Official Quant thread for CAT 2013

@YouMadFellow said:
Yes, I got the same. Wondering why the wrong solution is roaming everywhere around the internet , or may be we are terribly wrong @chillfactor Sir, Can you look into these two questions -> Quoted one, and the one where Red and Blue balls are arranged
I think 13 is correct for the red and blue balls question

f(n + 1) = f(n) + f(n - 1)
f(1) = 2 and f(2) = 3
f(3) = 5
f(4) = 8
f(5) = 13

which is the other question ??
is it the question involving no of factors of 4620?

I think 46 should be correct (all factors except 1 and 2)
@Aman.Malhotra said:
x^3+y^3 -xy
1
x^3+y^3(x+y)(x^2-xy+y^2)(x+y)((x-y)^2+xy)Now (x+y) can at max be 1 with x-y=0
@joyjitpal said:
Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?(a) White (b) Red (c) Red and White (d) Not possible to determine from a sample of one ball
c?
@Aman.Malhotra said:
x^3+y^3 -xy
x=y=1/2

1
IF it takes 40 days for 40 cows to graze a park, and 60 days for 30 cows to do the same , how many days will it take 20 cows to graze the park, given that grass is also growing everyday at the same rate ?
a) 240 days
b) 150 days
c) 180 days
d) 210 days
e) None of these
detailed solution woud be appreciated

@Aman.Malhotra said:
x^3+y^3 -xy
4) 1

to maximize x+y, put x=y in the above equation

you will get 2*x^3
(2x - 1)*x^2 maximum value of x can be 1/2 , so x+y=1 (for x>1/2, above equation will not satisfy)

please post the detailed solution

@Aman.Malhotra said:
x^3+y^3 -xy
Option 4
Put x=y in the equation
@hiteshkhurana82 said:
IF it takes 40 days for 40 cows to graze a park, and 60 days for 30 cows to do the same , how many days will it take 20 cows to graze the park, given that grass is also growing everyday at the same rate ?a) 240 daysb) 150 daysc) 180 daysd) 210 dayse) None of these detailed solution woud be appreciated
let z be initial amount of grass
let x be rate of grass growth/day, and y be rate of grazing/cow/day

40*y*40=40*x+z
==> 1600y=40x+z

30*y*60=60*x+z
==> 1800y=60x+z

200y=20x
==> x=10y,z=120x

20*d*x/10=d*x+120x
==> d=120


@hiteshkhurana82 said:
IF it takes 40 days for 40 cows to graze a park, and 60 days for 30 cows to do the same , how many days will it take 20 cows to graze the park, given that grass is also growing everyday at the same rate ?
a) 240 days
b) 150 days
c) 180 days
d) 210 days
e) None of these
detailed solution woud be appreciated

1600=a+40b
1800=a+60b
20b=200
b=10
a=1200
20*n=1200+10n
10n=1200
n=120??
@krum said:
not visible

A man borrows 6000at 5 % interest on reducing balance ,at the start of the year .If he pays 1200 at the end of each ear ,find the amount of loan outstanding at the beginning of the third year ???

please post the detailed solution for this ques

@gautam22 said:
average 8 aa jayegi......5 uske pas bacha pada hai 40 chal jayega usse......fir har point se min possible bharvata chalega kyunki price decrease kar raha hai vapis jaate hue.....
110/8*45+50*40/8+200*35/8=13950/8??????
thank u ans is correct
@gautam22 said:
average 8 aa jayegi......5 uske pas bacha pada hai 40 chal jayega usse......fir har point se min possible bharvata chalega kyunki price decrease kar raha hai vapis jaate hue.....
110/8*45+50*40/8+200*35/8=13950/8??????
it should be `150 instead of 110 i guess

hi

@krum said:
5600 ?
ans is 4155

Good eve

@gautam22 said:
y to maximize x+y put x=y......for ex: xy=3....1@krum


In that ques., we are putting x=y to simplify the equation..after that we can find maximum value of x that can satisfy that simplified equation..!!

@gautam22 said:
upar ques mein likha hai ki 5 litre bach gaya hai.....vo consider nahi karna kya?
karna hai mujhe usi point par confusion tha so i posted the ques ..as i was not able to get that point
@priyalli said:

A man borrows 6000at 5 % interest on reducing balance ,at the start of the year .If he pays 1200 at the end of each ear ,find the amount of loan outstanding at the beginning of the third year ???

please post the detailed solution for this ques

5% of 6000=300
after 1 year 6300-1200=5100
5%of 5100=255
after 2nd year=5355-1200=4155
hence 4155