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The mean of 8 numbers is 25. If 5 is subtracted from each number, what will be the new mean?

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: Assume the number to be a, b, c, d, e, f, g, h means this number is given. Then subtract 5 from each of these 8 numbers i.e. (a-5), (b-5) ……. (h-5) divide these new numbers formed by 8 to find the mean of the new number.

Complete step-by-step answer:
Mean of any number is given by,
\[\text{Mean=}\dfrac{a+b+c+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. n}}{n}\]
Mean is the average value of these 'n' numbers. Here, we are given with the mean of 8 numbers.
Let the number be a, b, c, d, e, f, g and h.
Mean of these 8 number would be,
\[\dfrac{a+b+c+d+e+f+g+h}{8}\]
Which is equal to 25, as per the question, so,
\[\dfrac{a+b+c+d+e+f+g+h}{8}=25\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (1)}\]
Naming it as equation (1)
Now, we are told to subtract each of these numbers by 5. So, the new number formed will be:
(a-5), (b-5), (c-5), (d-5), (e-5), (f-5), (g-5), (h-5)
Mean of these 8 new number will be:
\[\dfrac{\left( \text{a}-\text{5} \right)+\left( \text{b}-\text{5} \right)+\left( \text{c}-\text{5} \right)+\left( \text{d}-\text{5} \right)+\left( \text{e}-\text{5} \right)+\left( \text{f}-\text{5} \right)+\left( \text{g}-\text{5} \right)+\left( \text{h}-\text{5} \right)}{8}\]
This equation can be split into two: only variable and only numeral,
\[\begin{align}
  & \Rightarrow \dfrac{a+b+c+d+e+f+g+h-\left( 5+5+5+5+5+5+5+5 \right)}{8} \\
 & \Rightarrow \dfrac{a+b+c+d+e+f+g+h}{8}-\dfrac{40}{8} \\
\end{align}\]
From equation (1) we know the value of mean of old 8 number i.e. 25
\[\begin{align}
  & \Rightarrow 25-\dfrac{40}{8} \\
 & \Rightarrow 25-5=20 \\
\end{align}\]
Therefore, the mean of 8 new numbers formed by subtracting 5 from each of the old numbers is 20.

Note: Students need not to assume the numbers as done in the solution. Here, one can simply write sum in place of a+b+c+d+e+f+g+h. Students instead of subtracting 5 from each value might subtract from the whole sum just once, which is wrong.