Question

The sum of three consecutive multiples of 12 is 324. Find the greatest one among them.

Answer 1

Heya...
Here is your answer ----

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Let the first multiple be 12 (x)
Then, the second multiple = 12 (x+1)
And, the third multiple = 12 (x+2)

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ATQ -----
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=> 12 (x) + 12 (x+1) + 12 (x+2) = 324

=> 12x + 12x + 12 + 12x + 24 = 324

=> 36x + 36 = 324

=> 36x = 324 - 36

=> 36x = 288

=> x = 288÷36

=> x = 8
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=> First multiple = 12 (x) = 12 × 8 = 96

=> Second multiple = 12 (x+1) = 12 (8+1) = 12×9 = 108

=> Third multiple = 12 (x+2) = 12 (8+2) = 12×10 = 120
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HOPE IT HELPS...!!

Answer 2

heya!
here's your answer!
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as mentioned in the question,
let the first multiple of 12 be 12x
therefore ,

second consecutive multiple - 12( x + 1 )
third consecutive multiple - 12( x + 2 )

therefore,
according to question,
=> 12x + 12(x+1) + 12( x + 2 ) = 324
=> 12x + 12x + 12 + 12x + 24 = 324
=> 36x + 36 = 324
=> taking 36 common on left side,
=> 36(x+1) = 324
=> therefore,
=> x+1 = 324/36
=> x+1 = 9
=> x= 8

greatest multiple = 12(x+2) = 12(8+2)= 10 x 12= 120

first multiple = 12x = 12 x 8= 96

second multiple = 12(x+1) = 12 x 9 = 108

Hope this helps you:)