DS question a Day

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Re: DS 007

by rajs.kumar » Tue Nov 14, 2006 4:34 am
rajs.kumar wrote:DS 007

If △ denotes an operation, what is the value of (a△b) △c ?

1. a △ b = 5
2. 5 △ c = 3
Answer is C

(1) alone is insuff - no information about c
(2) alone is insuff - no information about a and b

(1) and (2) no information about △ but values after individual operations are provided

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Re: DS004

by Bharadwaj » Tue Nov 14, 2006 3:36 pm
ajith wrote:DS004

is |x+y| = 5?

1) |x| = 3
2) |y| = 2
Obviously we need both x and y. But both stmts are inadequate to answer the question..

Stmt1 : X = -3 or 3
Stmt2 : Y = -2, or 2

So |x+y| = 1 or 5

E is my answer...
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Re: DS 006

by kulksnikhil » Tue Nov 14, 2006 9:33 pm
rajs.kumar wrote:
rajs.kumar wrote:DS 006

13. If a, b, and c are integers, is the number 3(a + b) - c divisible by 3?
(1) a + b is divisible by 3
(2) c is divisible by 3
Answer is B.

(1) nothing known about C => insuff
(2) 3(a+b) and c are divisable => suff

I think there is some confusion here... the answer should be 'C' and not 'B'

Coz in (2) u r assuming even 3(a+b) is divisible by 3 which is given in (1) and not in B :D

Please correct me if I am wrong.

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Re: DS 006

by rajs.kumar » Wed Nov 15, 2006 1:48 am
kulksnikhil wrote: I think there is some confusion here... the answer should be 'C' and not 'B'

Coz in (2) u r assuming even 3(a+b) is divisible by 3 which is given in (1) and not in B :D

Please correct me if I am wrong.
the question stem states a, b and c are integers. Let say a + b = x, then is 3x divisable by 3?

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Re: DS 008

by rajs.kumar » Tue Nov 21, 2006 4:34 pm
rajs.kumar wrote:DS 008

If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x
(2) |x| > x
The question can be restated as -1 < x < 1

Stmt 1

x > 0 => x/x < x or x > 1
x < 0 => x/-x < x or x > -1 => insuff

Stmt 2

x > 0 => x > x (not true)
x < 0 => -x > x or x < 0 => insuff

(1) & (2) => x < 0 and x > -1 => -1 < x < 0 => suff

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Re: DS 008

by iraja » Wed Nov 22, 2006 1:29 pm
rajs.kumar wrote:
rajs.kumar wrote:DS 008

If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x
(2) |x| > x
The question can be restated as -1 < x < 1

Stmt 1

x > 0 => x/x < x or x > 1
x < 0 => x/-x < x or x > -1 => insuff

Stmt 2

x > 0 => x > x (not true)
x < 0 => -x > x or x < 0 => insuff

(1) & (2) => x < 0 and x > -1 => -1 < x < 0 => suff


I am bit confused with the answer.

Can't we try this approach??

Stmt 1

x/|x| < x

--> x/|x| - x < 0
--> x-x|x| /|x| < 0
--> x-x|x| < 0
--> x(1-|x|) < 0

--> 1-|x| < 0
1 < |x|

If you reverse the value, you can reverse the symbol

|x| > 1 So,It does prove that |x| is not less that 1.

2 is not suffient to answer. Thus B is the answer.

Is that right?

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DS

by mukul » Thu Dec 14, 2006 11:16 pm
I think what raj has supplied is pretty much the way to do it..its less twisted and simple...I would do it the same way

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Re: DS 008

by lidifantasy » Fri Jan 19, 2007 7:33 am
rajs.kumar wrote:
rajs.kumar wrote:DS 008

If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x
(2) |x| > x
The question can be restated as -1 < x < 1

Stmt 1

x > 0 => x/x < x or x > 1
x < 0 => x/-x < x or x > -1 => insuff

Stmt 2

x > 0 => x > x (not true)
x < 0 => -x > x or x < 0 => insuff

(1) & (2) => x < 0 and x > -1 => -1 < x < 0 => suff
I think the answer should be "A"
because according to(1),we can get that,
x/|x| < x----that means x/|x| equals to either 1 or -1, if the figar is positive, it
so we devide the whole integral set into:
x > 0 => x/x < x or x > 1
x < 0 => x/-x < x or x > -1
so |x|>1 it is suff.
(2)
x > 0 => x > x (not true)
x < 0 => -x > x or x < 0 => insuff

so choose A

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by navami » Mon Mar 14, 2011 5:05 am
ajith wrote:
ajith wrote:DS001

If x is positive, is x>3?

a) (x-1)^2 > 4
b) (x-2)^2 > 9



Answers In 3 days of time

When we solve a we get x>3 or x<-1(refer the above post for explanation)
since x is positive , x>3

So a is sufficient to answer

now conisdering b

(x-2)^2 > 9
=>(x-2+3)(x-2-3)>0
upon solving which we get x>5 or x<-1
since x is positive , x>5

So b is sufficient to answer

so the answer is D. Both alone are sufficient
A. On solving the equation, (x+1)^2 >4, we get (x-1)(x+3)>0

for this to be true:
either, both should be > 0 (positive)
(x-1)>0 and (x+3)>0 => x>1 and x>-3
or, both should be < 0 (positive)
(x-1)<0 and (x+3)<0 => x<1 and x<-3
so how can this be a sufficient condition?

B. Same doubt as in Option A
This time no looking back!!!
Navami

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by [email protected] » Sun Jul 31, 2011 4:09 pm
THE ANSWER IS D
FROM STATEMENT 1: (X-1) SQUARE > 4 MEANS EITHER X>3 OR X<-1. BUT FROM THE QUESTION IT WRITTEN THAT X IS A POSITIVE. SO X>3. SUFFICIENT.
FROM STATEMENT 2:(x-2)^2 > 9 MEANS THAT EITHER X>5 OR X<-1. BUT X IS POSITIVE . SO X>5. HENCE SUFFICIENT.
THUS ANSWER SHOULD BE D

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by naveen451 » Mon Aug 01, 2011 9:23 am
IMO D