Question

the average of the first 3 of 4 numbers is 16 and that of the last 3 is 15.if the sum of the of the first and last nuber is 13, what is the last nuber?​

Answer 1

let a, b, c and d be the four numbers

now,

$\frac{a + b + c}{3} = 16 \\ a + b + c = 48 \\ b + c = 48 - a\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i) \\ \\ \frac{b + c + d}{3} = 15 \\ b + c + d = 45 \\ 48 - a + d = 45 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: from \: (i) \\ 48 - (a - d) = 45 \\ a - d = 3 \\ \\ but \\ a + d = 13 \\ \\ (a - d) + (a + d) = 3 + 13 \\ 2a = 16 \\ a = 8 \\ then \\ d = 5$

Hence, the last number is 5