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Question 1 of 5
1. Question
Two persons A and B start simultaneously from two places c km apart and walk in the same direction. If A travels at the rate of p km/h and B travels at the rate of q km/h, then A has travelled before he overtakes B a distance of
Correct
Ans: b
Let A and B will meet after t h at point E.
Distance travelled by A = pt h
[because, distance = speed ´ time]
and distance travelled by B = qt h
According to the question,
pt = qt +C Þ pt – qt = c
t(p-q) = c
t=(c/p-q) …………………………………………………………(1)
\Distance travelled by A = pt = (pc/(p-q)) km [from Eq.(1)]
Incorrect
Ans: b
Let A and B will meet after t h at point E.
Distance travelled by A = pt h
[because, distance = speed ´ time]
and distance travelled by B = qt h
According to the question,
pt = qt +C Þ pt – qt = c
t(p-q) = c
t=(c/p-q) …………………………………………………………(1)
\Distance travelled by A = pt = (pc/(p-q)) km [from Eq.(1)]
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Question 2 of 5
2. Question
With a uniform speed, a car covers a distance in 8 h. Had the speed been increased by 4 km/h, the Same distance could have been covered in 7h and 30 min. What is the distance covered?
Correct
Ans: b
Let the distance between A and B be x km.
Case l Given, distance= xkm, speed= V km/h [let]
and time =8h
Because Speed = Distance/Time
Because V = x/8
Case 2 If speed = (V + 4) km/h
And time = 7 (1/2)h = 15/2 h
Because V + 4 = (x/(15/2))
V+4 = (2x/15) ⇒ (x/8) + 4 = 2x/15 [ From Eq. (1)]
(2x/15) – (x/8) = 4 ⇒ x/ 120 = 4
∴x = 480 km
Incorrect
Ans: b
Let the distance between A and B be x km.
Case l Given, distance= xkm, speed= V km/h [let]
and time =8h
Because Speed = Distance/Time
Because V = x/8
Case 2 If speed = (V + 4) km/h
And time = 7 (1/2)h = 15/2 h
Because V + 4 = (x/(15/2))
V+4 = (2x/15) ⇒ (x/8) + 4 = 2x/15 [ From Eq. (1)]
(2x/15) – (x/8) = 4 ⇒ x/ 120 = 4
∴x = 480 km
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Question 3 of 5
3. Question
A runs 5/3 times as fast as B. If A given B a start of 80 m, then how far must the winning post from the starting point be so that A and B might reach it at the same time?
Correct
Ans: a
Let the speed of B be x m/s.
∴Speed of A = 1(2/3)x = (5x/3)m/s
Ratio of speed of rates of A and B= 5x/3 : x = 5:3
Because 2m are gained in a race of 5 m.
Because 1 m are gained in a race of 5/2 m.
So, 80 m are gained in a race of (5/2) x 80= 200 m
Incorrect
Ans: a
Let the speed of B be x m/s.
∴Speed of A = 1(2/3)x = (5x/3)m/s
Ratio of speed of rates of A and B= 5x/3 : x = 5:3
Because 2m are gained in a race of 5 m.
Because 1 m are gained in a race of 5/2 m.
So, 80 m are gained in a race of (5/2) x 80= 200 m
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Question 4 of 5
4. Question
A train takes 10 s to cross a pole and 20 s to cross a platform of length 200 m. What is the length of the train?
Correct
Ans: d
Let length of the train be lm
∴Speed of train= (l/10)m/s
Distance covered by train to cross the platform of length
200 m= (l+ 200) m
Time taken to cross the platform = Distance/Speed
20 = (l+200)/(1/10)
(l/10)x20 = l+200
2l-1= 200
∴l = 200 m
Hence, length of the train is 200 m.
Incorrect
Ans: d
Let length of the train be lm
∴Speed of train= (l/10)m/s
Distance covered by train to cross the platform of length
200 m= (l+ 200) m
Time taken to cross the platform = Distance/Speed
20 = (l+200)/(1/10)
(l/10)x20 = l+200
2l-1= 200
∴l = 200 m
Hence, length of the train is 200 m.
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Question 5 of 5
5. Question
The distance between two points (A and B) is 110 km. X starts running from point A at a speed of 60 km/h and Y starts running from point Bat a speed of 40 km/h at the same time. They meet at a point C, somewhere on the line AB. What is the ratio of AC to BC?
Correct
Ans: a
Distance between two points= 110 km
Their relative speed = 60+ 40 = 100 km/h
Time after which they meet= Total distance/ Relative speed = 110/100 = 1.10 h
Distance covered by A in 1.10 h= AC = 60 x 1.10 = 66 km
Remaining distance = BC= 110- 66= 44 km
∴Required ratio = AC: BC= 66: 44= 3:2
Incorrect
Ans: a
Distance between two points= 110 km
Their relative speed = 60+ 40 = 100 km/h
Time after which they meet= Total distance/ Relative speed = 110/100 = 1.10 h
Distance covered by A in 1.10 h= AC = 60 x 1.10 = 66 km
Remaining distance = BC= 110- 66= 44 km
∴Required ratio = AC: BC= 66: 44= 3:2
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