Question

If the average of 9 consecutive positive integers is 55, then what is the largest integer?​

Answer 1

Let the first integer be X, then the largest integer will be X+8.

A/Q

$\frac {x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)} {9} = 55\\\\\sf \frac{9x+36}{9} =55 \\\\\sf \frac{9(x+4)}{9} = 55 \\\\\sf x+4=55 \\\\\sf x=51$

Then, largest integer will be x+8= 51+8 = 59

Answer 2

Answer:

59

Step-by-step explanation:

Let the numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6, x + 7, x + 8,

Then [x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7) + (x + 8)] / 9= 55

= 9x + 36 = 55 × 9

= 9x = 495 - 36

= 9x = 459

= x = 459/9

= x = 51

Largest number = x + 8 = 51 + 8 = 59

Hope it helps you.