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RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

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Question : 13 of 120

Marks: +1, -0

If

\frac{3}{x+y}+\frac{2}{x-y}=2

\frac{9}{x+y}-\frac{4}{x-y}=1

, then what is the value of

\frac{x}{y}

?

$\frac{3}{2}$
5
$\frac{2}{3}$
$\frac{1}{5}$

Solution:

Concept:

Use substitution to convert reciprocal terms into linear equations in

a

and

b

.

Explanation:

Let

a=\frac{1}{x+y}

and

b=\frac{1}{x-y}

.

The equations become

3a+2b=2

and

9a-4b=1

.

Multiply

3a+2b=2

by

2

:

6a+4b=4

.

Add with

9a-4b=1

:

15a=5

, so

a=\frac{1}{3}

.

Thus

\frac{1}{x+y}=\frac{1}{3}

, so

x+y=3

.

Put

a=\frac{1}{3}

in

3a+2b=2

:

1+2b=2

, so

b=\frac{1}{2}

.

Thus

\frac{1}{x-y}=\frac{1}{2}

, so

x-y=2

.

Solve

x+y=3

and

x-y=2

.

Adding gives

2x=5

, so

x=\frac{5}{2}

.

Subtracting gives

2y=1

, so

y=\frac{1}{2}

.

Therefore,

\frac{x}{y}=\frac{\frac{5}{2}}{\frac{1}{2}}=5

.

Answer:

\frac{x}{y}=5

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