[naga25french]

Originally Posted by iamnutty

what is the number of pairs of non zero integers whose sum is 72 more than their product is ?

a)0
b)1
c)2
d)3
e)4

Pls post method i do not knw answer

a+b = ab + 72

(a-1)(b-1) = -71

no solutions
whats the answer?

[protoplasm]

Originally Posted by uhhh...

yh we can.. knowing n+a-1>= n

n = 2, n+a-1= (5^3)*2
n= 2*5, n+a-1 = (5^2)*2..

two possible values

but this is wrong.

answer is 3 possible (a,n)

[simrat_sandy]

Originally Posted by bishoo123

Was any one able to solve it ? Could not get any clue.

we can do long division
I got -xy

[uhhh...]

Originally Posted by protoplasm

but this is wrong.

answer is 3 possible (a,n)

math cannot be wrong.. show me a flaw in my method.. thats not a plausible line of reasoning

[uhhh...]

Originally Posted by crack_2009

Let X = (√3 + √5)^222. If X is written in decimal form, what will be the first digit after the decimal point?

OPTIONS 1)5 2)6 3)7 4)8 5)9

(√3 + √5)^222 = integer (i) + fraction (f1).. 0 < f1 < 1

(√3 + √5)^222 = 2^111 * (4 + √15)^111

now

2^111 ( (4 + √15)^111 + (4 - √15)^111) = integer (I) .. binomial theorem

also 0 < 2(4 - √15) < 1 .. hence 0 < f2 = 2^111 * (4 - √15)^111 < 1

so I = i + f1 + f2

0 < f1 + f2 < 2

so only possible value of f1 + f2 = 1

or i = 2^111 ( (4 + √15)^111 + (4 - √15)^111) -1 = (√3 + √5)^222 + (√3 - √5)^222 - 1

or f1 = 1 - (√3 - √5)^222 .. now (√3 - √5)^222 is very small virtually 0

so the 1st decimal place of f1 = 9... option 5

edit:

ps: this bloody question cannot be asked in cat.. its so not like cat, took me 15 mins

[crack_2009]

Originally Posted by naga25french

a+b = ab + 72

(a-1)(b-1) = -71

no solutions
whats the answer?

we have (-70,2) and (2,-70)

[jinidgenie]

Originally Posted by simrat_sandy

let the distances be d1 and d2 that they travel before they meet each other
d1/d2 = (s1*k/s2*k)

d1/s2 =24 => s1*k/s2 =24 ...(1)

d2/s1 = s2*k/s1=54 ....(2)

from 2 k= 54*s1/s2

substituting k in 1

s1^2/s2^2 =4/9
so s1 is 2 and s2 is 3

=>d1= 72; d2= 108

so time taken by A to travel till meeting point
plus from meeting pt to Q
plus from Q till second meeting point

=36+54+36
so I'm getting 126
(hope I've not made any silly mistake typed it on the phone)

they meet at 12:06. their speeds are the in the ratio 2:3. Since distance PQ is constant, Sa(t+54/60)=Sb(t+24/60). where Sa, Sb are the speeds of a and b. Since both started at the same time, let t be the time they travelled before they met.

Solving the above eq, we get t=36 mins. Implies they first met at 10:36 and next meeting would be at 54 mins after this..which is 12:06

I hope this is correct..

[shanks4mba]

Originally Posted by naga25french

a+b = ab + 72

(a-1)(b-1) = -71

no solutions
whats the answer?

there will be two divisors...
hence 1 such solution...rt?

[shanks4mba]

Originally Posted by crack_2009

we have (-70,2) and (2,-70)

i guess there will be only one...in ur case u are takin ordered pairs...

[jinidgenie]

Originally Posted by brajesh.pandey

A travel from P to Q and B from Q to P at 10:00 am.
After meeting eachother in between A takes 54 minutes to reach Q and B takes 24 minutes for reaching P.

When will they meet again.Please provide a detailed solution. As for the answer sorry I do not have it (:

they meet at 12:06. their speeds are the in the ratio 2:3. Since distance PQ is constant, Sa(t+54/60)=Sb(t+24/60). where Sa, Sb are the speeds of a and b. Since both started at the same time, let t be the time they travelled before they met.

Solving the above eq, we get t=36 mins. Implies they first met at 10:36 and next meeting would be at 54 mins after this..which is 12:06