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Carmen bought several cans of soda. If each can was either a 75–cent

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Carmen bought several cans of soda. If each can was either a 75–cent [#permalink]

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New post 03 Sep 2017, 05:37
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Carmen bought several cans of soda. If each can was either a 75–cent can or a 90–cent can, how many 90–cent cans did Carmen buy?

(1) Carmen bought a total of 5 cans of soda.
(2) The total value of the cans of soda Carmen bought was 450 cents.

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Re: Carmen bought several cans of soda. If each can was either a 75–cent [#permalink]

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New post 03 Sep 2017, 11:45
C as first equation from statement would be x+y=5 which is insufficient as this would have different solutions. The second statement gives the equation 75x+90y=450 or simplfied 5x+3y=30 which again is insufficient. But combining the statements 2 equations and 2 variables give y=5 and hence sufficient.

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Carmen bought several cans of soda. If each can was either a 75–cent [#permalink]

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New post 04 Sep 2017, 11:13
Bunuel wrote:
Carmen bought several cans of soda. If each can was either a 75–cent can or a 90–cent can, how many 90–cent cans did Carmen buy?

(1) Carmen bought a total of 5 cans of soda.
(2) The total value of the cans of soda Carmen bought was 450 cents.


Let x =number of cans of 75

y =number of cans of 90

(1) Carmen bought a total of 5 cans of soda.

It could be:
x =1 and y =4
x =2 and y =3
Clearly Insufficient

(2) The total value of the cans of soda Carmen bought was 450 cents.

75x + 90y = 450

multiple of 10 + multiple of 10 = multiple of 10

1- To make 75x multiple of 10, then x must be even.

2- In DS question, neither conditions contradicts the other. So we can use combination of 5 can to reach answer. (Note: It does not mean that we combine them, rather we use it as clue)

Based on 2 points above, x must be 0 or 2 or 4

\(75 (4) + 90 (1)\neq{450}\).........Eliminate this possibility

\(75 (2) + 90 (3)\neq{450}\).........Eliminate this possibility

\(75 (0) + 90 (5)\neq{450}\).........Target

Sufficient

Answer: B
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Re: Carmen bought several cans of soda. If each can was either a 75–cent [#permalink]

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New post 04 Sep 2017, 18:35
Try x=6 which is even and u get y=0.

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Re: Carmen bought several cans of soda. If each can was either a 75–cent [#permalink]

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New post 30 Oct 2017, 05:06
Mo2men wrote:
Bunuel wrote:
Carmen bought several cans of soda. If each can was either a 75–cent can or a 90–cent can, how many 90–cent cans did Carmen buy?

(1) Carmen bought a total of 5 cans of soda.
(2) The total value of the cans of soda Carmen bought was 450 cents.


Let x =number of cans of 75

y =number of cans of 90

(1) Carmen bought a total of 5 cans of soda.

It could be:
x =1 and y =4
x =2 and y =3
Clearly Insufficient

(2) The total value of the cans of soda Carmen bought was 450 cents.

75x + 90y = 450

multiple of 10 + multiple of 10 = multiple of 10

1- To make 75x multiple of 10, then x must be even.

2- In DS question, neither conditions contradicts the other. So we can use combination of 5 can to reach answer. (Note: It does not mean that we combine them, rather we use it as clue)

Based on 2 points above, x must be 0 or 2 or 4

\(75 (4) + 90 (1)\neq{450}\).........Eliminate this possibility

\(75 (2) + 90 (3)\neq{450}\).........Eliminate this possibility

\(75 (0) + 90 (5)\neq{450}\).........Target

Sufficient

Answer: B



Good question,
i chose B and got it wrong.

Statement B
75x +90y = 450

x=6, y=0
x=0 y=5
hence insufficient.

1+2
x=0 and y=5
hence sufficient.
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Carmen bought several cans of soda. If each can was either a 75–cent [#permalink]

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New post 30 Oct 2017, 05:46
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Carmen bought several cans of soda. If each can was either a 75–cent can or a 90–cent can, how many 90–cent cans did Carmen buy?

(1) Carmen bought a total of 5 cans of soda.
(2) The total value of the cans of soda Carmen bought was 450 cents.
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Carmen bought several cans of soda. If each can was either a 75–cent   [#permalink] 30 Oct 2017, 05:46
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