Data Sufficiency:
If a, b, c, d and e are integers and p = 2^a * 3^b and q = 2^c * 3^d * 5^e, is p/q a terminating decimal?


(1) a > c

(2) b > d

Note: ^ represents raise to power
* represents Multiplication

The fraction will terminate if and only if the denominator has prime divisors only 2 and 5 or both.

Stmt1: insuff. The denominator may or may not have 3.

Stmt2: Suff. Denominator will have only 2 and 5.

Answer is 'B'

I am still not clear why only 2 and 5 and not 3? CAn you please elaborate?

If you reduce a fraction to its lowest terms, you are guaranteed to have a terminating decimal provided that the only prime factors in the denominator are 2 and/or 5. There are probably some rigorous proofs out there to show this, but you can just try some numerical examples to convince yourself of this property. (A calculator may be helpful.)

Rey

I still don't understand this question... Can someone please further clarify?

Rephrase the Question as:-
If a, b, c, d and e are integers and p = 2^a * 3^b and q = 2^c * 3^d * 5^e, is p/q a terminating decimal?

p/q= 2^a*3^b/ 2^c*3^d*5^e
= 2^a-c * 3^b-d * 5^-e

Is P/Q = terminating decimal? (which means p/q= 1.34656)

or in other words p/q not equal to values (like 10/3= 3.33333333333 which does not terminate)

Thanks

ok people, here it is, as succinctly as i can put it:
if you want a terminating decimal, you have to have only 2's and/or 5's in the denominator (after you factor everything into primes, and cancel all the primes that are common to the top and bottom).
this condition also works in converse: if you write a fraction whose denominator is a product of 2's and 5's only, it's guaranteed to be a terminating decimal - and if there's any other prime in there, it won't terminate.

--

in this question, then, you can rephrase the question as follows:
are there any 3's in the denominator? (yes --> not terminating; no --> terminating{
which is the same as
is b > d? (because all the 3's in the denominator will cancel if this is so, and not otherwise)

the rest follows.

Our number system is of base 10. Decimals in this system represents 10th part. When we say 5.8 it means 5 and 8/10. 5.86 = 5 & 8/10 & 6/100.
Thus a terminating decimal is always INTEGER/(10^n). If and only if a fraction can be converted to a form of integer divided by some power of 10 we will get a terminating decimal. This comes from the very basic fact that our number system is of base 10.

anything / 2 will be terminating. if we multiply numerator and denominator by 5 we will get a form INTEGER/10.
anything / 5 will terminate since we can multiply numerator and denominator by 2 we will get a form INTEGER/10.
Note that any 2 ^ (to the power of ) INT will also terminate as we can multiple by 5^ Same INT. ( 2^3 * 5^3 = 10^3)

Multiplying 3 with an integer will never give you 10 or 100 or any power of 10. why??? 10^n has only 2 & 5 as prime factors but if 3 * INT = 10000; then 10000 should have 3 as a factor, a condition not possible.
Similarly multiplying 7 or 9 or 11 or 15 etc with INT will never result to 10^n.
I hope now you are clear why 2's and 5's in Deno. will be terminating decimal and 3's 6's and 7's etc. will not.. Now read Ron's post for the DS question.
Finally the meaning of non-terminating decimals in this 10-base world -
3.33333..... is 3+ 3/10+ 3/100 + 3/1000 .... so on. This number in its decimal form of 3.3333..... can never be written exactly as INTEGER / 10^n. We can write if approximately as 33/10, 333/100 or 3333/1000, and so on but always we will always have some left over.

Well put, captain.

Rey