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The only contents of a parcel are 25 photographs and 30 negatives. Wha 10 Feb 2014, 00:32
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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 1/3 ounce.
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The only contents of a parcel are 25 photographs and 30 negatives. Wha 10 Feb 2014, 00:32
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SOLUTION

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 --> the total weight = 25*(3n) + 30n. Not sufficient.

(2) The total weight of 1 of the photographs and 2 of the negatives is 1/3 ounce --> p + 2n = 1/3. Not sufficient.

(1)+(2) We have two distinct linear equations with two unknowns: p = 3n and p + 2n = 1/3, thus we can solve for p and n and get the value of 25*(3n) + 30n. Sufficient.

Answer: C.
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The only contents of a parcel are 25 photographs and 30 negatives. Wha Updated on: 10 Feb 2014, 17:27
Originally posted by WoundedTiger on 10 Feb 2014, 10:32.
Last edited by WoundedTiger on 10 Feb 2014, 17:27, edited 1 time in total.
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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.

Sol: let the weight of 1 photograph be a ounce and that of negatives be b ounce so we need to find 25a+30b

St1 says a=3b is not enough cause weight of negative can be anything so not sufficient. Ans choice A &D ruled out

St2 says a+2b=1 ounce. It's difficult to get individual values

On combining we get a=3b and a+2b=1

We can solve for a and b

So ans is C

Just for information cause we need not find values
5b=1 so b=1/5 ounce and a=1/15 ounce
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The only contents of a parcel are 25 photographs and 30 negatives. Wha 10 Feb 2014, 16:11
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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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The only contents of a parcel are 25 photographs and 30 negatives. Wha 10 Feb 2014, 16:37
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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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The only contents of a parcel are 25 photographs and 30 negatives. Wha 29 Jun 2017, 14:53
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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.

\(p = 3n\)

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

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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The only contents of a parcel are 25 photographs and 30 negatives. Wha 09 May 2024, 09:17
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What if the question was 25 photographs and 50 negatives? Would statement two be sufficient then? Because for statement two to be sufficient we would have to assume all the photographs weigh the same and all the negatives weigh the same? so is it safe to assume so?­
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