Data Sufficiency

[Math Revolution GMAT math practice question]

a, b, c, d, e and f are integers. Is their median greater than their average (arithmetic mean)?

1) a < b < c < d < e < f
2) b-a=d-c=f-e

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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.

Since we have 6 variables (x, y and z) and 0 equations, E is most likely to be 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.

If a = 1, b = 2, c =5, d = 6, e = 7, f = 8, we have a median of 5.5 and an average of 29/6. The median is greater than the average, and the answer is "yes".

If a = 1, b = 2, c =3, d = 4, e = 5, f = 6, we have a median of 3.5 and an average of 3.5. The median is not greater than the average, and the answer is "no".

Since we don't have a unique solution, both conditions are not sufficient, when considered together.
.

Therefore, the answer is E.

Answer: E

In cases where 3 or more additional equations are required, such as for original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

Question : Is their median greater than their arithmetic mean (average).

Statement 1 = a<b<c<d<e<f
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
$$Average\ =\ \frac{Efx}{Ef\ }\ =\ \frac{\left(1\ +2\ +3\ +4\ +5\ +6\right)}{6}\ =\ \frac{21}{6\ }\ =\ 3.5$$
$$Median\ =\ \frac{\left(3\ +\ 4\ \right)}{2}\ =\ \frac{7}{2}\ =\ 3.5$$
The median is not greater than the average.
Statement 1 is NOT SUFFICIENT.

Statement 2 : b-a = d-c = f-e
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
2 - 1 = 4 - 3 = 6 - 5
1 = 1 = 1
= 1
Statement 2 is NOT SUFFICIENT.

In combining the 2 statement together we still have the same equation because there is no unique solution.
Thus both statement are not SUFFICIENT when considered together.
Option E is CORRECT.

deloitte247 wrote:Question : Is their median greater than their arithmetic mean (average).

Statement 1 = a<b<c<d<e<f
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
$$Average\ =\ \frac{Efx}{Ef\ }\ =\ \frac{\left(1\ +2\ +3\ +4\ +5\ +6\right)}{6}\ =\ \frac{21}{6\ }\ =\ 3.5$$
$$Median\ =\ \frac{\left(3\ +\ 4\ \right)}{2}\ =\ \frac{7}{2}\ =\ 3.5$$
The median is not greater than the average.
Statement 1 is NOT SUFFICIENT.

Statement 2 : b-a = d-c = f-e
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
2 - 1 = 4 - 3 = 6 - 5
1 = 1 = 1
= 1
Statement 2 is NOT SUFFICIENT.

In combining the 2 statement together we still have the same equation because there is no unique solution.
Thus both statement are not SUFFICIENT when considered together.
Option E is CORRECT.

Math Revolution provides VA method for DS questions.
VA suggests we should check both conditions together first, when we need 3 or more equations, which means E is most likely to be the answer.
If both conditions together are not sufficient, just choose E.
We don't need to check each of conditions one by one again. It is just time consuming.

deloitte247 wrote:Question : Is their median greater than their arithmetic mean (average).

Statement 1 = a<b<c<d<e<f
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
$$Average\ =\ \frac{Efx}{Ef\ }\ =\ \frac{\left(1\ +2\ +3\ +4\ +5\ +6\right)}{6}\ =\ \frac{21}{6\ }\ =\ 3.5$$
$$Median\ =\ \frac{\left(3\ +\ 4\ \right)}{2}\ =\ \frac{7}{2}\ =\ 3.5$$
The median is not greater than the average.
Statement 1 is NOT SUFFICIENT.

Statement 2 : b-a = d-c = f-e
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
2 - 1 = 4 - 3 = 6 - 5
1 = 1 = 1
= 1
Statement 2 is NOT SUFFICIENT.

In combining the 2 statement together we still have the same equation because there is no unique solution.
Thus both statement are not SUFFICIENT when considered together.
Option E is CORRECT.

Math Revolution provides VA method for DS questions.
VA suggests we should check both conditions together first, when we need 3 or more equations, which means E is most likely to be the answer.
If both conditions together are not sufficient, just choose E.
We don't need to check each of conditions one by one again. It is just time consuming.

deloitte247 wrote:Question : Is their median greater than their arithmetic mean (average).

Statement 1 = a<b<c<d<e<f
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
$$Average\ =\ \frac{Efx}{Ef\ }\ =\ \frac{\left(1\ +2\ +3\ +4\ +5\ +6\right)}{6}\ =\ \frac{21}{6\ }\ =\ 3.5$$
$$Median\ =\ \frac{\left(3\ +\ 4\ \right)}{2}\ =\ \frac{7}{2}\ =\ 3.5$$
The median is not greater than the average.
Statement 1 is NOT SUFFICIENT.

Statement 2 : b-a = d-c = f-e
If a = 1; b = 2; c = 3; d = 4; e = 5; f = 6.
2 - 1 = 4 - 3 = 6 - 5
1 = 1 = 1
= 1
Statement 2 is NOT SUFFICIENT.

In combining the 2 statement together we still have the same equation because there is no unique solution.
Thus both statement are not SUFFICIENT when considered together.
Option E is CORRECT.

Math Revolution provides VA method for DS questions.
VA suggests we should check both conditions together first, when we need 3 or more equations, which means E is most likely to be the answer.
If both conditions together are not sufficient, just choose E.
We don't need to check each of conditions one by one again. It is just time consuming.