For the first time SSC examination will be held in Computer Based pattern and managing time will be one of the important factor while solving the questions. So in order to make students familiar with such a situations we are providing questions in a time based manner , which will help students to manage the time properly.
1. In a 100m race, A runs at 5 km/h. A gives B a start of 8 m and still beats him by 8 sec. Find out the speed of B.
6.14 km/h
4.14 km/h
3.14 km/h
2.14 km/h
Solution:
1.(b) Time taken by A to cover 100 m =100÷(5×5/18)= 72 seconds.
∴ B covers (100 – 8) or, 92m in (72 + 8) or, 80 seconds.
∴ Speed of B = 92/80×18/5= 4.14 km/h
∴ B covers (100 – 8) or, 92m in (72 + 8) or, 80 seconds.
∴ Speed of B = 92/80×18/5= 4.14 km/h
2.In a Km race A can beat B by 80 m and B can beat C by 60 m. In the same race, A can beat C by:
135.2m
130.5 m
142 m
132.5 m
Solution:
2. (a) While A runs 1000 m,
B runs 1000 – 80 = 920 m
And while B runs 1000 m,
C runs 1000 – 60 = 940 m.
∴ While B runs 920 m;
C runs 940/1000×920= 4324/5 m
∴ While A runs 1000m,
C runs 4324/5 m
∴ A can beat C by 1000 -4324/5=676/5=135(1/5)m.
B runs 1000 – 80 = 920 m
And while B runs 1000 m,
C runs 1000 – 60 = 940 m.
∴ While B runs 920 m;
C runs 940/1000×920= 4324/5 m
∴ While A runs 1000m,
C runs 4324/5 m
∴ A can beat C by 1000 -4324/5=676/5=135(1/5)m.
3. In a Km race A beats B by 100 m and C by 200 m. By how many metre can B beat C in a race of 1350 m?
150 m
120 m
1200 m
210 m
Solution:
3. (a) While A runs 1000 m,
B runs 1000 – 100 = 900 m and
C runs 1000 – 200 = 800 m.
∴ While B run 900 m,
C runs 800 m.
∴ While B runs 1350 m;
C runs 800/900×1350=1200 m
∴ B can beat C by 1350 – 1200 = 150 m.
B runs 1000 – 100 = 900 m and
C runs 1000 – 200 = 800 m.
∴ While B run 900 m,
C runs 800 m.
∴ While B runs 1350 m;
C runs 800/900×1350=1200 m
∴ B can beat C by 1350 – 1200 = 150 m.
4.Two boys, A and B, runs at 4 1/2 and 6 Km an hour. A having 190 m start. The course being 1 km, B wins by a distance of:
60 m
65 m
45 m
75 m
Solution:
4. (a) Speeds (in m/sec) of A and B are 9/2×5/18=5/4 and 6×5/18=5/3 , respectively.
A has a start of 190 m. So A has to run 1000 – 190 = 810 m,
while B 1000 m. Time taken by B to cover 1000 m =3/5×1000=600 seconds.
In this time, A covers 5/4×600= 750 m.
So, B reaches the winning post while A remains 810 – 750 = 60 m behind.
∴ B wins by 60 m.
A has a start of 190 m. So A has to run 1000 – 190 = 810 m,
while B 1000 m. Time taken by B to cover 1000 m =3/5×1000=600 seconds.
In this time, A covers 5/4×600= 750 m.
So, B reaches the winning post while A remains 810 – 750 = 60 m behind.
∴ B wins by 60 m.
5.A and B runs a km race. If A gives B a start of 50 m, A wins by 14 sec and, if A gives B a start of 22 sec , B wins by 20 m. The time taken by A to run a km is?
100 sec
120 sec
105 sec
125 sec
Solution:
5. (a) Let, times (in sec) taken by A and B to run a Km, be x and y, respectively.
When B gets a start of 50 m, B runs. 1000 – 50 = 950m while A runs 1000 m.
∴ (950/100)y-x=14, i.e. 0.95y – x = 14 …(i) and
when B gets a start of 22 seconds,
A runs for (y – 22) seconds, while B runs for y seconds
∴ 1000- 1000/x (y-22)=20 i.e. 50y – 49x = 1100. …(ii)
Multiplying equation (i) by 49 and subtract from equation (ii)
3.45y = 414
∴ y = 120 sec ∴ (i)
⇒ x = 0.95 × 120 – 14 = 100 seconds.
When B gets a start of 50 m, B runs. 1000 – 50 = 950m while A runs 1000 m.
∴ (950/100)y-x=14, i.e. 0.95y – x = 14 …(i) and
when B gets a start of 22 seconds,
A runs for (y – 22) seconds, while B runs for y seconds
∴ 1000- 1000/x (y-22)=20 i.e. 50y – 49x = 1100. …(ii)
Multiplying equation (i) by 49 and subtract from equation (ii)
3.45y = 414
∴ y = 120 sec ∴ (i)
⇒ x = 0.95 × 120 – 14 = 100 seconds.
6.A runner runs 1 1/4 laps of a 5 lap race. What fractional part of the race remains to be run?
15/4
4/5
5/6
2/3
Solution:
6. (a) 5- 5/4=15/4.
7.A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds. B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet next at the starting point again?
46 minutes 12 seconds
45 minutes
42 minutes 36 seconds
26 minutes 18 seconds
Solution:
7. (a) L.C.M of 252, 308 and 198 = 2772 seconds = 46 × 60 + 12 = 46 min 12 seconds.
8. From a point on a circular track 5 Km long A, B and C started running in the same direction at the same time with speeds of 2 1/2 km/h, 3 km/h and 2 km/h respectively. Then on the starting point all three will meet again after?
30 hours
6 hours
10 hours
15 hours
Solution:
8. (c) A will reach at starting point in (5 ×2)/5=2 hours
B will reach at starting point in 5/3 hours
C will reach at starting point in 5/2 hours
Then, on the starting point all three will meet after the L.C.M. of 2, 5/3,5/2=10/1=10 hours.
B will reach at starting point in 5/3 hours
C will reach at starting point in 5/2 hours
Then, on the starting point all three will meet after the L.C.M. of 2, 5/3,5/2=10/1=10 hours.
9.Four runners started running simultaneously from a point on a circular track. They took 200 seconds, 300 seconds, 360 seconds and 450 seconds to complete one round. After how much time do they meet at the starting point for the first time?
1800 seconds
3600 seconds
2400 seconds
4800 seconds
Solution:
9. (a) Required = L.C.M. of 200, 300, 350 and 450 s = 1800 s
10.In a 100 m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 km/h, then the speed of Bimal is:
15.4 km/h
14.5 km/h
14.4 km/h
14 km/h
Solution:
10. (c) Time taken by Kamal to run 100 m =100/(18×5/( 18))=20 s
Therefore, time taken by Bimal to run 100m = 20 + 5 = 25 s
Hence, Bimal’s speed =100/25= 4 m/sec =(4 ×18)/5 km/h = 14.4 km/h
Therefore, time taken by Bimal to run 100m = 20 + 5 = 25 s
Hence, Bimal’s speed =100/25= 4 m/sec =(4 ×18)/5 km/h = 14.4 km/h
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