Answer
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Hint: We first find the sum of 100 numbers, subtract wrongly read numbers i.e., 192 & 33 from them and then we add correct numbers 92 & 83 to them and find the arithmetic mean
Complete step by step answer:
Arithmetic mean of (say) 5 numbers a, b, c, d, e is calculated by the expression $\dfrac{a+b+c+d+e}{5}=\dfrac{\text{sum}\ \text{of}\ \text{all}\ \text{the}\ \text{numbers}\ }{\text{No}\ \text{of}\ \text{numbers}}$
Let the 100 numbers are named as ${{a}_{1}},\ {{a}_{2}},\ {{a}_{3}},...............,\ {{a}_{100}}$
It is given that, arithmetic mean of 100 numbers are 89.05
i.e. $\dfrac{{{a}_{1}}+{{a}_{2}}+..............+{{a}_{100}}}{100}=89.05$
$\therefore {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.............{{a}_{100}}=89.05\times 100=8905$
It is given that two numbers 92 and 93 have been missed as 192. and 33 respectively
Let ${{a}_{99}}=192\ \text{and}\ {{a}_{98}}=33$ (you can take any two numbers as 192. & 33)
Now,
we subtract ${{a}_{99}}\ \text{and}\ {{a}_{98}}$ from total sum of 100 numbers
i.e. sum of remaining 98 numbers are $8905-192-33=8680$
Sum of remaining 98 numbers are 8680
Now, we add the correct numbers which are unfortunately missed to the sum of 98 numbers and then we find the arithmetic of correct 100 numbers.
i.e. Sum of 100 correct numbers $=8680+92+83$
$=8855$
Correct Arithmetic mean of 100 numbers
= sum of 100 correct numbers /100
=8855/100
= 88.55
Therefore, Option A is correct.
Note: In the given problem, we have to find the correct arithmetic mean of the numbers. We first need to subtract wrongly read numbers from the sum of total 100 numbers and then we have to add the correct numbers which are wrongly read to the sum of 100 numbers and then we have to find the correct arithmetic mean of those 100 numbers. You must be very careful in doing this as you might not subtract wrongly read numbers from the total sum of 100 numbers.
Complete step by step answer:
Arithmetic mean of (say) 5 numbers a, b, c, d, e is calculated by the expression $\dfrac{a+b+c+d+e}{5}=\dfrac{\text{sum}\ \text{of}\ \text{all}\ \text{the}\ \text{numbers}\ }{\text{No}\ \text{of}\ \text{numbers}}$
Let the 100 numbers are named as ${{a}_{1}},\ {{a}_{2}},\ {{a}_{3}},...............,\ {{a}_{100}}$
It is given that, arithmetic mean of 100 numbers are 89.05
i.e. $\dfrac{{{a}_{1}}+{{a}_{2}}+..............+{{a}_{100}}}{100}=89.05$
$\therefore {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.............{{a}_{100}}=89.05\times 100=8905$
It is given that two numbers 92 and 93 have been missed as 192. and 33 respectively
Let ${{a}_{99}}=192\ \text{and}\ {{a}_{98}}=33$ (you can take any two numbers as 192. & 33)
Now,
we subtract ${{a}_{99}}\ \text{and}\ {{a}_{98}}$ from total sum of 100 numbers
i.e. sum of remaining 98 numbers are $8905-192-33=8680$
Sum of remaining 98 numbers are 8680
Now, we add the correct numbers which are unfortunately missed to the sum of 98 numbers and then we find the arithmetic of correct 100 numbers.
i.e. Sum of 100 correct numbers $=8680+92+83$
$=8855$
Correct Arithmetic mean of 100 numbers
= sum of 100 correct numbers /100
=8855/100
= 88.55
Therefore, Option A is correct.
Note: In the given problem, we have to find the correct arithmetic mean of the numbers. We first need to subtract wrongly read numbers from the sum of total 100 numbers and then we have to add the correct numbers which are wrongly read to the sum of 100 numbers and then we have to find the correct arithmetic mean of those 100 numbers. You must be very careful in doing this as you might not subtract wrongly read numbers from the total sum of 100 numbers.
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