Question

The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is

Accepted Answer

70. Let original numbers be $a < b < c$. Average $= 28 \implies a + b + c = 84$. New numbers: $a' = a + 7$, $b' = b$, $c' = c - 10$. Sum of new numbers $= 84 + 7 - 10 = 81$. New mean $= \frac{81}{3} = 27$. Given new mean is 2 more than middle number $b$: $$27 = b + 2 \implies b = 25$$ New difference $c' - a' = 64$: $$(c - 10) - (a + 7) = 64 \implies c - a = 81$$ Since $b = 25$, $a + c = 84 - 25 = 59$. Solving $c - a = 81$ and $c + a = 59$: $$2c = 140 \implies c = 70$$ $$a = -11$$ Original largest number $c = 70$.

Finding largest number from modified mean conditions

Explanation

Practise this under exam conditions