Out of four numbers,the average of the first | vedclass

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(B) Let the $4$ numbers be $a, b, c, d$.The average of the first three numbers $(a, b, c)$ is $16$,so:$(a + b + c) / 3 = 16$$a + b + c = 48$  --- (Equ

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(B) Let the $4$ numbers be $a, b, c, d$.The average of the first three numbers $(a, b, c)$ is $16$,so:$(a + b + c) / 3 = 16$$a + b + c = 48$  --- (Equation $1$)The average of the last three numbers $(b, c, d)$ is $15$,so:$(b + c + d) / 3 = 15$$b + c + d = 45$  --- (Equation $2$)Subtracting Equation $2$ from Equation $1$:$(a + b + c) - (b + c + d) = 48 - 45$$a - d = 3$Given that the last number $d = 20$,substitute this value into the equation:$a - 20 = 3$$a = 23$Therefore,the first number is $23$.

Out of four numbers,the average of the first three is $16$ and that of the last three is $15$. If the last number is $20$,then the first number is:

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