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# The only contents of a parcel are 25 photographs and 30 nega

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Math Expert
Joined: 02 Sep 2009
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The only contents of a parcel are 25 photographs and 30 nega  [#permalink]

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10 Feb 2014, 00:32
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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.

Data Sufficiency
Question: 84
Category: Algebra Simultaneous equations
Page: 158
Difficulty: 600

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Re: The only contents of a parcel are 25 photographs and 30 nega  [#permalink]

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10 Feb 2014, 00:32
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.

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Re: The only contents of a parcel are 25 photographs and 30 nega  [#permalink]

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Updated on: 10 Feb 2014, 17:27
1
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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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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Re: The only contents of a parcel are 25 photographs and 30 nega  [#permalink]

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10 Feb 2014, 16:11
1
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 nega  [#permalink]

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10 Feb 2014, 16:37
1
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 nega  [#permalink]

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17 Feb 2014, 01:43
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.

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Re: The only contents of a parcel are 25 photographs and 30 nega  [#permalink]

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29 Jun 2017, 14:53
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

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Re: The only contents of a parcel are 25 photographs and 30 nega &nbs [#permalink] 29 Jun 2017, 14:53
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