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		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15/Homework_4&amp;diff=56516</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15/Homework 4</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15/Homework_4&amp;diff=56516"/>
		<updated>2010-10-20T07:52:56Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: /* Question 5 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== Question 1 ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan&#039;s mother? Tosh owns a cat, Bianca owns a frog. Jaela owns a parrot. Jun owns a snake. Suzan is the name of the frog. The cat is named Jun. The name by which they call the turtle is the name of the woman whose pet is Tosh. Finally, Suzan&#039;s mother&#039;s pet is Bianca.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Polya’s law is about ways to tackle a problem successfully so it should apply here as well. Step one: Read the question carefully. The question, when broken down, implicitly asks us two questions in place of one, because they both need to be answered before we can answer the original question. They are:&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1.	Who is Suzan’s mother?&amp;lt;br&amp;gt;&lt;br /&gt;
2.	What pet does Suzan’s mother own? (which is also the “ultimate” question)&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From reading the question we are already given a useful clue. That is, Suzan’s mother, must not be Suzan. We can forget about Suzan and narrow the list of five candidates down to four (though this doesn&#039;t prove particularly useful for the problem at hand as you will be able to see). We then work down the list of other clues, basically taking them, a better description wanting, at face value (and easy, too!). By simple logic we know that Suzan’s pet is a turtle (because it is the only pet left without an owner). So we get the following data in the form of a sorted table (Polya’s Step 2: Plan and map out your strategy in a chart, table, graph, etc.). &amp;lt;br&amp;gt;&lt;br /&gt;
[[File:Picture1.png]]&lt;br /&gt;
&lt;br /&gt;
We then try to find out the name of Suzan’s turtle. It is the name of the person whose pet is called Tosh. Tosh can be Jaela’s parrot or Jun’s snake, but not Tosh’s cat nor Bianca’s frog. (It cannot be Suzan’s turtle either unless the question is playing tricks with our head... A rather dumb trick that would be too, wouldn’t it?) So, if we express this in our table:&amp;lt;br&amp;gt;&lt;br /&gt;
[[File:Picture2.png]]&lt;br /&gt;
&lt;br /&gt;
So, is it Jaela or Jun? Do we have enough data to determine that yet? ( Step 1: Identify if enough data is given to solve the problem) Turns out we do. Jun cannot be Suzan’s turtle’s name, because we already know that Tosh’s cat is called Jun. Therefore, the turtle’s name must be Jaela.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That leaves two more unknowns on our table to be determined: the names of Jaela’s parrot and Jun’s snake. They can only be either Tosh / Bianca (because that’s the only two names left – simple!) To find the next clue, we re-visit the original clues – it is important to revisit the question since as you are reading anything, as more information emerges the original material seems to present new meanings as you read it again because now you can read into deeper layers of meaning with additional perspective. So, since the turtle’s name is also the name of the woman who’s pet is Tosh, and Jaela being the turtle’s name, we conclude that the name of Jaela’s parrot is Tosh. That leaves Jun’s pet, whose name must be Bianca.&amp;lt;br&amp;gt;&lt;br /&gt;
[[File:Picture3.png]]&lt;br /&gt;
&lt;br /&gt;
I believe you can now confidently answer the original question, which is “what pet belongs to Suzan’s mother”. We know from the beginning that Suzan’s mother’s pet is Bianca. When we look for Bianca in the table we’ve drawn up to tackle the problem, we can answer with a lot of confidence that Suzan’s mother is Jun, and perhaps more importantly (to answer the question directly), Suzan’s mother’s pet is a snake (&amp;quot;don&#039;t mess with him&amp;quot;)...apparently a boy snake with a girly name!&lt;br /&gt;
&lt;br /&gt;
== Question 2 ==&lt;br /&gt;
&lt;br /&gt;
Bohao, Tim, Dylan, Chan and Stewart&lt;br /&gt;
&lt;br /&gt;
5 players&lt;br /&gt;
&lt;br /&gt;
3 of them are right handed&lt;br /&gt;
2 of them are left handed&lt;br /&gt;
&lt;br /&gt;
3 of them are under 2m &lt;br /&gt;
2 of them are over 2m&lt;br /&gt;
&lt;br /&gt;
We are looking for the centre player who is left handed and also over 2m.&lt;br /&gt;
&lt;br /&gt;
-Tim or Chan must be right handed, because Dylan and Bohao are both right handed and Stewart is left handed.&lt;br /&gt;
-Bohao is over 2m tall so this means that either Dylan or Tim must be the same height as Chan and Stewart who are both under 2m.&lt;br /&gt;
-You can already see a common trend developing, the fact that we are trying to find both the variables for Tim&#039;s height as well as his handedness.&lt;br /&gt;
-This then narrows down to Dylan and TIm to be over 2m&lt;br /&gt;
-Dylan is right handed though&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
This leads us to the conclusion that Tim is the centre player because he is the only valid option, Dylan who is over 2m is right handed and therefor does not fit the description.&lt;br /&gt;
&lt;br /&gt;
== Question 3 ==&lt;br /&gt;
&lt;br /&gt;
== Question 4 ==&lt;br /&gt;
&#039;&#039;&#039;Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
1). First of all we sort out the details given by the question:&lt;br /&gt;
6 players-&amp;gt; P, C, J, S, L and F.&lt;br /&gt;
&lt;br /&gt;
Tournament= 5 days&lt;br /&gt;
&lt;br /&gt;
-&amp;gt;each player plays each of others ONCE.&lt;br /&gt;
&lt;br /&gt;
-&amp;gt;3 matches are played in 5 days&lt;br /&gt;
&lt;br /&gt;
Therefore, we know that 2 people play against each other in each day.&lt;br /&gt;
&lt;br /&gt;
1st day: &#039;&#039;&#039;C&#039;&#039;&#039; vs P  ( C wins)  J,S,L,F&lt;br /&gt;
&lt;br /&gt;
2nd day: &#039;&#039;&#039;C&#039;&#039;&#039; vs J  ( C wins)  P,S,L,F&lt;br /&gt;
&lt;br /&gt;
3rd day: &#039;&#039;&#039;J&#039;&#039;&#039; vs L  (J wins)   P,C,S,F&lt;br /&gt;
 &lt;br /&gt;
4th day: &#039;&#039;&#039;P&#039;&#039;&#039; vs S  (P wins)   C,J,L,F &lt;br /&gt;
&lt;br /&gt;
5th day: F vs ???&lt;br /&gt;
&lt;br /&gt;
Then, we use the info given, try to pair the unknown ones up:&lt;br /&gt;
&lt;br /&gt;
1st day: CP, ( JS, &#039;&#039;L&#039;&#039;F ) &lt;br /&gt;
&lt;br /&gt;
2nd day: CJ, ( PL, &#039;&#039;S&#039;&#039;F) &lt;br /&gt;
&lt;br /&gt;
3rd day: JL, ( &#039;&#039;P&#039;&#039;F, CS)&lt;br /&gt;
&lt;br /&gt;
4th day: PS, ( CL, &#039;&#039;J&#039;&#039;F)&lt;br /&gt;
&lt;br /&gt;
5th day: F havn&#039;t played with C according to the info above.&lt;br /&gt;
&lt;br /&gt;
Therefore, Fernanda is playing against &#039;&#039;&#039;Carla&#039;&#039;&#039; on the 5th day.&lt;br /&gt;
&lt;br /&gt;
== Question 5 ==&lt;br /&gt;
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.&lt;br /&gt;
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.&lt;br /&gt;
On Sunday, the barking of the neighbour&#039;s dog abruptly ended Homer&#039;s sleep.&lt;br /&gt;
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.&lt;br /&gt;
In fact, the salesman, the neighbour&#039;s dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.&lt;br /&gt;
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.&lt;br /&gt;
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer&#039;s vacation. On which day of his holiday was Homer actually able to sleep late?&lt;br /&gt;
First, We notice that:&lt;br /&gt;
Saturday --&amp;gt; Salesman&lt;br /&gt;
Sunday --&amp;gt; Dog&lt;br /&gt;
Monday --&amp;gt; Salesman/Construction&lt;br /&gt;
Tuesday --&amp;gt; ? Unknown?	&lt;br /&gt;
Wednesday - Salesman/Dog&lt;br /&gt;
Thursday --&amp;gt; ? Unknown?	&lt;br /&gt;
Friday --&amp;gt; ?unknown?&lt;br /&gt;
Saturday --&amp;gt; Construction&lt;br /&gt;
Sunday --&amp;gt; Dog&lt;br /&gt;
&lt;br /&gt;
By following the rule : “No one of the three noisemakers was quiet for three consecutive days” and “no pair of them made noise on more than one day”  Tuesday, Thursday and friday are all unknown. &lt;br /&gt;
As we can see we could put it like below. &lt;br /&gt;
&lt;br /&gt;
Saturday --&amp;gt; Salesman&lt;br /&gt;
Sunday --&amp;gt; Dog&lt;br /&gt;
Monday --&amp;gt; Salesman + Construction&lt;br /&gt;
Tuesday --&amp;gt;  Sleep in.&lt;br /&gt;
Wednesday - Salesman + Dog&lt;br /&gt;
Thursday --&amp;gt; Construction	&lt;br /&gt;
Friday --&amp;gt;  Salesman&lt;br /&gt;
Saturday --&amp;gt; Construction&lt;br /&gt;
Sunday --&amp;gt; Dog&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54514</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54514"/>
		<updated>2010-10-13T09:07:24Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: /* Question 11 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Danny Choi&lt;br /&gt;
* Peter David Dabrowski&lt;br /&gt;
* Allie Miller&lt;br /&gt;
* Anabelle Tory&lt;br /&gt;
* Vivian Zhang&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
Homework 3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 1&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1hour and 20min is identical to 80min, there is nothing else to explain.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 2&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things. &lt;br /&gt;
&lt;br /&gt;
Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 3&#039;&#039;&#039; ==&lt;br /&gt;
   &lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.&lt;br /&gt;
&lt;br /&gt;
First, you simply pick the box that marked &amp;quot;apple and oranges&amp;quot;. And it contend apple. Then you know that the box is only contain apples only. Now you know that the box that marked oranges cant not be the box that contain apply. So the &amp;quot;oranges box is contain apples and oranges. ==&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 4&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
Since I have no brother, then the blind fiddler must be my&#039;s sister.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 5&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 6&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&lt;br /&gt;
I must at least draw 4 times, that&#039;s because the 4th apple are the same as one of the first three  i draw.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 7&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;[[Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)pair of the same color&lt;br /&gt;
When we draw &#039;&#039;&#039;3&#039;&#039;&#039; socks without looking, we can be sure that there are a pair of the same color since there are only 2 kinds of color.&lt;br /&gt;
&lt;br /&gt;
ii)pair of different color&lt;br /&gt;
When we draw a total of &#039;&#039;&#039;41&#039;&#039;&#039; socks, we can be sure that we have two socks that are different coloured. On the 41st draw, there must be a different color sock since one of the color is all drawn out.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 8&#039;&#039;&#039; ==&lt;br /&gt;
Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&lt;br /&gt;
&lt;br /&gt;
Reuben was born in december 31. and spoke on january 1.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 9&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Assuming that the boat remains in the water as the tide rises, 10 rungs will be showing. This is because as the tide rises the boat will be levitated with the rise.&lt;br /&gt;
&lt;br /&gt;
It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 10&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 11&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
There is no worst player, because it is not possible at all. Since we notices that the worst player are the twin and the best player is same age and opposite sex of the twin. The woman, her older brother or her son, and her daughter  could be twins... Without further information, we could not figure whos who.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 12&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 13&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 14&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 15&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.&lt;br /&gt;
&lt;br /&gt;
There are 52 cards in total and &amp;lt;math&amp;gt; \tfrac{52}{2}=26 &amp;lt;/math&amp;gt; cards in each half. Half of all cards are red and half of all cards are black.&lt;br /&gt;
&lt;br /&gt;
In the first half there are &#039;&#039;x&#039;&#039; red cards and &#039;&#039;y&#039;&#039; black cards.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;+&#039;&#039;y&#039;&#039;=26&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;=26-&#039;&#039;y&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;y&#039;&#039;=26-&#039;&#039;x&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In the second half there are the red cards which are not in the first half 26-&#039;&#039;x&#039;&#039; which is equal to &#039;&#039;y&#039;&#039;. And there are the black cards which are not in the first half 26-&#039;&#039;y&#039;&#039; which is equal to &#039;&#039;x&#039;&#039;. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 16&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assuem that the family has &amp;quot;x&amp;quot; number of girls and &amp;quot;y&amp;quot; number of boys&lt;br /&gt;
&lt;br /&gt;
Then, according to the question we get two equations:&lt;br /&gt;
X - 1 = y&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
2(y-1)= X&lt;br /&gt;
&lt;br /&gt;
Let&#039;s solve these two equations by combining them together, then we get&lt;br /&gt;
&lt;br /&gt;
X=4&lt;br /&gt;
&lt;br /&gt;
Y=3&lt;br /&gt;
&lt;br /&gt;
Therefore, this family has 4 girls and 3 boys.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 17&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 18&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given, now we get 4 equations:&lt;br /&gt;
&lt;br /&gt;
1) y = x/3&lt;br /&gt;
2) z = (1/3)(x - x/3) &lt;br /&gt;
3) w = (1/3)[x – (1/3)(x - x/3) – x/3] &lt;br /&gt;
4) x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Plug the values into equation number 4 then we get--- (8/27)X =40 ---&amp;gt; X = 135&lt;br /&gt;
      &lt;br /&gt;
Therefore, there are 135 pennies at the start.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 19&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 20&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
One clock runs 5 minutes faster per hour while the other runs 5 minutes slower.&lt;br /&gt;
&lt;br /&gt;
After one hour, the two clocks will have a 10 minutes difference between each other.&lt;br /&gt;
&lt;br /&gt;
There are 60 minutes per hour, therefore 60/10 = 6.&lt;br /&gt;
&lt;br /&gt;
After 6 hours the two clocks will be one hour apart @ 6a.m.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 21&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 22&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 23&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 24&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)candle A = 6 hours&lt;br /&gt;
&lt;br /&gt;
ii)candle B = 3 hours&lt;br /&gt;
&lt;br /&gt;
So, Candle A takes twice more time to burn out than Candle B&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Candle A: burns 1/6 per hour&lt;br /&gt;
&lt;br /&gt;
Candle B: burns 1/3 per hour&lt;br /&gt;
&lt;br /&gt;
Let the time it takes to burn out in hours be X&lt;br /&gt;
&lt;br /&gt;
Then, Candle A = (6-x)/6&lt;br /&gt;
&lt;br /&gt;
 and Candle B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
According to the info we get from the question, we get ---&lt;br /&gt;
&lt;br /&gt;
2((x − 3) / 3) = (6 − x) / 6&lt;br /&gt;
&lt;br /&gt;
6(6 − 2x) = 3(6 − x)&lt;br /&gt;
&lt;br /&gt;
36 − 12x = 18 − 3x&lt;br /&gt;
&lt;br /&gt;
18 = 9x&lt;br /&gt;
&lt;br /&gt;
x = 2hrs&lt;br /&gt;
&lt;br /&gt;
Therefore, the time it takes for one of the candle to be half of the other candle is 2 hours.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 25&#039;&#039;&#039; ==&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54508</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54508"/>
		<updated>2010-10-13T09:02:45Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: /* Question 6 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Danny Choi&lt;br /&gt;
* Peter David Dabrowski&lt;br /&gt;
* Allie Miller&lt;br /&gt;
* Anabelle Tory&lt;br /&gt;
* Vivian Zhang&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
Homework 3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 1&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1hour and 20min is identical to 80min, there is nothing else to explain.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 2&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things. &lt;br /&gt;
&lt;br /&gt;
Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 3&#039;&#039;&#039; ==&lt;br /&gt;
   &lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.&lt;br /&gt;
&lt;br /&gt;
First, you simply pick the box that marked &amp;quot;apple and oranges&amp;quot;. And it contend apple. Then you know that the box is only contain apples only. Now you know that the box that marked oranges cant not be the box that contain apply. So the &amp;quot;oranges box is contain apples and oranges. ==&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 4&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
Since I have no brother, then the blind fiddler must be my&#039;s sister.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 5&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 6&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&lt;br /&gt;
I must at least draw 4 times, that&#039;s because the 4th apple are the same as one of the first three  i draw.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 7&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;[[Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)pair of the same color&lt;br /&gt;
When we draw &#039;&#039;&#039;3&#039;&#039;&#039; socks without looking, we can be sure that there are a pair of the same color since there are only 2 kinds of color.&lt;br /&gt;
&lt;br /&gt;
ii)pair of different color&lt;br /&gt;
When we draw a total of &#039;&#039;&#039;41&#039;&#039;&#039; socks, we can be sure that we have two socks that are different coloured. On the 41st draw, there must be a different color sock since one of the color is all drawn out.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 8&#039;&#039;&#039; ==&lt;br /&gt;
Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&lt;br /&gt;
&lt;br /&gt;
Reuben was born in december 31. and spoke on january 1.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 9&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Assuming that the boat remains in the water as the tide rises, 10 rungs will be showing. This is because as the tide rises the boat will be levitated with the rise.&lt;br /&gt;
&lt;br /&gt;
It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 10&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 11&#039;&#039;&#039; ==&lt;br /&gt;
 A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
There is no worst player, because it is not possible at all.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 12&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 13&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 14&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 15&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.&lt;br /&gt;
&lt;br /&gt;
There are 52 cards in total and &amp;lt;math&amp;gt; \tfrac{52}{2}=26 &amp;lt;/math&amp;gt; cards in each half. Half of all cards are red and half of all cards are black.&lt;br /&gt;
&lt;br /&gt;
In the first half there are &#039;&#039;x&#039;&#039; red cards and &#039;&#039;y&#039;&#039; black cards.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;+&#039;&#039;y&#039;&#039;=26&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;=26-&#039;&#039;y&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;y&#039;&#039;=26-&#039;&#039;x&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In the second half there are the red cards which are not in the first half 26-&#039;&#039;x&#039;&#039; which is equal to &#039;&#039;y&#039;&#039;. And there are the black cards which are not in the first half 26-&#039;&#039;y&#039;&#039; which is equal to &#039;&#039;x&#039;&#039;. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 16&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assuem that the family has &amp;quot;x&amp;quot; number of girls and &amp;quot;y&amp;quot; number of boys&lt;br /&gt;
&lt;br /&gt;
Then, according to the question we get two equations:&lt;br /&gt;
X - 1 = y&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
2(y-1)= X&lt;br /&gt;
&lt;br /&gt;
Let&#039;s solve these two equations by combining them together, then we get&lt;br /&gt;
&lt;br /&gt;
X=4&lt;br /&gt;
&lt;br /&gt;
Y=3&lt;br /&gt;
&lt;br /&gt;
Therefore, this family has 4 girls and 3 boys.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 17&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 18&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given, now we get 4 equations:&lt;br /&gt;
&lt;br /&gt;
1) y = x/3&lt;br /&gt;
2) z = (1/3)(x - x/3) &lt;br /&gt;
3) w = (1/3)[x – (1/3)(x - x/3) – x/3] &lt;br /&gt;
4) x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Plug the values into equation number 4 then we get--- (8/27)X =40 ---&amp;gt; X = 135&lt;br /&gt;
      &lt;br /&gt;
Therefore, there are 135 pennies at the start.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 19&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 20&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
One clock runs 5 minutes faster per hour while the other runs 5 minutes slower.&lt;br /&gt;
&lt;br /&gt;
After one hour, the two clocks will have a 10 minutes difference between each other.&lt;br /&gt;
&lt;br /&gt;
There are 60 minutes per hour, therefore 60/10 = 6.&lt;br /&gt;
&lt;br /&gt;
After 6 hours the two clocks will be one hour apart @ 6a.m.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 21&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 22&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 23&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 24&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 25&#039;&#039;&#039; ==&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54507</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54507"/>
		<updated>2010-10-13T09:01:57Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Danny Choi&lt;br /&gt;
* Peter David Dabrowski&lt;br /&gt;
* Allie Miller&lt;br /&gt;
* Anabelle Tory&lt;br /&gt;
* Vivian Zhang&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
Homework 3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 1&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1hour and 20min is identical to 80min, there is nothing else to explain.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 2&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things. &lt;br /&gt;
&lt;br /&gt;
Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 3&#039;&#039;&#039; ==&lt;br /&gt;
   &lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.&lt;br /&gt;
&lt;br /&gt;
First, you simply pick the box that marked &amp;quot;apple and oranges&amp;quot;. And it contend apple. Then you know that the box is only contain apples only. Now you know that the box that marked oranges cant not be the box that contain apply. So the &amp;quot;oranges box is contain apples and oranges. ==&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 4&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
Since I have no brother, then the blind fiddler must be my&#039;s sister.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 5&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 6&#039;&#039;&#039; ==&lt;br /&gt;
 Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&lt;br /&gt;
I must at least draw 4 times, that&#039;s because the 4th apple are the same as one of the first three  i draw.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 7&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;[[Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)pair of the same color&lt;br /&gt;
When we draw &#039;&#039;&#039;3&#039;&#039;&#039; socks without looking, we can be sure that there are a pair of the same color since there are only 2 kinds of color.&lt;br /&gt;
&lt;br /&gt;
ii)pair of different color&lt;br /&gt;
When we draw a total of &#039;&#039;&#039;41&#039;&#039;&#039; socks, we can be sure that we have two socks that are different coloured. On the 41st draw, there must be a different color sock since one of the color is all drawn out.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 8&#039;&#039;&#039; ==&lt;br /&gt;
Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&lt;br /&gt;
&lt;br /&gt;
Reuben was born in december 31. and spoke on january 1.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 9&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Assuming that the boat remains in the water as the tide rises, 10 rungs will be showing. This is because as the tide rises the boat will be levitated with the rise.&lt;br /&gt;
&lt;br /&gt;
It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 10&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 11&#039;&#039;&#039; ==&lt;br /&gt;
 A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
There is no worst player, because it is not possible at all.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 12&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 13&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 14&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 15&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.&lt;br /&gt;
&lt;br /&gt;
There are 52 cards in total and &amp;lt;math&amp;gt; \tfrac{52}{2}=26 &amp;lt;/math&amp;gt; cards in each half. Half of all cards are red and half of all cards are black.&lt;br /&gt;
&lt;br /&gt;
In the first half there are &#039;&#039;x&#039;&#039; red cards and &#039;&#039;y&#039;&#039; black cards.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;+&#039;&#039;y&#039;&#039;=26&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;=26-&#039;&#039;y&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;y&#039;&#039;=26-&#039;&#039;x&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In the second half there are the red cards which are not in the first half 26-&#039;&#039;x&#039;&#039; which is equal to &#039;&#039;y&#039;&#039;. And there are the black cards which are not in the first half 26-&#039;&#039;y&#039;&#039; which is equal to &#039;&#039;x&#039;&#039;. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 16&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assuem that the family has &amp;quot;x&amp;quot; number of girls and &amp;quot;y&amp;quot; number of boys&lt;br /&gt;
&lt;br /&gt;
Then, according to the question we get two equations:&lt;br /&gt;
X - 1 = y&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
2(y-1)= X&lt;br /&gt;
&lt;br /&gt;
Let&#039;s solve these two equations by combining them together, then we get&lt;br /&gt;
&lt;br /&gt;
X=4&lt;br /&gt;
&lt;br /&gt;
Y=3&lt;br /&gt;
&lt;br /&gt;
Therefore, this family has 4 girls and 3 boys.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 17&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 18&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given, now we get 4 equations:&lt;br /&gt;
&lt;br /&gt;
1) y = x/3&lt;br /&gt;
2) z = (1/3)(x - x/3) &lt;br /&gt;
3) w = (1/3)[x – (1/3)(x - x/3) – x/3] &lt;br /&gt;
4) x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Plug the values into equation number 4 then we get--- (8/27)X =40 ---&amp;gt; X = 135&lt;br /&gt;
      &lt;br /&gt;
Therefore, there are 135 pennies at the start.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 19&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 20&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
One clock runs 5 minutes faster per hour while the other runs 5 minutes slower.&lt;br /&gt;
&lt;br /&gt;
After one hour, the two clocks will have a 10 minutes difference between each other.&lt;br /&gt;
&lt;br /&gt;
There are 60 minutes per hour, therefore 60/10 = 6.&lt;br /&gt;
&lt;br /&gt;
After 6 hours the two clocks will be one hour apart @ 6a.m.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 21&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 22&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 23&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 24&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 25&#039;&#039;&#039; ==&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54504</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54504"/>
		<updated>2010-10-13T09:01:32Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Danny Choi&lt;br /&gt;
* Peter David Dabrowski&lt;br /&gt;
* Allie Miller&lt;br /&gt;
* Anabelle Tory&lt;br /&gt;
* Vivian Zhang&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
Homework 3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 1&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1hour and 20min is identical to 80min, there is nothing else to explain.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 2&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things. &lt;br /&gt;
&lt;br /&gt;
Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 3&#039;&#039;&#039; ==&lt;br /&gt;
   &lt;br /&gt;
   ==  One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.&lt;br /&gt;
&lt;br /&gt;
   First, you simply pick the box that marked &amp;quot;apple and oranges&amp;quot;. And it contend apple. Then you know that the box is only contain apples only. Now you know that the box that marked oranges cant not be the box that contain apply. So the &amp;quot;oranges box is contain apples and oranges. ==&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 4&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
Since I have no brother, then the blind fiddler must be my&#039;s sister.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 5&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 6&#039;&#039;&#039; ==&lt;br /&gt;
 Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&lt;br /&gt;
I must at least draw 4 times, that&#039;s because the 4th apple are the same as one of the first three  i draw.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 7&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;[[Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)pair of the same color&lt;br /&gt;
When we draw &#039;&#039;&#039;3&#039;&#039;&#039; socks without looking, we can be sure that there are a pair of the same color since there are only 2 kinds of color.&lt;br /&gt;
&lt;br /&gt;
ii)pair of different color&lt;br /&gt;
When we draw a total of &#039;&#039;&#039;41&#039;&#039;&#039; socks, we can be sure that we have two socks that are different coloured. On the 41st draw, there must be a different color sock since one of the color is all drawn out.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 8&#039;&#039;&#039; ==&lt;br /&gt;
Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&lt;br /&gt;
&lt;br /&gt;
Reuben was born in december 31. and spoke on january 1.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 9&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Assuming that the boat remains in the water as the tide rises, 10 rungs will be showing. This is because as the tide rises the boat will be levitated with the rise.&lt;br /&gt;
&lt;br /&gt;
It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 10&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 11&#039;&#039;&#039; ==&lt;br /&gt;
 A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
There is no worst player, because it is not possible at all.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 12&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 13&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 14&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 15&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.&lt;br /&gt;
&lt;br /&gt;
There are 52 cards in total and &amp;lt;math&amp;gt; \tfrac{52}{2}=26 &amp;lt;/math&amp;gt; cards in each half. Half of all cards are red and half of all cards are black.&lt;br /&gt;
&lt;br /&gt;
In the first half there are &#039;&#039;x&#039;&#039; red cards and &#039;&#039;y&#039;&#039; black cards.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;+&#039;&#039;y&#039;&#039;=26&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;=26-&#039;&#039;y&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;y&#039;&#039;=26-&#039;&#039;x&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In the second half there are the red cards which are not in the first half 26-&#039;&#039;x&#039;&#039; which is equal to &#039;&#039;y&#039;&#039;. And there are the black cards which are not in the first half 26-&#039;&#039;y&#039;&#039; which is equal to &#039;&#039;x&#039;&#039;. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 16&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assuem that the family has &amp;quot;x&amp;quot; number of girls and &amp;quot;y&amp;quot; number of boys&lt;br /&gt;
&lt;br /&gt;
Then, according to the question we get two equations:&lt;br /&gt;
X - 1 = y&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
2(y-1)= X&lt;br /&gt;
&lt;br /&gt;
Let&#039;s solve these two equations by combining them together, then we get&lt;br /&gt;
&lt;br /&gt;
X=4&lt;br /&gt;
&lt;br /&gt;
Y=3&lt;br /&gt;
&lt;br /&gt;
Therefore, this family has 4 girls and 3 boys.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 17&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 18&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given, now we get 4 equations:&lt;br /&gt;
&lt;br /&gt;
1) y = x/3&lt;br /&gt;
2) z = (1/3)(x - x/3) &lt;br /&gt;
3) w = (1/3)[x – (1/3)(x - x/3) – x/3] &lt;br /&gt;
4) x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Plug the values into equation number 4 then we get--- (8/27)X =40 ---&amp;gt; X = 135&lt;br /&gt;
      &lt;br /&gt;
Therefore, there are 135 pennies at the start.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 19&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 20&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
One clock runs 5 minutes faster per hour while the other runs 5 minutes slower.&lt;br /&gt;
&lt;br /&gt;
After one hour, the two clocks will have a 10 minutes difference between each other.&lt;br /&gt;
&lt;br /&gt;
There are 60 minutes per hour, therefore 60/10 = 6.&lt;br /&gt;
&lt;br /&gt;
After 6 hours the two clocks will be one hour apart @ 6a.m.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 21&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 22&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 23&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 24&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 25&#039;&#039;&#039; ==&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54503</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54503"/>
		<updated>2010-10-13T09:00:40Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: /* Question 6 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Danny Choi&lt;br /&gt;
* Peter David Dabrowski&lt;br /&gt;
* Allie Miller&lt;br /&gt;
* Anabelle Tory&lt;br /&gt;
* Vivian Zhang&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
Homework 3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 1&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1hour and 20min is identical to 80min, there is nothing else to explain.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 2&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things. &lt;br /&gt;
&lt;br /&gt;
Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 3&#039;&#039;&#039; ==&lt;br /&gt;
   &lt;br /&gt;
     One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.&lt;br /&gt;
&lt;br /&gt;
   First, you simply pick the box that marked &amp;quot;apple and oranges&amp;quot;. And it contend apple. Then you know that the box is only contain apples only. Now you know that the box that marked oranges cant not be the box that contain apply. So the &amp;quot;oranges box is contain apples and oranges.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 4&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
Since I have no brother, then the blind fiddler must be my&#039;s sister.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 5&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 6&#039;&#039;&#039; ==&lt;br /&gt;
 Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&lt;br /&gt;
I must at least draw 4 times, that&#039;s because the 4th apple are the same as one of the first three  i draw.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 7&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;[[Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)pair of the same color&lt;br /&gt;
When we draw &#039;&#039;&#039;3&#039;&#039;&#039; socks without looking, we can be sure that there are a pair of the same color since there are only 2 kinds of color.&lt;br /&gt;
&lt;br /&gt;
ii)pair of different color&lt;br /&gt;
When we draw a total of &#039;&#039;&#039;41&#039;&#039;&#039; socks, we can be sure that we have two socks that are different coloured. On the 41st draw, there must be a different color sock since one of the color is all drawn out.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 8&#039;&#039;&#039; ==&lt;br /&gt;
Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&lt;br /&gt;
&lt;br /&gt;
Reuben was born in december 31. and spoke on january 1.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 9&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Assuming that the boat remains in the water as the tide rises, 10 rungs will be showing. This is because as the tide rises the boat will be levitated with the rise.&lt;br /&gt;
&lt;br /&gt;
It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 10&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 11&#039;&#039;&#039; ==&lt;br /&gt;
 A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
There is no worst player, because it is not possible at all.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 12&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 13&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 14&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 15&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.&lt;br /&gt;
&lt;br /&gt;
There are 52 cards in total and &amp;lt;math&amp;gt; \tfrac{52}{2}=26 &amp;lt;/math&amp;gt; cards in each half. Half of all cards are red and half of all cards are black.&lt;br /&gt;
&lt;br /&gt;
In the first half there are &#039;&#039;x&#039;&#039; red cards and &#039;&#039;y&#039;&#039; black cards.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;+&#039;&#039;y&#039;&#039;=26&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;=26-&#039;&#039;y&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;y&#039;&#039;=26-&#039;&#039;x&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In the second half there are the red cards which are not in the first half 26-&#039;&#039;x&#039;&#039; which is equal to &#039;&#039;y&#039;&#039;. And there are the black cards which are not in the first half 26-&#039;&#039;y&#039;&#039; which is equal to &#039;&#039;x&#039;&#039;. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 16&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assuem that the family has &amp;quot;x&amp;quot; number of girls and &amp;quot;y&amp;quot; number of boys&lt;br /&gt;
&lt;br /&gt;
Then, according to the question we get two equations:&lt;br /&gt;
X - 1 = y&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
2(y-1)= X&lt;br /&gt;
&lt;br /&gt;
Let&#039;s solve these two equations by combining them together, then we get&lt;br /&gt;
&lt;br /&gt;
X=4&lt;br /&gt;
&lt;br /&gt;
Y=3&lt;br /&gt;
&lt;br /&gt;
Therefore, this family has 4 girls and 3 boys.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 17&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 18&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given, now we get 4 equations:&lt;br /&gt;
&lt;br /&gt;
1) y = x/3&lt;br /&gt;
2) z = (1/3)(x - x/3) &lt;br /&gt;
3) w = (1/3)[x – (1/3)(x - x/3) – x/3] &lt;br /&gt;
4) x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Plug the values into equation number 4 then we get--- (8/27)X =40 ---&amp;gt; X = 135&lt;br /&gt;
      &lt;br /&gt;
Therefore, there are 135 pennies at the start.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 19&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 20&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
One clock runs 5 minutes faster per hour while the other runs 5 minutes slower.&lt;br /&gt;
&lt;br /&gt;
After one hour, the two clocks will have a 10 minutes difference between each other.&lt;br /&gt;
&lt;br /&gt;
There are 60 minutes per hour, therefore 60/10 = 6.&lt;br /&gt;
&lt;br /&gt;
After 6 hours the two clocks will be one hour apart @ 6a.m.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 21&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 22&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 23&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 24&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 25&#039;&#039;&#039; ==&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54500</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54500"/>
		<updated>2010-10-13T08:57:54Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Danny Choi&lt;br /&gt;
* Peter David Dabrowski&lt;br /&gt;
* Allie Miller&lt;br /&gt;
* Anabelle Tory&lt;br /&gt;
* Vivian Zhang&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
Homework 3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 1&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1hour and 20min is identical to 80min, there is nothing else to explain.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 2&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things. &lt;br /&gt;
&lt;br /&gt;
Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 3&#039;&#039;&#039; ==&lt;br /&gt;
   &lt;br /&gt;
     One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.&lt;br /&gt;
&lt;br /&gt;
   First, you simply pick the box that marked &amp;quot;apple and oranges&amp;quot;. And it contend apple. Then you know that the box is only contain apples only. Now you know that the box that marked oranges cant not be the box that contain apply. So the &amp;quot;oranges box is contain apples and oranges.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 4&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
Since I have no brother, then the blind fiddler must be my&#039;s sister.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 5&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 6&#039;&#039;&#039; ==&lt;br /&gt;
 Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&lt;br /&gt;
I must at least draw 4 times. &lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 7&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;[[Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)pair of the same color&lt;br /&gt;
When we draw &#039;&#039;&#039;3&#039;&#039;&#039; socks without looking, we can be sure that there are a pair of the same color since there are only 2 kinds of color.&lt;br /&gt;
&lt;br /&gt;
ii)pair of different color&lt;br /&gt;
When we draw a total of &#039;&#039;&#039;41&#039;&#039;&#039; socks, we can be sure that we have two socks that are different coloured. On the 41st draw, there must be a different color sock since one of the color is all drawn out.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 8&#039;&#039;&#039; ==&lt;br /&gt;
Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&lt;br /&gt;
&lt;br /&gt;
Reuben was born in december 31. and spoke on january 1.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 9&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Assuming that the boat remains in the water as the tide rises, 10 rungs will be showing. This is because as the tide rises the boat will be levitated with the rise.&lt;br /&gt;
&lt;br /&gt;
It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 10&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 11&#039;&#039;&#039; ==&lt;br /&gt;
 A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
There is no worst player, because it is not possible at all.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 12&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 13&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 14&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 15&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.&lt;br /&gt;
&lt;br /&gt;
There are 52 cards in total and &amp;lt;math&amp;gt; \tfrac{52}{2}=26 &amp;lt;/math&amp;gt; cards in each half. Half of all cards are red and half of all cards are black.&lt;br /&gt;
&lt;br /&gt;
In the first half there are &#039;&#039;x&#039;&#039; red cards and &#039;&#039;y&#039;&#039; black cards.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;+&#039;&#039;y&#039;&#039;=26&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;=26-&#039;&#039;y&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;y&#039;&#039;=26-&#039;&#039;x&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In the second half there are the red cards which are not in the first half 26-&#039;&#039;x&#039;&#039; which is equal to &#039;&#039;y&#039;&#039;. And there are the black cards which are not in the first half 26-&#039;&#039;y&#039;&#039; which is equal to &#039;&#039;x&#039;&#039;. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 16&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assuem that the family has &amp;quot;x&amp;quot; number of girls and &amp;quot;y&amp;quot; number of boys&lt;br /&gt;
&lt;br /&gt;
Then, according to the question we get two equations:&lt;br /&gt;
X - 1 = y&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
2(y-1)= X&lt;br /&gt;
&lt;br /&gt;
Let&#039;s solve these two equations by combining them together, then we get&lt;br /&gt;
&lt;br /&gt;
X=4&lt;br /&gt;
&lt;br /&gt;
Y=3&lt;br /&gt;
&lt;br /&gt;
Therefore, this family has 4 girls and 3 boys.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 17&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 18&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given, now we get 4 equations:&lt;br /&gt;
&lt;br /&gt;
1) y = x/3&lt;br /&gt;
2) z = (1/3)(x - x/3) &lt;br /&gt;
3) w = (1/3)[x – (1/3)(x - x/3) – x/3] &lt;br /&gt;
4) x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Plug the values into equation number 4 then we get--- (8/27)X =40 ---&amp;gt; X = 135&lt;br /&gt;
      &lt;br /&gt;
Therefore, there are 135 pennies at the start.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 19&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 20&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
One clock runs 5 minutes faster per hour while the other runs 5 minutes slower.&lt;br /&gt;
&lt;br /&gt;
After one hour, the two clocks will have a 10 minutes difference between each other.&lt;br /&gt;
&lt;br /&gt;
There are 60 minutes per hour, therefore 60/10 = 6.&lt;br /&gt;
&lt;br /&gt;
After 6 hours the two clocks will be one hour apart @ 6a.m.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 21&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 22&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 23&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 24&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 25&#039;&#039;&#039; ==&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54494</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 15</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_15&amp;diff=54494"/>
		<updated>2010-10-13T08:53:27Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Danny Choi&lt;br /&gt;
* Peter David Dabrowski&lt;br /&gt;
* Allie Miller&lt;br /&gt;
* Anabelle Tory&lt;br /&gt;
* Vivian Zhang&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
Homework 3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 1&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1hour and 20min is identical to 80min, there is nothing else to explain.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 2&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since the policeman never had the opportunity to check the driver’s license, he wouldn’t have known that she did not have it on her. Therefore, that mistake was irrelevant to the question why the policeman didn’t stop her at her antics. There can be a few possibilities why the policeman didn’t stop the driver despite seeing her apparently breaking road etiquettes (failing to stop at a stop sign and going down a one-way street the wrong way). The policeman could have been preoccupied with another road incident and therefore didn’t manage to stop the female driver in time. Or it could be that execution of road policies was not usually strictly carried out in that vicinity (or city, or country). Or it could be that the woman was not driving at all- there isn’t any indication in the question that the woman was in fact conducting a vehicle when she did all those things. &lt;br /&gt;
&lt;br /&gt;
Reading the question “superficially”, as demonstrated by the solutions we have seen above, it is very easy to automatically assume that the woman broke the road codes while she was driving. We expect to be given the context in the beginning of any paragraph of texts. In this example, the first sentence seems to set the scene for the readers, “The lady driver did not have her driver’s license with her…”, to be immediately followed by a series of driving behaviours that we all know are breaking the road code. We link the lady and her driver’s license to the assumption that “she must be behind the wheel”. However, as one should be able to see, that piece of information was there to mislead instead of inform. I found that to understand the problem (Polya’s Step 1) really useful in thinking up solutions for this problem. Specifically it helped me to re-read the question carefully and especially not taking anything in it for granted. As we all know, a pedestrian who fails to stop at a stop sign (presumably those signs designed for vehicles), and “going” (walking) down a one way street the wrong way (wrong only for vehicles) does not get stopped by the police!&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 3&#039;&#039;&#039; ==&lt;br /&gt;
   &lt;br /&gt;
    One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
&lt;br /&gt;
   First, you simply pick the box that marked &amp;quot;apple and oranges&amp;quot;. And it contend apple. Then you know that the box is only contain apples only. Now you know that the box that marked oranges cant not be the box that contain apply. So the &amp;quot;oranges box is contain apples and oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 4&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
Since I have no brother, then the blind fiddler must be my&#039;s sister.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 5&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 6&#039;&#039;&#039; ==&lt;br /&gt;
 Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&lt;br /&gt;
I must at least draw 4 times. &lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 7&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;[[Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
i)pair of the same color&lt;br /&gt;
When we draw &#039;&#039;&#039;3&#039;&#039;&#039; socks without looking, we can be sure that there are a pair of the same color since there are only 2 kinds of color.&lt;br /&gt;
&lt;br /&gt;
ii)pair of different color&lt;br /&gt;
When we draw a total of &#039;&#039;&#039;41&#039;&#039;&#039; socks, we can be sure that we have two socks that are different coloured. On the 41st draw, there must be a different color sock since one of the color is all drawn out.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 8&#039;&#039;&#039; ==&lt;br /&gt;
Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&lt;br /&gt;
&lt;br /&gt;
Reuben was born in december 31. and spoke on january 1.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 9&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Assuming that the boat remains in the water as the tide rises, 10 rungs will be showing. This is because as the tide rises the boat will be levitated with the rise.&lt;br /&gt;
&lt;br /&gt;
It is easy to be distracted by the presence of actual figures in the problem. It would have been the easiest route to solve the problem – one has the figures, one simply needs to do some mathematical calculations. However, not only does that approach miss the point of the question totally, it also lacks creativity on the part of the problem-solver. Step 2 of the Polya method emphasises that creativity is one of the most important element (even before organisation and experience) one should adopt while planning for a strategy to tackle a problem.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 10&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 11&#039;&#039;&#039; ==&lt;br /&gt;
 A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
There is no worst player, because it is not possible at all.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 12&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 13&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 14&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 15&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.&lt;br /&gt;
&lt;br /&gt;
There are 52 cards in total and &amp;lt;math&amp;gt; \tfrac{52}{2}=26 &amp;lt;/math&amp;gt; cards in each half. Half of all cards are red and half of all cards are black.&lt;br /&gt;
&lt;br /&gt;
In the first half there are &#039;&#039;x&#039;&#039; red cards and &#039;&#039;y&#039;&#039; black cards.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;+&#039;&#039;y&#039;&#039;=26&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039;=26-&#039;&#039;y&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;y&#039;&#039;=26-&#039;&#039;x&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In the second half there are the red cards which are not in the first half 26-&#039;&#039;x&#039;&#039; which is equal to &#039;&#039;y&#039;&#039;. And there are the black cards which are not in the first half 26-&#039;&#039;y&#039;&#039; which is equal to &#039;&#039;x&#039;&#039;. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 16&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assuem that the family has &amp;quot;x&amp;quot; number of girls and &amp;quot;y&amp;quot; number of boys&lt;br /&gt;
&lt;br /&gt;
Then, according to the question we get two equations:&lt;br /&gt;
X - 1 = y&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
2(y-1)= X&lt;br /&gt;
&lt;br /&gt;
Let&#039;s solve these two equations by combining them together, then we get&lt;br /&gt;
&lt;br /&gt;
X=4&lt;br /&gt;
&lt;br /&gt;
Y=3&lt;br /&gt;
&lt;br /&gt;
Therefore, this family has 4 girls and 3 boys.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 17&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 18&#039;&#039;&#039; ==&lt;br /&gt;
&#039;&#039;&#039;[[Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start? ]]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar.&lt;br /&gt;
&lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given, now we get 4 equations:&lt;br /&gt;
&lt;br /&gt;
1) y = x/3&lt;br /&gt;
2) z = (1/3)(x - x/3) &lt;br /&gt;
3) w = (1/3)[x – (1/3)(x - x/3) – x/3] &lt;br /&gt;
4) x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Plug the values into equation number 4 then we get--- (8/27)X =40 ---&amp;gt; X = 135&lt;br /&gt;
      &lt;br /&gt;
Therefore, there are 135 pennies at the start.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 19&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 20&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 21&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 22&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 23&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 24&#039;&#039;&#039; ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== &#039;&#039;&#039;Question 25&#039;&#039;&#039; ==&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47987</id>
		<title>User:ChunHangChoi</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47987"/>
		<updated>2010-09-19T18:21:41Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Danny Choi.&lt;br /&gt;
&lt;br /&gt;
The Pythagoras&#039; Theorem&lt;br /&gt;
&lt;br /&gt;
 A² + B² = C² &lt;br /&gt;
 &lt;br /&gt;
One of the most significant math theorems was discover back in the early centrist, it theorem which named after a Greek mathematician. &lt;br /&gt;
Today’s the theory, was title  “the Pythagoras’ Theorem.” &lt;br /&gt;
It theory had more than three hundreds evident ways to prove it owns, the trigonometric is the one of the most popular way to prove it.   &lt;br /&gt;
As in school, many people leaned that how the “Pythagoras’ theorem” effects out daily life. We could use the theorem on buildings (height), arts (3D), and even locate the location (GPS). &lt;br /&gt;
One thing we could not doubt is that Pythagoras theorem is not just a math equation but it changes our world, through one theorem.&lt;br /&gt;
    →There will be two example shows below. &lt;br /&gt;
Before showing the example, the rule for Pythagoras’ theorem was that the right angel have to be 90 degree.&lt;br /&gt;
→Example 1.  If the information were only given the a (base) and b (height), how do we find the C (hypotenuse)? &lt;br /&gt;
If a =6 and b =8&lt;br /&gt;
Then we could use the equation A² + B² = C².&lt;br /&gt;
6² + 8² = C² →36 + 64 =C²→ 100=C²→C=10&lt;br /&gt;
→Example 2. There’s a ladder lean on the wall, and it Only given the a (base) and C (hypotenuse), how do we find B (height) then? &lt;br /&gt;
If b =8 and c =10&lt;br /&gt;
Then we could use the equation A² + B² = C².&lt;br /&gt;
A² + 8² = 10²→10² - 8² = A² →36= A² →A=6&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47984</id>
		<title>User:ChunHangChoi</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47984"/>
		<updated>2010-09-19T18:10:03Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Danny Choi.&lt;br /&gt;
&lt;br /&gt;
The Pythagoras&#039; Theorem&lt;br /&gt;
&lt;br /&gt;
 →A² + B² = C² &lt;br /&gt;
 &lt;br /&gt;
One of the most significant math theorems was discover back in the early centrist, it theorem which named after a Greek mathematician. &lt;br /&gt;
Today’s the theory, was title  “the Pythagoras’ Theorem.” &lt;br /&gt;
It theory had more than three hundreds evident ways to prove it owns, the trigonometric is the one of the most popular way to prove it.   &lt;br /&gt;
As in school, many people leaned that how the “Pythagoras’ theorem” effects out daily life. We could use the theorem on buildings (height), arts (3D), and even locate the location (GPS). &lt;br /&gt;
One thing we could not doubt is that Pythagoras theorem is not just a math equation but it changes our world, through one theorem.&lt;br /&gt;
    →There will be two example shows below. &lt;br /&gt;
Before showing the example, the rule for Pythagoras’ theorem was that the right angel have to be 90 degree.&lt;br /&gt;
→Example 1.  If the information were only given the a (base) and b (height), how do we find the C (hypotenuse)? &lt;br /&gt;
If a =6 and b =8&lt;br /&gt;
Then we could use the equation A² + B² = C².&lt;br /&gt;
6² + 8² = C² →36 + 64 =C²→ 100=C²→C=10&lt;br /&gt;
→Example 2. There’s a ladder lean on the wall, and it Only given the a (base) and C (hypotenuse), how do we find B (height) then? &lt;br /&gt;
If b =8 and c =10&lt;br /&gt;
Then we could use the equation A² + B² = C².&lt;br /&gt;
A² + 8² = 10²→10² - 8² = A² →36= A² →A=6&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47983</id>
		<title>User:ChunHangChoi</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47983"/>
		<updated>2010-09-19T18:02:27Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Danny Choi is my name.&lt;br /&gt;
&lt;br /&gt;
The Pythagoras&#039; Theorem&lt;br /&gt;
&lt;br /&gt;
→ A² + B² = C²&lt;br /&gt;
&lt;br /&gt;
    One of the most significant math theorems was discover back in the early centrist, it theorem which named after a Greek mathematician. Today’s the theory, was title  “the Pythagoras’ Theorem.” It theory had more than three hundreds evident ways to prove it owns, the trigonometric is the one of the most popular way to prove it.   As in school, many people leaned that how the “Pythagoras’ theorem” effects out daily life. We could use the theorem on buildings (height), arts (3D), and even locate the location (GPS). One thing we could not doubt is that Pythagoras theorem is not just a math equation but it changes our world, through one theorem.&lt;br /&gt;
&lt;br /&gt;
There will be two example shows below. &lt;br /&gt;
Before showing the example, the rule for Pythagoras’ theorem was that the right angel have to be 90 degree.&lt;br /&gt;
Example 1.  If the information were only given the a (base) and b (height), how do we find the C (hypotenuse)? &lt;br /&gt;
If a =6 and b =8&lt;br /&gt;
Then we could use the equation A² + B² = C².&lt;br /&gt;
6² + 8² = C²&lt;br /&gt;
36 + 64 =C²&lt;br /&gt;
        100=C²&lt;br /&gt;
            C=10&lt;br /&gt;
Example 2. There’s a ladder lean on the wall, and it Only given the a (base) and C (hypotenuse), how do we find B (height) then? &lt;br /&gt;
If b =8 and c =10&lt;br /&gt;
Then we could use the equation A² + B² = C².&lt;br /&gt;
A² + 8² = 10².&lt;br /&gt;
10² - 8² = A² &lt;br /&gt;
         36= A² &lt;br /&gt;
           A=6&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47982</id>
		<title>User:ChunHangChoi</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:ChunHangChoi&amp;diff=47982"/>
		<updated>2010-09-19T18:01:46Z</updated>

		<summary type="html">&lt;p&gt;ChunHangChoi: Created page with &amp;#039;My name is Danny Choi.&amp;#039;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;My name is Danny Choi.&lt;/div&gt;</summary>
		<author><name>ChunHangChoi</name></author>
	</entry>
</feed>