Question

Find two consecutive odd integers such that two-fifth of the smaller exceeds twoninth of the greater by 4.

Answer 1

Let the two consecutive no.s be x and x+2

According to the question-
2x/5 = 2/9( x+2) + 4
2x/5 = 2x/9 + 4/9 + 4
2x/5 - 2x/9 = 40/9
(18x - 10x)/45 = 40/9
8x = 200
x = 200/8 = 25
Therefore the two consecutive positive odd no.s are 25 and 27

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Answer 2

Hi ,

Let a and ( a + 2 ) are two

consecutive odd numbers

According to the problem given ,

Two - fifths of the smaller number

= 2a/5

Two - ninth's of the greater number

= 2a/9

Now ,

2a/5 - 2( a + 2 ) /9 = 4

Divide each term with 2 , we get,

a/5 - ( a + 2 ) /9 = 2

[ 9a- 5 ( a + 2 ) ]/45 = 2

9a - 5a - 10 = 90

4a = 90 + 10

4a = 100

a = 100/4

a = 25

Therefore ,

Required two consecutive odd

numbers are

a , a + 2

25 , 27

I hope this helps you.

:)