Question

if 1 is added to the numerator of a fraction, it becomes 1/5.If 1 is subtracted from the denominator ,it become1/7​

Answer 1

$\large\bf\underline {To \: find:-}$

• we need to find the fraction.

$\large\bf\underline{Given:-}$

• if 1 is added to the numerator of a fraction, it becomes 1/5.If 1 is subtracted from the denominator ,it become1/7.

$\huge\bf\underline{Solution:-}$

Let the numerator be x and denominator be y

❧ According to Question:-

$\star \underbrace{ \rm \: condition \: 1st \: }$

if 1 is added to the numerator of a fraction, it becomes 1/5

➛ (x + 1)/y = 1/5

➛ 5(x + 1) = y

➛ 5x + 5 = y .........1)

$\star \underbrace{ \rm \: condition \: 2nd \: }$

If 1 is subtracted from the denominator ,it become1/7

➛ x/y - 1 = 1/7

➛ 7x = y -1 .......2)

⏭️ putting value of y From 1) in 2)

➛ 7x = (5x + 5) - 1

➛ 7x = 5x + 5 - 1

➛ 7x - 5x = 4

➛ 2x = 4

➛ x = 4/2

❧ x = 2

✝️ putting value of x from 2) in 1)

➛5x + 5 = y

➛ 5 × 2 + 5 = y

➛ y = 10 + 5

➛ y = 15

Hence,

• ❥ The fraction is 2/15

━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

• let the numerator be x
• And denominator be y

Then,

$\sf→ \frac{x+1}{y}=\frac{1}{5}\\\sf→ 5(x+1)=y\\\sf→ 5x+5=y\\\sf→ 5x-y=-5---equ(1)$

Again,

$\sf→ \frac{x}{y-1}=\frac{1}{7}\\\sf→ 7x=y-1\\\sf→ 7x-y=-1--equ(2)$

• Subtracting equ(2) From equ(1)

we get,

$\sf→ 2x=4\\\sf→x=\frac{4}{2}\\\sf→ x=2$

• Keep value of x in equ(1)

$\sf→ 5x-y=-5\\\sf→ 5×2-y=-5\\\sf→ 10-y=-5\\\sf→ 10+5=y\\\sf→ y=15$

Hence,

The fraction will $\large\sf{\fbox{\red{\underline{\frac{2}{15}}}}}$