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A child bought candy that cost 85 cents. If the child only had quarters, dimes, and nickels, did he pay using at least two quarters? (quarter = $0.25, dime = $0.1 and nickel = $0.05)

1) The child used 8 coins.
2) The child used 5 of one type of coin.

OA E

Say the child uses x quarters, y dimes, and z nickels.

Thus, we have 25x + 10y + 5z = 85

=> 5x + 2y + z = 17 ---(1)

We have to determine whether x ≥ 2.

Let's take each statement one by one.

1) The child used 8 coins.

=> x + y + z = 8 ---(2)

Case 1: Say x = 1

Thus, from (2), we have y + z = 7 => y = 7 - z

From (1), we have 5x + 2y + z = 17 => 5*1 + 2(7 - z) + z = 17 => 5 + 14 - 2z + z = 17 => z = 2; thus, y = 5. Since y and z are positve integers, it's a possible solution. Since x < 2, the answer is no.

Case 2: Say x = 2

Thus, from (2), we have y + z = 6 => y = 6 - z

From (1), we have 5x + 2y + z = 17 => 5*2 + 2(6 - z) + z = 17 => 10 + 12 - 2z + z = 17 => z = 5; thus, y = 1. Since y and z are positve integers, it's a possible solution. Since x ≥ 2, the answer is yes.

No unique answer. Insufficient.

2) The child used 5 of one type of coin.

Case 1: Say x = 5

Since 5x + 2y + z = 17 and 5x = 5*5 = 25 > 17, x cannot be 5. Not a valid case.

Case 2: Say y = 5

Since 5x + 2y + z = 17 and 5x + 2*5 + z = 17 => 5x + z = 7

Case 2.a: Say x = 1; thus, from 5x + z = 7, we have 5x + z = 7 => 5*1 + z = 7 => z = 2. Since y and z are positve integers, it's a possible solution. Since x < 2, the answer is no.

x cannot be 2 since 5x = 5*2 = 10 > 7.

Case 3: Say z = 5

Since 5x + 2y + z = 17 and 5x + 2y + 5 = 17 => 5x + 2y = 12

Case 3.a: Say x = 2; thus, from 5x + 2y = 12, we have 5*2 + 2y = 12 => y = 1. Since y and z are positve integers, it's a possible solution. Since x ≥ 2, the answer is yes.

x cannot be 3 since 5x = 5*3 = 15 > 12.

No unique answer. Insufficient.

(1) and (2) together

Number of coins as per Case 2a = x + y + z = 1 + 5 + 2 = 8, a valid case as per Statement 1 condition

Since x < 2, the answer is no.

Number of coins as per Case 3a = x + y + z = 2 + 1 + 5 = 8, a valid case as per Statement 1 condition

Thus, only Case 2.a is valid and we have x < 2. Sufficient.

Since x ≥ 2, the answer is yes.

No unique answer. Insufficient.

The correct answer: E

Hope this helps!

-Jay
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