Arithmetic - Time, Speed & Distance - Previous Year CAT/MBA Questions
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What is X's average speed for the entire journey (to and fro)?
- (a)
176 mph
- (b)
180 mph
- (c)
165 mph
- (d)
Data insufficient
Answer: Option D
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Text Explanation :
We do not know the distance between Frankfurt and India and hence cannot calculate the average speed.
Hence, option (d).
Workspace:
Answer the next 2 questions based on the following information.
A thief, after committing the burglary, started fleeing at 12 noon, at a speed of 60 km/hr. He was then chased by a policeman X. X started the chase, 15 min after the thief had started, at a speed of 65 km/hr.
At what time did X catch the thief?
- (a)
3.30 p.m.
- (b)
3 p.m.
- (c)
3.15 p.m.
- (d)
None of these
Answer: Option C
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Text Explanation :
Since the policeman started 15 min late, in this time the thief would have already covered = 15 km. To
catch the thief, the policeman will have to make up for this distance of 15 km. Every hour the policeman is travelling (65 – 60) = 5 km more than the thief. Hence, to make up the distance of 15 km, he would take 3 hr.
Since policeman started at 12.15 p.m., he would catch the thief at 3.15 p.m.
Workspace:
If another policeman had started the same chase along with X, but at a speed of 60 km/hr, then how far behind was he when X caught the thief?
- (a)
18.75 km
- (b)
15 km
- (c)
21 km
- (d)
37.5 km
Answer: Option B
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Text Explanation :
Every hour the second policeman covers (65 – 60) = 5 km less than the first one. Since the first policeman catches the thief in 3 hr, in this time the second policeman will be (3 × 5) = 15 km behind.
Workspace:
Direction: Answer the questions based on the following information.
In a locality, there are five small cities: A, B, C, D and E. The distances of these cities from each other are as follows.
AB = 2 km AC = 2km AD > 2 km AE > 3 km BC = 2 km
BD = 4 km BE = 3 km CD = 2 km CE = 3 km DE > 3 km
If a ration shop is to be set up within 2 km of each city, how many ration shops will be required?
- (a)
2
- (b)
3
- (c)
4
- (d)
5
Answer: Option A
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Text Explanation :
If there is a shop at C, all A, B, C and D are within 2 km range. Another shop is needed for E.
Hence, 2 shops are required.
Workspace:
If a ration shop is to be set up within 3 km of each city, how many ration shops will be required?
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Answer: Option A
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Text Explanation :
If there is a shop at C; all A, B, D and E are within 3 km range. Hence, 1 shop is required.
Workspace:
In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and Chinmay will win a race of one and half miles, and what will be the final lead given by the winner to the loser? (One mile is 1,600 m.)
- (a)
Akshay, mile
- (b)
Chinmay, mile
- (c)
Akshay, mile
- (d)
Chinmay, mile
Answer: Option D
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Text Explanation :
In a mile race, Akshay can be given a start of 128 m by Bhairav. This means that Bhairav can afford to start after Akshay has travelled 128 m and still complete one mile with him. In other words, Bhairav can travel one mile, i.e. 1,600 m in the same time as Akshay can travel (1600 – 128) = 1,472 m. Hence, the ratio of the speeds of Bhairav and Akshay = Ratio of the distances travelled by them in the same time = = 25 : 32. Bhairav can give Chinmay a start of 4 miles. This means that in the time Bhairav runs 100 m, Chinmay only runs 96 m. So the ratio of the speeds of Bhairav and Chinmay = = 25 : 24.
Hence, we have B : A = 25 : 23 and B : C = 25 : 24. So A : B : C = 23 : 25 : 24. This means that in the time Chinmay covers 24 m, Akshay only covers 23 m. In other words, Chinmay is faster than Akshay. So if they race for miles = 2,400 m, Chinmay will complete the race first and by this time Aksahy would only complete 2,300 m. In other words, Chinmay would beat Akshay by 100 m = mile.
Workspace:
A man travels three-fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and return at a speed 5c in the same time, then
- (a)
- (b)
a + b = 99
- (c)
- (d)
None of these
Answer: Option C
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Text Explanation :
Let the total distance be x. So the man travels a distance at a speed 3a. Therefore, total time taken to travel this distance =
He then travels a distance at a speed 2b. Hence, time taken to travel this distance = So total time taken in going from A to B = Now he travels from B to A and comes back. So total distance travelled = 2x at an average speed 5c.
Hence, time taken to return = .
Since the time taken in both the cases remains the same, we can write
Therefore,
Workspace:
A man travels from A to B at a speed x km/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be
- (a)
3 hr
- (b)
6 hr
- (c)
2 hr
- (d)
4 hr
Answer: Option A
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Text Explanation :
Total time taken by the man to travel from A to D = 16 hr and total distance travelled = 36 km. The time that he would have taken had he not rested in between will be (16 – x – 2x) = (16 – 3x). But this time should be equal to the addition of the times that he takes to travel individual segments. This is given as:
Therefore, (16 - 3x).
So we get the equation 3x2 – 16x + 21 = 0. Solving this equation, we get x = 3 or x = . This should be the time for which he rested at B.
Workspace:
In a race of 200 m run, A beats S by 20 m and N by 40 m. If S and N are running a race of 100 m with exactly same speed as before, then by how many metres will S beat N?
- (a)
11.11 m
- (b)
10 m
- (c)
12 m
- (d)
25 m
Answer: Option A
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Text Explanation :
In the same time as A runs 200 m in the race, S runs 180 m and N runs 160 m.
In other words, in the same time as S runs 180 m, N runs 160 m.
So in the same time as S runs 100 m, N will run = 88.89 m.
Hence, in a 100 m race, S will beat N by (100 – 88.89) = 11.11 m.
Workspace:
I live X floors above the ground floor of a high-rise building. It takes me 30 s per floor to walk down the steps and 2 s per floor to ride the lift. What is X, if the time taken to walk down the steps to the ground floor is the same as to wait for the lift for 7 min and then ride down?
- (a)
4
- (b)
7
- (c)
14
- (d)
15
Answer: Option D
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Text Explanation :
Since I live X floors above the ground floor and it takes me 30 s per floor to walk down and 2 s per floor to ride the lift, it takes 30X s to walk down and 2X s to ride the lift after waiting 420 s.
⇒ 30X = 2X + 420 ⇒ X = 15.
Alternative method:
X > 14 as time taken to walk has to be greater than 7 min.
Workspace:
A man can walk up a moving ‘up’ escalator in 30 s. The same man can walk down this moving ‘up’ escalator in 90 s. Assume that his walking speed is same upwards and downwards. How much time will he take to walk up the escalator, when it is not moving?
- (a)
30 s
- (b)
45 s
- (c)
60 s
- (d)
90 s
Answer: Option B
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Text Explanation :
Let the length of the escalator be 90 ft.
(There is no loss of generality in making this assumption.)
Let the speed of the escalator be y ft per second and the man’s walking speed be x ft per second.
According to the question, we get
x + y
x - y
Adding the above equations, we get 2x = 4, i.e., x = 2.
∴ Time taken by the man to walk up the escalator
when it is not moving = or 45 s.
Workspace:
Two towns A and B are 100 km apart. A school is to be built for 100 students of town B and 30 students of Town A. Expenditure on transport is Rs. 1.20 per km per student. If the total expenditure on transport by all 130 students is to be as small as possible, then the school should be built at
- (a)
33 km from Town A.
- (b)
33 km from Town B
- (c)
Town A
- (d)
Town B
Answer: Option D
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Text Explanation :
Hence we find that the least expenditure will be incurred if the school is located in town B.
HINT:
Students please note that since there are more number of students from Town B, to minimise the total expenditure the school should be located as closer to town B as possible.
Workspace:
Shyam went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is times the distance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was half as much again as that from Chandigarh to Shimla. If the average speed for the entire journey was 49 kmph. What was the average speed from Chandigarh to Shimla?
- (a)
39.2 kmph
- (b)
63 kmph
- (c)
42 kmph
- (d)
None of these
Answer: Option C
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Text Explanation :
It is clear that the ratio of the distances between (Delhi- Chandigarh) : (Chandigarh-Shimla) = 3 : 4.
The ratio of the speeds between (Delhi-Chandigarh) : (Chandigarh-Shimla) = 3 : 2.
Let the distances be 3x and 4x respectively and speeds be 3y and 2y.
So the time taken will be and respectively.
Average speed =
Hence, y = 21. So the average speed from Chandigarh to Shimla = 2y = 42 kmph.
Workspace:
A and B walk from X to Y, a distance of 27 km at 5 kmph and 7 kmph respectively. B reaches Y and immediately turns back meeting A at Z. What is the distance from X to Z?
- (a)
25 km
- (b)
22.5 km
- (c)
24 km
- (d)
20 km
Answer: Option B
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Text Explanation :
Let they meet at a distance x kms from X.
So the total distance travelled by A = x at the speed of 5 kmph.
Total distance travelled by B = 27 + (27 – x) = (54 – x) at the speed of 7 kmph.
Time taken by A = .
Time taken by B =
Since they have met at the same time, they would have travelled for the same time. Hence or x = 22.5 kms.
Workspace:
The winning relay team in a high school sports competition clocked 48 minutes for a distance of 13.2 km. Its runners A, B, C and D maintained speeds of 15 kmph, 16 kmph, 17 kmph, and 18 kmph respectively. What is the ratio of the time taken by B to than taken by D?
- (a)
5 : 16
- (b)
5 : 17
- (c)
9 : 8
- (d)
8 : 9
Answer: Option C
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Text Explanation :
Since it is a relay race, all the runners ran the same distance.
Hence, for a same distance,
ratio of times =
Hence, ratio of times taken by B & D = 18 : 16 = 9 : 8.
Workspace:
Directions for next 3 questions:
Q started to move from point B towards point A exactly an hour after P started from A in the opposite direction. Q’s speed was twice that of P. When P had covered one-sixth of the distance between the points A and B, Q had also covered the same distance.
The point where P and Q would meet is
- (a)
Closer to A
- (b)
Exactly between A and B
- (c)
Closer to B
- (d)
P and Q will not meet at all
Answer: Option A
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Text Explanation :
It is clear that after a particular amount of time P and Q are equidistant from A and B respectively and speed of Q is twice the speed of P, therefore, in the remaining time distance moved by Q will be twice than P.
Hence, they would meet closer to A.
Hence, option (a).
Workspace:
How many hours would P take to reach B?
- (a)
2
- (b)
5
- (c)
6
- (d)
12
Answer: Option D
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Text Explanation :
Let the speed of P be x and the distance between A and B be d, so the speed of Q will be 2x.
According to the question, (1 + t)v = 2vt = (Let t be the travel time of Q)
⇒ t = 1, and d = 12v
Hence, the time taken by P to reach to B = = 12 hours.
Hence, option (d).
Workspace:
How many more hours would P (compared to Q) take to complete his journey?
- (a)
4
- (b)
5
- (c)
6
- (d)
7
Answer: Option C
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Text Explanation :
As P takes 12 hours to complete his journey, so Q moving with twice the speed of P will take hours to complete his journey.
Hence P will take 6 hours more than Q to complete the journey.
Hence, option (c).
Workspace:
A ship leave on a long voyage. When it is 18 miles from the shore, a seaplane, whose speed is ten times that of the ship, is sent to deliver mail. How far from the shore does the seaplane catch up with the ship?
- (a)
24 miles
- (b)
25 miles
- (c)
22 miles
- (d)
20 miles
Answer: Option D
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Text Explanation :
The separation between the ship and the seaplane is 18 miles.
Since the two are travelling in the same direction, the relative speed would be 9 times the speed of the ship (If speed of ship is x miles/hour, speed of the seaplane would be 10x and 10x – x = 9x).
Hence, to catch up with the ship, the seaplane would take hours.
Now, the ship covers x miles in an hour, so in hours it would cover 2 miles.
So when the seaplane catches up with the ship, it would be 18 + 2 = 20 miles from the shore.
Hence, option (d).
Workspace:
Every day Neera’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?
- (a)
1 hour
- (b)
50 minutes
- (c)
1/2 hour
- (d)
55 minutes
Answer: Option D
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Text Explanation :
Since her husband meets her mid way, the total time saved by him can be equally divided into time saved while going to station and that saved while returning home.
In other words, he saved 5 mins while going and 5 mins while coming.
So instead of usual time of 6:00 pm he must have met her at 5:55 pm.
So she must have walked for 55 mins.
Hence, option (d).
Workspace:
Each of these items has a question followed by two statements. As the answer,
Type 1, If the question can be answered with the help of statement I alone,
Type 2, If the question can be answered with the help of statement II alone,
Type 3, If both the statement I and statement II are needed to answer the question, and
Type 4, If the question cannot be answered even with the help of both the statements.
A train started from Station A, developed engine trouble and reached Station B, 40 minutes late. What is the distance between Stations A and B?
- The engine trouble developed after travelling 40 km from Station A and the speed reduced to 1/4th of the original speed.
- The engine trouble developed after travelling 40 km from station A in two hours and the speed reduced to ¼th of the original speed.
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Answer: Option B
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Text Explanation :
In order to solve the question, we need to know two things :
(a) the original speed of the train or the new speed of the train and
(b) at what distance from A or after how much time after leaving A the train brokedown.
The statement II provides both of these data viz. original speed = 20 kmph and distance from A = 40 kms. and hence only this is required to answer the question.
For e.g: If the distance between A & B is considered to be x, then time taken had it not broken down is x/20 hours.
The new time taken is [2 + (x – 40)/5] hours and we know that this time is 40 min. more than the original time.
The equation becomes : x/20 + 40/60 = [2 + (x – 40)/5], which can be easily solved to get value of x.
Hence, 2.
Workspace:
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