Question

the numerator amd denominator of a fraction are increased by 1 each, the fraction become 1/2 and if the numerator and denominator are deceased by 1 each, then the fraction become 1/4 . find the fraction?
A-2/3
B-0
C-2/5
D-5/ 8
​

Answer 1

2/5 all the options are wrong check it option c should be 2/5 not -2/5 ok

correct answer is 2/5

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Answer 2

Answer:-

The fraction is $\sf{\dfrac{2}{5}}$

Option (c) 2/5 (Answer)

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Explanation:-

$\sf{\star}$ Given:-

• When the denominator and numerator of a fraction are increased by 1 each the fraction becomes 1/2.
• When the denominator and numerator of a fraction are decreased by 1 each the fraction becomes 1/4.

$\sf{\star}$ To Find:-

The fraction

$\sf{\star}$ Assumption:-

Let the numerator be x and denominator be y.

$\sf{\star}$

1st case,

When the numerator and the denominator of a fraction is increased by 1 each, the fraction becomes 1/2.

Therefore,

$\sf{\dfrac{x+1}{y+1} = \dfrac{1}{2}}$

By cross-multiplication,

= $\sf{2(x+1) = 1(y+1)}$

= $\sf{2x+2 = y +1}$

= $\sf{2x - y = -2+1}$

= $\sf{2x - y = -1 \longrightarrow [i]}$

2nd case,

When the numerator and denominator of a fraction is decreased by 1 each the fraction becomes 1/4.

Therefore,

$\sf{\dfrac{x-1}{y-1} = \dfrac{1}{4}}$

By Cross-multiplication,

= $\sf{4(x-1) = 1(y-1)}$

= $\sf{4x - 4 = y-1}$

= $\sf{4x-y = 4-1}$

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

From eq.(i),

$\sf{2x-y = -1}$

= $\sf{-y = -1-2x}$

= $\sf{y = -(-1-2x)}$

= $\sf{y = 1+2x}$

=> $\sf{ y = 2x+1}$

Substituting the value of y in eq.[ii],

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

= $\sf{4x-(2x+1) = 3}$

= $\sf{4x-2x-1 = 3}$

= $\sf{2x = 3+1}$

= $\sf{2x = 4}$

= $\sf{x = \dfrac{4}{2}}$

= $\sf{x = 2}$

Putting the value of x in eq.[i],

$\sf{2x-y = -1}$

= $\sf{2\times 2 -y = -1}$

= $\sf{4-y = -1}$

= $\sf{-y = -4-1}$

= $\sf{-y = -5}$

= $\sf{y = -(-5)}$

= $\sf{y = 5}$

Now,

Numerator = x = 2

Denominator = y = 5

Therefore,

Fraction = $\sf{\dfrac{x}{y} = \dfrac{2}{5}}$

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