If a, b, c, d are integers, p =

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GMATBusters’ Quant Quiz Question -4

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If a, b, c, d are integers, p = 2^a*3^b and q = 2^c*5^d. Is p/q a terminating decimal?
1) a>c
2) b>d

GMATBusters · Answer

p/q is terminating if p, q are integers, and factors of denominator contain only multiple of 2 & 5.

p = 2^a*3^b and q = 2^c*5^d
hence, we are not concerned about the power of 2 & 5 but 3. because if 3 is in denominator it will make p/q non-terminating.
1) a>c
nothing about the power of 3, if power of 3 is negative, it will make p/q non-terminating.
Hence NOT Sufficient.
2) b>d
nothing about the power of 3, if the power of 3 is negative, it will make p/q non-terminating.
Hence NOT Sufficient.

Even after combining, we cannot find the power of 3.
Hence NOT Sufficient.

Answer E

Raxit85 · Answer

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

To get terminal decimal we may need 2 or 5 or both in denominator.

(1) a > c. Sufficient as we can have both 2 and 5 in in denominator.
(2) b > d. Sufficient as we can have both 2 and 5 in in denominator.

Ans D.

ashwini2k6jha · Answer

p/q=2^a-c*3^b*5^-d terminating?
state-1:Not sufficient
a>c
no information about b and d
stae:2 not sufficient 
b>d
nothing about a and can not decide conclusively about b and d
1+2:not sufficient answer E
let b=3 and d=2
b=-2 d=-3
in both cases denominator may or may not contain 3 or 5

Kinshook · Answer

Asked: If a, b, c, d are integers, p = 2^a*3^b and q = 2^c*5^d. Is p/q a terminating decimal?
p/q = 2^{a-c}*3^b*5^{-d}
p/q is a terminating decimal if b>=0; Since any powers of 2 & 5 result in terminating decimal.

1) a>c
Since a>c; a-c>0
Since b is unknown
NOT SUFFICIENT

2) b>d
If d<0; b may be <0 or 0 or >0; not sufficient
If d>=0; b>0; sufficient
But sign of d is unknown
NOT SUFFICIENT

(1) + (2)
1) a>c
Since a>c; a-c>0
2) b>d
If d<0; b may be <0 or 0 or >0; not sufficient
If d>=0; b>0; sufficient
But sign of d is unknown
NOT SUFFICIENT

IMO E

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Archit3110 · Answer

If a, b, c, d are integers, p = 2^a*3^b and q = 2^c*5^d. Is p/q a terminating decimal?
1) a>c
2) b>d
the ratio of x/y will be terminating when least form the denominator has NO prime factors other than 2 and 5
#1
a>c
no relation know whether c is -ve or +ve and d also not know insufficient
#2
b>d
no relation know whether d is -ve or +ve and c also not know insufficient
from 1&2
we do not know whether the values of c& d are +ve or -ve integers so insufficient
IMO E

uc26 · Answer

We are given that a,b,c,d are integers. We do not know if these are positive or negative integers. A reduced fraction can be said to be terminating decimal if the denominator (q in this case) is of the form  2^x 5^y . We know that  p=2^a*3^b  and  q=2^c*5^d . Notice that, p/q will be a terminating decimal only if b is non-negative since if b is negative, then the the denominator will contain some form of 3 which will make p/q a non-terminating decimal.

(1) a>c. We do not know the positive or negative nature of b. Not sufficient.

(2) b>d. We do not know if d is positive or negative. Consequently, b could be positive or negative. Not sufficient.

(1)+(2) Combining the statements, we still have no information about the positive or negative nature of b. Not sufficient.

Answer: E.

Tommy6e · Answer

Answer is E

If a, b, c, d are integers, p = 2^a*3^b and q = 2^c*5^d. Is p/q a terminating decimal?

The expression will be a terminating decimal if and only if its denominators will have powers of only 5 and/or 2.
We're told that a, b, c, and d are integers; however, we're not told they are positive.
The only option for this expression to not be a terminating decimal is when b is negative and thus 3 is in the denominator.

Statement 1:
1) a>c
This statement is irrelevant since both a and c decides how many 2's will or will not be in the denominator. This will not affect the answer.

Statement 2:
2) b>d
This keeps us with 2 scenarios
Scenario 1 - b is positive --> the expression is a terminating decimal
Scenario 2 - b is negative --> the expression is not a terminating decimal

Mixing both statements still keeps this open with the above two scenarios.

1BlackDiamond1 · Answer

If a, b, c, d are integers, p = 2^a*3^b and q = 2^c*5^d. Is p/q a terminating decimal?
1) a>c
Among 1/2, 1/3 and 1/5, only 1/3 is non terminating
The denominator q = 2^c*5^d needs to have 2 and 5 with positive powers for p/q to be a terminating decimal.
if a>c then the 2 in the denominator is eliminated and we are left with 5 in the denominator. now since d can be negative then the denominator becomes 1 and is divisible.
If c is negative we have a denominator with 3 or 3 and 5 which may or may not have a terminating decimal.

Statement I is insufficient.
2) b>d
Among 1/2, 1/3 and 1/5, only 1/3 is non terminating
Similar logic goes for Statement II if a, b, c, or d is negative then the terminating decimal does not hold true.

Statement II is insufficient.

Statement I and II together do not suffice that a, b, c, or d is nonnegative.
Hence answer is E

kapilagarwal · Answer

ANS : D EACH ALONE SUFFICIENT.

Posted from my mobile device

SR71Blackbird · Answer

The answer to this question is D

For a fraction to be terminating, the denominator should only have factors of 2 and/or 5 and no other factors. 
1/32 terminating (only factors of 2)
1/25 terminating (only factors of 5)
1/10 terminating (only factors of 2 and 5)
1/28 non-terminating (factors of 7 are present)

Statement 1: a >c c is =ve then 2 will be cancelled in denominator, 5 will be left over which is sufficient. c is -ve, 2 will go to nem. same situation.

Statement 2: b>d still doesn't change the conditions and the fraction will end (even if the powers are zero)

zs2 · Answer

If a, b, c, d are integers, p = 2^a*3^b and q = 2^c*5^d. Is p/q a terminating decimal?
1) a>c
2) b>d

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

Statement 1- 
a>c

let a=2, c=1, d=0, b=-1, putting values we get- fraction as 2/3 non terminating
let a=2, c=1, d=1, b=1, putting values we get- fraction as 6/5 terminating
not sufficient

Statement 2- 
let a=2, c=1, d=-1, b=-2, putting values we get- fraction as 50/3 non terminating
let a=2, c=1, d=1, b=2, putting values we get- fraction as 36/5 terminating
not sufficient

combining 1 and 2 --
cases taken in statement are similar to combined cases which were also not sufficient, Hence E is the answer

firas92 · Answer

For a fraction to be a terminating decimal, the denominator must have no other prime factors with positive powers than 2 and 5.

So in this case, the fraction will be a terminating decimal as long as b is not negative.

Neither statement tells us whether b is non negative

Answer is (E)

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lacktutor · Answer

If a, b, c, d are integers,  p = 2^a*3^b and  q = 2^c*5^d  Is p/q a terminating decimal?

(Statement1): a > c
(Statement2): b > d 
If you combine two statements, and they are not sufficient to solve the question, each statement alone is not sufficient.

Taken together 1&2, 
If a= 3, c= 1 ( a>c) and 
If b= 4, d =2 (b > d), then 
—>  p/q = 2^{3} * 3^{4} /2^{1}*5^{2} = 4*81/25 = 4*4*81/ 100= 1296/100= 12.96  ( terminating decimal)

If a =3, c=1 (a >c) and 
If b= —1, d= —2 (b >d), then 
—>  p/q= 2^{3}*3^{—1}/ 2^{1}* 5^{—2} = 4*25/3 = 100/3 = 33.333333  ( not terminating decimal) 
Insufficient

Answer (E)

Posted from my mobile device

jrk23 · Answer

i think "e" is the answer. As "b" can be negative also. and if b is negative than its not terminating.

So i will go with "e"

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If a, b, c, d are integers, p =

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