Question

❤Question❤

=> The sum of numerator and denominator of a fraction is three less
than twice the denominator. If each denominator and denominator is decreased
by 1, the fraction becomes 72. Find the fraction. (CBSE 2010-3M)

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Answer 1

Answer:

let the numerator be ' x '

and denominator be ' y '

$let \: the \: fraction \: be \: \frac{x}{y}$

$x + y + 3 = 2y$

$or \: x - y = - 3$

or x = y -3 _____ ( 1 )

$also \frac{ x - 1}{x - 1} = \frac{1}{2} \\ x - 1 = \frac{y - 1}{2} \\ x = \frac{y - 1}{2} + 1⇒2x = y - 1 + 2 = y + 1$

( or ) 2x - y = 1 ______ ( 2 )

substitute ( 1 ) in ( 2 )

$2(y - 3) - y = 1 \\ ⇒2y - 6 - y⇒y = 7$

put y = 7 in ( 1 ) we get x = 7 - 3 = 4

$∴x = 4, \: y = 7 \\ ∴the \: fraction \: is \: \frac{x}{y} = \frac{4}{7}$

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Explanation:

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Answer 2

Answer:

Let the numerator = x

and the denominator = y

So, the fraction = x/y According to the question,

Condition I:

x + y = 2y – 3

⇒ x + y – 2y = – 3

⇒ x – y = – 3 …(i)

Condition II:

⇒ 2(x – 1) = y – 1

⇒ 2x – 2 = y – 1

⇒ 2x – y = 1 …(ii)

On subtracting Eq. (i) from Eq. (ii),

we get 2x – y – x + y = 1 + 3

⇒ x = 4

On putting the value of x in Eq. (i),

we get

4 – y = – 3

⇒ y = 7 So,

the numerator is 4 and the denominator is 7 Hence,

the fraction is 4/7.