Official Quant thread for CAT 2013

MBA CAT 2024
21346
@amresh_maverick said:
@scrabbler how to solve this
Getting 3 (including last meeting) by some horrible method ...not sure....OA?

regards
scrabbler
21347
@scrabbler said:
Getting 3 (including last meeting) by some horrible method ...not sure....OA? regards scrabbler
OA only @jain4444 sir can tell. let us wait



21348
@amresh_maverick said:
OA only @jain4444 sir can tell. let us wait
Me to so-ing soon :(

regards
scrabbler
21349
There is a track on the outer boundary of a rectangular field ABCD (AB = 300m and BC = 200m). Initially Aman and Baman are at point A and B respectively. Both Aman and Baman start moving along the track in anti-clockwise manner at the same time with the same speed. If 'x' is the shortest possible distance between Aman and Baman at any instance then the minimum possible value of 'x' is (neglect the width of the track)
21350
@mohitjain said:
There is a track on the outer boundary of a rectangular field ABCD (AB = 300m and BC = 200m). Initially Aman and Baman are at point A and B respectively. Both Aman and Baman start moving along the track in anti-clockwise manner at the same time with the same speed. If 'x' is the shortest possible distance between Aman and Baman at any instance then the minimum possible value of 'x' is (neglect the width of the track)
x=150root2 ??
21351
@jain4444 3 times at the starting point..my approach was initially ratio of v's 29:19 they meet for 1st time after 29/10:19/10 i.e 1/10th away from starting pt. next meeting 28/10:18/10 ie 7/10th ahead of starting pt. thus for subsequent meetings pts are 4/10th ahead, AT ORIGIN,5/10th ahead,9/10th ahead,2/10th,4/10th,5/10th,5/10th,4/10th,2/10th,9/10th,5/10th,AT ORIGIN,4/10th,7/10th,9/10th,AT ORIGIN. now vijay's vel reduces to 0..so no of meetings=3.
21352
@mohitjain said:
There is a track on the outer boundary of a rectangular field ABCD (AB = 300m and BC = 200m). Initially Aman and Baman are at point A and B respectively. Both Aman and Baman start moving along the track in anti-clockwise manner at the same time with the same speed. If 'x' is the shortest possible distance between Aman and Baman at any instance then the minimum possible value of 'x' is (neglect the width of the track)
200?

regards
scrabbler
21353
Madhav, Munjal, Melkote bought a bike for Rs.45000. Munjal paid one-third of the total amount paid by Madhav and Melkote. Madhav paid half of the amount paid by Munjal and Melkote. How much did melkote pay for the bike?

guys can you please share the approach.

21354
@impluse said:
Madhav, Munjal, Melkote bought a bike for Rs.45000. Munjal paid one-third of the total amount paid by Madhav and Melkote. Madhav paid half of the amount paid by Munjal and Melkote. How much did melkote pay for the bike?guys can you please share the approach.
From 1
Madhav and Melkote , Munjal - 3x , x
4x=45k
x=munjal=11250

From 2
Munjal and Melkote , Madhav - 2y , y
3y=45k
2y=Munjal and Melkot=30k

so , melkote = 30000 - 11250=18750
21355

Two cars starting from the same time run in opposite directions along a circular track ,64 m in circumference.speed of car 1 doubles and car 2 halves every time they meet.If 16mph and 32 mph are the speeds of the cars,what will be the distance travelled by them when they meet for the 5th time ?

21356
@nole said:
Two cars starting from the same time run in opposite directions along a circular track ,64 m in circumference.speed of car 1 doubles and car 2 halves every time they meet.If 16mph and 32 mph are the speeds of the cars,what will be the distance travelled by them when they meet for the 5th time ?
Let the speed of the car 1, car 2 be S1,S2
When they meet for the 1st time,S1=16 mpm and S2= 32 mpm.
Ratio of their speeds = 1 : 2
∴Distance covered by Car 2 = (64 × 2)/3 = 42.67 meter When they meet for
the 2nd time,S 1= 32 mpm and S 2= 16 mpm.Ratio of their speeds = 2 : 1
∴Distance covered by Car 2 = (64/3) = 21.33 meter When they meet for
the 3rd time,S 1= 64 mpm and S 2= 8 mpm.Ratio of their speeds = 8 : 1
∴Distance covered by Car 2 = (64/9) = 7.11 meter When they meet for
the 4thtime,S 1= 128 mpm andS 2= 4 mpm.Ratio of their speeds = 32 : 1
∴Distance covered by Car 2 = (64/33) = 1.94 meter When they meet for
the 5th time,S 1= 256 mpm andS 2= 2 mpm.Ratio of their speeds = 128 : 1
∴Distance covered by Car 2= (64/129) = 0.496 ≈ 0.5 metre
∴Total distance covered by the Car 2 = (42.67 + 21.33 + 7.11 + 1.94 + 0.5) = 73.55metre
∴Total distance covered by Car 1 = Total distance covered by both the cars – Totaldistance covered by the Car 2= (5 × 64) – 73.55 = 320 – 73.55 = 246.45 metr
21357
@scrabbler said:
Getting 3 (including last meeting) by some horrible method ...not sure....OA? regards scrabbler
yes 3 it is

let track of length = 10 km

distance covered by A and V for 1st meet = 29 and 192nd meet = 28 and 19
3rd meet = 27 and 17
4th meet = 26 and 16 (this meet will be at starting point as sum of their distance covered became multiple of 10)

5th = 25 and 15
.
.
.
15th = 15 and 5 (2nd time when they meet at starting point)
16th = 14 and 4
.
.
19th = 10 and 1 (3rd meet)

21358
@amresh_maverick :: OA please?

There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at least one student is chosen from each state ?
21359
@Joey_Sharma said:
@amresh_maverick :: OA please?There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at least one student is chosen from each state ?
One from each in 3.3.3 ways
The remaining in 6C2
So 27.6C2 = 405
Whats the OA
21360
@catahead said:
One from each in 3.3.3 waysThe remaining in 6C2So 27.6C2 = 405Whats the OA
@Joey_Sharma said:
@amresh_maverick :: OA please?There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at least one student is chosen from each state ?
OA : 108

(3 ,1 ,1) -------- 1*3C1*3C1 *3 =27
(2,2, 1) --------- 3C2*3C2*3C1*3=81

81+27=108 ans
21361
In how many ways can 2310 be expressed as a product of 3 factors ?? with soln. pls
21362
@catahead said:
One from each in 3.3.3 waysThe remaining in 6C2So 27.6C2 = 405Whats the OA
@Joey_Sharma said:
@amresh_maverick :: OA please?There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at least one student is chosen from each state ?
OA : 108

(3 ,1 ,1) -------- 1*3C1*3C1 *3 =27
(2,2, 1) --------- 3C2*3C2*3C1*3=81

81+27=108 ans
21363
calculate remainder when 7219485697/72?
21364
how many integer values are there for x and y such that 4x+7y=3? where x
21365
@Calvin4ever said:
In how many ways can 2310 be expressed as a product of 3 factors ?? with soln. pls
25+15+1 = 41