d The average of 9 observations was 9

# Arithmetic MCQ 10 (Average observation)

Arithmetic MCQ 10 : The average of 9 observations was 9, that of the 1st  5 being 10 and that of the last 5 being 8. What was the 5th observation?

a) 9                 b) 8                              c) 7                              d) 6

Solution:

Let the observations are ​$$x_1, x_2,x_3,x_4,x_5,x_6,x_7,x_8 and x_9$$​. It is given that their average is 9.

Since average is taken by taking sum of entries divided by number of entries,

Hence,

$9=\frac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8 + x_9}{9}$

This gives;

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8 + x_9=81 —–(1)$

Given that average of first 5 entries is 10 i-e;

$10=\frac{x_1+x_2+x_3+x_4+x_5}{5}$

or

$x_1+x_2+x_3+x_4+x_5=50 ——(2)$

Also given that average of last five observations is 8 i-e;

$8= \frac{x_5+x_6+x_7+x_8 + x_9}{5}$

or

$x_5+x_6+x_7+x_8 + x_9=40—-(3)$

from eq 2 we get;

$x_1+x_2+x_3+x_4=50-x_5 —(A)$

from eq 3 we get;

$x_6+x_7+x_8 + x_9=40-x_5 —- (B)$

Putting values from eq A and B in eq 1;

$(x_1+x_2+x_3+x_4)+x_5+(x_6+x_7+x_8 + x_9)=81$

$(50-x_5)+x_5+(40-x_5)=81$

Removing brackets we get ;

$50-x_5+x_5+40-x_5=81$

Simplifying we get;

$90-x_5=81$

Hence,

$x_5 = 90-81=9$

So the fifth observation is 9 — option  a is correct !

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