Ken has a group of coins worth $6.70. He has four times as many nickels as dimes, and twenty fewer quarters than nickels. How many quarters does he have?
A) 19
B) 18
C) 16
D) 15
E) 14
Ken has a group of coins worth $6.70. He has four times as m
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- AbhishekRyu
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Let x = the NUMBER of dimesAbhishekRyu wrote:Ken has a group of coins worth $6.70. He has four times as many nickels as dimes, and twenty fewer quarters than nickels. How many quarters does he have?
A) 19
B) 18
C) 16
D) 15
E) 14
So, 4x = the NUMBER of nickels
And 4x - 20 = the NUMBER of quarters
One dime is worth $0.1, so 0.10x = the total VALUE of the dimes (in dollars)
Likewise, (0.05)(4x) = the total VALUE of the nickels (in dollars)
And (0.25)(4x - 20) = the total VALUE of the quarters (in dollars)
The total VALUE of all coins is $6.70
So, we can write: 0.10x + (0.05)(4x) + (0.25)(4x - 20) = 6.70
Simplify to get: 0.10x + 0.20x + x - 5 = 6.70
Simplify again to get: 1.3x - 5 = 6.70
Add 5 to both sides to get: 1.3x = 11.70
Solve: x = 11.70/1.3 = 9
So, there are 9 DIMES, 36 nickels and 16 quarters.
Answer: C
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Let's let Q = the number of quarters, D = the number of dimes, and N = the number of nickels that Ken has. We can create the money value equation as:AbhishekRyu wrote:Ken has a group of coins worth $6.70. He has four times as many nickels as dimes, and twenty fewer quarters than nickels. How many quarters does he have?
A) 19
B) 18
C) 16
D) 15
E) 14
0.25Q + 0.1D + 0.05N = 6.7
25Q + 10D + 5N = 670
5Q + 2D + N = 134
We also know that Ken has has four times as many nickels as dimes, and twenty fewer quarters than nickels. We can create two additional equations, as follows:
N = 4D
N/4 = D
and
N - 20 = Q
Substituting into the money value equation, we have:
5(N - 20) + 2(N/4) + N = 134
5N - 100 + N/2 + N = 134
6N + N/2 = 234
Multiplying by 2, we have:
12N + N = 468
13N = 468
N = 36, so he has 16 quarters
Alternate Solution:
We know that
N = 4D
and
N - 20 = Q
We see that N is a multiple of 4 and since 20 is also a multiple of 4, Q must be a multiple of 4 also.
Looking at our answer choices, only choice C is a multiple of 4, so C must be the correct choice.
Answer: C
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