My approach would be to resolve the fractions into prime factors. In this case whichever number has the largest denominator is the smallest number.

After resolving into prime factors cancel out the common terms ...

A) 3^2 * 5^6 * 2^2

B) 2 ^4 * 5^7

C) 5^7 * 2^3

D) 2^2 * 5^7

E) 3 * 2^4 *5^6

Then cancelling 5^6 and 2^2 in all the terms..
We know right away D and E are smaller than B.
Then between B and A, B is greater.
Then evaluate B and E. E is greater of all. Hence 1/E will be the smallest fraction.

rahul.s wrote:Which of the following is the smallest number?

OA: E

I'd like to learn some approaches to tackle this problem. I spent a good 5 minutes calculating!

arrange the denomn in terms of (5^5)*x

1/5^5*180
1/5^5*80
1/5^5*200
1/5^5*100
1/5^5*240

clearly E is smallest

could you please elaborate???

thephoenix wrote:

rahul.s wrote:Which of the following is the smallest number?

OA: E

I'd like to learn some approaches to tackle this problem. I spent a good 5 minutes calculating!

arrange the denomn in terms of (5^5)*x

1/5^5*180
1/5^5*80
1/5^5*200
1/5^5*100
1/5^5*240

clearly E is smallest

bhumika.k.shah wrote:could you please elaborate???

thephoenix wrote:

rahul.s wrote:Which of the following is the smallest number?

OA: E

I'd like to learn some approaches to tackle this problem. I spent a good 5 minutes calculating!

arrange the denomn in terms of (5^5)*x

1/5^5*180
1/5^5*80
1/5^5*200
1/5^5*100
1/5^5*240

clearly E is smallest

a)1/900*5^4=1/180*5^5
b)1/80*5^5
c)1/5000*5^3=1/1000*5^4=1/200*5^5
likewise converting everything in terms of 5^5
d)2/40*5^5=2/200*5^5=1/100*5^5
e)1/30000*5^2=1/6000*5^3=1/1200*5^4=1/240*5^5

now each deno is in the form 5^5 * x so it depends on x the larger the x the smaller the no. is which is E

thanks for the explanations rahul and phoenix