There are two benches in an office.The capacity of these two benches are 4 and 3 persons respectively.How many methods are possible for making a sitting arrangement of 7 persons in these two benches.
A person goes to office either by car, jeep, bus or train. The probabilities of which being 2/7, 3/7, 1/7 ,1/7 respectively. The probability that he reaches office later if he takes jeep, bus, car or train is 2/9, 1/9, 5/9, 2/9 respectively. If he reaches office in time, then the probability that he travelled by car is
the probability that he travelled by car is 2/7
The probability that he reaches office later if he takes car is 5/9
So probability that he reaches office in time by car =(2/7)*(4/9)= 8/63
Similarly probability for jeep = (3/7)*(7/9) = 21/63
Similarly probability for bus = (1/7)*(8/9) = 8/63
Similarly probability for train = (1/7)*(7/9) = 7/63
so required probability = (8/63)/(8/63)+(21/63)+(8/63)+(7/63)= 8/44 = 2/11
Two trains are moving with speeds 40 km/hr and 50 km/hr in same direction. Train with speed 40 km/hr is 30 km ahead of the second train.
Total path required for making a meeting of the two trains is
OPtion
1) 90 km
2) 100 km
3) 110 km
4) 120 km
5) 130 km
6) 140 km
7) 150 km
8) 160 km
9) 170 km
10) 180 km
Solution
3 hrs are required for meeting , so total path required = 50*3 = 150 km
What will be the probability of selecting 2 integers with same unit's place in such a way that the unit's place of the product of these 2 integers would be same as that of the given integers