### Question 2

SSC-CGL 2018 June 4 Shift 1

The average of twelve numbers is 42. The last five numbers have an average of 40 and the first four numbers have an average of 44. The sixth number is 6 less than the fifth number and 5 less than the seventh number. What will be the average of the 5th and 7th numbers?

- 44
- 44.5
- 43
- 43.5

### Solution 1

Sum of first 4 nos. = 4 x (their average)

∴ sum of first four = $\displaystyle 4 \times 44 = 176$

Let 6-th = x

So fifth = (x + 6) and seventh = (x + 5)

Sum of 5th, 6th, 7th is: $\displaystyle 3x + 11$

Sum of last 5 nos. = 5 x (their average)

∴ sum of last five = $\displaystyle 5 \times 40 = 200$

Twelve numbers have been given with an average of 42. So we can use the basic math to write an equation.

$\displaystyle 42 = \frac{176 + (3x + 11) + 200}{12}$

Solving for x we get x = 39, and finally, the fifth as (39 + 5 = 44) and sixth as (39 + 6 = 45)

Hence average of 44 and 45 = 44.5 Answer!

### Solution 2

Use the principle that the sum of deviations of each number from the average is zero.

Take the last five numbers each equal to 40. Sum of deviations from the average (i.e., 42) is (40 - 42) x 5 = -10.

Similarly, the deviations of the first four = (44 - 42) x 4 = +8

Total deviation of the fifth, sixth and seventh numbers should be +2 so that combined deviation of all numbers is 0.

Hence, we should have a total deviation of (x + 6 - 42) + (x - 42) + (x + 5 - 42) = 2

Solving, x = 39. Hence fifth and seventh numbers will be (x + 6) and (x + 5), i.e., 45 and 44, the answer is 44.5.

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### Solution 3

Group like this: first four are 44, fifth = x + 6, sixth x, seventh x + 5, eighth 40, and last four 40.

The average of first and last four balances to 42.

The average of 5-th, 6-th, 7-th, 8-th must also be 42.

$\displaystyle \frac{(x + 6) + x + (x + 5) + 40}{4} $ $\displaystyle = 42$

Whence, x = 39 and finally $\displaystyle 44.5 \:\underline{Ans}$

### Solution 4

Fifth and seventh numbers are (x + 6) and (x + 5), hence consecutive. Their average cannot be a whole number. So (a) and (c) are rejected.

Also, the sum of fifth, sixth and seventh numbers should be even (can you see why?).

Try option (d), i.e, 43.5. If this option is the answer, then fifth is 44, sixth will be 38 and seventh 43. Their sum is odd. Hence, (d) is rejected.

Hence, only (b) is possible!

This Blog Post/Article "(solved)Question 2 SSC-CGL 2018 June 4 Shift 1" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.