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

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Ken has a group of coins worth $6.70. He has four times as many nickel  [#permalink]

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New post 02 Oct 2018, 11:16
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Question Stats:

88% (01:51) correct 13% (00:04) wrong based on 8 sessions

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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

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Re: Ken has a group of coins worth $6.70. He has four times as many nickel  [#permalink]

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New post 02 Oct 2018, 14:45
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Abhi077 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
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Re: Ken has a group of coins worth $6.70. He has four times as many nickel  [#permalink]

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New post 02 Oct 2018, 14:52
Abhi077 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 be the number of nickels, Y be the number of dimes, and Z be the number of quarters


Four times as many nickels as dimes can be translated into (1)

x = 4y (1)
z = x - 20 (2) which can be rearranged to be z + 20 = x


6.70 = 0.05x+0.1y+0.25z
6.70 = 0.05*4y+0.1y+0.25(4y-20)

6.70 = 0.2y+0.1y+1y - 5
11.70 = 1.3y

Y = 9
X = 36
Z = 36 - 20 = 16 << number of quarters.

Another approach would be to pick a number after setting up the initial equations (1) and (2)

If we select Z = 16 then x would be 36

If we multiply 16 * 0.25 = 4
And 36 * 0.05 = 1.8

X = 36 then y is 9
9*0.1 = 0.9

1.8+0.9+4 = 6.70.

Answer choice C

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Re: Ken has a group of coins worth $6.70. He has four times as many nickel  [#permalink]

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New post 02 Oct 2018, 15:13
Quote:
Ken has a group of coins - nickels, dimes and quarters only - 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 N be the number of nickels, D the number of dimes, Q the number of quarters and let us consider CENTS as the "problem´s unit".

Doing so, we have:

\(N = 4D\,\,\,\,\, \Rightarrow \,\,\,\,5N = 20D\)

\(Q = N - 20\,\,\,\)

\(? = Q = N - 20\)

\(5N + 10D + 25Q = 670\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,2} \,\,\,10N + 5N + 50\left( {N - 20} \right) = 2 \cdot 670\)

\(65N = 2 \cdot 670 + 1000\, = 20\left( {67 + 50} \right)\,\,\,\,\mathop \Rightarrow \limits^{:\,\,5\,\,} \,\,\,\,13N = 4\left( {9 \cdot 13} \right)\,\,\,\,\,\, \Rightarrow \,\,\,N = 36\)

\(? = 16\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Ken has a group of coins worth $6.70. He has four times as many nickel  [#permalink]

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New post 07 Oct 2018, 19:36
Abhi077 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
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Re: Ken has a group of coins worth $6.70. He has four times as many nickel &nbs [#permalink] 07 Oct 2018, 19:36
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Ken has a group of coins worth $6.70. He has four times as many nickel

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