Quantitative Aptitude Ques 1891
Question: Which of the following equations are equivalent?
I. $ {{( \frac{1}{2}M+\frac{2}{3}N )}^{2}} $ II. $ \frac{4}{9}N^{2}+\frac{1}{4}M^{2}+\frac{2}{3}MN $ III. $ ( \frac{M}{2}+\frac{2}{3}N )( \frac{1}{2}M-\frac{2}{3}N ) $ IV. $ \frac{1}{4}{{( M+\frac{4}{3}N )}^{2}} $
Options:
A) II and III
B) I and IV
C) I and II
D) I and III
E) I, II and IV
Show Answer
Answer:
Correct Answer: E
Solution:
- Simplifying .all the equations, I. $ {{( \frac{1}{2}M+\frac{2}{3}N )}^{2}}=\frac{1}{4}M^{2}+\frac{4}{9}N^{2}+\frac{2}{3}MN $ II. $ \frac{4}{9}N^{2}+\frac{1}{4}M^{2}+\frac{2}{3}MN=\frac{1}{4}M^{2}+\frac{4}{9}N^{2}+\frac{2}{3}MN $ III. $ ( \frac{M}{2}+\frac{2}{3}N )( \frac{1}{2}M-\frac{2}{3}N ) $ $ =\frac{1}{4}M^{2}+\frac{1}{3}MN-\frac{1}{3}MN-\frac{4}{9}N^{2}=\frac{1}{4}M^{2}-\frac{4}{9}N^{2} $ IV. $ \frac{1}{4}{{( M+\frac{4}{3}N )}^{2}}=\frac{1}{4}[ M^{2}+\frac{16}{9}N^{2}+\frac{8}{3}MN ] $ $ =\frac{1}{4}M^{2}+\frac{4}{9}N^{2}+\frac{2}{3}MN $ From the above four solutions, we find that I, II and IV are equivalent.
BETA
