Question

A fraction becomes 4 when 1 is added to both numerator and denominator and it becomes 7 when 1 is subtracted from both numerator amd denominator . Then find the numerator and denominator of given fraction

Answer 1

Answer:

$\large Fraction =\dfrac{15}{3}$

Step-by-step explanation:

Let the numerator and denominator of  fraction be x and y respectively.

Fraction = $\dfrac{x}{y}$

In first case when 1 is added to both it become 4

$\large \dfrac{x+1}{y+1}=4 \\\\\\\large x+1=4y+4\\\\\\\large x=4y+3 \ ...( i)$

In second case when 1 is subtracted from both it become 7.

$\large \dfrac{x-1}{y-1}=7\\\\\\\large x-1=7y-7\\\\\\\large x=7y-6 \ ...( ii)$

From ( i ) and  ( ii ) we have

4 y + 3 = 7 y - 6

7 y - 4 y = 6 + 3

3 y = 9

y = 3

Now put  y = 3 in ( i )

x = 4 y + 3

x = 12 + 3

x = 15

Thus we get  Fraction = $\dfrac{15}{3}$

Verification:

when 1 is added to both it become 4

$\large Fraction = \dfrac{15+1}{3+1}=\dfrac{16}{4}=4$

when 1 is subtracted from both it become 7.

$\large Fraction = \dfrac{15-1}{3-1}=\dfrac{14}{2}=7$

Hence verified.

Answer 2

• Let the numerator be N and denominator be D

Fraction = $\dfrac{N}{D}$

$\textbf{Case 1)}$

When 1 is added to both it become 4.

• A.T.Q.

=> $\dfrac{N\:+\:1}{D\:+\:1}$ = $\dfrac{4}{1}$

Cross-multiply them

=> N + 1 = 4(D + 1)

=> N + 1 = 4D + 4

=> N = 4D + 3 _______ (eq 1)

____________________________

$\textbf{Case 2)}$

When 1 is subtracted from both (N and D) it becomes 7

• A.T.Q

=> $\dfrac{N\:-\:1}{D\:-\:1}$ = $\dfrac{7}{1}$

Cross-multiply them

=> N - 1 = 7(D - 1)

=> N - 1 = 7D - 7

=> N = 7D - 6

=> (4D + 3) = 7D - 6 [From (eq 1)]

=> 4D + 3 = 7D - 6

=> 4D - 7D = - 6 - 3

=> - 3D = - 9

=> 3D = 9

=> D = 3

• Put value of D in (eq 1)

=> N = 4(3) + 3

=> N = 12 + 3

=> N = 15

______________________________

Fraction = $\dfrac{15}{3}$

___________ $\bold{[ANSWER]}$