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Phil has a certain number of coins. He has 10 more dimes...

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by Poisson » Fri Aug 26, 2016 3:49 am
Phil has a certain number of coins in his pocket. He has 10 more dimes than nickels and 22 more quarters than dimes. If Phil has a total of $13.00 in change, how many coins does he have in total?

A. 42
B. 36
C. 60
D. 72
E. 81

Could someone please show me the most efficient method to answer this question? Thanks so much
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by MartyMurray » Fri Aug 26, 2016 4:29 am
Poisson wrote:Could someone please show me the most efficient method to answer this question? Thanks so much
Work with one variable, the number of nickels, N.

Number Of Nickels: N

Number Of Dimes: N + 10

Number Of Quarters: N + 10 + 22 = N + 32

Set up equation.

N(5) + (N + 10)(10) + (N + 32)(25) = 1300

5N + 10N + 100 + 25N + 800 = 1300

40N + 900 = 1300

40N = 400

N = 10

N + 10 = 20

N + 32 = 42

Total Coins = 72

The correct answer is D.
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by Brent@GMATPrepNow » Fri Aug 26, 2016 4:39 am
Poisson wrote:Phil has a certain number of coins in his pocket. He has 10 more dimes than nickels and 22 more quarters than dimes. If Phil has a total of $13.00 in change, how many coins does he have in total?

A. 42
B. 36
C. 60
D. 72
E. 81
I typically assign a variable to the smallest quantity.
Also, in these situations, I like to deal with cents (rather than dollars), so I can avoid using decimals.

Let n = # of nickels. This means 5n = VALUE of all of the nickels (in CENTS)

Phil has 10 more dimes than nickels ....
So, n + 10 = # of dimes. This means 10(n +10) = VALUE of all of the dimes(in CENTS)

Phil has 22 more quarters than dimes
So, (n + 10) + 22 = # of quarters. Simplify to get: n + 32 = # of quarters
This means 25(n +32) = VALUE of all of the quarters(in CENTS)

Phil has a total of $13.00 in change
So, (VALUE of nickels) + (VALUE of dimes) + (VALUE of quarters) = 1300 CENTS
Or... 5n + 10(n +10) + 25(n +32) = 1300
Expand to get: 5n + 10n + 100 + 25n + 800 = 1300
Simplify: 40n + 900 = 1300
Solve to get: n = 10

Since n = # of nickels, there are 10 nickels
Since n + 10 = # of dimes, there are 20 dimes
Since n + 32 = # of quarters, there are 42 quarters

10 + 20 + 42 =72
Answer: D

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by GMATGuruNY » Fri Aug 26, 2016 7:46 am
Poisson wrote:Phil has a certain number of coins in his pocket. He has 10 more dimes than nickels and 22 more quarters than dimes. If Phil has a total of $13.00 in change, how many coins does he have in total?

A. 42
B. 36
C. 60
D. 72
E. 81
He has 10 more dimes than nickels and 22 more quarters than dimes.
Least possible case:
N=1, D=1+10 = 11, Q=11+22=33.
In this case:
Total number of coins = 1+11+33 = 45.
Total value of the coins = 1*5 + 11*10 + 33*25 = 940.

Required total value = 1300 cents.
Since 1300-940 = 360, the value in blue must increase by 360 cents.
Every additional set of 3 coins -- 1 nickel, 1 dime, 1 quarter -- will increase the total value of the coins by 40 cents.
Since 360/40 = 9, adding nine 3-coin sets to the 45 coins above will increase the total value of the coins from 940 to 1300:
45 + 9*3 = 72 coins.

The correct answer is D.
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by Matt@VeritasPrep » Thu Sep 01, 2016 6:01 pm
I think there's an easier way.

Start with 0 nickels. If we have 0 nickels, we've got 10 dimes and 32 quarters, for a total of $9.

Every time we add a nickel, we must also add a dime and a quarter, since we need to have the same distance between them. Every nickel we add thus contributes 5¢ + 10¢ + 25¢, or 40¢. We need to add $4, so this requires 40¢ * 10, or 10 of each coin.

We started with 42 coins and added 10 more of each, or 30 more, so we end with 72.
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