EgmatQuantExpert wrote:
Hi Guys,
Please find below a practice question based on similar lines as the
OG question in this thread:
If \(a\) is a multiple of 5, \(b\) is a multiple of 8 and \(c\) is a multiple of 10 with \(c\) having no common factor other than 10 with the product \(ab\), then \(\frac{a^3*b^2}{c^2}\) must be a multiple of which of the following?
I. 20
II. 40
III. 160 (A) I only
(B) I & III only
(C) II & III only
(D) I & II only
(E) I, II & III
Please post your analysis along with the answers.
The OA and the solution will be posted on May 1, 2015.
Regards
Harsh
The correct answer is Option DGivenWe are given three numbers a, b, c such that a is a multiple of 5, b is a multiple of 8 and c is a multiple of 10 with c having no common factors with the product ab other than 10. We are asked to find \(\frac{(a^3*b^2)}{c^2}\) is a multiple of which of 20, 40 and 160.
ApproachWe are given that a is a multiple of 5, so a = 5x.
Since b is a multiple of 8, b = \(2^3\)y
Also, c is a multiple of 10. So, we can write c = 2*5z. Now, we are given that the only factor that c has in common with the product ab is 10 (which is 2*5). The important inference to draw from this information is that z contains no 2 or 5.
We will use the above expressions to find out if \(\frac{(a^3*b^2)}{c^2}\) is a multiple of which of 20, 40 and 160.
Working OutSubstituting the values of a, b and c, we can write \(\frac{(a^3*b^2)}{c^2}\) = \(\frac{((5x)^{3}*(2^{3}*y)^2)}{(2*5z)^2}\) = \(\frac{(5*2^{4}*x^{3}*y^{2})}{z^2}\) which can be simplified to \(\frac{(80x^{3}y^{2})}{z^2}\) . So, we can say that \(\frac{(a^{3}*b^{2})}{c^2}\) will definitely be a factor of 20 and 40.
Answer (D)