Official Quant thread for CAT 2013

MBA CAT 2024
28685
@vishcat said:
Guys help me solve this--->A box contains two coins. One coin has a head on both sides. The other has a head on one side and a tail on the other. A coin is selected from the box at random and then one of its faces is observed at random. If this face is head, what is the probability that the other face is also a head
use bayes theorem
28686
@ananyboss said:
Hi all, Easy question though but I am stuck :/Q. A Mixture of 125 ltr of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% ?Can it be solved by allegation ??? Please help with approachi need to understand the concept !
25+x/125+x = 25/100

2500+100x = 3125 +25x
75 x = 625 ,
x = 125/3 litres ~ 41.66 litres:-)
28687
@albiesriram said:
25+x/125+x = 25/1002500+100x = 3125 +25x75 x = 625 , x = 125/3 litres ~ 41.66 litres
The Answer given is -- 8.33
28688
@ananyboss said:
The Answer given is -- 8.33
6.25/.75=25/3=8.33
28689
@rahulgaur666 said:
6.25/.75=25/3=8.33
Can u elaborate a little ? :/


28690
@ananyboss said:
Hi all, Easy question though but I am stuck :/Q. A Mixture of 125 ltr of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% ?Can it be solved by allegation ??? Please help with approachi need to understand the concept !
125lit=>100 lit wine and 25 lit water
water added=x
25+x=1/4 * (125+x)
x=25/3


28691
@ananyboss said:Hi all, Easy question though but I am stuck :/Q. A Mixture of 125 ltr of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% ?


its simple....first of all content of water and wine is :

Wine water

100(80%) 25(20%)

now, 100(75%)


=> 100= (75/100)x

x= 133.33

=> water added is 8.33. [133.33-100-25].
28692
@ananyboss said:
Can u elaborate a little ? :/
25+x/125+x=.25
25+x=31.25+.25x
.75x=6.25
x=8.33
28693
@hesse said:
@ananyboss said:Hi all, Easy question though but I am stuck :/Q. A Mixture of 125 ltr of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% ?its simple....first of all content of water and wine is : Wine water 100(80%) 25(20%) now, 100(75%)=> 100= (75/100)xx= 133.33=> water added is 8.33. [133.33-100-25].
Got it !!!! We work on the solute which is constant and not changing 😃 Whereas quantity of water is changing so working on that will not give right answer... Thanks
28694
@ananyboss said:
Hi all, Easy question though but I am stuck :/Q. A Mixture of 125 ltr of wine and water contains 20% water. How much water must be a concept !
125/x = 15/1
x= 125/15= 8.33
28695

a^4+b^4+c^4=? formula

28696
@albiesriram said:
25+x/125+x = 25/1002500+100x = 3125 +25x75 x = 625 , x = 125/3 litres ~ 41.66 litres

lol.. 625/75 = 8.33 only . I Badly need to do some calculation exercises . Doing hell a lot of silly mistakes. . Had a gut feeling, how could the quantity be that larger while solving , but still went on with it.

@ananyboss said:
The Answer given is -- 8.33
28697
@IIM-A2013 said:
@chillfactor what is fermats theorem
http://en.wikipedia.org/wiki/Fermat's_Last_Theorem
@nole said:
@chillfactor cos(ADC) = -7/24 ? kaise aya ?
If x + y = 180, then

Cos(y) = Cos(180 - x) = -Cos(x)
28698

let x,y be two positive numbers such that x+y=1.then the minimum value of(x+1/x)^2+(y+1/y)^2 is

a 12 b 20 c 12.5 d 13.3 e none

28699

"@IIM-A2013

let x,y be two positive numbers such that x+y=1.then the minimum value of(x+1/x)^2+(y+1/y)^2 is
a 12 b 20 c 12.5 d 13.3 e none"

12.5

28700
@hesse excellent tell approach
28701
@IIM-A2013 nvm misread the question
28702
@anantn its 12.5
28703

@IIM-A2013

whenever we have such question in which we have to minimize or maximize we have to put values of both x and y to be same(acc to the condn. like in this case we can not have same value for mazimizing.)

so x+y =1 => 2x=1 or x=y=1/2

and after putting this value we will get minimum value to be 12.5
28704
I'm driving my car of mass 1000 kg at 100 km/hr when I accidentally hit the rear end of a truck of mass 5000 kg traveling in the same direction at 90 km/hr. My car and the truck lock together and move as a single object. Let F1 be the force the truck exerts on my car and let F2 be the force my car exerts on the truck. What is F1/F2 during the collision?

Details and assumptions
You may ignore friction and air resistance.
The problem can be taken to be 1-d.
F1 and F2 are not the magnitudes of the forces, remember, but the actual forces themselves.