21. A and B run a race of 2000m. First A gives B a start of 200m and beats him by 30 sec. Next, A gives B a start of 3 minutes and is beaten by 1000m. find the time in minutes in which A and B can run the race separately.
Sol:
make speed of B equal in both cases
\(\frac{{1800}}{{t + 30}} = \frac{{2000}}{{\frac{t}{2} + 180}}\)
\(\frac{9}{{t + 30}} = \frac{{20}}{{t + 360}}\)
20t + 600 = 9t + 3240
11t = 2640
t = 240 sec = 4 min ⟵ Time taken by A Answer
& time taken by B = \(\frac{t}{2} + 180 + \frac{{240}}{2} + 180\) = 300 sec = 5 min Answer
22. A and B run a race of 3 km. A gives B a heaed-start of 400m and beats him by 30 sec. While coming back, A gives B a leadw of 2.5 min and gets beaten by 500m. What is the difference between the times in minutes in which A and B can run the race for one side separately.
Sol:
∴ \(\frac{{2600}}{{t + 30}} = \frac{{3000}}{{\frac{{5t}}{6} + 150}}\)
\(\frac{{13}}{{t + 30}} = \frac{{90}}{{5t + 900}}\)
90t + 2700 = 65t + 11700
25t = 9000
t = 360 secc = 6 min ⟵ Time taken by A
& time taken by B = \(\frac{{5t}}{6} + 150\)
= \(\frac{{5 \times 360}}{6} + 150\)
= 450 sec
= 7.5 min
∴ required time difference = 7.5 – 6 = 1.54 min Answer
23. In a 150m race, A wins over B by 35m and in the same race C can give a start of 25m to B. By how much distance can A give a start to C, so that A beats C by 5m ?
Sol:
A beats C by 12m & since it is given that A beats C by 5m so A gives a start of 12 – 5 = 7m to C Answer
24. In a 1000m race A gives B a start of 40m and beats him by 19 sec. If A gives a start of 30 sec then B beats A by 40 m. Find the ratio of speeds of A and B.
Sol:
Speed of B = Speed of B
\(\frac{{960}}{{t + 19}} = \frac{{1000}}{{\frac{{24t}}{{25}} + 30}}\)
\(\frac{{24}}{{t + 19}} = \frac{{625}}{{24t + 750}}\)
625t + 19 × 625 = 576t + 24 × 750
49t = 6125t = 125 sec
∴ \(\frac{{Speed\;of\;A}}{{Speed\;of\;B}} = \frac{{\frac{{1000}}{t}}}{{\frac{{960}}{{t + 19}}}} = \frac{{t + 19}}{t} \times \frac{{25}}{{24}}\)
= \(\frac{{144}}{{125}} \times \frac{{25}}{{24}}\)
= 6 : 5 Answer
25. In a race of 100m A can give a start of 20m to B and a start of 40m to C. How much start can B give to C in a 100m race.
Sol:
∴ B gives a start of 1m in 4m race
∴ in 100m race start will be \(\frac{1}{4} \times 100\) = 25m Answer
26. A can beat B in a 100 m race by 10 m. B can beat C in a 100 m race by 10 m. What is the ratio of time taken by A, B and C to complete the race.
Sol:
27. In a game of 100 points, A can give B 10 points and C 20 points. Then how many points B can give to C ?
Sol:
28. If the ratio of speed of A and B is 5 : 9 and A loses the race by 48 m, what is the length of race track ?
Sol:
Since time is constant so ratio of distance will be same as ratio of speeds.
29. The ratio of time taken by A and B to run a certain distance is 2 : 3 and A wins by 100 m. Then find the length of the race track.
Sol:
Since distance is constant
∴ ratio of speed ∝ \(\frac{1}{{ratio\;of\;time}}\)
30. In a 480 m race, A gave B a head-start of 6 seconds and still won the race by 10 sec. The ratio of the speed of A & B are 3 : 1. How long will B take to cover a race of 1000 m ?
Sol:
Since distance is constant
∴ Speed ∝ \(\frac{1}{{Time}}\)
⟹ \(\frac{1}{t} = \frac{3}{{t + 16}}\)
⟹ t = 8
∴ required time = \(\frac{{8 + 16}}{{480}} \times 1000\) = 50 sec Answer