Ex 2.2, 8 Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.Let first integer be x
Second integer = x + 1
3rd integer = (x + 1) + 1
= x + 2
Now,
Multiply 1st number by 2 = 2x
Multiply 2nd number by 3 = 3(x + 1)
Multiply 3rd number by 4 = 4(x + 2)
Rough
Integers are −2, −1, 0, 1, 2, …
Consecutive integers are 1, 2, 3
Difference between Consecutive Integers = 1
If number starting from 1, then
First number = 1
Second number = 1 + 1 = 2
Third number = 2 + 1 = 3
Given, sum of these three is 74
2x + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
2x + 3x + 4x + 3 + 8 = 74
9x + 11 = 74
9x = 74 − 11
9x = 63
x = 63/9
x = 7
∴ First integer = x = 7
Second integer = x + 1 = 7 + 1 = 8
Third integer = x + 2 = 7 + 2 = 9

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.