Question

The sum of three consecutive integers is 12 more than the smallest of the integers. Find all integers

Answer 1

Correct question :

The sum of three consecutive integers is 12 more than twice the smallest of the integers. Find all the integers .

Answer :

9 , 10 , 11

Solution :

Here ,

It is given that , the sum of three consecutive integers is 12 more than the smallest of the integers .

Thus ,

Let the three consecutive integers be ;

x , (x + 1) , (x + 2) .

Now ,

According to the question ;

=> x + (x + 1) + (x + 2) = 2x + 12

=> 3x + 3 = x + 12

=> 3x - 2x = 12 - 3

=> x = 9

=> x = 9

Thus ,

1st integer = x = 9

2nd integer = x + 1 = 9 + 1 = 10

3rd integer = x + 2 = 9 + 2 = 11

Hence ,

The required integers are ;

9 , 10 , 11 .

Answer 2

Correct Question : The sum of three consecutive integers is 12 more than twice the smallest of the integers. Find all integers.

Let the integers be x , x + 1 and x +2 .

Sum of three consecutive integers = 3x + 3

According to the question ,

   →      3x + 3 = 12 + 2x

   →      3x - 2x = 12 - 3

   →       x = 9

The numbers are 9 , 10 and 11 .

Hope it helps u ....... :)