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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  [#permalink]

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New post 14 Jun 2016, 14:09
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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

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

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New post 14 Jun 2016, 16:06
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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:

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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 14 Jun 2016, 16:01
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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:
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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
1
x+y = 16
y=16-x

10x+25y=235
2x+5y=47
2x+80-5x=47
-3x=-33
x=11
y=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 02 Dec 2016, 06:48
3
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
1
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]

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New post 17 Mar 2018, 04:46
pushpitkc niks18 Hatakekakashi
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Quote:
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.


Let x: no coins with 10 cents as face value
Let y: no coins with 25 cents as face value

I am clear about forming x + y = 16

How can I be sure to form which of two below algebraic equations?

0.1 x + 0.25 y = 2.35 OR

0.25 x + 0.1 y = 2.35

Since I am getting y = 5 (or x= 11) in first case and (x=5 or y=11) in second case.
What mistake I am making since I must get an unique solution from answer choice in PS.
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A collection of 16 coins, each with a face value of either 10 cents or  [#permalink]

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New post 17 Mar 2018, 04:53
adkikani wrote:
pushpitkc niks18 Hatakekakashi
amanvermagmat

Quote:
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.


Let x: no coins with 10 cents as face value
Let y: no coins with 25 cents as face value

I am clear about forming x + y = 16

How can I be sure to form which of two below algebraic equations?

0.1 x + 0.25 y = 2.35 OR

0.25 x + 0.1 y = 2.35

Since I am getting y = 5 (or x= 11) in first case and (x=5 or y=11) in second case.
What mistake I am making since I must get an unique solution from answer choice in PS.


Hi adkikani

in your assumption x represents 10 cents or 0.1 dollars, so total contribution from x coins will be 0.1x

similarly y represents 25 cents or 0.25 dollars, so total contribution from y coins=0.25y

Hence total amount =0.1x+0.25y=2.35

and total number of coins =16 i.e x+y=16
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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 17 Mar 2018, 04:58
adkikani wrote:
pushpitkc niks18 Hatakekakashi
amanvermagmat

Quote:
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.


Let x: no coins with 10 cents as face value
Let y: no coins with 25 cents as face value

I am clear about forming x + y = 16

How can I be sure to form which of two below algebraic equations?

0.1 x + 0.25 y = 2.35 OR

0.25 x + 0.1 y = 2.35

Since I am getting y = 5 (or x= 11) in first case and (x=5 or y=11) in second case.
What mistake I am making since I must get an unique solution from answer choice in PS.


Adding to what niks18 said

If you have confusion regarding two choices

Always verify

If x=5 y=11

Substitute in the equation

And if the answer equals 2.35 that's the answer

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

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New post 18 Mar 2018, 07:47
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 know x + y = 16 ----- I

x = 16 - y -----(II)

Now, 0.10x + 0.25y = 2.35

0.10(16 - y) + 0.25y = 2.35

1.6 - 0.10y +0.25y = 2.35

1.6 - 0.15y = 2.35

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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New post 07 Apr 2018, 07:02
GMATPrepNow wrote:
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:


GMATPrepNow, How and When do you we know that we can solve a word problem (or any other problem for that matter) by using Testing "The Answer Choices Method"
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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 08 Apr 2018, 06:58
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FANewJersey wrote:
GMATPrepNow wrote:
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:


GMATPrepNow, How and When do you we know that we can solve a word problem (or any other problem for that matter) by using Testing "The Answer Choices Method"


For most word problems, you should at least consider the possibility of testing the answer choices. HOWEVER this doesn't necessarily mean that testing the answer choices will be the best/fastest approach.

I'd say that, in most cases, testing the answer choices is not the best approach. That said, it's a great strategy if you can't think of any other ways to solve the question.

In other situations, testing the answer choices is impossible. For example:

In how many ways can we arrange the letters in the word FAST?
A) 4
B) 8
C) 12
D) 16
E) 24

As you can see, there is no way to actually test the answer choices for this question.

I hope that helps.

Cheers,
Brent
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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 Apr 2018, 05:38
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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 23 May 2019, 19:47
Let's assume x are the coins valued at 0.10 and y are the coins valued at 0.25. We want to find value of y.

First, remove decimal bits and form algebraic equations:

10x + 25y = 235 --> equation 1
x+y = 16 --> equation 2
x= 16-y --> equation 3

Second, insert equation 3 into equation 1:

10 (16-y) + 25y = 235
160 - 10y + 25y = 235
160 + 15y = 235
15y= 75
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] 23 May 2019, 19:47
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