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by AbeNeedsAnswers » Sun Jul 02, 2017 11:19 am

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Answers

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B

C

D

E

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Difficulty—

How else can I solve this besides algebra?

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

B
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by Brent@GMATPrepNow » Sun Jul 02, 2017 11:33 am
AbeNeedsAnswers wrote:How else can I solve this besides algebra?

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

B
Let D = the NUMBER of 10-cent coins
Let Q = the NUMBER of 25-cent coins

Notice that the VALUE of Q 25-cent coins = ($0.25)Q
For example, the VALUE of six 25-cent coins = ($0.25)6 = $1.50
And the VALUE of ten 25-cent coins = ($0.25)10 = $2.50
etc
Likewise, the VALUE of D 10-cent coins = ($0.10)D

The collection has 16 coins
We can write: D + Q = 16

The collection has a total value of $2.35
So, (0.10)D + (0.25)Q = 2.35

So, we have the following system:
(0.10)D + (0.25)Q = 2.35
D + Q = 16

Take the top equation and multiply both sides by 10 to get:
D + 2.5Q = 23.5
D + Q = 16

Subtract the bottom equation from the top to get:
1.5Q = 7.5
Solve: Q = 5

Answer: B

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by GMATGuruNY » Sun Jul 02, 2017 1:11 pm
AbeNeedsAnswers wrote:How else can I solve this besides algebra?

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 PLUG IN THE ANSWERS, which represent the number of 25-cent coins.
When the correct answer choice is plugged in, the total amount of money will be 235 cents.

Answer choice D: 9 25-cent coins, implying 7 10-cent coins, for a total of 16 coins
Total amount of money = (9*25) + (7*10) = 295 cents.
Since the total amount is TOO GREAT, fewer 25-cents coins are needed.
Eliminate D and E.

Answer choice B: 5 25-cent coins, implying 11 10-cent coins, for a total of 16 coins
Total amount of money = (5*25) + (11*10) = 235 cents.
Success!

The correct answer is B.
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by [email protected] » Sun Jul 02, 2017 8:32 pm
Hi AbeNeedsAnswers,

TESTing THE ANSWERS (the approach that Mitch used) is a great way to approach this question. When questions deal with 'limited options', then you can also use a bit of 'brute force' - and the relationships that exist between the numbers - to get to the correct answer.

Here, since there are only 16 coins - and we're dealing with only 10-cent coins and 25-cent coins - there are only 16 possible combinations of coins. We have an exact dollar figure to work with ($2.35), so we can simply work through enough options and consider whether they're 'too big' or 'too small.'

To start, what would the total be if you had 8 of each coin?
8(.10) + 8(.25) = .80 + 2.00 = $2.80

This is clearly 'too high', so to decrease this value we need there to be more 10s and fewer 25s. Since the difference in value is..

.25 - .10 = .15

We know that exchanging one 25-cent piece for one 10-cent piece will decrease the total value by $0.15. Since we have to get from $2.80 down to $2.35, we need a decrease of $0.45. So how many 'increments' of $0.15 will it take to get to $0.45....? Three.

Thus, if we have...
8 + 3 = 11 10-cent pieces and
8 - 3 = 5 25-cent pieces

We'll end up with $2.35. The question asks for the number of 25-cent pieces....

Final Answer: B

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by Jay@ManhattanReview » Wed Jul 05, 2017 12:22 am
AbeNeedsAnswers wrote:How else can I solve this besides algebra?

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 have to make 235 cents out of 16 coins, which are 10 cents or 25 cents.

If you decide to plug-in the option values, you should choose smart values to plug-in. You need not plug-in values randomly; it may cost you more time. In the GMAT, if the answers are in numerals, they would be arranged either in an ascending or a descending order.

B and D strategy:

1. Plug-in the value of option B for the number of 25-cent coins. If the result is equal to 235, the answer is option B, if the result is more than 235, the correct answer is option A; however, if the result is less than 235, try option D.

2. Plug-in the value of option D for the number of 25-cent coins. If the result is equal to 235, the answer is option D, if the result is more than 235, the correct answer is option C; however, if the result is less than 235, the correct answer is option E.

Let's apply this.

Option B: 5
Say the number fo 25-cent coins is 5, thus, the number fo 10-cent coins is 16 - 5 = 11.

This makes (5 x 25 + 11 x 10) = 125 + 110 = 235 cents.

Since the result equals the desired value, 235, option B is the correct answer.

Hope this helps!

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by Matt@VeritasPrep » Wed Jul 05, 2017 12:29 am
Let's try it without algebra, as requested. (This is also faster than most of the methods above, I think.)

Start with the average. $2.35 / 16 ≈ 15¢, which is 5¢ away from a dime and 10¢ away from a quarter.

By that logic, we should have about twice as many dimes as quarters, since the average is twice as far from a quarter as it is from a dime.

The closest such possibility with sixteen coins is 11 dimes and 5 quarters ... which = $2.35, touchdown!

How's that? :D
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by Matt@VeritasPrep » Wed Jul 05, 2017 12:34 am
Another approach too!

Let's start by assuming we only have quarters. If we have 9 of them, we've got $2.25, so 9 quarters + 1 dime = $2.35.

From there, we can trade quarters in for 2.5 dimes to try to get to 16 coins. If we trade 2 quarters for 5 dimes, we now have 7 quarters + 6 dimes ... not enough coins.

But one more trade gets us there! 2 more quarters for 5 more dimes leaves us 5 quarters and 11 dimes, which gives us 16 coins and hey, whowouddathunkit, a purse of $2.35.
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by Matt@VeritasPrep » Wed Jul 05, 2017 12:36 am
One last approach (that isn't as good).

Start with an easy number: 16 dimes. That gives you $1.60. From there, trade in one dime for one quarter until you get to $2.35.

Every time you exchange a dime for a quarter, you'll net 15¢. To get to $2.35, you need to make ($2.35 - $1.60) / 15¢ => 5 exchanges. So you'll be left with 11 dimes and 5 quarters.

There you have it ... my 3¢ :)
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by santhosh_katkurwar » Sat Sep 30, 2017 1:13 am
What if international students dont know 1 dollar = 100 cents?
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by Brent@GMATPrepNow » Sun Oct 01, 2017 12:08 pm
santhosh_katkurwar wrote:What if international students dont know 1 dollar = 100 cents?
This is an official GMAT question, so it appears that you need to know that $1 = 100 cents.

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