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!

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!

can you prove this by picking negative numbers ?
eg b=-2, d=-3

why e should be one [e =1]

why e should be one [e =1]

why e should be one [e =1]

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

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

Thnx for explaining the case when b,d are negative. It helps.

B is the OA.

Thnx for explaining the case when b,d are negative. It helps.

B is the OA.[/quote]

I just got this problem on a practice test and just FYI, the question stem says "a,b,c,d,e are non-negative integers"

Therefore, there is no need to explain the negative for B,D.