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A collection of 16 coins, each with a face value of either 10 cents or

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A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?

A) 3
B) 5
C) 7
D) 9
E) 11
[Reveal] Spoiler: OA

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AbdurRakib wrote:
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?

A) 3
B) 5
C) 7
D) 9
E) 11


One solution is to start TESTING answer choices.

A) 3
If there are 3 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 13 coins must be dimes (worth $0.10 each)
3 quarters = $0.75 and 13 dimes = $1.30
So the TOTAL value = 0.75 + $1.30 = $2.05
NO GOOD - we want a total of $2.35

B) 5
If there are 5 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 11 coins must be dimes (worth $0.10 each)
5 quarters = $1.25 and 11 dimes = $1.10
So the TOTAL value = 1.25 + $1.10 = $2.35
BINGO!

Answer:
[Reveal] Spoiler:
B

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A collection of 16 coins, each with a face value of either 10 cents or [#permalink]

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AbdurRakib wrote:
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?

A) 3
B) 5
C) 7
D) 9
E) 11


Another approach is to use algebra:

Let Q = # of quarters ($0.25)
So, 16 - Q = # of dimes ($0.10)

We're told that: (value of quarters) + (value of dimes) = $2.35
So, we get: 0.25Q + 0.10(16 - Q) = 2.35
Expand: 0.25Q + 1.6 - 0.1Q = 2.35
Simplify: 0.15Q + 1.6 = 2.35
0.15Q = 0.75
Q = 5

Answer:
[Reveal] Spoiler:
B


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Re: A collection of 16 coins, each with a face value of either 10 cents or [#permalink]

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New post 15 Jun 2016, 00:52
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25x + 10(16-x) = 235
15x = 75
x = 5

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Re: A collection of 16 coins, each with a face value of either 10 cents or [#permalink]

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New post 06 Aug 2016, 01:47
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x+y = 16
y=16-x

10x+25y=235
2x+5y=47
2x+80-5x=47
-3x=-33
x=11
y=5

B

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Re: A collection of 16 coins, each with a face value of either 10 cents or [#permalink]

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AbdurRakib wrote:
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?

A) 3
B) 5
C) 7
D) 9
E) 11


We can let the number of 10-cent coins = d and the number of 25-cent coins = q. We are given that there are 16 total coins; thus, d + q = 16.

We are also given that the total face value is $2.35, thus:

0.1d + 0.25q = 2.35

10d + 25q = 235

Isolating d in our first equation, we have d = 16 - q. We can substitute 16 - q for d in our second equation and we have:

10(16 - q) + 25q = 235

160 - 10q + 25q = 235

15q = 75

q = 5

There are 5 coins with a face value of 25 cents.

Answer: B
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Re: A collection of 16 coins, each with a face value of either 10 cents or [#permalink]

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New post 09 Mar 2017, 16:04
x - number of 10-cent coins
y - number of 25-cent coins

x+y=16
x*0.10 + y*0.25 = 2.35
x*0.10 + y*0.10 + y*0.15 = 2.35
0.10*(x+y) + y*0.15 = 2.35
0.15*y = 2.35-1.60
y = 5

Answer B

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Re: A collection of 16 coins, each with a face value of either 10 cents or [#permalink]

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New post 13 Sep 2017, 22:50
let x be 10 cents coin and y be 25 cents coin
then
10x+25y = 235.................(i)
x+y = 16 .................(ii)

(ii) * 10
10x+10Y = 160 ............(iii)

then (i) - (iii)
15 y = 75
y=5
Hence Answer B
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Re: A collection of 16 coins, each with a face value of either 10 cents or   [#permalink] 13 Sep 2017, 22:50
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