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Asifpirlo
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Sue and Nancy wish to buy a snack. They combine their money Updated on: 25 Aug 2013, 05:56
Originally posted by Asifpirlo on 24 Aug 2013, 15:49.
Last edited by Bunuel on 25 Aug 2013, 05:56, edited 1 time in total.
Edited the question.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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25% (medium)

Question Stats:

83% (02:33) correct 17% (02:06) wrong based on 84 sessions

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Sue and Nancy wish to buy a snack. They combine their money and find they have $4.00, consisting of quarters, dimes, and nickels. If they have 35 coins and the number of quarters is half the number of nickels, how many quarters do they have?

(A) 5
(B) 10
(C) 20
(D) 3
(E) 6
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B
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Bunuel
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Sue and Nancy wish to buy a snack. They combine their money 25 Aug 2013, 06:00
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Asifpirlo
Sue and Nancy wish to buy a snack. They combine their money and find they have $4.00, consisting of quarters, dimes, and nickels. If they have 35 coins and the number of quarters is half the number of nickels, how many quarters do they have?

(A) 5
(B) 10
(C) 20
(D) 3
(E) 6
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q + d + n = 35;
The number of quarters is half the number of nickels --> n = 2q --> q + d + 2q = 35;

25q + 10d + 5n = 400 --> 25q + 10d + 5*2q = 400 --> 5q + 2d + 2q = 80.

Solving q + d + 2q = 35 and 5q + 2d + 2q = 80 gives q = 10.

Answer: B.
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Sue and Nancy wish to buy a snack. They combine their money 25 Aug 2013, 15:34
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Bunuel
Asifpirlo
Sue and Nancy wish to buy a snack. They combine their money and find they have $4.00, consisting of quarters, dimes, and nickels. If they have 35 coins and the number of quarters is half the number of nickels, how many quarters do they have?

(A) 5
(B) 10
(C) 20
(D) 3
(E) 6

q + d + n = 35;
The number of quarters is half the number of nickels --> n = 2q --> q + d + 2q = 35;

25q + 10d + 5n = 400 --> 25q + 10d + 5*2q = 400 --> 5q + 2d + 2q = 80.

Solving q + d + 2q = 35 and 5q + 2d + 2q = 80 gives q = 10.

Answer: B.
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This is an easy problem but 25q + 10d + 5n = 400 is the equation i have to use quickly.
you used algebra but i often try with this sort problem via COUNTING method and i kill time by doing so... :(
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Sue and Nancy wish to buy a snack. They combine their money 25 Aug 2019, 07:20
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Alternate Method: Test the answer choices.

Total: $4.00 = 400 cents

Number of Coins: 35

Types of coins:

    Nickels - 5 cents

    Dimes - 10 cents

    Quarters - 25 cents

Number of quarters is half the number of nickels.

Number of Quarters Choices: 3, 5, 6, 10, 20

Execution:

20 Quarters = 20 x 25 = 500 cents. Too big. Eliminate 20.

For every 1 quarter there are 2 nickels.

1 Quarter + 2 Nickels = 35 cents. An Odd Number x 35 Cents = A Number Ending in 5. Dimes = 10 Cents. No number ending in 5 can be combined with numbers ending in 10 to add up to 400. Eliminate all odd numbers of quarters: 3, 5.

Left with 6 and 10.

6 x 35 = 210. Need 19 dimes to get to 400. Already have 6 x 1 Quarter + 6 x 2 Nickels = 18 coins. 19 + 18 > 35.

The only possible number of quarters left is 10.

Confirm:

Quarters and Nickels: 10 x 25 + 20 x 5 = 350

Need 50 more to get to 400.

Dimes: 5 x 10 = 50

10 + 20 + 5 = 35 Coins

The correct answer is (B).
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Sue and Nancy wish to buy a snack. They combine their money 14 Jan 2026, 03:49
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