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Mark has fewer than 9 coins. Each coin is either a quarter

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by AAPL » Thu Dec 27, 2018 5:32 am

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Magoosh

Mark has fewer than 9 coins. Each coin is either a quarter (worth $0.25) or a dime (worth $0.10). How many dimes does Mark have?

1) The total value of Mark's coins is $1.70.
2) Mark has three times as many quarters as he has dimes.

OA A
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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Thu Dec 27, 2018 11:15 pm
AAPL wrote:Magoosh

Mark has fewer than 9 coins. Each coin is either a quarter (worth $0.25) or a dime (worth $0.10). How many dimes does Mark have?

1) The total value of Mark's coins is $1.70.
2) Mark has three times as many quarters as he has dimes.

OA A
say Mark has x numbers of quarters and y numbers of dimes.

Thus, we have

2 ≤ x + y < 9

We have to find out the value of y.

Let's take each statement one by one.

1) The total value of Mark's coins is $1.70.

0.25x + 0.10y = 1.70

5x + 2y = 43

y = (34 - 5x)/2

y = 17 - (5x)/2

Since x and y are integers, we must have x a multiple of 2.

To ensure that x + y < 9, only one value of x is valid, which is x = 6, thus, y = 2. Sufficient.

2) Mark has three times as many quarters as he has dimes.

x = 3y

To ensure that x + y < 9, we have y = 1 and x =3 or y = 2 and x = 6. No unique value of y. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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by Jay@ManhattanReview » Thu Dec 27, 2018 11:15 pm
AAPL wrote:Magoosh

Mark has fewer than 9 coins. Each coin is either a quarter (worth $0.25) or a dime (worth $0.10). How many dimes does Mark have?

1) The total value of Mark's coins is $1.70.
2) Mark has three times as many quarters as he has dimes.

OA A
say Mark has x numbers of quarters and y numbers of dimes.

Thus, we have

2 ≤ x + y < 9

We have to find out the value of y.

Let's take each statement one by one.

1) The total value of Mark's coins is $1.70.

0.25x + 0.10y = 1.70

5x + 2y = 43

y = (34 - 5x)/2

y = 17 - (5x)/2

Since x and y are integers, we must have x a multiple of 2.

To ensure that x + y < 9, only one value of x is valid, which is x = 6, thus, y = 2. Sufficient.

2) Mark has three times as many quarters as he has dimes.

x = 3y

To ensure that x + y < 9, we have y = 1 and x =3 or y = 2 and x = 6. No unique value of y. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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by GMATGuruNY » Fri Dec 28, 2018 9:11 am
AAPL wrote:Magoosh

Mark has fewer than 9 coins. Each coin is either a quarter (worth $0.25) or a dime (worth $0.10). How many dimes does Mark have?

1) The total value of Mark's coins is $1.70.
2) Mark has three times as many quarters as he has dimes.
Statement 1:
To MINIMIZE the number of coins so that a sum of 170 cents is yielded by at most 8 coins, we must MAXIMIZE the number of quarters.
Since 7*25 = 175 cents, the number of quarters must be less than 7.
170 cents will be yielded by 6 quarters (totaling 150 cents) and 2 dimes (totaling 20 cents), for a total of 8 coins and 170 cents.
Using fewer quarters will increase the number of coins beyond 8, violating the condition that Mark has fewer than 9 coins.
Implication:
No other combination of quarters and dimes is possible.
Thus, the number of dimes = 2.
SUFFICIENT.

Statement 2:
Case 1: 1 dime and 3 quarters, for a total of 4 coins
Case 2: 2 dimes and 6 quarters, for a total of 8 coins
Since the number of dimes can be different values, INSUFFICIENT.

The correct answer is A.
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