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10 Feb 2014, 00:32
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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 1/3 ounce.
(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.
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.
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.
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.
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
(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
