Question

In a fraction sum of numerator and denominator is equal to 60. If 2 is added to the numerator and 2 is subtracted from denominator . The value of fraction becomes 1/3. Find the Fraction.

Answer 1

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Sum of numerator and denominator is 60
• If 2 is added in the numerator and 2 is subtracted from denominator the fraction becomes ⅓ .

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Original fraction = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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• let the numerator be x
• let the denominator be y

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Acc. to the 1st statement :-

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x + y = 60 ----- ( i )

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Acc. to the 2nd statement :-

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$\sf : \implies\:{\frak{ \dfrac{x + 2 }{y - 2 } = \dfrac{1}{3} }}$

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$\sf : \implies\:{\frak{ 3 ( x + 2 ) = y - 2 }}$

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$\sf : \implies\:{\frak{ 3x + 6 = y - 2 }}$

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$\sf : \implies\:{\frak{ 3x - y = - 2 - 6 }}$

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$\sf : \implies\:{\frak{ 3x - y = -8 }}$ ----- ( ii )

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• Adding eq ( i ) and ( ii )

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3x - y + x + y = 60 - 8

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3x + x = 52

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4x = 52

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x = 52 / 4

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x = 26 / 2

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x = 13

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• putting value of x in eq ( i )

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x + y = 60

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13 + y = 60

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y = 60 - 13

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y = 47

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• $\sf : \implies\:{\frak{ Required \: fraction = \dfrac{13 }{47 } }}$

Answer 2

Given

Sum of numerator and denominator is 60

If 2 is added in the numerator and 2 is subtracted from denominator the fraction becomes ⅓ .

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To Find:-

Original fraction = ??

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Solution:-

let the numerator be "x"

let the denominator be "y"⠀⠀⠀⠀

Acc. to the 1st statement :-

x + y = 60 --------- ( i )

Acc. to the 2nd statement :-

=> x+2/n-2 = 1/3

=> 3(x+6)= n-2

=> 3x-n= -2-6

=> 3x-n = -8 ----------(ii)

Adding eq ( i ) and ( ii )

=> 3x - y + x + y = 60 - 8

=> 3x + x = 52

=> 4x = 52

=> x = 52 / 4

=> x = 26 / 2

=> x = 13

putting value of x in eq ( i )

=> x + y = 60

=> 13 + y = 60

=> y = 60 - 13

=> y = 47

Answer:-

13/47 is the required fraction.

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