Speed Time And Distance - Study Mode
[#276] A ship leaves on a long voyage. When it is 18 miles from shore, a sea plane, whose speed is ten times that of the ship, is sent to deliver mail. How far from the shore does the sea plane catch up with the ship?
Correct Answer
(D) 20 miles
Explanation
Solution: A _______18miles_____B ___x(let)___C
Let the speed of the ship be 1 mile/h and then speed of the plane becomes 10 miles/h. Let time t is taken to cover the distance x from point B to point C by ship. Hence, in time t plane will cover (18 + x) miles. Now, $$frac{{ ext{x}}}{1} = frac{{18 + { ext{x}}}}{{10}}$$ Or, 10x = 18 + x Or, x = 2 miles. Hence, ship will cover 2 miles in 2 hours and plane will cover 20 miles in same time. Short method: Relative speed = (10 - 1) = 9 miles/h. hence, time taken to catch up the ship is 2 hours so distance covered by the ship = 20 miles.
[#277] Two planes move along a circle of circumference 1.2 km with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction, one plane overtakes the other every 60 seconds. Find the speed of the slower plane.
Correct Answer
(B) 0.03 km/s
Explanation
Solution: The sum of speeds would be 0.8 m/s (relative speed in opposite direction). Also if we go by option (B) the speeds will be 0.03 and 0.05 m/s respectively. At this speed the overlapping would occur in every 60 second. Alternate : Let their speeds be x m/sec and y m/sec respectively. Then, $$eqalign{
& frac{{1200}}{{x + y}} = 15 cr
& Rightarrow x + y = 80.....(i) cr} $$ And, $$eqalign{
& frac{{1200}}{{x - y}} = 60 cr
& Rightarrow x - y = 20.....(ii) cr} $$ Adding (i) and (ii), we get : 2x = 100 or x = 50 Putting x = 50 in (i), we get : y = 30 Hence, speed of slower plane : = 30 m/sec = 0.03 km/sec
[#278] Two joggers left Delhi for Noida simultaneously. The first jogger stopped 42 min later when he was 1 km short of Noida and the other one stopped 52 min later when he was 2 km short of Noida. If the first jogger jogged as many kilometers as the second, and the second as kilometers as first, the first one would need 17 min less than the second. Find the distance between Delhi and Noida?
Correct Answer
(B) 15 km
Explanation
Solution: $$eqalign{
& { ext{Speed of first Jogger}} cr
& = {frac{{ {x - 1} }}{{42}}} imes 60,{ ext{kmph}} cr
& { ext{Speed of }},{2^{nd}},{ ext{jogger}} cr
& = {frac{{ {x - 2} }}{{52}}} imes 60,{ ext{kmph}} cr
& { ext{Then}}, cr
& {frac{{x - 2}}{{{s_b}}}} - {frac{{x - 1}}{{{s_b}}}} cr
& cr} $$ Now, we use option checking method which gives us that option (B) is correct.
[#279] An ant moved for several seconds and covered 3 mm in the first second and 4 mm more in each successive second than its predecessor. If the ant had covered 1 mm in the first second and 8 mm more in each successive second, then the difference between the path it would cover during the same time and actual path would be more than 6 mm but less than 30 mm. find the time for which the ant moved (in seconds).
Correct Answer
(B) 4 s
Explanation
Solution: 3 + 7 + 11 + 15 + . . . . . (1)
1 + 9 + 17 + 25 + . . . . . (2)
The condition is satisfied for the 4 seconds in 2 nd journey, ant covered 17 m more than 1 st one.
[#280] A racetrack is in the form of a right triangle. The longer of the legs of track is 2 km more than the shorter of the legs (both these legs being on a highway). The start and end points are also connected to each other through a side road. The escort vehicle for the race took the side road and rode with a speed of 30 km/h and then covered the two intervals along the highway during the same time with a speed of 42 km/h. find the length of the race track.
Correct Answer
(A) 14 km
Explanation
Solution: The given conditions are met on a Pythagoras triplet 6, 8, 10. Since, the racetrack only consists of the legs of the right angle triangle the length must be, = 6 + 8 = 14 km.