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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Arman Akram
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Anonymous
I am a science graduate with bachelor degree in mathematic and physics as major subjects and MSc in mathematics!
But I went through the old css papers of the said subjects, it is difficult to score high marks, So, I am looking for some other good scoring subjects!