Question

find two consecutive odd numbers such that two fifths of the smaller number exceeds two ninths of the larger by 4.

Answer 1

Given :-

➪ ⅖ of a number (smaller) exceeds ²⁄₉ of another number (larger) by 4

To find :-

➪ Those consecutive odd numbers

Solution :-

➪ Let the required numbers be (x) and (x + 2)

➪ Then, according to the question :-

➪ ⅖ (x) - ²⁄₉ (x + 2) = 4

➪ 18x - 10 (x + 2) = (45 × 4)     (L.C.M of 9 and 5 (45) is transferred)

➪ 18x - 10x - 20 = 180

➪ 8x = 200

➪    x = ²⁰⁰⁄₈

➪    x = 25

The value of x is 25

Hence, (x) = 25 and (x + 2) = 25 + 2 = 27

∴ The required numbers are 25 and 27

Verification :-

➪ ⅖ (x) - ²⁄₉ (x + 2) = 4

➪ ⅖(25) - ²⁄₉(27) = 4

➪ 10 - 6 = 4

➪ 4 = 4

➪ L.H.S = R.H.S

Answer 2

Question :

Find two consecutive odd numbers such that two fifths of the smaller number exceeds two ninths of the larger by 4

Solution :

Let one odd number be ' 2n + 1 '

This is smallest odd number .

Other consecutive odd number be ' 2n + 3 '

This is largest odd number .

A/c , " Two fifths of the smaller number exceeds two ninths of the larger by 4 "

$:\implies \sf \dfrac{2}{5}\ (2n+1)=\dfrac{2}{9}\ (2n+3)+4$

$:\implies \sf \dfrac{2}{5}(2n+1)= \dfrac{2(2n+3)+36}{9}$

$:\implies \sf 9(2n+1)=5(2n+3)+ 90$

$:\implies \sf 18n+9=10n+15+90$

$:\implies \sf 18n-10n =105-9$

$:\implies \sf 8n=96$

$:\implies \sf \pink{n=12}\ \; \bigstar$

First consecutive smallest odd number :

                   = 2n + 1

                   = 2(12) + 1

                   = 24 + 1

                   = 25  $\green{\bigstar}$

Second consecutive largest odd number :

                   = 2n + 3

                   = 2(12) + 3

                   = 24 + 3

                   = 27  $\orange{\bigstar}$

★ ═════════════════════ ★

Alternative : You may solve this question by taking ' x ' as smallest consecutive odd number and ' x + 2 ' as biggest consecutive odd number .