Question

The numerator and the denominator of a fraction are in the ratio 3: 2. If 3 is added to the numerator
and 2 is subtracted from the denominator, a new fraction is formed whose value is 9/4. Find the original
fraction.




Please don't post wrong answer s or I will report​

Answer 1

Let us consider a positive integer a

Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that

a = 3b + r……………………………(1)

where r = 0,1,2,3…..

Case 1: Consider r = 0

Equation (1) becomes

a = 3b

On squaring both the side

a2 = (3b)2

a2 = 9b2

a2 = 3 × 3b2

a2 = 3m

Where m = 3b2

Case 2: Let r = 1

Equation (1) becomes

a = 3b + 1

Squaring on both the side we get

a2 = (3b + 1)2

a2 = (3b)2 + 1 + 2 × (3b) × 1

a2 = 9b2 + 6b + 1

a2 = 3(3b2 + 2b) + 1

a2 = 3m + 1

Where m = 3b2 + 2b

Case 3: Let r = 2

Equation (1) becomes

a = 3b + 2

Squaring on both the sides we get

a2 = (3b + 2)2

a2 = 9b2 + 4 + (2 × 3b × 2)

a2 = 9b2 + 12b + 3 + 1

a2 = 3(3b2 + 4b + 1) + 1

a2 = 3m + 1

where m = 3b2 + 4b + 1

∴ square of any positive integer is of the form 3m or 3m+1.

Hence proved.

Answer 2

S O L U T I O N :

Let the ratio of numerator & denominator be r .

• Numerator = 3r
• Denominator = 2r

$\boxed{\bf{The\:original\:fraction\:become = \frac{3r}{2r} }}$

$\underline{\underline{\tt{Accoding\:to\:the\:question\::}}}$

$\mapsto\tt{\dfrac{Numerator + 3}{Denominator -2} = \dfrac{9}{4} }$

$\mapsto\tt{\dfrac{3r+ 3}{2r-2} = \dfrac{9}{4} }$

$\mapsto\tt{4(3r + 3) = 9(2r - 2) \:\: \underbrace{\sf{cross-multiplication}}}$

$\mapsto\tt{12r + 12 = 18r - 18}$

$\mapsto\tt{12r-18r = - 18-12}$

$\mapsto\tt{-6r = -30}$

$\mapsto\tt{r=\cancel{-30/-6}}$

$\mapsto\bf{r=5}$

Now,

• Numerator = 3r = 3 × 5 = 15
• Denominator = 2r = 2 × 5 = 10

Thus;

$\boxed{\bf{The\:original\:fraction\:become = \frac{3r}{2r} =\frac{15}{10} }}$