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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Phil has a certain number of coins. He has 10 more dimes...
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MartyMurray
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Work with one variable, the number of nickels, N.Poisson wrote:Could someone please show me the most efficient method to answer this question? Thanks so much
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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Brent@GMATPrepNow
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I typically assign a variable to the smallest quantity.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
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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GMATGuruNY
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He has 10 more dimes than nickels and 22 more quarters than dimes.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
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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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.
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.
