Official Quant thread for CAT 2013

@maroof10 said:
@staaalinnn I have doubt in Q4No. of steps when the escalator is stationary is30 + 15s = 20 + 20ss= 2No. of steps= 30 + 15*2 =60next...

So now Tarun moves at twice the speed of Ram, i.e., 4 steps/sec. So, his relative speed is (4-2) = 2 steps/sec.
And relative speed of Ram is (2+2) = 4 steps/sec.
Let they meet in time t.
2*t + 4*t = 60 => t = 10
Hence steps covered by Tarun to meet Ram is 4*10 = 40 steps.

@staaalinnn Understood....ThankU
@sbharadwaj The answer gives is 3.
k + 1/k
So, k+1/k+1=3

Thanks.....
@maroof10 said:
@sbharadwaj The answer gives is 3.k + 1/k So, k+1/k+1=3Thanks.....
It doesn't say max/min value of x right.? How can we conclude it to be 3.?
@chandrakant.k
only 3

Try this Time & Work Problem:
Q) There are a certain number of taps installed to fill a certain tank. The taps are numbered 1, 2, 3, ... A particular tap numbered n (n>= 4) can fill the tank in (n^2 - 3*n) minutes. The first three taps don't work at all. If all the taps from the fourth onward are opened simultaneously, then find the least time in which the tank can be filled.
ANS: 18/11 minutes
Level: Easy
@scrabbler said:
Should be 8/9?regardsscrabbler
do explain...don't just post the answer
@sbharadwaj the options are 1>2 2>3 3>4 4>1
so, value 3 is the obvious choice..
@x2maverickc said:
let's say the prob he know the ans is x and y denotes the prob of him not knowing the ans.x+y=1in case of x----> prob he gets it correct=1in case of y----> prob he gets it correct=1/4thus x + y/4=2/3solve the two equations x=5/9.OA?
no its 8/9
@iLoveTorres said:
is the answer 5/6?
no its 8/9
@iLoveTorres said:
is the answer 5/6?
no its 8/9
@staaalinnn said:
Try this Time & Work Problem:Q) There are a certain number of taps installed to fill a certain tank. The taps are numbered 1, 2, 3, ... A particular tap numbered n (n>= 4) can fill the tank in (n^2 - 3*n) minutes. The first three taps don't work at all. If all the taps from the fourth onward are opened simultaneously, then find the least time in which the tank can be filled. ANS: 18/11 minutesLevel: Easy
Summation 1/n(n-3)
1/n(n-3)=1/3(1/(n-3) - 1/n)
=1/3 (1/1 -1/4 +1/2 -1/5 +1/3-1/6+1/4-1/7+.... infinity) since more no of taps will only help
=1/3(1+1/2+1/3)
=1/3((6+3+2)/6)
=11/18

18/11 minutes?
@Subhashdec2 said:
Summation 1/n(n-3)1/n(n-3)=1/3(1/(n-3) - 1/n)=1/3 (1/1 -1/4 +1/2 -1/5 +1/3-1/6+1/4-1/7+.... infinity) since more no of taps will only help=1/3(1+1/2+1/3)=1/3((6+3+1)/6)=10/1818/10 minutes?
Its 18/11.. check ur sum..its 1/3((6+3+2)/6) not 1/3((6+3+1)/6)
@staaalinnn said:
Its 18/11.. check ur sum..its 1/3((6+3+2)/6) not 1/3((6+3+1)/6)
sorry my bad
There are1120 students in a school namely 'Junior King II'. The ratio of the number of boys to the number of girls is (n + 1) : n, where n is a natural number. Each of the students play exactly one of the five games – Badminton, Table Tennis, Cricket, Football and Volleyball. One-fifth of the total students play football whereas one-fifth of the girls play Badminton. The ratio of the number of boys to girls who play Football is 5 : 2. The number of girls who play Cricket is 80% of that of the boys. The number of boys who play Badminton is 70. The number of boys who play Volleyball is 50% more than that of girls. The number of boys who play Table Tennis is equal to that of girls, which in turn, is equal to 80.

How many girls play Cricket?
A. 80 B. 96 C. 120 D. 64 E. None of the above

85. What is the difference between the number of boys who play Cricket and Volleyball?
A. 10 B. 30 C. 70 D. 80 E. 100
@Subhashdec2 said:
7x and 3x7x/2 -9x/4=3.751.25x=3.75x=321 boys and 9 girlsboys have 10.5rsgirls have 6.75rstotal money 17.25it has to be distributed in the ratio 2:12/3 *17.25 =11.51/3*17.25 =5.75girls have to reduce 1rsby giving .75 paisa coin and taking .50 coin a reduction of .25 is madetotal =1/.25=4
yar yahi doubt hai...if girls give two coins to the boys i.e 1.5 rs and boy gives 1 coin to girl i.e 50 paise to bhi to condition satisy ho rahi he na?

@staaalinnn how to solve T =1/4 +1/10 + 1/18 + 1/28 +....
@maroof10 said:
@staaalinnn how to solve T =1/4 +1/10 + 1/18 + 1/28 +....
T = 1/(1*4) + 1/(2*5) + 1/(3*6) + 1/(4*7) + ....

T(n) = 1/(n*(n+3)) = (1/3) * [ 1/n - 1/n + 3]

Sum = (1/3) * [ 1 - 1/4 + 1/2 - 1/5 + 1/3 - 1/6 + 1/4 - 1/7 + ..... ] = (1/3)*[1 + 1/2 + 1/3]
= (1/3)*(11/6) = 11/18 ?
@maroof10 said:
@staaalinnn how to solve T =1/4 +1/10 + 1/18 + 1/28 +....
multiply and divide by 3...
then write like this

1/3(3/1*4 + 3/5*2 + 3/3*6...)

1/3(1/1-1/4 + 1/2-1/5 +1/3-1/6+1/4-1/7....)
terms will get cancel out..
u'll get..
18/11..
@vbhvgupta said:
A bag contains 15 balls of the same size. Each ball is of a single colour, white, red or blue. How many red balls are there in the bag?Statement 1:The probability of drawing a red ball is the same as that of drawing a blue ballStatement 2:The probability of randomly drawing a white ball from the bag is 20%
Is answer :
Combining both the statements we can solve?