Question

consider a proper fraction such that the numerator is 2 less than the denominator. if the numerator and denominator are each increased by 1, then the ratio of the numerator and denominator of the new fraction is 4:5. determine the original fraction

Answer 1

Answer:-

Let the fraction be x/y.

Given:

Numerator is 2 less than the denominator.

⟶ Numerator = denominator - 2

⟶ x = y - 2 -- equation (1)

And,

If numerator and denominator are increased by 1 , the ratio of numerator and denominator is 4 : 5.

According to the above condition,

$\longrightarrow \sf \: \dfrac{x + 1}{y + 1} = \dfrac{4}{5}$

Substitute the value of x from equation (1).

$\longrightarrow \sf \: \frac{y - 2 + 1}{y + 1} = \frac{4}{5} \\ \\ \longrightarrow \sf \:5(y - 1) = 4(y + 1) \\ \\ \longrightarrow \sf \:5y - 5 = 4y + 4 \\ \\ \longrightarrow \sf \:5y - 4y = 4 + 5 \\ \\ \longrightarrow \boxed{ \sf \:y = 9}$

Substitute the value of y in equation (1)

⟶ x = y - 2

⟶ x = 9 - 2

⟶ x = 7

∴ The original fraction x/y is 7/9

Answer 2

Let the fraction be x/y.

Given:

Numerator is 2 less than the denominator.

⟶ Numerator = denominator - 2

⟶ x = y - 2 -- equation (1)

And,

If numerator and denominator are increased by 1 , the ratio of numerator and denominator is 4 : 5.

According to the above condition,

\longrightarrow \sf \: \dfrac{x + 1}{y + 1} = \dfrac{4}{5}⟶

y+1

x+1

=

5

4

Substitute the value of x from equation (1).

\begin{gathered} \longrightarrow \sf \: \frac{y - 2 + 1}{y + 1} = \frac{4}{5} \\ \\ \longrightarrow \sf \:5(y - 1) = 4(y + 1) \\ \\ \longrightarrow \sf \:5y - 5 = 4y + 4 \\ \\ \longrightarrow \sf \:5y - 4y = 4 + 5 \\ \\ \longrightarrow \boxed{ \sf \:y = 9}\end{gathered}

⟶

y+1

y−2+1

=

5

4

⟶5(y−1)=4(y+1)

⟶5y−5=4y+4

⟶5y−4y=4+5

⟶

y=9

Substitute the value of y in equation (1)

⟶ x = y - 2

⟶ x = 9 - 2

⟶ x = 7

∴ The original fraction x/y is 7/9