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UPSC NDA I 2015 Mathematics Solved Paper

Question : 30 of 120

Marks: +1, -0

2p+3q=18

4p^2+4pq-3q^2-36=0

(2p+q)

Solution:

Consider the following equations,

2p+3q=18

(2p+3q)^2=324

4p^2+12pq+9q^2=324

...(1)

4p^2+4pq-3q^2=36

...(2)

Subtracting equation 2 from 1, we get

1 - 2

\Rightarrow 8pq+12q^2 =288

\Rightarrow 4q(2p+3q)=288

\Rightarrow 4q(18)=288

\Rightarrow q=\frac{288}{72}

By substituting the value of ‘q’ so obtained, we get

2p+3q=18

2p+3(4)=18

2p+12=18

2p=6

p=3

\therefore p=3,q=4

Hence the required value of the given expression is

2p+q

=2(3)+4

=10

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