Question 62

The average of a set of 15 numbers is 41. Five numbers 21, 24, 39, 47 and 97 are removed from the above set. What will the average of the remaining 10 numbers?

Solution

The average of a set of 15 numbers is 41.

Sum of a set of 15 numbers = $$41\times15$$

= 615 Eq.(i)

Five numbers 21, 24, 39, 47 and 97 are removed from the above set.

Sum of these five removed numbers from the set = 21+24+39+47+97

= 228 Eq.(ii)

Sum of the remaining 10 numbers = Eq.(i)-Eq.(ii)

= 615-228

= 387

Average of the remaining 10 numbers = $$\frac{387}{10}$$

= 38.7

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