UPSC NDA I 2015 Mathematics Paper

Concept:Use substitution from the linear equation into the quadratic equation to solve for one variable, then compute the required expression.Explanation:Given: $2p + 3q = 18$ and $4p^{2} + 4pq - 3q^{2} - 36 = 0$.From the first equation, write $2p = 18 - 3q$.Substitute this into the second equation. Notice that $4p^{2} = (2p)^2$ and $4pq = 2 \cdot (2p) \cdot q$.So the second equation becomes: $(18 - 3q)^2 + 2(18 - 3q)(q) - 3q^2 - 36 = 0$.Expand: $(324 + 9q^2 - 108q) + (36q - 6q^2) - 3q^2 - 36 = 0$.Combine like terms: $324 - 36 = 288$ for constants; $9q^2 - 6q^2 - 3q^2 = 0$ for $q^2$ terms; $-108q + 36q = -72q$ for $q$ terms.Thus, $288 - 72q = 0$, so $72q = 288$ and $q = 4$.Now $2p = 18 - 3(4) = 18 - 12 = 6$.Therefore, $2p + q = 6 + 4 = 10$.Answer:$10$