Question

The ratio of the numerator to the denominator of a fraction is 2 to 3. If both the numerator and the denominator are increased by 2, the fraction becomes 3/4 . What is the original fraction? Which of the following systems of equations can be used to solve the problem?
A. n + 4 = 3 and d + 5 = 4
B. 3n - 2d = 0 and 4n + 2 = 3d + 2
C. 3n = 2d and 4n + 8 = 3d + 6

Answer 1

Answer:

The fraction is $\frac{4}{6}$

Option (B) is correct.

$3n-2d=0$  and $4n+8=3d+6$

Step-by-step explanation:

Let n denotes numerator and d denotes denominator of the fraction.

Given : The ratio of the numerator to the denominator of a fraction is 2 to 3.

That is $\frac{n}{d}=\frac{2}{3}$

Cross multiply , we get,

$3n=2d$

Or, $3n-2d=0$   ........(1)

Also, given : . If both the numerator and the denominator are increased by 2, the fraction becomes 3/4

That is $\frac{n+2}{d+2}=\frac{3}{4}$

Cross multiply , We get,

$4{n+2}={3}{d+2}$

$4n+8=3d+6$  .........(2)

$3d-4n=2$  .......(3)

Thus, from (1) and (2) , option (B) follows.

Solving equation (1) and (3) to get the original fraction using elimination method,

3n - 2d = 0    ............(1)

and 3d - 4n = 2       .........(3)

Multiply equation (1) by 3 , we get ,

9n - 6d = 0  ..........(4)

Multiply equation (3) by 2 , we get ,

-8n + 6d = 4 ..........(5)

Adding (4) and  (5) , we get,

9n - 6d -8n + 6d = 4 + 0

⇒ n = 4

Put n = 4 in (1), we get

3n - 2d = 0 ⇒ 3(4) - 2d = 0 ⇒ 12 -2d = 0  ⇒  12 = 2d ⇒ d = 6

so, the numerator is 4 and  the denominator is 6.

Thus, the fraction is $\frac{4}{6}$