Question

The sum of three consecutive multiples of 11 is 636. Find these multiples. ​

Answer 1

Let one multiple be x

2nd multiple = x + 11

And 3rd multiple = x + 22

A.T.Q

$x + (x + 11) + (x + 22) = 636$

$x + x + 11 + x + 22 = 636$

$3x + 33 = 636$

$3x = 636 - 33 = 603$

$x = \frac{603}{3} = 201$

Thus multiples are 201 , 212 and 223

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Answer 2

Three consecutive multiples are 201 , 212 and 223

Given:

The sum of 3 consecutive multiples of 11 is 636.

To Find:

Solution:

Consider the:

• One multiple as - x
• 2nd multiple = x + 11
• 3rd multiple = x + 22

So,

→ x + (x + 11) + (x + 22) = 636 x + (x + 11) +(x + 22) = 636

→ x + x + 11 + x + 22 = 636x + x+11 + x + 22 = 636

→ 3x + 33 = 6363x + 33 = 636

→ 3x = 636 - 33 = 6033x = 636 - 33 = 603

→ x = $\sf \dfrac{603}{3}$ = 201