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There are 3 kinds of gifts A, B and C in a box. The number

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There are 3 kinds of gifts A, B and C in a box. The number

by Max@Math Revolution » Tue Dec 17, 2019 11:38 pm
[GMAT math practice question]

There are 3 kinds of gifts A, B and C in a box. The number of gifts A, B and C are a, b and c, respectively. The prices of A, B and C are $3, $2 and $1. The total price of all gifts in the box is $48. What is the total price of gift A?

1) a < b < c
2) a, b and c are all even numbers.
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by SampathKp » Wed Dec 18, 2019 2:58 am
Max@Math Revolution wrote:[GMAT math practice question]

There are 3 kinds of gifts A, B and C in a box. The number of gifts A, B and C are a, b and c, respectively. The prices of A, B and C are $3, $2 and $1. The total price of all gifts in the box is $48. What is the total price of gift A?

1) a < b < c
2) a, b and c are all even numbers.
A B C
a . b . c
$3 . $2 . $1
Total price of all gifts in a box is $ 48
i.e.

3a+2b+c = 48 is the equation given in the question

Now in (1) Take a=1, b=2 and get value of c in the above equation c= 48- (3+4) = 41
Take a=2, B=3 to get values of c in the above equation c = 48- (6+6) = 36

So we will get multiple values of a satisfying given equation. Clearly (1) by itself is not sufficient to answer the question.

Now in (2) Take a=2, b=2 to get value of c from above equation c = 48 - (6+4) = 40
take a=2, b=4 , then c =34

So we will get multiple values of a satisfying the given equation. Clearly (2) by itself is not sufficient to answer the question.

Combining (1) and (2) a<b<c and a, b and c are all even numbers check for a=2, b=4 to get value of c in the equation c =34.
take a=2, b=6 , c = 30. Clearly combining (1) and (2) is not sufficient to answer the question.

Hence Answer is E, both statement together are NOT sufficient to answer the question.
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by Max@Math Revolution » Thu Dec 19, 2019 10:44 pm
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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.

We have 3a + 2b + c = 48 from the original condition.

Since we have 3 variables (a, b and c) and 1 equation, 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)
If a = 2, then we have 2b + c = 42 and we have c = 34 when we have b = 4.
If a = 4, then we have 2b + c = 36 and we have c = 24 when we have b = 6.
Thus, a =2, b = 4, c = 34 and a = 4, b = 6, c = 24 are solutions.

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the answer.
Answer: E

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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