Question

The mean of eight numbers in a set is 41. The mean of two numbers in the set is 29. Calculate the mean of the remaining 6 numbers

Answer 1

Step-by-step explanation:

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$\large\pink{ {\boxed { Question :-}}}$

⊙ The mean of eight numbers in a set is 41. The mean of two numbers in the set is 29. Calculate the mean of the remaining 6 numbers.

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$\Huge\boxed {{ \orange{Answer :-}}}$

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$\implies {a+b+c+d+e+f+g+h}=328$

$\implies\frac{a + b + c + d + e + f + g + h }{8} =41$

$\implies \frac{g + h}{2}=29$

$g + h=58$

$So  \implies a+b+c+d+e+f+g+h=328 Becomes...$

$\implies \: a+b+c+d+e+f+58=328$

$\implies \: a+b+c+d+e+f=270$

$\implies \: Mean \: of \: these \: six \: numbers...$

$\implies \frac{a + b + c + d + e + f + g}{6}= \frac{270}{6}= \pink{\boxed {45}}$

$\green{ \boxed{ \implies{45}}}$

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Explanation:

The mean is that single value that if you multiply it by the count of values used to derive it you end up with their sum.

8 values with mean of 41

8 x 41 = 328 <-sum of all values.

2 values with mean of 29

2 x 29 = 58

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So the remaining 6 values must sum to 328 - 58 = 270

Thus the mean of the remaining 6 is 270÷6=45

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