Official Quant thread for CAT 2013

MBA CAT 2024
18426
@Logrhythm said:
nah...it is given that they complete the work in 2 days...
i took ram does 1 unit on first day...that means second day he must do 2 units...
so in 2 days he does 3 units...
now since saleem takes 40% less time to complete the work done by ram in 2 days (units)
=> 2*60/100 = in 6/5 days he does 3 units...
hence in 2 days he does -> 3*5*2/6 = 5 units
this implies that the total work would have been 3+5 = 8 units (as ram and saleem together took 2 days to complete the job)

this is what i had done.......now let me knw if something is wrong...
yea this much is fine man
u misunderstood the meaning of their efficiencies remaining constant
the normal efficiency of saleem is 1 unit in .6 days
in 1 day he does 5/3 units of work
18427
@Subhashdec2 said:
yea this much is fine manu misunderstood the meaning of their efficiencies remaining constantthe normal efficiency of saleem is 1 unit in .6 daysin 1 day he does 5/3 units of work
ohhh i though same efficiency meant 1 unit for both
han then it wld be 3 days...
1*3+5/3*3 = 8 units in 3 days...
18428

52^52^52/7


Find the remainder
18429
@NiTTis said:
52^52^52/7Find the remainder
e(7) is 6..
52^52%6
6=3*2
52^52%2=0
52^52%3=1
2k=3k+1 ie 4
52^4%7 = 3^4%7 = 4...
18430
@NiTTis said:
52^52^52/7Find the remainder
4??
18431
@Logrhythm said:
e(7) is 6..52^52%66=3*252^52%2=052^52%3=12k=3k+1 ie 452^4%7 = 3^4%7 = 4...
can u name the method used
or which is the best method i can use to find remainder and last 2 digits
18432

52^52^52 mod 7

first we calculate 52^52 mod 6 so that it can be expressed in the form of 6k+r

52^52 mod 6 = 4^52 mod 6 = 4
hence
52^(6k+4) mod 7 = 52^6k mod 7 X 52 ^4 mod 7

now 52^6k mod 7=1
and 52^4 mod 7 = 3^4 mod 7 =4
Hence 4 mod 7 =4

@NiTTis
18433
@NiTTis said:
can u name the method usedor which is the best method i can use to find remainder and last 2 digits
i have used the concept of euler's....u can google it...u wld find a link by totalgadha...wo padh lena...u'll be able to understand
18434

two ships sail in a fog towards each other with the same speed. when they are 4 km apart the captains decelerate the engines for 4 minutes with a decceleration rate of 0.1m/s^2, and the ships continue sailing with the speed attained.for what range of values of the initial speed v will the ships avoid collision?

a 0

18435
ABCD is a rectangle with BC = a units and DC = 3 a units. The perpendicular dropped from point A meets BD at point F. The diagonals AC and BD intersect at point G. What is the area (in squareunits) of Ž”AFG?(a) [root(3) x a2]/12(b) [root(3) x a2]/6(c) [root(3) x a2]/10(d) [root(3) x a2]/8Plz give answer along with explanation.
OA:d
@Exodia
approach please...
18436
@NiTTis said:
52^52^52/7Find the remainder
4?
E(7) = 6
52^6k mod 7 = 1
52^52 mod 6 = 4
52^4 mod 7 = 4
18437
Find the remainder when 57! is divided by 61.(a)1,(b)0,(c)60,(d)51,(e)Cannot be determined
18438
@raopradeep said:
Find the remainder when 57! is divided by 61.(a)1,(b)0,(c)60,(d)51,(e)Cannot be determined
60! mod 61=-1
57!*58*59*60 mod 61 =-1
57!*-6 mod 61 =-1
-6R = 61b-1
55R=61b-1
R=-10 satisfy
which is equivalent to 51
18439
@raopradeep said:
Find the remainder when 57! is divided by 61.(a)1,(b)0,(c)60,(d)51,(e)Cannot be determined
51 ?

59! mod 61 = 1
59*58*57! mod 61 = 1
(-2)*(-3)*57! mod 61 = 1
6*x = 61k + 1
k = 5
6*x = 61*5 + 1
x = 51
18440
@raopradeep said:
Find the remainder when 57! is divided by 61.(a)1,(b)0,(c)60,(d)51,(e)Cannot be determined
59!%61 = 1
(59*57)*x%61 = 1
6*x%61 = 1
x = 51...
18441
@Logrhythm
18442
@Logrhythm said:
59!%61 = 1(59*57)*x%61 = 16*x%61 = 1x = 51...
explain how 59!%61=1
18443

find the sum 1/2 + 2^2/2^2 + 3^2/2^3 + 4^2/2^4+............

4
5
6
7
18444
@raopradeep said:
explain how 59!%61=1
wilson's theorem

(prime number -1)! mod (prime number) = (prime number-1) mod (prime number)
(prime number - 2)! mod (prime number) = 1 mod (prime number)
18445
@Logrhythm said:
59!%61 = 1(59*57)*x%61 = 16*x%61 = 1x = 51...
kal to time & work chal raha tha ye no's kab shuru hua ??