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. 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. 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.

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.

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. 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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