Question

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100 Point Question!!

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

Answer 1

Let the smaller integer be x

Let the larger integer be x + 2

Given, two-fifth of the smaller exceeds two-ninth of the greater by 4



According to the question,

$\dfrac{2x}{5}$ = $\dfrac{2}{9}$ × x + 2 + 4

Taking LCM of the denominators of the right side, we get

$\dfrac{2x}{5}$ = $\dfrac{2x \: + \: 4 \: + \: 36}{9}$


By cross multiplying we get,

2x × 9 = 5(2x + 40)

18x = 10x + 200

18 - 10x = 200

8x = 200

x = $\dfrac{200}{8}$

x = 25


Smaller integer => x => 25

Lager integer => x + 2 => 25 + 2 => 27


$\therefore{}$ The integers are 25 and 27 respectively

Answer 2

Let the smaller number be x.

Then the larger number be x + 2.

Given that two - filth of the smaller exceeds two - ninth of the greater by 4.

⇒ (2/5)x = 2/9(x + 2) + 4

⇒ (2x)/5 = (2x + 4)/9 + 4 * 45

⇒ (2x)/5 = (2x + 4)/9 + (180)

⇒ 9(2x) = 5(2x + 4) + 180

⇒ 18x = 10x + 20 + 180

⇒ 18x = 10x + 200

⇒ 8x = 200

⇒ x = 200/8

⇒ x = 25.

Therefore,

⇒ Smaller number = 25

⇒ Larger number = 27.

Hope it helps!