[linksuresh]

Originally Posted by HYgoldking

1) Two different positive numbers a and b differ from their reciprocals by 1. Find a+b
A) 1 B) √6 c) √5 D) 3 E) None of these.
A) 1

consider, b>a>0
hence 1/a-1/b = 1
=> b-a = ab
=> a = b/(b+1)

assume b=1 .. hence a=1/2
so a+b = 3/2

so what is the flaw ?

[dare2]

Originally Posted by shivam_01

well in that case also the numbers will sum to even only i guess

no boss.. in that case it'll always be odd..

[dare2]

Originally Posted by linksuresh

consider, b>a>0
hence 1/a-1/b = 1
=> b-a = ab
=> a = b/(b+1)

assume b=1 .. hence a=1/2
so a+b = 3/2

so what is the flaw ?

i think the question is a - 1/a =1, b - 1/b =1

if this case is then,

a + b = (2ab + a + b)/ab

still it cant be 1.

@Hygoldking are u sure that the answer is 1?

[prakharc]

Originally Posted by HYgoldking

1) Two different positive numbers a and b differ from their reciprocals by 1. Find a+b
A) 1 B) √6 c) √5 D) 3 E) None of these.
A) 1

3) How many subsets of the set {1,2,3…,30} have the property that the sum of all elements is more than 232?
A) 2^30 B) 2^29 C) 2^29 -1 D) 2^29+1 E ) None of these

E ) None of these

4) Let the point P be (4,3). Choose a point Q on y=x line and another point R on the line y=0 such that sum of lengths PQ+QR+PR is minimum. Find this minimum length.
A) 5√2 B) 3+3√2 C) 5 D) 4 E) 10
c) 5

5) Let f be a function defined on odd natural numbers which return natural numbers such that f(n+2)=f(n)+n
and f(1)=0 . Then f(201)?
A) 10000 B)20000 C) 40000 D) 2500 E) None of these

A) 10000

7) The product of n numbers is n and their sum is 0. Then n is always divisible by
A) 3 B) 4 C) 5 D) 2 E) None of the foregoing

B) 4

Are we discussing Implex's test here right now....?? Nice test if someone has not attempted should first attempt the paper then discuss.......

[HYgoldking]

Originally Posted by dare2

i think the question is a - 1/a =1, b - 1/b =1

if this case is then,

a + b = (2ab + a + b)/ab

still it cant be 1.

@Hygoldking are u sure that the answer is 1?

a - 1/a =1

a^2-1=a

So,

a^2-a-1=0

b^2-b-1=0

Roots are (1 +- (5)^1/2)/2

a+b = Sum of the roots

(1 + (5)^1/2)/2 + (1 - (5)^1/2)/2

(1+1)/2 = 1

[prakharc]

Originally Posted by dare2

i think the question is a - 1/a =1, b - 1/b =1

if this case is then,

a + b = (2ab + a + b)/ab

still it cant be 1.

@Hygoldking are u sure that the answer is 1?

a=1/a +1 => a^2 - a -1 =0 ..... 'a' and 'b' are roots of this equaton...as they are the only possible solution so a+b= sum of roots= 1

[vivek417]

Originally Posted by dare2

i think the question is a - 1/a =1, b - 1/b =1

if this case is then,

a + b = (2ab + a + b)/ab

still it cant be 1.

@Hygoldking are u sure that the answer is 1?

My take:-

2 cases:-

x- (1/x) = 1
x^2 -x-1 = 0

Hence x = (1+5^0.5)/2......the other one can't b taken as a & b r positive numbers

Next:- (1/x) - x = 1
x^2+x-1 = 0

Solve it to get x= (-1+5^0.5)/2....negative one can't b taken

Hence a+b = 5^0.5.......sum of the above 2 numbers

Thanks
Vivek

[dare2]

Originally Posted by prakharc

a=1/a +1 => a^2 - a -1 =0 ..... 'a' and 'b' are roots of this equaton...as they are the only possible solution so a+b= sum of roots= 1

@hygoldking, @prakharc
thanks.. dont know y quadratic equation didnt come in my mind.. tried 2 add them directly..

well i think i need a small break.. time 4 a fag..

[vivek417]

Originally Posted by prakharc

a=1/a +1 => a^2 - a -1 =0 ..... 'a' and 'b' are roots of this equaton...as they are the only possible solution so a+b= sum of roots= 1

How can u take both the roots to be a & b? a & b are both positive numbers....

a^2 - a -1 =0 yields.....a = (1+-(5)^0.5)/2

(1-5^0.5)/2 is a negative number...u can't take this into consideration....

Thanks
Vivek

[dare2]

Originally Posted by prakharc

Are we discussing Implex's test here right now....?? Nice test if someone has not attempted should first attempt the paper then discuss.......

which test are you talking about?? can i have the link of that?