can you prove this by picking negative numbers ?shanrizvi wrote:The best way to do this is by picking numbers.
Is p/q = (2^a.3^b)/(2^c.3^d.5^e) a terminal decimal?
e=1
Statement 1: a>c
a=4, c=3
b=2, d=2
p/q = 2/5 (terminating decimal)
Lets pick different values for b and d. b=3, d=4
p/q = 2/15 (not a terminating decimal)
Hence, statement 1 is insufficient.
Statement 2: b>d
b=4, d=3
a=2, b=2
p/q = 3/5 (terminating decimal)
lets take a=2, b=3
p/q = 3/10 (terminating decimal)
Statement 2 is sufficient. Answer: B!
Thnx for explaining the case when b,d are negative. It helps.sreak1089 wrote:Ans is B IMO.
stmt # 1: a > c Irrespective of whether a > c or a < c, any power of 2 would result in a terminating decimal. Same goes with powers of 5. Since we have powers of 3 in the expression, which can yield a non-terminating decimial, hence stmt #1 is NOT SUFFICIENT.
stmt # 2: b > d => Whether the expression is terminating or not depends on what (b-d) evaulates to. Three possible cases exist for (b-d)
b +ve d +ve => (b-d) +ve (since b > d)
Hence expression is terminating.
b +ve d -ve => (b-d) +ve (since b-d = b+|d|)
Hence expression is terminating.
b -ve d -ve => (b-d) +ve (since b-d = |d|-|b| &&
|d| > |b|)
Hence expression is terminating.
Hence stmt # 2 is SUFFICIENT.
Answer is B
