Ken has a group of coins worth $6.70. He has four times as m

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

Let x = the NUMBER of dimes
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

Cheers,
Brent

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

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:

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