Official Quant thread for CAT 2013

MBA CAT 2024
15723
@The_Loser said:
21600 = 2^5 * 3^3 * 5^2here from 2 we can consider 3 value. These are 2^0, 2^2, 2^4from 3 we can consider two vales. These are 3^0, 3^2from 5 we can consider two vales. These are 5^0, 5^2Total values = 3 * 2 * 2 = 12
Best method 😃 Thanks
15724

Q => N persons sitting around round table. Find the odd against two specified persons sitting next to each other.

1) N+1/2
2) N-3/2
3) N+3/2
4) N+2/3

15725
There are 2 candles which can burn for 7hrs and 8hrs respectively. The two candles are lit
simultaneously and initially the first candle was twice the length of the second candle. How
many hours after the candles are lit, will the two candles be of equal length?
a. 5.66 hours b. 6.22 hours c. 5.33 hours d. 5.83 hours
15726
@joyjitpal said:
There are 2 candles which can burn for 7hrs and 8hrs respectively. The two candles are litsimultaneously and initially the first candle was twice the length of the second candle. Howmany hours after the candles are lit, will the two candles be of equal length?a. 5.66 hours b. 6.22 hours c. 5.33 hours d. 5.83 hours
6.22?
15727
@The_Loser said:
Q => N persons sitting around round table. Find the odd against two specified persons sitting next to each other.1) N+1/22) N-3/23) N+3/24) N+2/3


2.....explanation


first of all....odds=(number of cases of an event happening)/(number of cases of that event not happening)


now...total cases---(n-1)!

total cases where those two are together....


= 2*(n-2)!

total where they aren't=

(n-1)!- 2*(n-2)!= (n-2)!(n-3)



odds= (n-2)!(n-3)/2*(n-2)!

=n-3/2






15728
@joyjitpal said:
There are 2 candles which can burn for 7hrs and 8hrs respectively. The two candles are litsimultaneously and initially the first candle was twice the length of the second candle. Howmany hours after the candles are lit, will the two candles be of equal length?a. 5.66 hours b. 6.22 hours c. 5.33 hours d. 5.83 hours
6.22?
15729
@adwaitjw said:
6.22?
approach batao bhai OA nahi hai
15730
@joyjitpal said:
There are 2 candles which can burn for 7hrs and 8hrs respectively. The two candles are litsimultaneously and initially the first candle was twice the length of the second candle. Howmany hours after the candles are lit, will the two candles be of equal length?a. 5.66 hours b. 6.22 hours c. 5.33 hours d. 5.83 hours
6.22 hours
15731
@joyjitpal said:
approach batao bhai OA nahi hai
Suppose second candle is of 56 unit length (LCM :P)
So first candle will be 112 unit length.
So after 6 hrs.
part of first candle remained 16 unit length
part of second candle remained 14 unit length.
i.e. still first candle is longer. i.e. it has not matched the length of second candle yet. So ans greater than 6 hrs. Only one option. :P
15732
If f(x)*f(1/x) = f(x) + f(1/x) and f(3) = 28, then f(4) is:
a. 63 b. 65 c. 17 d. None of these
15733
@joyjitpal said:
There are 2 candles which can burn for 7hrs and 8hrs respectively. The two candles are litsimultaneously and initially the first candle was twice the length of the second candle. Howmany hours after the candles are lit, will the two candles be of equal length?a. 5.66 hours b. 6.22 hours c. 5.33 hours d. 5.83 hours
consider the length of 1st candle be 2x, 2nd candle be = x.
V (1st) = 2x/7 = v1
V (2nd) = x/8 = v2

now consider it as 2 persons moving in same direction as 2 candles are burning in same direction.
so time to catch the 2nd candle via 1st one = Dist bw them/ relative speed
= x
--------------------
2x/7 - x/8
Time = 6.22 hr

15734
@adityaknsit said:
2.....explanationfirst of all....odds=(number of cases of an event happening)/(number of cases of that event not happening)now...total cases---(n-1)!total cases where those two are together....= 2*(n-2)!total where they aren't=(n-1)!- 2*(n-2)!=(n-2)!(n-3)odds= (n-2)!(n-3)/2*(n-2)!=n-3/2
we need odds against i think
15735

for first speed is 2h/7 and for second is h/8
now relative speed 9h/56
distance to travel for equal length is 'h'

so its 56/9..
@joyjitpal
15736

what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real number

a)11
B)13
c)12
d)14
15737
@raopradeep said:
what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)14
b

put x equal to 1
15738
@joyjitpal said:
we need odds against i think
that is what i've taken...


actually as i have taken, the non-happening of an event (them not sitting together) is the event.
15739
@adityaknsit said:
that is what i've taken...actually as i have taken, the non-happening of an even (them not sitting together) is the event.
oh understood
15740
@raopradeep said:
what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)14
x=2, f =14
x=1, f=13
x=-11, f=25.
so, OA = 13 - B
15741
@raopradeep said:
what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)14
take it on a number line.
Points are = -11, 1 ,2.

So minimum value would be there at 1, this would be 13.
Option - B
15742
@joyjitpal said:
If f(x)*f(1/x) = f(x) + f(1/x) and f(3) = 28, then f(4) is:a. 63 b. 65 c. 17 d. None of these
...koi batado....