Question

the sum of three numbers is 182, the second is 6 times bigger than the first and the third is 14 times bigger than the second number, what are those numbers?​

Answer 1

The first number is a , the second number is b , the third number is c

( 0 < a < b < c )

The sum of three numbers is 182

⇒ a + b + c = 182 (1)

The second is 6 times bigger than the first

⇒ b = 6a (2)

The third is 14 times bigger than the second number

⇒ c = 14b = 14.6a = 84a (3)

(1),(2),(3): ⇒ a + 6a + 84a = 182

⇒ 91a = 182

⇒ a = 2

⇒ b = 6a = 6.2 = 12

⇒ c = 14b = 14.12 = 168

Those numbers are 2 , 12 , 168

ok done. Thank to me :>

Answer 2

The numbers are
`2,12,168`

Explanation:

Let the numbers be
`a,b,c`

We have

The sum of three numbers is
`182`

`a+b+c=182-----eqn1`

The second is
`6` times bigger than the first

`b=6a-----eqn2`

The third is
`14` times bigger than the second number

`c=14b-------eqn3`

Substituting in eqn 3

`c=14* 6a=84a ----eqn4`

Substituting in eqn 1

`a+6a+84a=182\\\\91a=182\\\\a=2`

Substituting in eqn 2

`b=6*2=12`

Substituting in eqn 3

`c=14*12=168`

The numbers are
`2,12,168`