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word problems equations...

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by topspin360 » Sat Jul 07, 2012 2:08 pm
how do we setup equations for the following question. I was able to do it by plugging numbers but that took a LONG time. trying to find a short cut route...


Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and half-dollars (50¢). On Monday, Billy bought one candy for less than a dollar and paid for it with exactly four coins (i.e., he received no change). On Tuesday, he bought two of the same candy and again paid with exactly four coins. On Wednesday, he bought three of the candies, on Thursday four of the candies, and on Friday five of the candies; each day he was able to pay with exactly four coins. Which of the following could be the price of one candy in cents?

8¢
13¢
40¢
53¢
66¢
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by GMATGuruNY » Sat Jul 07, 2012 2:27 pm
topspin360 wrote:how do we setup equations for the following question. I was able to do it by plugging numbers but that took a LONG time. trying to find a short cut route...


Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and half-dollars (50¢). On Monday, Billy bought one candy for less than a dollar and paid for it with exactly four coins (i.e., he received no change). On Tuesday, he bought two of the same candy and again paid with exactly four coins. On Wednesday, he bought three of the candies, on Thursday four of the candies, and on Friday five of the candies; each day he was able to pay with exactly four coins. Which of the following could be the price of one candy in cents?

8¢
13¢
40¢
53¢
66¢
We can plug in the answers, which represent the price of one candy.
Since four of the 5 coins are each a multiple of 5, the correct answer almost certainly will also be a multiple of 5.

Answer choice C: 40 cents
One candy: 10+10+10+10.
Two candies, with a total cost of 80: 25+25+25+5.
Three candies, with a total cost of 120: 50+50+10+10.
Four candies, with a total cost of 160: 50+50+50+10.
Five candies, with a total cost of 200: 50+50+50+50.
Success!

The correct answer is C.

Note the following:
Because we identified the most likely of the 5 answer choices. we had to try only ONE of them -- a very efficient way to solve the problem.
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by eagleeye » Sat Jul 07, 2012 2:56 pm
topspin360 wrote:how do we setup equations for the following question. I was able to do it by plugging numbers but that took a LONG time. trying to find a short cut route...


Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and half-dollars (50¢). On Monday, Billy bought one candy for less than a dollar and paid for it with exactly four coins (i.e., he received no change). On Tuesday, he bought two of the same candy and again paid with exactly four coins. On Wednesday, he bought three of the candies, on Thursday four of the candies, and on Friday five of the candies; each day he was able to pay with exactly four coins. Which of the following could be the price of one candy in cents?

8¢
13¢
40¢
53¢
66¢
This is a reasoning problem hence it requires some non-linear thinking.

Observe that we have 1 cent and all the other options are multiple of 5. Also note that the max. no. of coins in each case is 4. Here's where the problem gives us a clue. If we use all of the 1 cent coins, the max. amount we can make is 4 cents itself. So the number of 1 cent coins is definitely less than 4.
I will use the following two facts to eliminate wrong choices.
a. The 5cent, 10 cent, 25 cent, 50 cent coins all give us a number which either ends in a 5 or a 10.
b. We cannot have 4 one cent coins.

Together, they tell us that we can't have the following option: there are no numbers which end in 4 or 9. (Even*5+4 = number ending in 4 and Odd*5+4 = number ending in 9).

Let's check for the numbers which can end up in 4 or 9:

8*3=24. ends in 4. Wrong.
13*3 = 39 ends in 9. Wrong
53*3 = ends in 9. Wrong
66*4 = ends in 4. Wrong

Hence 40 is the only possible option. 40 is the answer. :)
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by Anurag@Gurome » Sat Jul 07, 2012 8:09 pm
Refer to the post here >> https://www.beatthegmat.com/is-there-a-f ... tml#481985
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by topspin360 » Sun Jul 08, 2012 8:44 pm
Thanks guys, makes perfect sense. I guess some of these problems require a very sharp eye to be actively looking for these kind of patterns.
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