Question - Is x < y?
1. x - y + 1 < 0
2. x - y - 1 < 0
How would you go about solving this one?
Is x < y? from DS 1000 series
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Ans A here.
From stem 1 : x-y+1<0>x-y<-1 i.e (x-y) negative i.e. x<y ;sufficient
From stem 2 : x-y-1<0>x-y<1 here it could be 0, more than zero or even negative.So not sufficient.
From stem 1 : x-y+1<0>x-y<-1 i.e (x-y) negative i.e. x<y ;sufficient
From stem 2 : x-y-1<0>x-y<1 here it could be 0, more than zero or even negative.So not sufficient.
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Hi,amitansu wrote:Ans A here.
From stem 1 : x-y+1<0>x-y<-1 i.e (x-y) negative i.e. x<y ;sufficient
From stem 2 : x-y-1<0>x-y<1 here it could be 0, more than zero or even negative.So not sufficient.
I have a couple of more questions, which I have not answered correctly. Are you happy to answer more questions?
Vikas
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1. S is a set of integers such thatamitansu wrote:Go ahead and post 'em....
Amit
i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is –4 in S?
(1) 1 is in S.
(2) 2 is in S.
2. If x and y are integers, is xy + 1 divisible by 3?
(1) When x is divided by 3, the remainder is 1.
(2) When y is divided by 9, the remainder is 8.
3. If x≠0, is |x| < 1?
(1) x^2<1
(2) |x| < 1 / x
Thanks, for helping out Amit. Highly appreciate it.
Vikas
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Vikas,
Just a small request : Post one q per thread.So that it would be better to refer.
However for q i can you put it in a better way ? i couldn't follow it !!
For q 2 :
Ans would be E.
From 1 : x=3q+1 (q quotient here)
different values of x are 4,7,10,13 ......but no idea about y so insufficient
From 2: y=9q+8
so different values of y would be 17,26,35,44....still no idea about x . so insufficient.
Combining both also we have many combinations of xy which wouldn't verify whether (xy+1) is divisible by 3 oe not.Insufficient
For q 3 :
From 1 : x^2<1> +0r- (x)<1 which is mod(x)<1 here x being a factor.So sufficient.
From 2: mod(x)<1>+or-(x)<1/x (x is a factor here) but with -x being a factor would contradict the value.For example let x =1/2
so 1/2<1>1/2<2; but is -1/2 < 1/-(1/2) no.
However we are asked about whether mod(x)<1 ?
and from above we knew, yes definitely.So suficient.
Ans D.
Just a small request : Post one q per thread.So that it would be better to refer.
However for q i can you put it in a better way ? i couldn't follow it !!
For q 2 :
Ans would be E.
From 1 : x=3q+1 (q quotient here)
different values of x are 4,7,10,13 ......but no idea about y so insufficient
From 2: y=9q+8
so different values of y would be 17,26,35,44....still no idea about x . so insufficient.
Combining both also we have many combinations of xy which wouldn't verify whether (xy+1) is divisible by 3 oe not.Insufficient
For q 3 :
From 1 : x^2<1> +0r- (x)<1 which is mod(x)<1 here x being a factor.So sufficient.
From 2: mod(x)<1>+or-(x)<1/x (x is a factor here) but with -x being a factor would contradict the value.For example let x =1/2
so 1/2<1>1/2<2; but is -1/2 < 1/-(1/2) no.
However we are asked about whether mod(x)<1 ?
and from above we knew, yes definitely.So suficient.
Ans D.
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Amit,amitansu wrote:Vikas,
Just a small request : Post one q per thread.So that it would be better to refer.
However for q i can you put it in a better way ? i couldn't follow it !!
For q 2 :
Ans would be E.
From 1 : x=3q+1 (q quotient here)
different values of x are 4,7,10,13 ......but no idea about y so insufficient
From 2: y=9q+8
so different values of y would be 17,26,35,44....still no idea about x . so insufficient.
Combining both also we have many combinations of xy which wouldn't verify whether (xy+1) is divisible by 3 oe not.Insufficient
For q 3 :
From 1 : x^2<1> +0r- (x)<1 which is mod(x)<1 here x being a factor.So sufficient.
From 2: mod(x)<1>+or-(x)<1/x (x is a factor here) but with -x being a factor would contradict the value.For example let x =1/2
so 1/2<1>1/2<2; but is -1/2 < 1/-(1/2) no.
However we are asked about whether mod(x)<1 ?
and from above we knew, yes definitely.So suficient.
Ans D.
Thanks again. I'll post one question per thread. For question 2 the correct answer is given 'C'. Your answer to question 3 is correct. As for question 1, I am starting a new thread for it.
Vikas
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1. S is a set of integers such that
i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is –4 in S?
(1) 1 is in S.
(2) 2 is in S.
Hey Amit,
I have posted the q on a new thread, but I thought to give you some explanation on the q here. From above, the set should be something like -
S = {a, -a, b, ab}
1 is in S = {1, -1, b, 1 * b}
not sufficient to say whether -4 is in the set
2 is in S = {2, -2, b, 2b}
not sufficient to say whether -4 is in S
To my surprise the answer is 'B'. Let me know, if the explanation is of any help to you.
Vikas
i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is –4 in S?
(1) 1 is in S.
(2) 2 is in S.
Hey Amit,
I have posted the q on a new thread, but I thought to give you some explanation on the q here. From above, the set should be something like -
S = {a, -a, b, ab}
1 is in S = {1, -1, b, 1 * b}
not sufficient to say whether -4 is in the set
2 is in S = {2, -2, b, 2b}
not sufficient to say whether -4 is in S
To my surprise the answer is 'B'. Let me know, if the explanation is of any help to you.
Vikas
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(a) says that 1 is in the set. It does not say that only 1 is in the set. So the ans is -4 may or may not be in the set hence insufficient.
mehravikas wrote:1. S is a set of integers such that
i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is –4 in S?
(1) 1 is in S.
(2) 2 is in S.
Hey Amit,
I have posted the q on a new thread, but I thought to give you some explanation on the q here. From above, the set should be something like -
S = {a, -a, b, ab}
1 is in S = {1, -1, b, 1 * b}
not sufficient to say whether -4 is in the set
2 is in S = {2, -2, b, 2b}
not sufficient to say whether -4 is in S
To my surprise the answer is 'B'. Let me know, if the explanation is of any help to you.
Vikas
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What about point B {2 is in S}netigen wrote:(a) says that 1 is in the set. It does not say that only 1 is in the set. So the ans is -4 may or may not be in the set hence insufficient.
mehravikas wrote:1. S is a set of integers such that
i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is –4 in S?
(1) 1 is in S.
(2) 2 is in S.
Hey Amit,
I have posted the q on a new thread, but I thought to give you some explanation on the q here. From above, the set should be something like -
S = {a, -a, b, ab}
1 is in S = {1, -1, b, 1 * b}
not sufficient to say whether -4 is in the set
2 is in S = {2, -2, b, 2b}
not sufficient to say whether -4 is in S
To my surprise the answer is 'B'. Let me know, if the explanation is of any help to you.
Vikas
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Vikas,
Reply to Question 2> Answer should be C and not E.
This is how:
X can be a number like 4, 7, 10, 13, 16, 19 etc ..
Y can be 17, 26 etc ..
1. is not sufficient.
2. this is also not sufficient because no relation to x.
Now combining both the statements,
if a number of form 3x+1 is multiplied by number of form 9x-1, it will always be divisible by 3 because the +1 will offset -1, try any of the above written combinations.
Hence answer is C.
Reply to Question 2> Answer should be C and not E.
This is how:
X can be a number like 4, 7, 10, 13, 16, 19 etc ..
Y can be 17, 26 etc ..
1. is not sufficient.
2. this is also not sufficient because no relation to x.
Now combining both the statements,
if a number of form 3x+1 is multiplied by number of form 9x-1, it will always be divisible by 3 because the +1 will offset -1, try any of the above written combinations.
Hence answer is C.
Thanks
Airan
Airan
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Yes, for Q 2 ans is "C" .
I just missed it !!
by pluggin the values of p and q we can confirm it !
Now for Q 1. i guess, since it says if some no. is there, then it's negation would also be there and as per 2 condition : if there are two nos. then their multiplication would also be there...
so, if 2 is there then -2 would also be there as per condition 1.
Again, if 2 and -2 are two existing nos. then their multiplication i.e. -4 should be there.So it meets the condition.
I just missed it !!
by pluggin the values of p and q we can confirm it !
Now for Q 1. i guess, since it says if some no. is there, then it's negation would also be there and as per 2 condition : if there are two nos. then their multiplication would also be there...
so, if 2 is there then -2 would also be there as per condition 1.
Again, if 2 and -2 are two existing nos. then their multiplication i.e. -4 should be there.So it meets the condition.