Statement I: Both \(x\) and \(y\) are consecutive even multiples of \(32\).
So, if \(x = 32(2k) = 64k\), then \(y = 32(2k + 2) = 64(k + 1)\)
Product of x and \(y = 64k(64k + 64) = 4096k(k + 1)\)
We do not know the value of \(‘k’\), so we cannot find the product.
Thus, Statement I alone is not sufficient to answer the question.
Statement II: The LCM of x and \(y\) is \(3584\).
Let x and \(y\) be \(‘ha’\) and \(‘hb’\), respectively, where \(‘h’\) is the \(HCF\) of \(x\) and \(y\), and \(‘a’\) and \(‘b’\) are co-primes.
Thus, \(hab = 3584\)
Product of \(x\) and \(y\) = \(ha(hb) = h^2(ab) = 3584\;h\)
We do not know the value of \(‘h’\), so we cannot find the product of \(x\) and \(y\).
Using statements I and II together,
\(LCM \) of \(x\) and \(y\) = \(64k(k + 1) = 3584\)
or, \(k(k + 1) = 56\)
or, \(k = 7\)
Now, the product of \(x\) and \(y\), i.e., \(4096k(k + 1)\) can be calculated.
Thus, both the statements are necessary to answer the question.
Hence, option \(A\) is the correct answer.
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |