Question

the sum of three numbers is 92. the second number is three times the first number. the third number exceeds the second number by 8 . find the numbers

Answer 1

Let the 1st number be x.

Given that the second number is 3 times the first number = 3x.

Given that the third number exceeds the second number by 8. = 3x + 8.

Given that sum of three numbers = 92.

x + 3x + 3x + 8 = 92

7x + 8 = 92

7x = 92 - 8

7x = 84

x = 84/7

x = 12.


Then 2nd number = 3x

                               = 3 * 12

                               = 36.


Then 3rd number = 3x + 8

                              = 3(12) + 8

                              = 36 + 8

                              = 44.


Therefore the numbers are 12,36,44.


Verification:

x + 3x + 3x + 8 = 92

12 + 36 + 44 = 92

48 + 44 = 92

92 = 92.


Hope this helps!

Answer 2

Let the 1st number be p

Given that the second number is 3 times the first number = 3p

Given that the third number exceeds the second number by 8. = 3p + 8.

Given that sum of three numbers = 92.

p + 3p + 3p + 8 = 92

7p+ 8 = 92

7p = 92 - 8

7p= 84

p = 84/7

p = 12.


Then 2nd number = 3p

                               = 3 * 12 

                               = 36.


Then 3rd number = 3p + 8 

                              = 3(12) + 8 

                              = 36 + 8

                              = 44.


Therefore the numbers are 12,36,44.


Verification:

p + 3p + 3p+ 8 = 92

12 + 36 + 44 = 92

48 + 44 = 92

92 = 92.

p =92