Question

Find two consecutive odd numbers such that one fifth of the smaller exceeds one ninth of the greater by 2

Answer 1

1st: 2x-1

2nd: 2x+1

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Equation:

(1/5)(2x-1) = (1/9)(2x+1) + 2

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Multiply thru by 45 to get:

9(2x-1) = 5(2x+1) + 90

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18x - 9 = 10x + 95

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8x = 104

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x = 13

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1st: 2x-1 = 25

2nd: 2x+1 = 27

Answer 2

Answer:

25 and 27

Step-by-step explanation:

Let two consecutive natural numbers = x, x + 1

∴ One – fourth of the smaller = x/4

One – fifth of the greater = (x + 1)/5

According to the statement :

x/4 = ((x + 1)/5) + 1 ⇒ x/4 – (x + 1)/5 = 1

⇒ (5x – 4 (x + 1))/20 = 1 ⇒ ((5x – 4x – 4)/20) = 1

⇒( x – 4)/20 = 1

⇒ x – 4 = 20

(Cross – multiplying)

⇒ x = 20 + 4 ⇒ x = 24

∴ x + 1 = 24 + 1 = 25

Two consecutive numbers are 24 and 25