Three wheels can complete respectively 60, 36, 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?
(a) 5/2 seconds
(b) 5/3 seconds
(c) 5 seconds
(d) 7.5 seconds
(e) None of these
Answer: (c)
Let the three wheels be A, B &C which can complete 60, 36, 24 revolutions per minute respectively. Then,
A makes 1 rev. per sec
B makes \(\displaystyle \frac{6}{{10}}\) rev. per sec
C makes \(\displaystyle \frac{4}{{10}}\) rev. per sec
In other words A, B and C take \(\displaystyle 1,\frac{5}{3}\And \frac{5}{2}\) seconds to complete one revolution.
L.C.M of \(\displaystyle 1,\frac{5}{3}\And \frac{5}{2}\) = \(\displaystyle \frac{{L.C.Mof1,5,5}}{{H.C.Fof1,3,2}}=5\)
Hence, after every 5 seconds the red spots on all the three wheels touch the ground
A train after travelling 150 km meets with an accident and then proceeds with 3/5 of its former speed and arrives at its destination 8 h late. Had the accident occurred 360 km further, it would have reached the destination 4 h late. What is the total distance travelled by the train?
If x is the original speed, then difference in time
\(\displaystyle \frac{{360}}{{\frac{{3x}}{5}}}-\frac{{360}}{x}=4\) Solving, we get x=60 km/hr If T is the total hrs, then \(\displaystyle 60T=150+60\times \frac{3}{5}\times \left( {T+8-2.5} \right)\)
Solving, we get T=14.5 hrs and distance =60 \(\displaystyle \times \)14.5=870 km
A 180-metre long train crosses another 270- metre long train running in the opposite direction in 10.8 seconds. If the speed of the first train is 60 kmph, what is the speed of the second train in kmph?
(a) 80
(b) 90
(c) 150
(d) Can’t be determined
(e) None of these
Answer: (b)
Relative speed of two trains = \(\displaystyle \frac{{180+270m}}{{10.8}}m/s=\frac{{4500}}{{108}}m/s\)
Deepa rides her bike at an average speed of 30 km/hr and reaches her destination in 6 hrs. Hema covers the same distance in 4 hrs. If Deepa increases her average speed by 10 km/hr and Hema increases her average speed by 5 km/hr, what would be the difference in their time taken to reach the destination?
(a) 54 minutes
(b) 1 hour
(c) 40 minutes
(d) 45 minutes
(e) None of these
Answer: (a)
Average Speed of Deepa = 30 kmph.
Time taken to reach the destination = 6 hours.
So, total distance of destination = 6 \(\displaystyle \times \)30 = 180 Km.
Time taken to covered 180 km by Hema = 4 hours.
Speed of Hema = 180 /4 = 45 kmph.
Deepa’s new average speed = 30 +10 = 40 kmph.
Time Taken = 180/40 = 4.5 hours.
Hema’s new average speed = 45+5 = 50 Kmph.
Time taken by Hema = 180/50 = 3.6 hours. Time difference = 4.5 – 3.6 = 0.9 hours = 0.9 \(\displaystyle \times \) 60 = 54 minutes
A man starts going for morning walk every day. The distance walked by him on the first day was 2 km. Every day he walks half of the distance walked on the previous day. What can be the maximum total distance walked by him in his lifetime?
(a) 4 km
(b) 20 km
(c) 8 km
(d) Data inadequate
(e) None of these
Answer: (a)
Distance walked by man = \(\displaystyle 2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{{16}}…\infty \)
