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Carla
 
 

Online Word Problems #5

by Carla Wed Apr 18, 2007 7:44 pm

Online Question Bank : Word Problems : #5

The solution presented two methods :

(1) to try each answer choice for 1 piece of candy, then for 2, etc.. - I tried this method but was too slow and just ended up guessing.

(2) "remainder and divisibility" - I wonder if you could just explain this to me in greater detail.

I generally find problems related to remainder to be challenging and maybe some help on this problem could help me.
Guest
 
 

by Guest Sun Apr 22, 2007 11:58 am

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¢
Guest
 
 

by Guest Mon Apr 23, 2007 11:34 am

I tried substitution strategy..
1) 8->5/1/1/1; 16->5/5/5/1; 24->10/5/1/X
2) 13->10/1/1/1; 26->10/10/5/1; 39:25/10/5/X
3)40-10/10/10/10; 80->50/10/10/10; 120->50/50/10/10; 160:50/50/50/10 sufficient;
-- did not go any further...
Carla
 
 

Thanks

by Carla Mon Apr 23, 2007 8:46 pm

The substitution method worked for me as well.. but it was very slow. The solution suggests a different method with a trick based on remainders and divisibility - did you have the chance to look at that? I found it a bit hard to follow...

here it is.. what do you think?/....



This problem could also have been solved using divisibility and remainders. Notice that all of the coins are multiples of 5 except pennies. In order to be able to pay for a certain number of candies with exactly four coins, the total price of the candies cannot be a value that can be expressed as 5x + 4, where x is a positive integer. In other words, the total price cannot be a number that has a remainder of 4 when divided by 5. Why? The remainder of 4 would alone require 4 pennies.

We can look at the answer choices now just focusing on the remainder when each price and its multiples are divided by 5:




Price per candy Remainder Remainder Remainder Remainder
when price for when price for when price for when price for
candy is 2 candies is 3 candies is 4 candies is
divided by 5 divided by 5 divided by 5 divided by 5


8 3 1 4 1
13 3 1 4 2
40 0 0 0 0
53 3 1 4 2
66 1 2 3 4


The only price for which none of its multiples have a remainder of 4 when divided by 5 is 40¢.

Notice that not having a remainder of 4 does not guarantee that exactly four coins can be used; however, having a remainder of 4 does guarantee that exactly for coins cannot be used!

The correct answer is C.
mww7786
 
 

Hey.

by mww7786 Mon Apr 23, 2007 9:44 pm

Carla,

Good. I see that your trying to get on top this. i've been after the GMAT stuff for a while now. 3 + months. I have the PR online class, and the manual. I highly recommend the kaplan 800 book. It's not that long of a read, but very very worth your time.

I assumed that your a student since I saw your posting. Anyways, I just wanted to link up with you and say hi for extra motivational reasons. it's always to a study pal as well.

Keep in touch,
Mark West
Tallahassee, FL
StaceyKoprince
ManhattanGMAT Staff
 
Posts: 9359
Joined: Wed Oct 19, 2005 9:05 am
Location: Montreal
 

ManhattanGMAT Word Translations bank #5

by StaceyKoprince Tue Apr 24, 2007 2:40 am

Another really tough one. Please note that you will not be able to do every problem on the test and part of your study task is to figure out which problems to do completely and which problems to do as educated guesses and then move on. This one may not be worth your time unless you're looking for a 750+.

The first day he buys 1 candy, the second day 2 candies, the third day 3 candies, and so on. So if we say the price he pays the first day is x, then he pays 2x the second day, 3x the third day, and so on.

He uses exactly 4 coins each time. If he uses any combination of nickels, dimes, quarters, and half-dollars (but NO pennies), then the amount will add up to a multiple of 5, since all of those denominations are multiples of 5. So:
sum of nickels, dimes, quarters, and half-dollars ONLY => will be a multiple of 5
sum of nickels, dimes, quarters, half-dollars and exactly 1 penny => will be (a multiple of 5) + 1
sum of nickels, dimes, quarters, half-dollars and exactly 2 pennies => will be (a multiple of 5) + 2
sum of nickels, dimes, quarters, half-dollars and exactly 3 pennies => will be (a multiple of 5) + 3
sum of nickels, dimes, quarters, half-dollars and exactly 4 pennies => will be (a multiple of 5) + 4

But we can't use that last option b/c Billy only gets to use 4 coins total - so all 4 coins can't be pennies.

If I take (a multiple of 5) + 4 and divide that sum by 5, I will get a remainder of 4 (those 4 pennies). Since this is not a valid option (because I can't make all 4 coins pennies), any choice that uses this is not the correct choice.

Then see the chart in the solution (I can't reproduce charts here) that tests the answer choices. As soon as one choice shows a remainder of 4 (indicating the need to make all 4 coins pennies), I can cross it off.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep