Question

The sum of numerator and denominator of a
fraction is 8. If 3 is added to both numerator
3
and denominator the fraction becomes
4.
Find the original fraction.​

Answer 1

Correct Question:-

The sum of numerator and denominator of a fraction is 8. If 3 is added to both numerator and denominator the fraction becomes 3/4. Find the original fraction.

Given:-

• Sum of numerator and denominator of a fraction = 8
• If 3 is added to both numerator and denominator the fraction becomes 3/4

To Find:-

• The original fraction

Assumption:-

• Let the numerator be x
• and denominator be y

Solution:-

ATQ,

Sum of both numerator and denominator is 8

Hence,

x + y = 8

=> x = 8 - y $\longrightarrow$ [i]

Now,

Also it is given that,

When 3 is added to both numerator and denominator the fraction becomes 3/4

Hence,

$\sf{\dfrac{x+3}{y+3} = \dfrac{3}{4}}$

= $\sf{4(x + 3) = 3(y+3)}$

= $\sf{4x+12 = 3y + 9}$

= $\sf{4x - 3y = -12+9}$

= $\sf{4x - 3y = -3 \longrightarrow[ii]}$

Putting the value of x in eq.[ii] from eq.[i]

= $\sf{4x - 3y = -3}$

= $\sf{4(8-y)-3y = -3}$

= $\sf{32-4y-3y = -3}$

= $\sf{32-7y = -3}$

= $\sf{-7y = -32-3}$

= $\sf{-y = \dfrac{-35}{7}}$

= $\sf{y = 5}$

Putting the value of y in eq.[i]

$\sf{x = 8-y}$

= $\sf{x = 8-5}$

= $\sf{x=3}$

Now,

The original fraction as follows:-

$\sf{\dfrac{x}{y} = \dfrac{3}{5}}$

Therefore the original fraction is $\sf{\dfrac{3}{5}}$

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