edited by
4,615 views
0 votes
0 votes

Let $a, b, c, d$ and $e$ be integers such that $a = 6b = 12c,$ and $2b = 9d = 12e.$ Then which of the following pairs contains a number that is not an integer?

  1. $\left[ \frac{a}{27}, \frac{b}{e} \right] $
  2. $\left[ \frac{a}{36}, \frac{c}{e} \right] $
  3. $\left[ \frac{a}{12}, \frac{bd}{18} \right] $
  4. $\left[ \frac{a}{6}, \frac{c}{d} \right] $
edited by

1 Answer

Best answer
1 votes
1 votes
Answer (D)  :- Here $\frac{c}{d}$ could not be an integer

Given a = 6b = 12c

Let  $\frac{a}{12}$ = $\frac{b}{2}$ = c = k

a=12k

b=2k

c=k...............................i

Similarly,

2b = 9d = 12e

Let $\frac{b}{18}$ = $\frac{d}{4}$ =$\frac{c}{3}$ =k

b=18k

d=4k

e=3k..............................ii

 from i and ii we get

a=108k

b=18k

c=9k

d=4k

e=3k

 Now, putting in option we get D) as ans.
selected by

Related questions

0 votes
0 votes
1 answer
1
go_editor asked May 5, 2016
521 views
If three positive real numbers $x, y, z$ satisfy $y – x = z – y$ and $x y z = 4,$ then what is the minimum possible value of $y?$$2^{\frac{1}{3}}$$2^{\frac{2}{3}}$$2^...
0 votes
0 votes
0 answers
2
0 votes
0 votes
1 answer
5
go_editor asked May 4, 2016
1,088 views
Let $x$ and $y$ be positive integers such that $x$ is prime and $y$ is composite. Then,$y – x$ cannot be an even integer$xy$ cannot be an even integer.$\frac{x+y}{x}$ c...