one more :)
What is the value of (x - y)^4?
(1) The product of x and y is 7.
(2) x and y are integers.
How can answer to this be option E i.e. OA
I think this is incorrect option, unless u puys have some wise thoughts to share ;)
If x and y are positive, is x^3 > y?
(1) (x)^1/2 > y
(2) x > y
one more :)
What is the value of (x - y)^4?
(1) The product of x and y is 7.
(2) x and y are integers.
C.From second we can conclude x,y is either 1 or 7.Being positive power , answer will alswyas be same.
guy with guts SaysC.From second we can conclude x,y is either 1 or 7.Being positive power , answer will alswyas be same.
that's what even I marked.. but OA is option E... :cheerio:
I am sure OA is wrong.. until unless someone can prove both of us wrong today..
How can answer to this be option E i.e. OA
I think this is incorrect option, unless u puys have some wise thoughts to share ;)
If
x and y are positive, is x^3> y?
(1) (
x)^1/2 > y
(2) x > y
You are right.Answer cannt be E.It should be B.
However in the question , if X^1/3 is asked , the answer will be E
that's what even I marked.. but OA is option E... :cheerio:
I am sure OA is wrong.. until unless someone can prove both of us wrong today..
yaar you are getting so may wrong OA's.Chakkar kya hey?material is authentic?
bcoz Even sometimes you may be getting wrong but bcoz of faulty choice the answer may be right.
yaar you are getting so may wrong OA's.Chakkar kya hey?material is authentic?
bcoz Even sometimes you may be getting wrong but bcoz of faulty choice the answer may be right.
ha haa.. yeah today I have decided to kill ur time along with mine by posting wrong OA..
material is some PDF.. I dunno wat.. but if u wud notice I have posted numerous DS back2back.. reaso being my answer did not match with OA and I could not figure out why.. instead of douting the material I thought of confirming myself first..
watever man..
If
vmt 0, is v^2m^3t^-4> 0?
(1)
m > v^2
(2) m > t^-4
best luck ;)
I think VMT>0 is given , the question can be rearranged as-
(v*m*t)^2*m/t^6
1.m>v2-cant say
2.m>t^-4 , so m>t^-6- so B is right anwer.
ha haa.. yeah today I have decided to kill ur time along with mine by posting wrong OA..
material is some PDF.. I dunno wat.. but if u wud notice I have posted numerous DS back2back.. reaso being my answer did not match with OA and I could not figure out why.. instead of douting the material I thought of confirming myself first..
watever man..
Thats the right approach u are following.And even its good for the PG junta.
You are right.Answer cannt be E.It should be B.
However in the question , if X^1/3 is asked , the answer will be E
good that someone agreed that E cannot be the answer..
BTW how come B.. if sq. root of x is greater than y, i.e. (x)^1/2 > y, then we have x > y... which is sufficient
so answer should be D and not B.. isnt it..??..
If
vmt 0, is v^2m^3t^-4> 0?
(1)
m > v^2
(2) m > t^-4
best luck ;)
guy with guts SaysI think VMT>0 is given , the question can be rearranged as-
(v*m*t)^2*m/t^6
1.m>v2-cant say
2.m>t^-4 , so m>t^-6- so B is right anwer.
Did u consider any decimal values.. try again, u might get answer different than B.. ;)
that's what even I marked.. but OA is option E... :cheerio:
I am sure OA is wrong.. until unless someone can prove both of us wrong today..
Hi Varun,
The answer should be E.
Please find my explanation:
If x and y are positive, is x^3> y?
(1) (x)^1/2 > y
(2) x > y
Consider two cases:
When x=4, y=1.
These values will suffice the condition 1.
and condition 2. and the answer to the question will be yes x^3 > y.
If one uses these kind of value, he will make the mistake that both from 1 and 2 the statement can be answered. So the option D.
But consider the case:
x=1/4, y=1/5.
These values suffices condition 1. But x^3 is not >y.
These values suffices condition 2. But x^3 is not >y.
So statement 1 and statement 2 cannot answer the question. x^3 may be or may not be greater than y depending upon the values of x and y.
So the answer is E.
good that someone agreed that E cannot be the answer..
BTW how come B.. if sq. root of x is greater than y, i.e. (x)^1/2 > y, then we have x > y... which is sufficient
so answer should be D and not B.. isnt it..??..
No.Try with x=.2 and y=.4.
I think the answer is indeed E only.
I have posted the answer and explanation above. The answer should be E.
that's what even I marked.. but OA is option E... :cheerio:
I am sure OA is wrong.. until unless someone can prove both of us wrong today..
Varun, the numbers may be fraction also. So you cannot determine the numbers from the two statements.
So the answer should be E.
I think VMT>0 is given , the question can be rearranged as-
(v*m*t)^2*m/t^6
1.m>v2-cant say
2.m>t^-4 , so m>t^-6- so B is right anwer.
V^2.M^3.T^-4 = V^2.M^3/T^4.
Irrespective of the value of v and t being positive or negative, the value of V^2 and T^4 will always be positive. So the value of the function depend on M. If M is positive, the function will be positive else it will be negative.
From 1, we can conclude that m will be greater than zero as v^2 will always be positive.
From 2, we can conclude that m will be greater then zero as 1/t^4 will always be positive.
So the answer is D. Using either of the statements, the question can be answered.
one more :)
What is the value of (x - y)^4?
(1) The product of x and y is 7.
(2) x and y are integers.
guy with guts SaysC.From second we can conclude x,y is either 1 or 7.Being positive power , answer will alswyas be same.
that's what even I marked.. but OA is option E... :cheerio:
I am sure OA is wrong.. until unless someone can prove both of us wrong today..
Hi Varun,
The answer should be E.
Please find my explanation:
If x and y are positive, is x^3> y?
(1) (x)^1/2 > y
(2) x > y
Consider two cases:
When x=4, y=1.
These values will suffice the condition 1.
and condition 2. and the answer to the question will be yes x^3 > y.
If one uses these kind of value, he will make the mistake that both from 1 and 2 the statement can be answered. So the option D.
But consider the case:
x=1/4, y=1/5.
These values suffices condition 1. But x^3 is not >y.
These values suffices condition 2. But x^3 is not >y.
So statement 1 and statement 2 cannot answer the question. x^3 may be or may not be greater than y depending upon the values of x and y.
So the answer is E.
Amar and Mr.Gutsy.. I believe both of u are getting confused with so many questions and posts.. incidentally both of u are talking sense only, knowingly or unknowingly
Amar.. I was talking about some other ques I guess whn I said option E was wrong.. but anyway, thanks for such eleborate explanation for the other ques.. appreciate it 😉
Hey crack this..
If x and y are consecutive odd integers, what is the sum of x and y?
(1) The product of x and y is negative.
(2) One of the integers is equal to 1.
one more :)
What is the value of (x - y)^4?
(1) The product of x and y is 7.
(2) x and y are integers.
i think both statement need to solve example
but if u have put What is the value of (x - y)^3? then it will more interesting...
Hey crack this..
If x and y are consecutive odd integers, what is the sum of x and y?
(1) The product of x and y is negative.
(2) One of the integers is equal to 1.
Ans A
St 1 : product is -ve, hence one is +ve and other -ve, so nos are -1 and 1 ...Sufficient
St 2 : if one integer is -1, other could be 1 or -3, not sufficient ..