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Andrew bought pizzas for his swim team. Pepperoni pizzas cos

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Andrew bought pizzas for his swim team. Pepperoni pizzas cos

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Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and combination pizzas cost $17 and he bought only pepperoni or combination pizzas. He spent a total of $184 on pizzas. How many pizzas did he buy?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16

OA A

Source: Princeton Review

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BTGmoderatorDC wrote:
Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and combination pizzas cost $17 and he bought only pepperoni or combination pizzas. He spent a total of $184 on pizzas. How many pizzas did he buy?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16

OA A

Source: Princeton Review
Letting p = the number of pepperoni pizzas bought and c = the number of combination pizzas bought, we can create the equation:

13p + 17c = 184

13p = 184 - 17c

p = (184 - 17c)/13

We can see we can buy at most 10 combination pizzas. (If the value of c were more than 10, then the number of pepperoni pizzas bought would be negative.)

If c = 10, then p = (184 - 170)/13 = 14/13, which is not an integer.

If c = 9, then p = (184 - 153)/13 = 31/13, which is not an integer.

If c = 8, then p = (184 - 136)/13 = 48/13, which is not an integer.

If c = 7, then p = (184 - 119)/13 = 65/13 = 5, which is an integer.

Therefore, a total of 7 + 5 = 12 pizzas are bought.

Answer: A

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scott@targettestprep.com



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BTGmoderatorDC wrote:
Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and combination pizzas cost $17 and he bought only pepperoni or combination pizzas. He spent a total of $184 on pizzas. How many pizzas did he buy?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16
13x + 17y = 184.

Multiples of 13:
13, 26, 39, 52, 65, 78, 91, 104...
Multiple of 17:
17, 34, 51, 68, 85, 102, 119....

The values in blue sum to 184.
Since 65/13 = 5 and 119/17 = 7, the number of pizzas = 5+7 = 12.

The correct answer is A.

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