Question

The sum of three numbers is 32. If the second number is four more than the first while the third is twice the second, find the numbers.

Answer 1

Answer:

First number - 5, Second number - 9 and Third number 18

Step-by-step explanation:

Let first number be x

Second number - 4 + x

Third number - 2(4+x)

x + 4+x + 2(4+x) = 32

x + 4+x + 8+2x = 32

4x + 12 = 32

4x = 32 - 12

4x = 20

x = 20/4

x = 5

So, x = 5

4+x = 4+5 = 9

2(4+x) = 8 + 10 = 18

Answer 2

Given: Sum of three numbers = 32

           Second number is 4 more than first

           Third number is twice the second number

To find: The numbers

Let: The numbers be X, Y and Z

Solution: According to the given question,

Sum of three numbers is 32

i.e., X + Y + Z = 32                                   ...(1)

also, second number is 4 more than the first number

i.e., Y = X + 4                                           ...(2)

and third number is twice the second number

i.e., Z = 2Y                                               ...(3)

putting values of Y and Z from equation (2) and (3) into equation (1)

⇒ X + (X + 4) + (2Y) = 32

Putting value of Y from equation (2) again in the above equation

⇒ X + X + 4 + 2(X + 4) = 32

⇒ 2X + 4 + 2X + 8 = 32

⇒ 4X + 12 = 32

⇒ 4X = 32 - 12 = 20

⇒ X = 20/4 = 5

Putting value of X in equation (2)

⇒ Y = 5 + 4 = 9

Putting value of Y in equation (3)

⇒ Z = 2 X 9 = 18

Therefore, the three numbers are 5, 9 and 18.