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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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