Question

Two consecutive odd integers are such that one third the smaller one is greater than one seventh of the larger one by 6 find the numbers

Answer 1

let the smaller no be x

other no = x+2

A/Q

x/3=(x+2)/7 + 6

x/3 - (x+2)/7 = 6

7x-3x-6/21=6

4x-6=126

4x=132

x=33

therefore numbers are 33 and 35

Answer 2

Given :

Two consecutive odd integers are such that one third the smaller one is greater than one seventh of the larger one by 6

To Find :

find the numbers

Solution:

Let the two odd consecutive inetegers be x and x+2

We are given that one third the smaller one is greater than one seventh of the larger one by 6

So, $\frac{1}{3}x-\frac{1}{7}(x+2)=6\\\frac{7x-3x-6}{21}=6\\\frac{4x-6}{21}=6\\4x-6=21 \times 6$

4x-6=126

4x=132

x=33

x+2=33+2=35

Hence 33 and 35 are two odd consecutive integers