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Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and

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Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and  [#permalink]

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New post 17 May 2018, 12:07
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Question Stats:

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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

Any different approach on this apart from hit and trial. Please help.
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Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and  [#permalink]

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New post Updated on: 18 May 2018, 07:42
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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

Pepperoni = P
Combo = C

13P + 17C = 184

P & C each have to be integers. The max C can be is 10 since anything more would give value for 17C greater than 184. So I would find corresponding values of P starting from C = 10, C = 9, C = 8...etc and stop once I get to a pair that works.


C = 10 ---> 13P = 14 ---> P isn't an integer

C = 9 -----> 13P = 21 ----> P isn't an integer

C = 8 ---> 13P = 48 ----> P isn't an integer

C = 7 ---> 13P = 65 -----> P = 5 ---> WE DONE.

C = 7
P = 5
----> C + P = 12

Answer: A
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Originally posted by arosman on 17 May 2018, 14:16.
Last edited by arosman on 18 May 2018, 07:42, edited 1 time in total.
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Re: Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and  [#permalink]

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New post 18 May 2018, 07:00
SidR wrote:
Any different approach on this apart from hit and trial. Please help.


We have only one equation with 2 variables. The added information that helps find a unique value for number of Pepperoni and Combination pizza is the fact that both numbers have to be non-negative integers.

The only way to find the answer is through iteration as shown by arosman
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Re: Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and  [#permalink]

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New post 08 Jun 2018, 08:52
SidR

Question mentions "he bought only pepperoni or combination pizzas."
Means only one category
And 184 is divided neither by 13 nor 17.so that's not possible

Could you please check question and confirm whether my understanding is correct or am I missing something

Posted from my mobile device
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Re: Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and  [#permalink]

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New post 08 Jun 2018, 09:38
push12345 wrote:
SidR

Question mentions "he bought only pepperoni or combination pizzas."
Means only one category
And 184 is divided neither by 13 nor 17.so that's not possible

Could you please check question and confirm whether my understanding is correct or am I missing something

Posted from my mobile device


The question isn't saying that either he bought ALL pepperoni or he bought ALL combination pizzas. It's saying that each pizza is one of either pepperoni or combination.
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Re: Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and  [#permalink]

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New post 04 Oct 2018, 19:47
SidR 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


Letting p = the number of pepperoni pizzas bought and c = the number of combination pizzas bought, wee 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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Re: Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and  [#permalink]

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New post 05 Oct 2018, 15:20
Hi All,

We're told that Andrew bought pizzas for his swim team; pepperoni pizzas cost $13 each, combination pizzas cost $17 each and he bought only pepperoni or combination pizzas (spending a total of $184 on pizzas). We're asked for the TOTAL number of pizzas he bought. This question can be solved in a number of different ways, but it has a great 'pattern-matching' shortcut that get you to the correct answer without too much math.

To start, it's worth noting that the combined cost of 1 pepperoni pizza and 1 combination pizza is $13 + $17 = $30. With a total of $184 of pizzas, (6)($30) = $180 is fairly close to that total. Notice that it's $4 less than what we need it to be... which is exactly the difference in price between a combination pizza and a pepperoni pizza.

So, if we buy 6 pepperoni pizzas and 6 combination pizzas, we'll spend $180 in total. If we 'swap' one pepperoni pizza for one combination pizza, then we'll end up spending 4 MORE dollars, for a total of $184. That would be 5 pepperoni pizzas and 7 combination pizzas --> a total of 12 pizzas.

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Re: Andrew bought pizzas for his swim team. Pepperoni pizzas cost $13 and &nbs [#permalink] 05 Oct 2018, 15:20
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