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:

Here are some videos that cover the concepts required to answer this question:



_________________

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:
_________________

25x + 10(16-x) = 235
15x = 75
x = 5

x+y = 16
y=16-x

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

B

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
_________________

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

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

pushpitkc niks18 Hatakekakashi
amanvermagmat



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

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

Posted from my mobile device
_________________

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)
_________________

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
_________________

Thank you GMATPrepNow

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

25=5*5
235=5*47
GCF=5 => answer B
Can check:
5*25=125
(16-5)*10=110
125+110=235

Given: 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.

Asked: How many of the coins have a face value of 25 cents?

Let the number of 10 cents coin and 25 cents coins be x & y respectively
10x + 25y = 235
x + y = 16

Q. y = ?

x = 16 - y
10 (16-y) + 25y = 235
160 + 15y = 235
15y = 235 - 160 = 75
y = 5

IMO B
_________________

\(\frac{a}{10} + \frac{(16-a)}{4 }= \frac{235}{100}\)

Or, \(\frac{10a+25(16-a)}{100}= \frac{235}{100}\)

Or, \(10a+400 - 25a = 235\)

Or, \(-15a = -165\)

So, a = 11 and thus there are 5 no of 25 cents coins, Answer must be (B)
_________________

\(x= 10\) cents coin

\(y= 25\) cents coin

\(x+y=16; x=16-y\)

\(0.10x+0.25y=2.35\)

\(10x+25y=235\)

\(10(16-y)+25y=235\)

\(160-10y+25y=235\)

\(15y=75\)

\(y=5 \)

The answer is \(B\)
_________________