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The costs of equities of type A and type B (in dollars) are two differ

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The costs of equities of type A and type B (in dollars) are two differ [#permalink]

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New post 18 May 2017, 13:55
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The costs of equities of type A and type B (in dollars) are two different positive integers. What is the total cost of 2 equities of type A and 3 equities of type B?

(1) 4 type A equities and 5 type B equities together cost 27 dollars.
(2) Cost of a type A equity plus the Cost of a type B equity is 6 dollars.

Source: Nova GMAT
[Reveal] Spoiler: OA

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Re: The costs of equities of type A and type B (in dollars) are two differ [#permalink]

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New post 19 May 2017, 09:58
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Find out 2A+3B?

A = Cost of 1 Equity of Type A
B = Cost of 1 Equity of Type B

(1) 4 type A equities and 5 type B equities together cost 27 dollars.

4A+5B = 27

The only possible values of A & B which satisfy the above is (3,3)
Therefore, 2A+3B = 6+9 = 15
This statement is sufficient.



(2) Cost of a type A equity plus the Cost of a type B equity is 6 dollars.

This statement is not sufficient, as there are mutiple combinations of A and B

Answer is A
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Re: The costs of equities of type A and type B (in dollars) are two differ [#permalink]

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New post 19 May 2017, 11:33
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(May I please suggest that you have a look at the question. The question gives the right answer similar to OA when you remove the word 'different' from the main question stem. Instead of saying A and B are two 'different' positive integers, the question should say that A and B are two positive integers. Or maybe I am missing something)

Let cost per equity of type A be A, and cost per equity of type B be B.
We are given that both A and B are positive integers. And we have to find 2A + 3B

Statement 1. 4A + 5B = 27.
We should not just assume that one equation will not give us any solution - we need to consider the fact that both A and B are positive integers and then do some trial/error to see which all positive integer values of A & B will satisfy the above equation.

We can write 4A = 27-5B or A = (27-5B)/4
Start substituting positive integer values of B and write down the cases where A also comes to be a positive integer.
for B=1, A=22/4 (not integer)
for B=2, A=17/4 (not integer)
for B=3, A=12/4 = 3 (integer)
for B=4, A=7/4 (not integer)
for B=5, A=2/4 (not integer)

No need to substitute further values because if B>5, A will be negative which is not the given case.
So this statement gives ONLY one set of positive integer values of A & B which are 3 & 3 respectively.
We can put this is 2A + 3B to get our answer. Sufficient

Statement 2. A+B = 6
Not sufficient to find 2A+3B.

Hence answer is A.
Re: The costs of equities of type A and type B (in dollars) are two differ   [#permalink] 19 May 2017, 11:33
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