If y = (6+x)^2/(6-x)^2 is not a perfect square integer and x is less than 6 and greater than 0; th

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14 Dec 2009 find the value of Y

If y = (6+x)^2/(6-x)^2 is not a perfect square integer and x is less than 6 and greater than 0; then the value of y is

OPtion

1) 100/9
2) 16/3
3) 81/8
4) 49/25
5) 20
6) 1/2
7) 121
8) 25
9) 25/16
10) 24

Solution

On taking x = 1, 2, 3, 4, 5 respectively, we get
y = 49/25, 4, 9, 25, 121
Only 49/25 is not a perfect square integer, hence is the solution.

Correct Option: 4 Best Solution (10)

Dr.Dipin Singh   14 years AGO

y = (6+x)^2/(6-x)^2
6+x and 6-x should be relatively prime .

Only in case of x=1, 7 and 5 are prime.

so putting x=1, we get
y= 7^2/5^2= 49/25 .. option 4

Like? Yes (1) | No   

Pargat Pal Singh   14 years AGO

y = (6+x)^2/(6-x)^2

only value of x in range 1-5 to satisfy given condition is x=1.
then y = (7/5)^2=49/25

Like? Yes (1) | No   

Dolly   14 years AGO

Putting x=1
we get

y = (6+x)^2/(6-x)^2 = 49/25 which is not a perfect square integer.

so option 4.

Like? Yes (1) | No   

Aman   14 years AGO

When we put x=1 in the expression for y,

y= 7^2/5^2=49/25

Like? Yes (1) | No   

Arneet   14 years AGO

y = (6+x)^2/(6-x)^2

when x=1

y= 7^2/5^2=49/25

Like? Yes (1) | No   

Mehul MKDP   14 years AGO

IF X = 1 Y= 1.96
OTHERS ARE PERFECT SQ.

49/15 = 1.96

Like? Yes (1) | No   

Karan   14 years AGO

when x=1,


y= 7^2/5^2=49/25

Like? Yes (1) | No   

Dundi   14 years AGO

(6+1)^2/(6-1)^2 = 49/25 balance all will be perfect square....

Like? Yes (1) | No   

rajendra   14 years AGO

if x=1 then y=49/25 which is not an integer........
but if put other values of x between 0 to 6 then yhe value of y will be an integer....

Like? Yes (1) | No   

sujit   14 years AGO

Numbers that are NOT integers are fractions, decimal numbers and complex numbers.

Put all the integers between 1-5 and test.

Like? Yes (1) | No