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

Answer:

The numbers are $2,12,168$

Step-by-step 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

$c=14\times 6a=84a ----eqn4$

Substituting in eqn 1

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

Substituting in eqn 2

$b=6\times2=12$

Substituting in eqn 3

$c=14\times12=168$

The numbers are $2,12,168$

Answer 2

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 :>