Official Quant thread for CAT 2013

MBA CAT 2024
22747
@amresh_maverick said:
Second and third digit mein 0 ka option bhi hai to 1/6th of casesye to detail mein samjhana
3 digit numbers hai...
it can of the form 1xx, x1x or xx1

1xx -> both x can be filled in 6 ways -> 6*6 = 36 numbers...
x1x -> 1st x mein 0 nahi aa sakta so -> 5*6 = 30 numbers
xx1 -> 5*6 = 30

total = 96...
22748
@Logrhythm said:
a=b=c=1/3(1/9+1/3)*3 = 4/3...??i took the expression as ((a/3) + a) if it is (a/(3+a)) then it wld be 3/10....
could you plz explain the approach ? how did u equate a=b=c =1/3??
22749

If f(f) = f(t-1) + f(t+1) , find Min value of K such that f(t+k) +f(t) =0

22750
@amresh_maverick said:
Second and third digit mein 0 ka option bhi hai to 1/6th of casesye to detail mein samjhana
Second digit could be 0, 1, 2, 3, 4, 5. All are equiprobable since all are cycling continuously. So each one is in second position equal number of times = 1/6th of the 180 numbers = 30 times.

Sometimes symmetry is a great help in PnC :D

regards
scrabbler
22751
@pathetic said:
could you plz explain the approach ? how did u equate a=b=c =1/3??
to find the max value in such cases, always keep a,b,c,d....z as close as possible...thumb rule hai...it comes from the concept of am >= gm >= hm
22752
@amresh_maverick said:
options are 3644556072
55??
22753
@amresh_maverick said:
If f(f) = f(t-1) + f(t+1) , find Min value of K such that f(t+k) +f(t) =0

considering t>0
f(1)=f(0)+f(2)
f(2)=f(1)+f(3)

f(1)=f(0)+f(3)+f(1)
f(0)+f(3)=0

k=3?
22754
@amresh_maverick said:
If f(f) = f(t-1) + f(t+1) , find Min value of K such that f(t+k) +f(t) =0
f(0) + f(3) = 0

for t>0 , k = 3??
22755

did we do this one:
1024*2065 + 2564*2653 + 5216*5645 is in octal form ........ find number of digit 6 that comes after multiplication ??

22756

No of points with integral coordinates inside the region x+2y =0 and y>=0

22757
@amresh_maverick said:
No of integral sols of |x-3| + |y-5| =11
44 ?

x = -8 , y = 0
x = -7 , y = 4,6
.......
............
.............
..........
x = 14 , y = 5
total solutions = 1 + 7*2 + 2 + 13*2 + 1 = 44
22758
@amresh_maverick said:
No of integral sols of |x-3| + |y-5| =11
|x-k| + |y-p| = x

always have 4x integral solutions...so 44...
22759
@Logrhythm said:
|x-k| + |y-p| = xalways have 4x integral solutions...so 44...
bhai logic??

22760
@amresh_maverick said:
No of points with integral coordinates inside the region x+2y =0 and y>=0
x+2y+c=50
x+a+c=50
no of solutions=52!/50!2! = 26*51

divide it by 2 since half cases will have a odd =13*51=510+153=663??
22761
@amresh_maverick said:
If f(f) = f(t-1) + f(t+1) , find Min value of K such that f(t+k) +f(t) =0
3 ?

f(t) = f(t - 1) + f(t + 1)
f(t + 1) = f(t) + f(t + 2)
adding these two

f(t + 2) + f(t - 1) = 0
f(t + 3) + f(t) = 0
so k = 3
22762
@amresh_maverick said:
No of points with integral coordinates inside the region x+2y =0 and y>=0
650??
22763
@saurav205 said:
bhai logic??
Points on a square of side xrt2...at distance of rt2...

regards
scrabbler
22764
@abhijit90 said:
sir , is single sitting degree valid for cat
wrong thread??
BTW what does single sitting degree mean??
22765
Pavan, Mohan and Sohan were three friends. Pavan bought a cornetto, which is a large ice-cream in the shape of a perfect cone. He first had the top of it (by volume) and then gave it to Mohan. Mohan then had the top (again by volume) of the remaining part and gave the rest to Sohan. The level of the ice-cream then was 4 cm from the bottom, i.e., the vertex. Find the height of the initial cone (in cm).
8.1
9.9
9
7.2
22766
@amresh_maverick said:
No of points with integral coordinates inside the region x+2y =0 and y>=0
x=0,y=0,1,2,3,...,25 -> 26 points..
x=1,y=0,1,2,3,...,24 -> 25 points..
x=2,y=0,1,2,3,...,24 -> 25 points..
x=3,y=0,1,2,3,...,23 -> 24 points..
.
.
.
x=47,y=1 -> 2 points
x=48,y=1 -> 2 points
x=49,y=0 -> 1 point
x=50,y=0 -> 1 point

total = 26 + 2(1+2+3+...+25) = 676 points??