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

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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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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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

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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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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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

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
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