Latest Elitmus Aptitude Question SOLUTION: if (a+b)^2=25 and (a-b)^2=1 1)a is prime 2)modulus of a>2 and a is a natural number a) b) c) d)

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(#M40030664) ELITMUS Numerical Ability question data sufficiency Keep an EYE

if (a+b)^2=25 and (a-b)^2=1
1)a is prime
2)modulus of a>2 and a is a natural number
a)
b)
c)
d)

Asked In Elitmus abc (11 years ago)

Unsolved Read Solution (18)

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Read Solution (18) :

RAKESH  (11 years ago)

(a+b)^2=25 so (a+b)=+-5
(a-b)^2=1 so (a-b)=+-1
solve these equations
a+b=+5,a-b=+1 then a=3,b=2
a+b=+5,a-b=-1 then a=2,b=3
a+b=-5,a-b=-1 then a=-3,b=-2
a+b=-5,a-b=+1 then a=-2,b=-3
so a=-2,-3,2,3
1)a is prime is incorrect(negative no. cant be prime)
2)modulus of a>2 and a is a natural number is incorrect(-2,-3 is not natural)
both are incorrect statement

Is this solution Helpfull? Yes (121) | No (6)  

Priyansh   (10 years ago)

here a can take 4 values i.e. 3, -3, 2, -2,,, on solving the equations with all the possibilities,,, now considering the frst case a is prime thus a can be either 2 or 3 as -ve numbers cant be prime so two value exists,,, but not giving the unique solution,,,, now considering the second case |a|>2 thus |-3|>2 and also |3|>2 thus again two values exists,,,, now considering both the statements a is prime and |a|>2 here we getting only one value is |3|>2 and also 3 is a prime number thus it can be answered using both

Is this solution Helpfull? Yes (8) | No (2)  

pradeep singh  (10 years ago)

(a+b)^2=25 and (a-b)^2=1
by solving a=3,2,-3,-2
1) a is prime then a= 2,3
2) A) |a|> 2 i.e. a= 3,-3

B) a is natural number i.e. a=2,3
from A) AND B) a=3
Statement 2) alone is sufficient

Is this solution Helpfull? Yes (8) | No   

dilbar  (11 years ago)

either statement is sufficient

Is this solution Helpfull? Yes (4) | No (9)  

Mahesh  (11 years ago)

a can be either -2, -3, 2 or 3
So, none of the conclusion are true

The possibilities for (a,b) are (2,3)(3,2)(-2,-3)(-3,-2)

so, none of the conclusions are definete

Answer is none follows

Is this solution Helpfull? Yes (2) | No (4)  

shruti  (11 years ago)

@mahesh : -3,-2 are not natural no.

Is this solution Helpfull? Yes (2) | No   

vigneshbabu  (10 years ago)

since a can take negative values like -2,-3
it cant be prime as well as natural.
so both statements are incorrect

Is this solution Helpfull? Yes (1) | No   

Guddu kumar  (10 years ago)

by solving above two equation we will get four set of values
(a=13,b=12),(a=12,b=13),(a=-12,b=-13),(a=-13,b=12)
again a is prime no and a is also natural no so
a can have only one value out of 4 ie a=13
so b=12..........

Is this solution Helpfull? Yes (1) | No (6)  

rajeev kumar jaiswal  (10 years ago)

(a+b)^2=25========>a^2 + b^2 + 2ab=25...(1)
(a=b)^2=1=========>a^2 + b^2 - 2ab=1....(2)
solve above 2 eqn we get
a^2+ b^2 = 13 in that eqn we put a=3 and b=2 get the value....and anssss c part

Is this solution Helpfull? Yes (1) | No   

vinod  (11 years ago)

a is prime

Is this solution Helpfull? Yes | No (2)  

ABHINAV SHRIVASTAVA  (11 years ago)

a is prime

Is this solution Helpfull? Yes | No (2)  

nishtha mathur  (11 years ago)

a is a prime

Is this solution Helpfull? Yes | No (2)  

arsh khan  (11 years ago)

a is a prime number

Is this solution Helpfull? Yes | No (1)  

ananya sharan  (10 years ago)

a is prime

Is this solution Helpfull? Yes | No (3)  

anu maurya  (10 years ago)

both are incorrect as a=2,-2,3,-3
(which is nt a prime no. n a natural no.)

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Harshit shah  (10 years ago)

both are sufficient
from 1st statement we get a=1,3,-1,-3
and in second statement it is given that modulus of a>2 so a=3 or-3 and it is also given that a is natural no. so a=3

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AMIT KUMAR  (10 years ago)

a is a prime

Is this solution Helpfull? Yes | No (1)  

monika kumari  (5 years ago)

The question is:
f (a+b)^2=25 and (a-b)^2=1
1)a is prime number.
2) |a|>2 and a is an integer

Is this solution Helpfull? Yes | No   

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