Data Sufficiency

If v,w,x,y are non-negative integers, and a=3w2v b=5z3y2x is a/b a terminating decimal?

(1) w>y
(2) v>x

so far, only 6 out of 85 people answered correctly

Please help.

Just a try.

A fraction can be terminating only if the denominator is 2 and/or 5 , else it is recurring.

so none of the statements alone or combined tell us anything about what the denominators x,y,z are hence answer should be (e) none of the statements are sufficient.

xxpatzz wrote:If v,w,x,y are non-negative integers, and a=3w2v b=5z3y2x is a/b a terminating decimal?

(1) w>y
(2) v>x

so far, only 6 out of 85 people answered correctly

Please help.

I'm assuming that w, v, x, y and z are exponents and that the question is asking:

Is (3^w)(2^v) / (5^z)(3^y)(2^x) a terminating decimal?

Dividing by a power of 2 results in a terminating decimal:
1/2 = .5
1/2² = .25
1/2³ = .125
etc.

Dividing by a power of 5 results in a terminating decimal:
1/5 = .2
1/5² = .04
1/5³ = .008
etc.

Dividing by a power of 3 results in a non-terminating decimal:
1/3 = .33333...
1/3² = .11111...
1/3³ = .037037037...

Thus, we need to know whether (3^w)(2^v) / (5^z)(3^y)(2^x) will require dividing by a power of 3.
If there are more 3's in the denominator than in the numerator, then (3^w)(2^v) / (5^z)(3^y)(2^x) will require dividing by a power of 3.
Since the number of 3's depends on the sizes of the exponents, the question can be rephrased:

Is y>w?

Statement 1: w>y.
Thus, we know that (3^w)(2^v) / (5^z)(3^y)(2^x) will NOT require dividing by a power of 3, indicating that (3^w)(2^v) / (5^z)(3^y)(2^x) will be a terminating decimal.
Sufficient.

Statement 2: v>x.
No information about whether y>w.
Insufficient.

The correct answer is A.

Mom4MBA wrote:Just a try.

A fraction can be terminating only if the denominator is 2 and/or 5 , else it is recurring.

so none of the statements alone or combined tell us anything about what the denominators x,y,z are hence answer should be (e) none of the statements are sufficient.

Hi,
No. 2 and 5 are the only prime numbers as denominator which result in terminating.
Every denominator that can be written in 2^a*5^b will result in terminating number, where a and b are non negative integers.
Example:20 ->2^2*5^1

So...this 2 and 5 denom rule is universal , should I note this down ...or there are any expcetions too ?

ArpanaAmishi wrote:So...this 2 and 5 denom rule is universal , should I note this down ...or there are any expcetions too ?

Hi,
Yes. Whenever the denominator is of that form , it will be terminating. No exceptions.

If v,w,x,y are non-negative integers, and a=3w2v b=5z3y2x is a/b a terminating decimal?

(1) w>y
(2) v>x

so far, only 6 out of 85 people answered correctly

people will definitely give you wrong answer if you give them wrong question. you should have mentioned the powers properly. Had you written properly, it wasn't a tough question.
a=3^w 2^v b=5^z 3^y 2^x

GMATGuruNY wrote:

xxpatzz wrote:If v,w,x,y are non-negative integers, and a=3w2v b=5z3y2x is a/b a terminating decimal?

(1) w>y
(2) v>x

so far, only 6 out of 85 people answered correctly

Please help.

I'm assuming that w, v, x, y and z are exponents and that the question is asking:

Is (3^w)(2^v) / (5^z)(3^y)(2^x) a terminating decimal?

Dividing by a power of 2 results in a terminating decimal:
1/2 = .5
1/2² = .25
1/2³ = .125
etc.

Dividing by a power of 5 results in a terminating decimal:
1/5 = .2
1/5² = .04
1/5³ = .008
etc.

Dividing by a power of 3 results in a non-terminating decimal:
1/3 = .33333...
1/3² = .11111...
1/3³ = .037037037...

Thus, we need to know whether (3^w)(2^v) / (5^z)(3^y)(2^x) will require dividing by a power of 3.
If there are more 3's in the denominator than in the numerator, then (3^w)(2^v) / (5^z)(3^y)(2^x) will require dividing by a power of 3.
Since the number of 3's depends on the sizes of the exponents, the question can be rephrased:

Is y>w?

Statement 1: w>y.
Thus, we know that (3^w)(2^v) / (5^z)(3^y)(2^x) will NOT require dividing by a power of 3, indicating that (3^w)(2^v) / (5^z)(3^y)(2^x) will be a terminating decimal.
Sufficient.

Statement 2: v>x.
No information about whether y>w.
Insufficient.

The correct answer is A.

Unfortunately, w v z y x are not exponents, I just double checked the original question.

Mom4MBA wrote:

If v,w,x,y are non-negative integers, and a=3w2v b=5z3y2x is a/b a terminating decimal?

(1) w>y
(2) v>x

so far, only 6 out of 85 people answered correctly

people will definitely give you wrong answer if you give them wrong question. you should have mentioned the powers properly. Had you written properly, it wasn't a tough question.
a=3^w 2^v b=5^z 3^y 2^x

Nope, I just double checked the original question ,they are not exponents.

xxpatzz wrote:

GMATGuruNY wrote:

xxpatzz wrote:If v,w,x,y are non-negative integers, and a=3w2v b=5z3y2x is a/b a terminating decimal?

(1) w>y
(2) v>x

so far, only 6 out of 85 people answered correctly

Please help.

I'm assuming that w, v, x, y and z are exponents and that the question is asking:

Is (3^w)(2^v) / (5^z)(3^y)(2^x) a terminating decimal?

Dividing by a power of 2 results in a terminating decimal:
1/2 = .5
1/2² = .25
1/2³ = .125
etc.

Dividing by a power of 5 results in a terminating decimal:
1/5 = .2
1/5² = .04
1/5³ = .008
etc.

Dividing by a power of 3 results in a non-terminating decimal:
1/3 = .33333...
1/3² = .11111...
1/3³ = .037037037...

Thus, we need to know whether (3^w)(2^v) / (5^z)(3^y)(2^x) will require dividing by a power of 3.
If there are more 3's in the denominator than in the numerator, then (3^w)(2^v) / (5^z)(3^y)(2^x) will require dividing by a power of 3.
Since the number of 3's depends on the sizes of the exponents, the question can be rephrased:

Is y>w?

Statement 1: w>y.
Thus, we know that (3^w)(2^v) / (5^z)(3^y)(2^x) will NOT require dividing by a power of 3, indicating that (3^w)(2^v) / (5^z)(3^y)(2^x) will be a terminating decimal.
Sufficient.

Statement 2: v>x.
No information about whether y>w.
Insufficient.

The correct answer is A.

Unfortunately, w v z y x are not exponents, I just double checked the original question.

If v, w, x, y, and z are not exponents:

a/b = (3w2v)/(5z3y2x) = (vw)/(5xyz).

The following combinations satisfy both statements:
w=2, y=1, v=2, x=1, z=1.
a/b = (vw)/(5xyz) = (2*2)/(5*1*1*1) = .8.

w=2, y=1, v=2, x=1, z=12.
a/b = (vw)/(5xyz) = (2*2)/(5*1*1*12) = .0666666...

Since in the first case a/b is a terminating decimal and in the second case a/b is not a terminating decimal, the correct answer is E.

I still suspect that the variables are supposed to represent exponents. Otherwise, the question would need to specify that x, y and z are non-ZERO so that a/b is always defined. The word non-NEGATIVE seems more appropriate for exponents; this restriction guarantees that the denominator will not include fractions.

learning is if denominator can be expressed in terms of 2^x * 5^y. then it is terminating decimal.
else it will be always a non terminating decimal with 3^w and 7^k coming into picture.