Question

34. The denominator of a rational number is greater than its numerator by 7. If the
numerator is increased by 4 and the denominator is decreased by 8 the new
rational number becomes 8/3. Find the original rational n
number. plzz help in this ques ​

Answer 1

Answer:

Let the fraction be qp

So q=p+7

q−6p+17=2

⇒p+7−6p+17=2

⇒p+1p+17=2

⇒p+17=2p+2

⇒p=15

q=15+7=22

So qp=2215.

Answer 2

☯ Let numerator of a rational no. be p and the denominator be q.

⠀⠀⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Denominator of a rational no. is greater than its Numerator by 7

$:\implies\sf q = p + 7\qquad\qquad\bigg\lgroup\bf eq.\;(1)\bigg\rgroup$

If the numerator is increased by 4 and the denominator is decreased by 8 the new rational number becomes 8/3.

$:\implies\sf \dfrac{p + 4}{q - 8} = \dfrac{8}{3}$

$:\implies\sf 3(p + 4) = 8(q - 8)$

$:\implies\sf 3p + 12 = 8q - 64$

$:\implies\sf 3p - 8q = - 64 - 12$

$:\implies\sf - 5p - 8(p + 7) = - 64 - 12$

$:\implies\sf - 5p - 8p - 56 = - 76$

$:\implies\sf - 13p = - 76 + 56$

$:\implies\sf - 13p = - 20$

$:\implies{\boxed{\sf{\purple{p = \dfrac{20}{13}}}}}\;\bigstar$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Now, Putting value of p in eq. (1),

$:\implies\sf q = \dfrac{20}{13} + 7$

$:\implies\sf q = \dfrac{20 + 91}{13}$

$:\implies{\boxed{\sf{\purple{q = \dfrac{111}{13}}}}}\;\bigstar$

Therefore required rational number is,

$:\implies\sf \dfrac{ \frac{20}{13}}{ \frac{111}{13}}\qquad\qquad\bigg\lgroup\bf Fraction = \dfrac{p}{q}\bigg\rgroup$

$:\implies\sf \dfrac{20}{ \cancel{13}} \times \dfrac{ \cancel{13}}{111}$

$:\implies{\boxed{\sf{\pink{ \dfrac{20}{111}}}}}\;\bigstar$

$\therefore$ Hence, the required rational number is 20/111 which is non terminating but repeating.