Arithmetic

Mean, Median & Mode

Medium

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

Entered answer:

✅ Correct Answer: 70

When we have three numbers and need to track changes, the key is to set up clear variables and use the given conditions systematically.

We call our three distinct numbers $a$ , $b$ , and $c$ where $a < b < c$ (smallest to largest).

The average of three numbers is 28, so:

$\frac{a + b + c}{3} = 28$

$a + b + c = 84$

This gives us our first equation connecting all three numbers.

After the changes:

New smallest = $a + 7$ (original smallest increased by 7)

New middle = $b$ (middle number stays the same)

New largest = $c - 10$ (original largest reduced by 10)

The problem states "the order remains unchanged" - this means $a + 7 < b < c - 10$ , confirming our operations don't change which number is smallest, middle, or largest.

The new average equals the middle number plus 2:

$\frac{(a + 7) + b + (c - 10)}{3} = b + 2$

$\frac{a + 7 + b + c - 10}{3} = b + 2$

$\frac{a + b + c - 3}{3} = b + 2$

Since we know $a + b + c = 84$ :

$\frac{84 - 3}{3} = b + 2$

$\frac{81}{3} = b + 2$

$27 = b + 2$

$b = 25$

Since $b = 25$ and $a + b + c = 84$ :

$a + 25 + c = 84$

$a + c = 59$ ...(Equation I)

The difference between the new largest and new smallest is 64:

$(c - 10) - (a + 7) = 64$

$c - 10 - a - 7 = 64$

$c - a - 17 = 64$

$c - a = 81$ ...(Equation II)

We have:

Equation I: $a + c = 59$

Equation II: $c - a = 81$

$(a + c) + (c - a) = 59 + 81$

$2c = 140$

$c = 70$

Substituting back into Equation I:

$a + 70 = 59$

$a = -11$

Therefore, the largest number in the original set is 70.

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