Official Quant thread for CAT 2013

MBA CAT 2024
21266
@scrabbler said:
Absolutely! Medians are good fun.Try this:Consider a triangle ABC such that AB = 11, AC = 12 and BC = 13. D, E and F are points on AC such that BD is an altitude, BE an angle bisector and BF a median. Find the lengths DF and EF.(Also try generalising this to any triangle with sides k-1, k and k+1 with the three lines drawn to the line of length k. Something curious I noticed on a long journey when I ran out of better things to do )regardsscrabbler
here is the solution that i came to:

ATDH.
21267
@mohitjain said:
2)A, B, C are the vertices of a triangle of area 60 cm2. Let AD be the median from A on BC and BY be the median from B on AD. If BY is extended to meet AC in E, what is the area of triangle AYE?
@scrabbler do correct if the method is wrong. :)
21268
If f(x) = 9^x/(9^x + 3)
then what would be the value of f(1/882) + f(2/882) + f(3/882) + ... + f(881/882)
21269
@Logrhythm said:
If f(x) = 9^x/(9^x + 3) then what would be the value of f(1/882) + f(2/882) + f(3/882) + ... + f(881/882)
is it 441
21270
@Logrhythm said:
If f(x) = 9^x/(9^x + 3) then what would be the value of f(1/882) + f(2/882) + f(3/882) + ... + f(881/882)
441
21271
@Logrhythm said:
If f(x) = 9^x/(9^x + 3) then what would be the value of f(1/882) + f(2/882) + f(3/882) + ... + f(881/882)
Here,

Taking first,
f(1/882) + f(881/882) =

[9^(1/882)]/[9^(1/882) + 3] + [9^(881/882)]/[9^(881/882) + 3]
Taking LCM we get,

=> 2*9^({1+881}/882) + 3*{9^(1/882)} + 3*{9^(881/882)}/ 9^({1+881}/882) + 3*3 + 3*{9^(1/882)} + 3*{9^(881/882)}

=> 1

I don't know how to represent it any other way!!:P

Hence, Taking 441 Pairs - 1 + 1 + 1 + ... = 441!!
21272
@Logrhythm said:
If f(x) = 9^x/(9^x + 3) then what would be the value of f(1/882) + f(2/882) + f(3/882) + ... + f(881/882)
441??
21273

N=202 x 20002 x 200000002 x 20000000000000002 x 20000000000.....2(31 zeros). The sum of the digits in this multiplication will be
1) 112
2) 160
3) 144
4) None of these

21274
@mohitjain @joyjitpal @hanushanand @viewpt

The answer is 440.5

f(x) = 9^x/(3 + 9^x)
=> f(1-x) = 9^(1-x)/(9^(1-x) + 3) = {(9/9^x)/(9/9^x) + 3} = 9/(3*9^x + 9) = 3/(3+9^x)
=> f(x) + f(1-x) = 9^x/(3 + 9^x) + 3/(3+9^x) = 1
so, f(1) + f(881/882) = f(2/882) + f(880/882) = .... = 1 [440 such pairs]...
and middle term -> f(441/882) = f(1/2) = 1/2
hence, sum = 440+1/2 = 440.5

21275
@nyancat
explain how?
21276
@Logrhythm said:
@mohitjain@joyjitpal@hanushanand@viewptThe answer is 440.5 f(x) = 9^x/(3 + 9^x) => f(1-x) = 9^(1-x)/(9^(1-x) + 3) = {(9/9^x)/(9/9^x) + 3} = 9/(3*9^x + 9) = 3/(3+9^x)=> f(x) + f(1-x) = 9^x/(3 + 9^x) + 3/(3+9^x) = 1so, f(1) + f(881/882) = f(2/882) + f(880/882) = .... = 1 [440 such pairs]...and middle term -> f(441/882) = f(1/2) = 1/2hence, sum = 440+1/2 = 440.5
21277
The four numbers a (1) 4 (2) 35/6 (3) 15/2 (4) 7 (5) 11
21278
@Exodia said:
N=202 x 20002 x 200000002 x 20000000000000002 x 20000000000.....2(31 zeros). The sum of the digits in this multiplication will be1) 1122) 1603) 1444) None of these
Mera to 1024 coming!! [EDIT]

5*32 = 160!!
21279
@Exodia said:
N=202 x 20002 x 200000002 x 20000000000000002 x 20000000000.....2(31 zeros). The sum of the digits in this multiplication will be1) 1122) 1603) 1444) None of these
21280
@joyjitpal said:
The four numbers a
ab ac ad bc bc and cd are 6 diff pairs...
a+b = 1
a+c = 2
b+c = 3

this gives a=0,b=1 anc c=2
now since a+d=4

hence d can only be 4...

21281
@Exodia said:
N=202 x 20002 x 200000002 x 20000000000000002 x 20000000000.....2(31 zeros). The sum of the digits in this multiplication will be1) 1122) 1603) 1444) None of these
160
21282
@joyjitpal said:
The four numbers a
OA 1
21283
@joyjitpal said:
The four numbers a (1) 4 (2) 35/6 (3) 15/2 (4) 7 (5) 11
21284
In how many ways can a first prize, a second prize and four identical third prizes be
awarded to a group of 15 people?
21285
Two cars starting from the same point at the same time run in opposite direction along a circular track,64 meter in circumference.Speed of Car1 doubles and Car 2 halves every time they meet.If 16 mpm and 32 mpm are the speed of the cars,what will be the distance travelled by each when they meet for the fiifth time?
a)258.13 m and 61.87m
b)263.55m and 56.45m
c)271.18m and 48.82m
d)246.45m and 73.55m