Question

the sum of numerator and denominator of a certain fraction is 10 if 1 is subtracted from both the numerator and denominator the fraction is decreased by 2 by 10 find the fraction

Answer 1

ola!!

refer to attachment ⤴⤴⤴

Answer 2

» The sum of numerator and denominator of a certain fraction is 10.

• Let Numberator be N and Denominator be D.

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

A.T.Q.

=> N + D = 10

=> N = 10 - D _______ (eq 1)

______________________________

» If 1 is subtracted from both the numerator and denominator the fraction is decreased by 2/10.

A.T.Q.

=> $\dfrac{N\:-\:1}{D\:-\:1}\:=\:\dfrac{N}{D}\:-\:\dfrac{2}{10}$

=> $\dfrac{N\:-\:1}{D\:-\:1}\:=\:\dfrac{10N\:-\:2}{10D}$

Cross-multiply them

=> 10D(N - 1) = (D - 1)(10N - 2)

=> 10ND - 10D = D(10N - 2) -1(10N - 2)

=> 10ND - 10D = 10ND - 2D - 10N + 2

=> 10ND - 10ND - 10D + 2D = - 10N + 2

=> - 8D = - 10N + 2

=> - 4D = - 5N + 1

=> - 4D = -(10 - D) + 1 [From (eq 1)]

=> - 4D = - 10 + D + 1

=> - 4D - D = - 9

=> - 5D = - 9

=> D = 9/5

Put value of D in (eq 1)

=> N = 10 - 9/5

=> N = (50 - 9)/5

=> N = 41/5

___________________________

Fraction = $\dfrac{N}{D}$ = $\dfrac{41}{5}$ × $\dfrac{5}{9}$

=> $\dfrac{41}{9}$

__________ [ ANSWER ]

__________________________

✡ Verification :

We have N = 10 - D as a Equation.

From above calculations we have N = 41/5 and D = 9/5

Put value of N and D in above equation.

→ 41/5 = 10 - 9/5

→ 41/5 = (50 - 9)/5

→ 41/5 = 41/5

______________________________