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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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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.
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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 06 Jun 2017, 15:20
SajjadAhmad wrote:
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


1)
When there are problems involving integer quantities and the equation with proportion and total value is provided, the first test to do is identify the couple of factors:

4A + 5B = 27
4 5
8 10
12 15
16 20
20 25
24

Testing all the possible combination, the only possible solution is 4a = 12 and 5b=15 --> now I know a and b --> 1 IS SUFFICIENT

2) not suff
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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 06 Jul 2017, 17:03
SajjadAhmad wrote:
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


We can let A = the cost of equity A and B = the cost of equity B. We must determine the value of 2A + 3B.

Statement One Alone:

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

We can create the following equation:

4A + 5B = 27

4A = 27 - 5B

A = (27 - 5B)/4

Since A must be an integer, we need to determine what B must be to make 27 - 5B a multiple of 4.

We see that the only value for B that will allow 27 - 5B to be a multiple of 4 is when B = 3. Thus:

A = (27 - 5(3))/4

A = (27 - 15)/4

A = 12/4

A = 3

Since we have determined that A = 3 and B = 3, statement one is sufficient to answer the question.

Statement Two Alone:

The cost of a type A equity plus the cost of a type B equity is 6 dollars.

We can create the following equation:

A + B = 6

Since we cannot determine a value for A or B, statement two alone is not sufficient to answer the question.

Answer: 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 07 Jul 2017, 22:55
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SajjadAhmad wrote:
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


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Let \(a\) and \(b\) be the costs of equities of type A and B, respectively.
The question asks what the value of \(2a + 3b\) is.

There are 2 variables and 0 equation. By Variable Approach method, C is the answer most likely. However, since this question is about the area of integers, which is one of the key question areas, we need to check A or B by Common Mistake Type 4A . Key question areas are integers, statistics, inequalities, absolute values and probability.

Condition 1) : \(4a + 5b = 27\).
Since \(a\) and \(b\) are positive integers, we have a unique solution \(a = 3\) and \(b = 3\).
Thus this condition is sufficient.

Condition 2) : \(a + b = 6\).
This equation has many solutions, for example, \(a = 1\), \(b = 6\) or \(a = 2\), \(b = 5\).
Thus this condition is not sufficient.

Therefore, the answer is A.


Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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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 08 Jul 2017, 16:59
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I may understand the problem incorrectly, but it is mentioned that the costs are two different positive integers, while the solution is true only when the costs are 3 and 3.
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The costs of equities of type A and type B (in dollars) are two differ   [#permalink] 08 Jul 2017, 16:59
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