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Re: The only contents of a parcel are 25 photographs and 30 negatives. Wha [#permalink]
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The only contents of a parcel are 25 photographs and 30 negatives. What is the total weight, in ounces, of the parcel's contents?
(1) The weight of each photograph is 3 times the weight of each negative.
(2) The total weight of 1 of the photographs and 2 of the negatives is i ounce.

What is 25P+30N?
Statement 1: 1P=3N. Not sufficient
Statement 2: 1P+2N = 1 ounce. Not sufficient

C. We can solve using both statements.
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Re: The only contents of a parcel are 25 photographs and 30 negatives. Wha [#permalink]
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Theres a little mistake. St(2) should say: The total weight of 1 of the photographs and 2 of the negatives is 1/3 ounce.

Lets call the weight of the photographs "P" and negatives "N"

St(1) dosent give us much. Just that \(P=3N\). Insufficient.
St(2) tells us that \(P+2N=1/3\). Also Insufficient.

Putting them together we get: \(3N+2N=1/3\)
So we can find N and P

Ans is C.
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Re: The only contents of a parcel are 25 photographs and 30 negatives. Wha [#permalink]
The only contents of a parcel are 25 photographs and 30 negatives. What is the total weight, in ounces, of the parcel's contents?

(1) The weight of each photograph is 3 times the weight of each negative.

\(p = 3n\)

\(25(3n) + 30n = 75n + 30n = 105n\)

As we are not aware of the value of n we cannot determine the total weight.

Hence, (1) =====> is NOT SUFFICIENT

(2) The total weight of 1 of the photographs and 2 of the negatives is 1/3 ounce.

\(1p + 2n = \frac{1}{3}\)

We still cannot determine the value of total weight as we are not aware of the values of p & n

Hence, (2) =====> is NOT SUFFICIENT

Combining (1) & (2)

Substituting the value obtained from (1) in (2) should give us total weight.

Hence, cobnined (1) & (2) =====> is SUFFICIENT

Hence, Answer is C
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Re: The only contents of a parcel are 25 photographs and 30 [#permalink]
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Re: The only contents of a parcel are 25 photographs and 30 [#permalink]
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