Fred has a jarful of nickels, dimes, and quarters, in the ra

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Fred has a jarful of nickels, dimes, and quarters, in the ratio of 2:5:10, respectively. If the total value of these coins is $15.50, how many dimes are in Fred's jar?

A. 17
B. 25
C. 50
D. 85
E. 155

OA B

Source: Princeton Review

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by [email protected] » Thu Dec 27, 2018 9:51 am
Hi All,

We're told that Fred has a jarful of nickels, dimes, and quarters, in the ratio of 2:5:10, respectively and the total value of these coins is $15.50. We're asked for the number of DIMES in Fred's jar. This question is based on ratio rules - and ratios are all about multiples - so you can do the math in a number of different ways. Here's how you can use a ratio pattern to your advantage.

For every 2 nickels, there are 5 dimes and 10 quarters, so we can start by calculating the value of that group (1 nickel = $0.05, 1 dime = $0.10 and 1 quarter = $0.25)..

(2)($0.05) + (5)($0.10) + (10)($0.25) =
$0.10 + $0.50 + $2.50 =
$3.10

The number of coins has to increase in that same ratio, so the total will be a MULTIPLE of $3.10

$15.50/$3.10 = 5 'groups' of coins (described above). With 5 groups of 5 DIMES, we have a total of 25 dimes.

Final Answer: B

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Rich
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by Scott@TargetTestPrep » Sun Feb 10, 2019 7:28 am
BTGmoderatorDC wrote:Fred has a jarful of nickels, dimes, and quarters, in the ratio of 2:5:10, respectively. If the total value of these coins is $15.50, how many dimes are in Fred's jar?

A. 17
B. 25
C. 50
D. 85
E. 155

OA B

Source: Princeton Review
The ratio of coins is N : D : Q = 2x : 5x : 10x.

We can create the equation for the money value of the coins, in cents, as:

5(2x) + 10(5x) + 25(10x) = 1550

10x + 50x + 250x = 1550

310x = 1550

x = 5

So there are 5(5) = 25 dimes in the jar.

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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