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An electronics distributor receives $75 for every DVD player he ships

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An electronics distributor receives $75 for every DVD player he ships  [#permalink]

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New post 18 Sep 2019, 00:33
1
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

81% (01:28) correct 19% (01:31) wrong based on 36 sessions

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An electronics distributor receives $75 for every DVD player he ships and $125 for every television he ships. How many DVD players did he ship last month?

(1) Last month the number of DVD players shipped was 40 less than twice the number of televisions shipped.

(2) Last month the distributor shipped a total value of $24,500 in DVD players and television sets.

Source: McGraw-Hill's GMAT (1-23)

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Re: An electronics distributor receives $75 for every DVD player he ships  [#permalink]

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New post 25 Sep 2019, 09:36
An electronics distributor receives $75 for every DVD player he ships and $125 for every television he ships. How many DVD players did he ship last month?

(1) If d is the number of DVD players shipped and t is the number of TVs shipped, then we get :

d= 2t-40 from (1) ; Can't find d with t unknown.
NOT SUFFICIENT.

(2) The total sale from both DVDs and TVs was $24,500.

We can form a simultaneous equation 75d + 125t = 24,500.
When simplified we get, 3d+5t = 980

However, we can't find d here as well as t is unknown.
NOT SUFFICIENT.

Combining (1) and (2), we can solve the equation by substituting the equation obtained in (1) in the simultaneous equation to arrive at d.

|Ans: (C)
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Re: An electronics distributor receives $75 for every DVD player he ships  [#permalink]

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New post 25 Sep 2019, 13:04
SajjadAhmad wrote:
An electronics distributor receives $75 for every DVD player he ships and $125 for every television he ships. How many DVD players did he ship last month?

(1) Last month the number of DVD players shipped was 40 less than twice the number of televisions shipped.

(2) Last month the distributor shipped a total value of $24,500 in DVD players and television sets.

Source: McGraw-Hill's GMAT (1-23)


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Assume \(d\) is the number of DVD players shipped and \(t\) is the number of TV shipped.

Since we have 2 variables (\(d\) and \(t\)) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
We have \(d = 2t - 40\) from condition 1) and \(75d + 125t = 24500\) from condition 2).
\(75d + 125t = 24500\)
\(⇔ 75(2t-40) + 125t = 24500\)
\(⇔ 275t - 3000 = 24500\)
\(⇔ 275t = 27500\)
\(⇔ t = 100\)
Then we have \(75d + 12500 = 24500\) or \(75d = 12000\) and \(d = 160\).
Since both conditions together yield a unique pair of solution \(t = 100\) and \(d = 160\), they are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since \(d = 2t - 40\) does not yield a unique solution obviously, condition 1) is not sufficient.

Condition 2)
\(75d + 125t = 24500\)

If \(d = 10\), then we have \(5t = 4900 - 30 = 4870\) and \(t = 974\).
If \(d = 20\), then we have \(5t = 4900 - 60 = 4840\) and \(t = 968\).
We found two pairs of solutions \((d,t) = (10,974)\) or \((d,t) = (20,968)\).
Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: An electronics distributor receives $75 for every DVD player he ships   [#permalink] 25 Sep 2019, 13:04
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