Latest Elitmus Aptitude Question SOLUTION: IF x and y are real then is x positive? 1)x^2m and y^2n have same sigh given both m and n are positive integers. 2)Both x^m and Y^n are positive, Given m and n are consecutive inte

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(#M40018513) ELITMUS Numerical Ability question Data Sufficiency Keep an EYE

IF x and y are real then is x positive?
1)x^2m and y^2n have same sigh given both m and n are positive integers.
2)Both x^m and Y^n are positive, Given m and n are consecutive integers.
DATA SUFFICIENCY

Asked In Elitmus Sumit Rawal (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

Shubham Gupta  (10 years ago)

Both the statements are not suffient.,.
1. x^2m will always be postive.
2. if m and n are consecutive integers,we cant conclude that x is positive.
for ex-
x=-3 and m=2,n=3
if this case -3 ^ 2 will be positive whereas we can see that its not positive

Is this solution Helpfull? Yes (32) | No (3)  

jaspreet  (10 years ago)

Both r nt sufficient.first is of no use.in second eg.x is -2, m is 4, then xto power m is +, also if x is 2 we get the same rsult.how can v determine it den.y can b anything too.

Is this solution Helpfull? Yes (8) | No (4)  

Glued Brain  (10 years ago)

The 2nd statement alone is enough to answer this question and not both the statements alone because, (+x)^2m will also give a +ve number and (-x)^2m will also give a positive no. And how can you conclude whether x is +ve or _ve from the first statement. All you can conclude is X^2m is +ve and not x itself

Is this solution Helpfull? Yes (4) | No (7)  

SAMBIT KUMAR GHOSH  (10 years ago)

statement2 is sufficient to answer this question.

Is this solution Helpfull? Yes (4) | No (3)  

Sumit Rawal  (10 years ago)

acc to vidushi answer it can be concluded that both the statements are alone sufficient to answer the question.since any no to even power is always positive that 1 will hold true in every case

Is this solution Helpfull? Yes (3) | No (2)  

Nitish  (10 years ago)

since in 2 it is written x^m and y^n are positive
means x^n is positive
in only one condition when x is postive

Is this solution Helpfull? Yes (3) | No (1)  

Sumit Rawal  (10 years ago)

So can Glued can we conclude that statement 2 alone is sufficient to answer the question but not statement 1???

Is this solution Helpfull? Yes (3) | No   

Krishanu chatterjee  (10 years ago)

both r not sufficient as if x=neg no the 1 cond equals x^2m which is square which will make it positive
while if x is pos it too remains positive
while in 2nd condition m is not told if it is even or odd if even we cant determine the no x is positive or negative
so it is data insufficient

Is this solution Helpfull? Yes (3) | No (2)  

Vidushi  (10 years ago)

let m=3 and n=4(consecutive nos. given)

bcoz;x^m and y^n are +ve this implies x cant be negative;thus x is positive

Is this solution Helpfull? Yes (2) | No (5)  

Ravindra  (10 years ago)

can't be say anything

both statement is not sufficient

Is this solution Helpfull? Yes (1) | No   

Glued Brain  (10 years ago)

I suppose so..

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Nitish  (10 years ago)

jaspreet is right in this sense

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Anjali Sadhukhan  (8 years ago)

yahpp..its right..both are insufficient to answer.

Is this solution Helpfull? Yes | No (5)  

Cp Choudhary  (8 years ago)

right @nitish nd @vidushi............only 2nd statement is sufficient to give answer
nd we can not answer by 1st statement

Is this solution Helpfull? Yes | No (1)  

satyaki  (6 years ago)

both are not sufficient as in second case (-2)^2 gives A +ve answer but x is not +ve....means x^m is positive but x is not

Is this solution Helpfull? Yes | No   

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