Question

In a fraction if numerator is increased by 2 and denominator is increased by 3 it becomes a 3 upon 4 and if numerator is decreased by 3 and denominator is decreased by 6 it becomes 4 upon 3.. find the sum of the numerator and denominator


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Answer 1

AnswEr:-

Sum of numerator and denominator = 16

$\rule{200}{1}$

1st case:-

• Numerator increased by 2 and denominator increased by 3 ,fraction = 3/4

Let numerator be N & denominator be D.

•°• Fraction = N/D

ATQ:-

⇒ (N + 2)/(D + 3) = 3/4

⇒ 3(D + 3) = 4(N + 2)

⇒ 3D + 9 = 4N + 8

⇒ 3D = 4N + 8 - 9

⇒ 3D = 4N - 1

⇒ D = (4N - 1)/3 - (Eq.1)

◗ 2nd case:-

• Numerator decreased by 3 and denominator decreased by 6,fraction = 4/3

ATQ:-

⇒ (N - 3)/(D - 6) = 4/3

⇒ (N - 3)/[(4N - 1)/3 - 6] = 4/3

⇒ (N - 3)/[(4N - 1 - 18)/3] = 4/3

⇒(N - 3)/[(4N - 19)/3] = 4/3

⇒ (N - 3) × [3/(4N - 19)] = 4/3

⇒ (3N - 9)/(4N - 19) = 4/3

⇒ 4(4N - 19) = 3(3N - 9)

⇒ 16N - 76 = 9N - 27

⇒ 16N - 9N = -27 + 76

⇒ 7N = 49

⇒ N = 49/7

⇒ N = 7

•°• Numerator = 7

• Putting value of N in (Eq.1):-

⇒ D = [4(7)-1]/3

⇒ D = 28 - 1/3

⇒ D = 27/3

⇒ D = 9

•°• Denominator = 9

Therefore,fraction = 7/9

$\rule{200}{1}$

Sum of numerator and denominator:-

➩ Sum = 7 + 9

➩ Sum = 16

$\therefore\underbrace{\textsf{Sum \: of \: numerator \: \& \: denominator = {\textbf{16 }}}}$

Answer 2

$\huge\bold\red{\underline{Solution:}}$

$\bold\green{\underline{Let:}}$

The numerator of fraction be X

The denominator of fraction be Y

$\bold\green{\underline{Find:The\:sum\:of\: numerator\:and\: denominator}}$

$\bold\blue{\underline{Case-1}}$

$\bold\green{\underline{Given:}}$

Numerator is increased by 2 and denominatoris increased by 3.It become 3/4

⟹X+2/Y+3=3/4

⟹4X+8=3Y+9

⟹4X-3Y=9-8

⟹4X-3Y=1

$\bold\purple{\boxed{4X-3Y=1-------(1)}}$

$\bold\green{\underline{Case-2}}$

The numerator is decreased by 3 and denominator is decreased by 6.It become 4/3

⟹X-3/Y-6=4/3

⟹3X-9=4Y-24

⟹3X-4Y=-24+9

⟹3X-4Y=-15

$\bold\orange{\boxed{3X-4Y=-15-------(2)}}$

$\bold\red{\underline{A.T.Q}}$

⟶4X-3Y=1

⟶3X-4Y=-15

In EQ 1 multiple by 3 and EQ 2 multiple by 4

than subtracting eachother

⟶12X-9Y=3

⟶12X-16Y=-60

⟶7Y=63

⟶Y=63/7

⟶Y=9

Now, putting the value of y in EQ 1

⟶4X-3Y=1

⟶4X-3*9=1

⟶4X-27=1

⟶4X=27+1

⟶4X=28

⟶X=7

Hence,

The sum of numerator and denominator=7+9=16

$\bold\orange{\boxed{Sum\:of\:N\:and\:D=16}}$