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(#M40037244) INFOSYS Numerical Ability question ask in infosys exam Keep an EYE

x^2 - y^2=16 and xy=16 so find out x+y

Asked In Infosys Rajneesh Rahi (10 years ago)

Solved ashish Read Solution (7)

Is this Puzzle helpful?   (48)   (16) Submit Your Solution Algebra

Read Solution (7) :

ashish  (10 years ago)

x^2-y^2=16
(x+y)(x-y)=16
so 16 comes in following table
1*16,2*8,4*4
using 2*8 equation
x+y=8 and x-y=2 so x is 5 or 3 and y will be 3 or 5
so answer is 8...
not sure check it out

Is this solution Helpfull? Yes (15) | No (43)  

Mukesh Nayal  (10 years ago)

sum will be
4(sqrt(5)+1)/sqrt(2*sqrt(5)-1)

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

Raghvendra Prajapati  (10 years ago)

8
x^2-y^2=16
4*4 1*16,2*8
x+y=8 and x-y=2
so ans 8

Is this solution Helpfull? Yes (3) | No (9)  

prashant nagrurkar  (10 years ago)

is d ans 16?

Is this solution Helpfull? Yes | No (2)  

Nishi  (8 years ago)

answer is 8

Is this solution Helpfull? Yes | No (4)  

Abhishek Satpathy  (7 years ago)

x^2 - y^2 = (x+y)(x-y) = 16;
xy = 16;
16 = 1 x 16 = 8 x 2 = 4 x 4;
So we have to assume x and y such that the value of xy remains 16.
If we take { (x+y) = 8 } and { (x-y) = 2 },
8 = 1 x 8 = 4 x 4 = 2 x 4;
If we take x=4 and y=4 then,
x+y = 8.
Basically, we have to prove x+y = 8 both sides.

Final Answer is 8.

Is this solution Helpfull? Yes | No (5)  

souradeep saha  (5 years ago)

x^2-y^2=16
(x+y)(x-y)=16

(x+y)^2(x-y)^2=256
(x+y)^2{(x+y)^2-4xy}=256
(x+y)^4-64(x+y)^2=256
a^2-64a=256 //(x+y)^2=a//
a=32+16(sqrt(5))
x+y=8.24(app)
x-y=1.94(app)
x=5.09
y=3.14
x!=5 && y!=3 if so then xy!=16
x!=4 && y!=4 if so then x^2 - y^2==0

Is this solution Helpfull? Yes | No (1)  

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