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How many coins

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by coolhabhi » Wed Nov 23, 2016 4:11 am
If 378 coins consist of dollars, 50 cent and 25 cent coins whose values are in the ratio of 13 : 11 : 7, the number of 50 cent coins will be

(A) 128 (B) 132 (C) 133 (D) 136 (E) 138

OE: B
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by MartyMurray » Wed Nov 23, 2016 4:37 am
For the same value of 1 dollar you need 2 50 cent coins or 4 25 cent coins.

Since we are looking for an answer in terms of number of coins, convert the value ratio to a number of coins ratio.

Dollars remains the same. 13

To get the 50 cent coins ratio number, multiply 11 by 2. 22

To get the 25 cent coins ratio number, multiply 7 by 4. 28

So the coins ratio is 13:22:28.

Add the numbers in the ratio to get a total divisor for 378.

13 + 22 + 28 = 63

To find out how many times the ratio has to be multiplied to show the exact number of each type of coin, divide.

378/63 = 6

Calculate the number of 50 cent coins.

6 x 22 = 132

The correct answer is B.
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by DavidG@VeritasPrep » Wed Nov 23, 2016 4:58 am
coolhabhi wrote:If 378 coins consist of dollars, 50 cent and 25 cent coins whose values are in the ratio of 13 : 11 : 7, the number of 50 cent coins will be

(A) 128 (B) 132 (C) 133 (D) 136 (E) 138

OE: B

If the value of the 50-cent coins is a multiple of 11, the number of 50-cent coins must also be a multiple of 11. (If you had $11 worth of 50-cent coins, you'd have 22 coins. If you had $22 worth of 50-cent coins you'd have 44 coins, etc.) Look at the answers. The only multiple of 11 is B
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by Matt@VeritasPrep » Fri Nov 25, 2016 3:11 am
Another approach:

Let's say we have x half dollars.

Since dollars are worth twice as much, we have (1/2) * (13/11) * x of them. ((1/2) for the double value, (13/11) for the ratio of dollars to half dollars.)

Since quarters are worth half as much, we have 2 * (7/11) * x of them.

x + (13/22)x + (14/11)x = 378

x = 132
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by Matt@VeritasPrep » Fri Nov 25, 2016 3:13 am
I do think David's approach is the way this is meant to be solved, though. The answers only have one multiple of 11, which can't be a coincidence, and any computational approach is meant to punish you for not looking for the Easter egg hidden in the answers. :)
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