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Five DS questions for practice today. Try !

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Five DS questions for practice today. Try !

by neilcao » Mon Jan 24, 2011 8:09 am
No.1 The probability for A to happen is 40%, for B to happen is 65%. The probability for any one of them to happen?

（1）A and B are independent
（2）The probability for A and B both not to happen is 22%

For this question, we can tell from choice 2 that the probability for A and B both to happen is 78%, how could we deep further?

No.2 n is positiver integer and n >=0 ï¼› n has remainder 1 when divided by 2, which one of the following is correct?
(1) n >0
(2) n = 2^n
(3) 3^n = (-3)^n

I am thinking that n could be 3, 5, 7, 9, 11..... the answer is only n > 0 is correct?


No3. Is 3 a^2+b^2+c^2 = (a+b+c)^2, a, b, c are all positive integer.
(1) a^2=b^2=4
(2) 1/a+1/b+1/c=0

For choice 1, I think we can try some numbers, and it is not sufficient. For choice 2, could that be a,b,c are all 0, so it is sufficient?


No.4. 15n, n is positive integer, can 15n be divided by 6 ?
(1)n can be divided by 3ï¼›(2)n can be divided by 2

For choice 1, it is not sufficient, e.g.15 x 6= 90. can be divided by 6, but if n= 3, 45 cannot be divided by 6
For choice 2, it is sufficient I suppose ?


No.5 N >10，can we determine the value of n?
(1) the tenth number of n =1/n
(2) the hundredth of n = 1/n

Need some ideas here.
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by Anurag@Gurome » Mon Jan 24, 2011 12:18 pm
neilcao wrote:2. n is positiver integer and n >=0 ï¼› n has remainder 1 when divided by 2, which one of the following is correct?
(1) n >0
(2) n = 2^n
(3) 3^n = (-3)^n


When a positiver integer has remainder 1 when divided by 2, the integers is an odd integer.

Now,
  • 1. n > 0 is true as n is a positive integer.
    2. n = 2^n --> An even integer --> FALSE
    3. 3^n = (-3)^n --> Only possible if n is even --> FALSE
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by Rahul@gurome » Mon Jan 24, 2011 12:31 pm
neilcao wrote:3. Is a² + b² + c² = (a + b + c)², a, b, c are all positive integer.
(1) a² = b² = 4
(2) 1/a + 1/b + 1/c = 0
Statement 1: a² = b² = 4
Thus, a = b = ±2
Hence, (a + b) = -4 OR 0 OR 4
=> (a + b + c) = (c - 4) OR c OR (c + 4)
=> (a + b + c)² = (c - 4)² OR c² OR (c + 4)²

Now, a² + b² + c² = 4 + 4 + c² = 8 + c²

If a = b = 2, (a + b + c)² = (c + 4)² = (c² + 8c + 16) = c² + 8 = a² + b² + c² for c = -1
If a = -2 and b = 2, (a + b + c)² = c² ≠ c² + 8 = a² + b² + c² for any value of c

Not sufficient

Statement 2: 1/a + 1/b + 1/c = 0
Implies (ab + bc + ca) = 0
Now, (a + b + c)² = (a² + b² + c² + 2ab + 2bc + 2ac) = (a² + b² + c²)

Sufficient

The correct answer is B.
Last edited by Rahul@gurome on Mon Jan 24, 2011 10:27 pm, edited 2 times in total.
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by Rahul@gurome » Mon Jan 24, 2011 9:20 pm
neilcao wrote:4. If n is positive integer, can 15n be divided by 6 ?
(1) n can be divided by 3
(2) n can be divided by 2
Statement 1: n can be divided by 3.
Now, if n is divisible by 2 --> 15n is divisible by 6
But, if n is not divisible by 2 --> 15n is not divisible by 6

Not sufficient

Statement 2: n can be divided by 2.
Now, n is divisible by 2 and 15 is divisible by 3.
Hence, their product 15n is divisible by 2*3 = 6.

Sufficient

The correct answer is B.
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by Rahul@gurome » Mon Jan 24, 2011 9:35 pm
neilcao wrote:5. If n is a positive integer that is less than 10, what is the value of n?

1) n is the tenths digit in the decimal representation of 1/n.
2) n is the hundredths digit in the decimal representation of 1/n.
I believe the original question is more like the above.

Now, as n < 10, n can be any of the following
  • n = 1 --> 1/n = 1.0
    n = 2 --> 1/n = 0.5
    n = 3 --> 1/n = 0.3333...
    n = 4 --> 1/n = 0.25
    n = 5 --> 1/n = 0.2
    n = 6 --> 1/n = 0.1666...
    n = 7 --> 1/n = 0.14...
    n = 8 --> 1/n = 0.125
    n = 9 --> 1/n = 0.1111....
Statement 1: Only possible value of n is 3.

Sufficient

Statement 2: Possible values of n are 3 and 6.

Not sufficient

The correct answer is A.
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by kvcpk » Mon Jan 24, 2011 9:48 pm
neilcao wrote:No3. Is 3 a^2+b^2+c^2 = (a+b+c)^2, a, b, c are all positive integer.
(1) a^2=b^2=4
(2) 1/a+1/b+1/c=0

For choice 1, I think we can try some numbers, and it is not sufficient. For choice 2, could that be a,b,c are all 0, so it is sufficient?
Need some ideas here.
Are you sure about this question?

if a,b,c are positive, I dont think 1/a + 1/b + 1/c can ever be 0.
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by Rahul@gurome » Mon Jan 24, 2011 10:07 pm
kvcpk wrote:Are you sure about this question?

if a,b,c are positive, I dont think 1/a + 1/b + 1/c can ever be 0.
You're correct.
Ideally the question should declare a, b, c as non-zero integers.
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by kvcpk » Mon Jan 24, 2011 10:19 pm
Rahul@gurome wrote:
kvcpk wrote:Are you sure about this question?

if a,b,c are positive, I dont think 1/a + 1/b + 1/c can ever be 0.
You're correct.
Ideally the question should declare a, b, c as non-zero integers.
Hi Rahul,

I checked your post above. Even if a,b,c are non zero integers,
1/a + 1/b + 1/c =0 doesnt mean a+b+c =0
eg: a=2,b=-4,c=-4
ab+bc+ca=0 would be right.

you might need to edit that.
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by Rahul@gurome » Mon Jan 24, 2011 10:29 pm
@kvcpk : Thanx for pointing it out. Edited the reply.
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