Question

Three consecutive integers are such that when they are taken in increasing order and multiplied by 2,3and 4 respectively, they add up to 74.Find these numbers.​

Answer 1

Answer:

the numbers are 7 , 8 and 9

Step-by-step explanation:

let the three consecutive integers be x , ( x + 1 ) and ( x + 2  )

the first number must be multiplied by 2

so it is x * 2 = 2x

the second number must be multiplied by 3

so it is ( x + 1 ) * 3 = 3x + 3

the third number must be multiplied by 4

so it is ( x + 2 ) * 4  = 4x + 8

but it is given that 2x + ( 3x + 3 ) + ( 4x + 8 ) = 74

=> 2x + 3x + 3 + 4x + 8 = 74

=> 9x + 11 = 74

=> 9x = 74 - 11 = 63

=> x = 63 / 9

= 7

the three numbers are x , ( x + 1 ) and ( x + 2 )

= 7 , ( 7 + 1 ) and ( 7 + 2 )

= 7 , 8 , 9

therefore the 3 numbers are 7 , 8  and 9

Answer 2

Answer:-

$\sf{The \ numbers \ are \ 7, \ 8 \ and \ 9.}$

Given:

• Three consecutive integers are such that when they are taken in increasing order and multiplied by 2,3 and 4 respectively. They add up to 74.

To find:

• The number.

Solution:

Let the constant be x.

$\sf{\therefore{Three \ consecutive \ numbers \ are:-}}$

(x-1), x and (x+1)

According to the given condition.

$\sf{2(x-1)+3x+4(x+1)=74}$

$\sf{\therefore{2x-2+3x+4x+4=74}}$

$\sf{\therefore{9x=74-2}}$

$\sf{\therefore{9x=72}}$

$\sf{\therefore{x=\frac{72}{9}}}$

$\boxed{\sf{\therefore{x=8}}}$

$\sf{\therefore{Numbers \ are:}}$

$\sf{(8-1), \ 8 \ and \ (8+1)}$

$\sf{\therefore{Numbers \ are:}}$

$\sf{7, \ 8 \ and \ 9.}$

$\sf\purple{\tt{\therefore{The \ numbers \ are \ 7, \ 8 \ and \ 9.}}}$