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Time and distance is a very easy topic and has limited variety of questions. I will take up questions from each variety and will solve them with shortest method possible.
Let the distance travelled on foot be X km. Then distance travelled on bicycle will be Rs. (80 - X)
X/8 + (80 - X)/16 = 7
After forming this equation, don't solve it for X. Just observe, X/8 denotes that X is a multiple of 8 (although you can't be 100% sure, but SSC generally likes whole numbers). That means answer is either 32 km or 48 km. Put X = 32 km in the equation and check if it satisfies the equation. It does!
Answer : (A)
This is a commonly asked question. Just remember -
Time taken to cross a telephone pole = Time taken by the train to cover a distance equal to its length
Speed = 54 km/hr = 54 * 5/18 m/sec = 15 m/sec
Time taken by the train to cover a distance equal to its length = 300/15 = 20 sec
Answer : (A)
When the objects are moving in opposite direction, Relative velocity = Speed of A + Speed of B
When the objects are moving in the same direction, Relative velocity = Speed of A - Speed of B
Let the length of each train be L metres.
Here, the trains are moving in the same direction, hence relative velocity = Speed of train A - Speed of Train B = 10 km/hr = 10 * 5/18 m/sec = 25/9 m/sec
We know, Distance = Speed * Time
Here, Distance = Sum of the length of both the trains, i.e., 2L
Speed = Relative velocity
Time = Time taken by the faster train to overtake the slower train
So, 2L = 36 * 25/9
L = 50 m
Answer : (A)
This question is similar to Q. No. 2. We just have to replace the Speed of the train, with Relative Velocity.
Let the length of the fast train be L.
Trains are running in the same direction, hence Relative Velocity = 40 - 20 = 20 km/hr = 50/9 m/sec
L = 50/9 * 5 = 250/9 metres
Answer : (C)
This question is similar to the above question. But it has a twist! The
two trains are not moving at the same time. First train is starting at 7
am, while second train is starting at 8 am.
Method 1 (Relative Velocity)
Remember! To apply the relative velocity formula in such questions, we will first have to make the trains move at the same time. How?
Let the distance between the trains be X km
The first train takes 4 hours (7 am to 11 am) to move from A to B
Hence speed of the first train = X/4
The second train takes 3.5 hours (8 am to 11:30 am) to move from B to A
Hence speed of the second train = X/3.5
From 7 am to 8 am (1 hour), only the first train is moving. Distance travelled by the first train in 1 hour = X/4 * 1 = X/4
Distance left between the two trains = X - X/4 = (3/4)X
Relative velocity of the trains = X/4 + X/3.5
Time taken by the trains to cross one another = (Distance between them) / (Relative velocity)
= (3/4)X / (X/4 + X/3.5)
= 1.40 hours
= 1 hour 24 minutes
So the trains will cross one another at 8 am + 1 hr 24 min = 9 : 24 am
Answer (D)
Method 2 (Equate the distance)
Let the trains cross each other in t hours
First train starts early, so it will travel for complete t hours. But second train starts 1 hour after the first train, so it will travel for (t - 1) hours
Distance travelled by first train in t hours + Distance travelled by second train in (t - 1) hours = X
X/4 * t + X/3.5 * (t - 1) = X
t = 2.4 hours = 2 hours 24 minutes
But this time will be added to 7 a.m.
So, Answer = 7 a.m. + 2 hours 24 minutes = 9:24 a.m.
This question is similar to the above question.
Chor-Sipahi questions are best tackled with Relative Velocity.
Relative Velocity of the Thief and the Owner = 50 - 40 = 10 km/hr
Distance travelled by the thief in half an hour (from 1:30 p.m. to 2 p.m.) = 0.5 * 40 = 20 km
Now the distance between the owner and the thief is 20 km.
So time taken to catch the thief = Distance between them / Relative velocity = 20/10 = 2 hours
He will catch the thief at 2 p.m. + 2 hours = 4 p.m.
Answer : (A)
Method 1 (Relative velocity)
Let P and Q meet in t hours
Distance travelled by P in t hours = 20t (because the speed of P is 20 km/hr)
Distance travelled by Q in t hours = 30t (because the speed of Q is 30 km/hr)
Given, 30t - 20t = 36 km
or t = 3.6 hours
So P and Q meet after 3.6 hours
Relative velocity of P and Q = 20 + 30 = 50 km/hr
Distance between the two places = Time taken by P and Q to meet * Relative velocity = 50 * 3.6 = 180 km
Answer : (A)
Method 2
Let P and Q meet at point M. P has to travel X km to reach M and Q has to travel Y km.
Time taken by P to reach M = Time taken by Q to reach M
X/20 = Y/30
X/Y = 2/3
Given, Y - X = 36 km
Divide the whole equation with Y
1 - X/Y = 36/Y
1 - 2/3 = 36/Y [Put X/Y = 2/3]
Y = 108 km
Hence X = Y - 36 = 72 km
Distance between the two places = X + Y = 108 + 72 km = 180 km
Answer : (A)
Apply the direct formula
Average speed for the complete journey = 2XY/(X + Y) = 2*20*30/50 = 24 km/hr
Answer : (D)
Q. 11) A car travels from P to Q at a constant speed. If its speed is increased by 10 kmph, it
would've taken 1 hour less to cover the distance. It would’ve taken further 45 minutes lesser if the
speed was further increased by 10 kmph. The distance between P and Q is ?
Let the distance between P and Q be X km. Let the car travels this distance at the speed of S km/hr in T
hours.
Then, S*T = (S + 10)(T - 1)
ST = ST – S + 10T – 10
10T = S + 10 … (1)
Similarly, ST = (S+20)(T-1.75)
For more, check-out: http://sschacks.blogspot.in/
ST = ST – 1.75S + 20T – 35
20T = 1.75S + 35 … (2)
Solve (1) and (2). You will get S = 60 km/hr, T = 7 hours
X = ST = 60*7 = 420 km
Distance between P and Q = 420 km
#dailyquizadda #KeepLearning
Time and distance is a very easy topic and has limited variety of questions. I will take up questions from each variety and will solve them with shortest method possible.
Q. 1)
Let the distance travelled on foot be X km. Then distance travelled on bicycle will be Rs. (80 - X)
X/8 + (80 - X)/16 = 7
After forming this equation, don't solve it for X. Just observe, X/8 denotes that X is a multiple of 8 (although you can't be 100% sure, but SSC generally likes whole numbers). That means answer is either 32 km or 48 km. Put X = 32 km in the equation and check if it satisfies the equation. It does!
Answer : (A)
Q. 2)
This is a commonly asked question. Just remember -
Time taken to cross a telephone pole = Time taken by the train to cover a distance equal to its length
Speed = 54 km/hr = 54 * 5/18 m/sec = 15 m/sec
Time taken by the train to cover a distance equal to its length = 300/15 = 20 sec
Answer : (A)
Q. 3)
If two objects A and B are moving at a given speed and we are asked
"when will A overtake B" or "When will the police catch the thief", we
use the concept of Relative Velocity. It's very simpleWhen the objects are moving in opposite direction, Relative velocity = Speed of A + Speed of B
When the objects are moving in the same direction, Relative velocity = Speed of A - Speed of B
Let the length of each train be L metres.
Here, the trains are moving in the same direction, hence relative velocity = Speed of train A - Speed of Train B = 10 km/hr = 10 * 5/18 m/sec = 25/9 m/sec
We know, Distance = Speed * Time
Here, Distance = Sum of the length of both the trains, i.e., 2L
Speed = Relative velocity
Time = Time taken by the faster train to overtake the slower train
So, 2L = 36 * 25/9
L = 50 m
Answer : (A)
Q. 4)
This question is similar to Q. No. 2. We just have to replace the Speed of the train, with Relative Velocity.
Let the length of the fast train be L.
Trains are running in the same direction, hence Relative Velocity = 40 - 20 = 20 km/hr = 50/9 m/sec
L = 50/9 * 5 = 250/9 metres
Answer : (C)
Q. 5)
You can solve such questions with two methods -
Method 1 (Relative Velocity)
Relative velocity of A and B = 6 + 4 = 10 km/hr
They have to cover a distance to 20 km.
Hence they will meet in 20/10 = 2 hours
So if they start at 7 a.m., they will meet at 9:00 a.m.
Answer : (C)
Method 2 (Equate the distance)
Let A and B after t hours
Then distance covered by A in t hours + Distance covered by B in t hours = 20
t*4 + t*6 = 20
or t = 2 hours
Q. 6)
Method 1 (Relative Velocity)
Remember! To apply the relative velocity formula in such questions, we will first have to make the trains move at the same time. How?
Let the distance between the trains be X km
The first train takes 4 hours (7 am to 11 am) to move from A to B
Hence speed of the first train = X/4
The second train takes 3.5 hours (8 am to 11:30 am) to move from B to A
Hence speed of the second train = X/3.5
From 7 am to 8 am (1 hour), only the first train is moving. Distance travelled by the first train in 1 hour = X/4 * 1 = X/4
Distance left between the two trains = X - X/4 = (3/4)X
Relative velocity of the trains = X/4 + X/3.5
Time taken by the trains to cross one another = (Distance between them) / (Relative velocity)
= (3/4)X / (X/4 + X/3.5)
= 1.40 hours
= 1 hour 24 minutes
So the trains will cross one another at 8 am + 1 hr 24 min = 9 : 24 am
Answer (D)
Method 2 (Equate the distance)
Let the trains cross each other in t hours
First train starts early, so it will travel for complete t hours. But second train starts 1 hour after the first train, so it will travel for (t - 1) hours
Distance travelled by first train in t hours + Distance travelled by second train in (t - 1) hours = X
X/4 * t + X/3.5 * (t - 1) = X
t = 2.4 hours = 2 hours 24 minutes
But this time will be added to 7 a.m.
So, Answer = 7 a.m. + 2 hours 24 minutes = 9:24 a.m.
Q. 7)
This question is similar to the above question.
Chor-Sipahi questions are best tackled with Relative Velocity.
Relative Velocity of the Thief and the Owner = 50 - 40 = 10 km/hr
Distance travelled by the thief in half an hour (from 1:30 p.m. to 2 p.m.) = 0.5 * 40 = 20 km
Now the distance between the owner and the thief is 20 km.
So time taken to catch the thief = Distance between them / Relative velocity = 20/10 = 2 hours
He will catch the thief at 2 p.m. + 2 hours = 4 p.m.
Answer : (A)
Q. 8)
Method 1 (Relative velocity)
Let P and Q meet in t hours
Distance travelled by P in t hours = 20t (because the speed of P is 20 km/hr)
Distance travelled by Q in t hours = 30t (because the speed of Q is 30 km/hr)
Given, 30t - 20t = 36 km
or t = 3.6 hours
So P and Q meet after 3.6 hours
Relative velocity of P and Q = 20 + 30 = 50 km/hr
Distance between the two places = Time taken by P and Q to meet * Relative velocity = 50 * 3.6 = 180 km
Answer : (A)
Method 2
Time taken by P to reach M = Time taken by Q to reach M
X/20 = Y/30
X/Y = 2/3
Given, Y - X = 36 km
Divide the whole equation with Y
1 - X/Y = 36/Y
1 - 2/3 = 36/Y [Put X/Y = 2/3]
Y = 108 km
Hence X = Y - 36 = 72 km
Distance between the two places = X + Y = 108 + 72 km = 180 km
Answer : (A)
Q. 1)
Average speed for the complete journey = 2XY/(X + Y) = 2*20*30/50 = 24 km/hr
Answer : (D)
Q. 2)
Let the total distance be 100 km
Average Speed = Total Distance/Total Time
Total time = 70/20 + 10/25 + 20/8 = 3.5 + 0.4 + 2.5 = 6.4
Average Speed = 100/6.4 = 15.625 m.p.h
Q. 3)
This is again a very frequently asked question. Let the distance of his school be X km.
(Time taken to reach the school at 3 km/hr) - (Time taken to reach the school at 4 km/hr) = (10 + 10) minutes or 1/3 hours
X/3 - X/4 = 1/3
Hence X = 4 km
Answer : (B)
Direct Formula
Distance = S1*S2/(S1 - S2) * Time difference
S1 = 4 km/hr, S2 = 3 km/hr, Time Difference = 10 - (-10) = 20 minutes or 1/3 hours
Distance = 4*3/(4-3) * 1/3 = 4 km
Note: In the above formula, while calculating the time difference, "late" time is written with negative sign.
Q. 4)
Method 1
Let time taken by second runner = t. So time taken by first runner = t + 32/60 = t + 8/15
Since distance is constant, hence speed and time are inversely proportional
S2/S1 = T1/T2
16/15 = (t + 8/15)/t
16/15 = 1 + 8/15t
1/15 = 8/15t
t = 8 hours
So second runner takes 8 hours to cover the distance with a speed of 16 km/hr
Hence distance = 8*16 = 128 km
Method 2
Let the distance be X km. Then,
X/15 - X/16 = 32/60
Solve for X
X = 128 km
Answer: (A)
Method 3 (Direct formula)-
In such questions you can use the same formula you used for Q. (3)
Distance = S1*S2/(S1 - S2) * Time difference
Distance = 16*15/(16 - 15) * (32/60) = 16 * 15 * 32/60 = 128 km
Q. 5)
A man rows down a river 15 km in 3 hrs.
Hence, Downstream Speed(v) = 15/3 = 5 km/hr
Similarly, Upstream Speed(u) = 15/7.5 = 2 km/hr
v = Rate in still water + Rate of stream
u = Rate in still water - Rate of stream
Add the above 2 equations-
Rate in still water = (v + u)/2 = (5 + 2)/2 = 3.5 km/hr
Answer: (C)
Q. 6)
We have in the above question-
Speed of the current = (v - u)/2
u = 36/6 = 6 km/hr
v = 48/6 = 8 km/hr
Speed of the current = (8 - 6)/2 = 1 km/hr
Answer: (D)
Q. 7)
Let the distance be X km. Let he takes 't' time downstream, then he will take '2t' time upstream.
Downstream speed(v) = X/t
Upstream speed(u) = X/2t
Speed of the boat in still water/Speed of the current = (v + u)/(v - u) = (X/t + X/2t)/(X/t - X/2t)
= 3/2 : 1/2
= 3 : 1
Answer : (B)
Direct formula-
So, Speed of the boat in still water/Speed of the current = (2t + t)/(2t - t) = 3 : 1
Q. 8)
Given,
24/u + 28/v = 6 or 12/u + 14/v = 3 ... (1)
30/u + 21/v = 6.5 ... (2)
The best way to solve (1) and (2) is by eliminating a variable.
Multiply equation (1) by 3
36/u + 42/v = 9 ... (3)
Multiply equation (2) by 2
60/u + 42/v = 13 ... (4)
Subtract equation (3) from (4)
24/u = 4
u = 6 km/hr
Put u = 6 in equation (1)
v = 14 km/hr
Speed of the boat in still water = (u + v)/2 = (6 + 14)/2 = 10 km/hr
Answer: (D)
Q. 9) Two guns were fired from the same place at an interval
of 13 minutes but a person in a train approaching the place hears the
second shot 12 mins 30 seconds after the first. Find the speed of the
train(approx) supposing that sound travels at 330 m/s.
A. 40 B. 47 C. 55 D. 60
Distance travelled by sound in 30 sec = Distance travelled by train in 12 min 30 sec
Let the speed of the train be X m/sec
Distance travelled by sound in 30 sec = 330*30 metres
Distance travelled by train in 12 min 30 sec (750 sec) = X*750
330*30 = X*750
X = 13.2 m/sec = 13.2 * 18/5 km/hr = 47.52 km/hr
Answer: 47 km/hr
Explanation
When you hear the gun shot, that means the sound has travelled to your ears.
First consider a simple scenario when the train is not moving. When the
two shots are fired from A, a person sitting in the train will hear them
at an interval of 13 minutes only. The sound travels the distance from A
to B.
Now let us consider the scenario when the train is moving from B to A.
When the first shot is fired, the sound will travel from A to B and the
person sitting inside the train will hear it instantly. Now when the
second shot is fired after 13 minutes, the sound would not have to
travel from A to B, because the person sitting inside the train is not
at B any more. He has moved from position B to X. Hence the sound only
needs to travel from A to X.
Hence in this case, the person is hearing the shot after 12 minutes 30
seconds. Instead of travelling for 13 minutes (from A to B), now the
sound is travelling only for 12 min 30 sec (from A to X). Hence we can
say,
AB = Distance travelled by sound in 13 minutes
AX = Distance travelled by sound in 12 minutes 30 seconds
XB = Distance travelled by sound in 30 seconds ... (1)
After 12 minutes 30 seconds, the sound moves from A to X and also the train moves from B to X.
BX = Distance travelled by train in 12 minutes 30 seconds ... (2)
Hence from (1) and (2) we can say-
Distance travelled by sound in 30 sec = Distance travelled by train in 12 min 30 sec
Q. 10) Two guns were fired from the same place at an interval of 10
minutes and 30 seconds, but a person in a train approaching the place
hears second shot 10 minutes after the first. The speed of train (in
km/hr), supposing that sound travels at 330m/s is:
A. 19.8 B. 58.6 C. 59.4 D. 111.8
Distance travelled by sound in 30 sec = Distance travelled by train in 10 minutes (600 sec)
330*30 = X*600
X = 16.5 m/sec or 59.4 km/hr
Answer: (C)
Q. 11) A car travels from P to Q at a constant speed. If its speed is increased by 10 kmph, it
would've taken 1 hour less to cover the distance. It would’ve taken further 45 minutes lesser if the
speed was further increased by 10 kmph. The distance between P and Q is ?
Let the distance between P and Q be X km. Let the car travels this distance at the speed of S km/hr in T
hours.
Then, S*T = (S + 10)(T - 1)
ST = ST – S + 10T – 10
10T = S + 10 … (1)
Similarly, ST = (S+20)(T-1.75)
For more, check-out: http://sschacks.blogspot.in/
ST = ST – 1.75S + 20T – 35
20T = 1.75S + 35 … (2)
Solve (1) and (2). You will get S = 60 km/hr, T = 7 hours
X = ST = 60*7 = 420 km
Distance between P and Q = 420 km
#dailyquizadda #KeepLearning
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