Arithmetic - Time, Speed & Distance - Previous Year CAT/MBA Questions

Answer: Option B

Explanation :

Let the time taken to meet after starting be t mins.

Time taken by A to reach Y after meeting = (10 – t) mins
Time taken by B to reach X after meeting = 9 mins

⇒ t² = (10 - t) × 9
⇒ t² +9t – 90 = 0
⇒ (t + 15)(t - 6) = 0
⇒ t = 6 (-15 is rejected)

Total time taken by B to reach Y = 6 + 9 = 15 mins.

Hence, option (b).

Answer: Option A

Explanation :

Let the speed of the slower ship be x km/hr.
Distance travelled by it in 2 hours = 2x kms

Speed of faster ship = (x + 6) km/hr.
Distance travelled by it in 2 hours = 2(x + 6) kms

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⇒ (2x)² + (2(x + 6))² = (60)2 
⇒ 4x² + 4(x²  12x + 36) = 3600
⇒ 8x² + 48x + 144 = 3600
⇒ 8x² + 48x - 3456 = 0
⇒ x² + 6x - 432 = 0
⇒ (x + 24)(x - 18) = 0
⇒ x = 18 (-24 is rejected)

∴ Speed of slower person is 18 km/hr

Hence, option (a).

Answer: 60

Explanation :

Let speed of Moody and escalator be ‘m’ and ‘e’ steps/sec and total number of steps in the escalator be ‘N’.

We will assume that Moody is going in the same direction as the escalator. If that is not the case value of e will come out to be negative.

Moody takes 30 seconds at normal speed ⇒ N = 30m + 30 ×e   …(1)

Moody takes 20 seconds at twice the speed ⇒ N = 20 × 2m + 20 ×e   …(2)

4 × (1) – 3 × (2)
N = 60e
⇒ N/e = 60

∴ If someone was standing on the escalator, he/she will take 60 seconds to climb the escalator.

Hence, 60.

Answer: Option A

Explanation :

When travelling towards each other, the two cars take 1.5 hours to meet.

⇒ The distance travelled by slower car in this case till they meet = 1.5 × 60 = 90 kms.

Hence, option (a).

Answer: Option C

Explanation :

Let the length of the smaller train be L meters.
Also, let the speeds of faster and slower trains be ‘f’ and ‘s’ m/s.

Given, f - s = 6 × 5/18 = 5/3 m/s
⇒ f = s + 5/3   …(1)

Faster train crosses a lamp post in 12 seconds.
∴ 160/f = 12
⇒ f = 40/3

∴ s = f – 5/3 = 35/3

The two trains cross each other in 14 seconds.
∴ (160 + L)/(f + s) = 14
⇒ L = 14(f + s) – 160
⇒ L = 14 × 25  – 160
⇒ L = 350 – 160 = 190

Hence, option (c).

Answer: Option B

Explanation :

Let the lengths of trains A and B be ‘A’ and ‘B’ and their speeds be ‘a’ and ‘b’ respectively.

Time taken for front end of A to cross rear end of B = 46 seconds

⇒ 46 = $\frac{B}{a+b}$

⇒ B = 46a + 46b   …(1)

Total time taken for rear ends of the two train to cross each other = (46 + 69) = 115 seconds

⇒ 115 = $\frac{A+B}{a+b}$

⇒ A + B = 115a + 115b   …(2)

(2) - (1)

⇒ A = 69a + 69b   …(3)

Now, $\frac{A}{B}$ = $\frac{69(a+b)}{46(a+b)}$ = $\frac{69}{46}$ = $\frac{3}{2}$

Hence, option (b). 

Answer: 8

Explanation :

Let the length of the track be ‘L’ meter and speeds of Mira and Amal be ‘m’ and ‘a’ meters/minute respectively.

When they walk in opposite direction, they meet in 3 minutes.
∴ 3(m + a) = L   
⇒ 3m + 3a = L   …(1)

When they walk in same direction, Amal completes 3 rounds more than Mira. Hence, Amal completes 3L meters more than Mira.
∴ 45a – 45m = 3L   …(2)

(1) × 15 - (2)

⇒ 90m = 12L

⇒ L/m = 7.5 

Hence, it takes Mira 7.5 minutes to complete one round.

Hence, in 60 minutes she will cover = 60/7.5 = 8 rounds.

Hence, 8.

Answer: Option B

Explanation :

Let the speed and length of train be ‘s’ m/s and ‘t’ m respectively.

Speeds of two persons is 2 kmph and 4 kmph i.e., 5/9 m/s and 10/9 m/s

Train passes the first person in 90 seconds.

∴ $\frac{t}{s-\frac{5}{9}}$ = 90

⇒ t = 90s – 50   …(1)

Train passes the second person in 100 seconds.

∴ $\frac{t}{s-\frac{10}{9}}$ = 100

⇒ t = 100s – 1000/9   …(2)

(1) = (2)

⇒ 90s – 50 = 100s – 1000/9

⇒ s = 55/9 m/s and t = 500 m

∴ Time taken to cross an electric pole = $\frac{500}{55/9}$ = $\frac{4500}{55}$ = $\frac{900}{11}$ = 81.81 ≈ 82 seconds.

Hence, option (b).

Answer: Option B

Explanation :

Let the time taken for them to meet after starting from opposite ends be t mins.

Also, let speeds of Car 2 be b kmph.

​​​​​​​

Distance AM = 60 × t = b × 20   …(1)

Distance BM = b × t = 60 × 45   …(2)

(1) × (2)

⇒ t² = 900

⇒ t = 30 minutes

From (2) we get,

60 × 30 = b × 20

⇒ b = 90 kmph

Hence, option (b).

Answer: Option D

Explanation :

Let the distance between home and office be d kms.

∴ $\frac{d}{8}-\frac{d}{15}=\frac{35}{60}=\frac{7}{12}$

⇒ d = 10 kms

If he leaves home at 9:10 a.m. and reaches office at 10 a.m., i.e., time taken = 50 minutes = 5/6 hours

∴ Required speed = $\frac{10}{5/6}$ = 12 kmph.

Hence, option (d).

Answer: Option C

Explanation :

Since the train travelled at 1/3rd of its speed, it will take thrice the usual time.

Hence, if the normal time taken for train is t minutes, now it will take 3t minutes.

∴ 3t – t = 30 

⇒ t = 15 minutes.

While coming back train travels at usual speed for 5 minutes and then takes 4 minutes break.

∴ If the train has to come back at scheduled time, it needs to cover the remaining distance in (15 – 5 – 4 =) 6 minutes. Train usually takes 10 minutes to travel this distance.

∴ The time reduced to 6/10 = 3/5th hence speed should become 5/3 times.

∴ % increase = $\frac{\frac{5}{3}-1}{1}$ × 100 = 66.66% ≈ 67%

Hence, option (c).

Answer: Option B

Explanation :

Let their speeds be a, b and c respectively and the length of the race be D.

Car A beats car B by 45 kms

∴ $\frac{a}{b}=\frac{D}{D-45}$   …(1)

Car B beats car C by 50 kms

∴ $\frac{b}{c}=\frac{D}{D-50}$   …(2)

Car A beats car C by 90 kms

∴ $\frac{a}{c}=\frac{D}{D-90}$   …(3)

   (1) × (2) = (3)

⇒ $\frac{D}{D-45}\times \frac{D}{D-50}=\frac{D}{D-90}$

⇒ $\frac{D}{(D-45)(D-50)}=\frac{1}{D-90}$

⇒ D² – 95D + 2250 = D² – 90D

⇒ 5D = 2250

⇒ D = 450

Hence, option (b).

Answer: Option D

Explanation :

Let the time taken for them to meet after starting from opposite ends be t mins.

Also, let speeds of Ram and Rahim be a and b respectively.

​​​​​​​

Distance AM = a × t = b × 4   …(1)

Distance BM = b × t = a × 1   …(2)

(1) × (2)

⇒ t² = 4

⇒ t = 2

From (1) we get,

a × 2 = b × 4

⇒ a : b = 4 : 2 = 2 : 1

Hence, option (d).

Answer: Option A

Explanation :

Time taken by Ram to reach the starting point A = $\frac{2\times \pi \times 0.1}{15}=\frac{\pi }{75}$

Time taken by Rahim to reach the starting point A = $\frac{2\times \pi \times 0.02}{5}=\frac{\pi }{125}$

Both of them will meet only at A since they are running in different circles which meet at A.

Both of them will reach the starting point after LCM$\left[\frac{\pi }{75},\frac{\pi }{125}\right]$ = $\frac{\pi }{25}$

∴ Number of rounds completed by Ram when both of them meet again = $\frac{\frac{\pi }{25}}{\frac{\pi }{75}}$ = 3

Hence, (a).

Answer: Option C

Explanation :

Let the distance from A to B = d and that from B to C = 3d.

Speed of train T₁ = s from A to B and 2s from B to C.

Total time taken by T₁ = $\frac{d}{s}+\frac{3d}{2s}=\frac{5d}{2s}$

Speed of train T₂ = 2s from A to B and s from B to C.

Total time taken by T₂ = $\frac{d}{2s}+\frac{3d}{s}=\frac{7d}{2s}$

∴ Ratio of time taken by T₁ and T₂ = $\frac{5d}{2s}:\frac{7d}{2s}$ = 5 : 7

Hence, option (c).

Answer: Option B

Explanation :

Train A starts at 9 a.m. whereas Train B starts at 10:30 a.m.

Distance travelled by A till 10:30 a.m. = 40 × 1.5 = 60 kms.

At 10:30 a.m. remaining distance between A and B = 90 – 60 = 30 kms.

Time taken for both of them to meet now = 30/(40 + 20) = 0.5 hours i.e., 30 mins.

∴ Both trains meet half after 10:30 a.m. i.e., at 11 a.m.

Hence, option (b).

Answer: Option B

Explanation :

Let the distance from home to office be d kms.

∴ $\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$

⇒ d = 28 kms.

∴ Normal time taken = $\frac{28}{40}$ = 0.7 hour = 42 minutes.

Now, she covers 2/3^rd of this distance in 1/3rd time i.e., in 14 minutes.

She now takes rest for 8 minutes.

∴ Time elapsed = 14 + 8 = 22 minutes.

Time remaining = 20 minutes.

In 20 minutes she needs to travel 1/3^rd of the distance i.e., 28/3 kms.

∴ Time taken = $\frac{28/3}{20/60}$ = 28 kmph.

Hence, option (b).

Answer: Option D

Explanation :

Speeds of Anil, Sunil and Ravi, which are 15 kmph, 10 kmph and 8 kmph respectively.

Length of the track is 3 km.

Time taken by Anil to complete one round = 3/15 × 60 = 12 minutes.

Time taken by Sunil to complete one round = 3/10 × 60 = 18 minutes.

Both Anil and Sunil will reach at the starting points after LCM[12, 18] = 36 minutes.

The distance covered by Ravi in 36 minutes (= 0.6 hours) = 8 × 0.6 = 4.8 km.

Hence, option (d).

Answer: Option D

Explanation :

First car starts at 10 am : Let the speed and time taken be a and t respectively.

Second car starts at 11 am : Let the speed be b. Time taken = t − 1.

Required percentage = [(b − a)/a] × 100 = [(b/a) − 1] × 100

Required percentage will be highest when b/a is highest.

Now, since distances covered by both cars are same, so D = at = b(t − 1)

∴ b/a = t/(t − 1) = 1/[1 − (1/t)]

(b/a) is maximum when t is minimum.

t_min = 6 (given)

∴ b/a = 6/5.

Maximum required percentage = [(6/5) − 1] × 100 = 20%.

Hence, option (d).

Answer: Option A

Explanation :

Let time taken by Amal be 3t, so time taken with each speed = 3t/3 = t.  

Let total distance travelled be 3d. Hence, Bimal travels d at each of the 3 different speeds.

For Amal : 3d = 10t + 20t + 30t = 60t.

∴ d = 20t.

Time taken by Bimal = (d/10) + (d/20) + (d/30) = (11d/60) = 11t/3.

Required percentage = [{(11t/3)/3t} − 1] × 100 = (2/9) × 100 = 22.22 ≈ 22%.

Hence, option (a).

Answer: 880

Explanation :

Let x be the length of the racecourse. 

The first horse beat the second by 11 metres and the third by 90 metres.

∴ Distances travelled by the first, second and third horse are x, x − 11 and x − 90 respectively. 

The second horse beat the third by 80 metres.

Distances travelled by the second and third horse are x and x − 80 respectively.

Ratio of speeds of second and third horse is constant which is equal to the ratio of the distances travelled by the second and third horse.

∴ (x − 11)/(x − 90) = x/(x − 80)

⇒ x² - 91x + 880 = x² - 90x

⇒ x = 880

Hence, 880.

Answer: Option C

Explanation :

Let the radius of A and B be a and b respectively. (a = 30 and b = 40)  

Distance travelled by any wheel D = (Circumference C) × (Number of revolutions N)

∴ D ∝ R × N (∵ C ∝ Radius R)

∴ N ∝ 1/R

∴ N_a/N_b = b/a = 40/30 = 4/3   ...(1)   

It is given that N_a = N_b + 5000 ...(2)

Solving (1) and (2), we get; N_a = 20000 and N_b = 15000.

D = C_a × N_a = 2πa × N_a = v_b × (3/4)   [where v_b is the velocity of B]

∴ 2π × (30 × 10^-5) × 20000 = v_b × (3/4)

∴ v_b = 16π.

Hence, option (c).

Answer: Option C

Explanation :

It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am

Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 11:00 am

If they leave one by one every minute, the 45^th motorcyclist would have left by 10:45 am to reach B at 11:00 am.

Thus, time taken by one motorcyclist to reach B from A = 15 minutes.

Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am

So, the last motorcyclist should have left A by 10:15 am

Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B

Hence, option (c).

Answer: 48

Explanation :

Let the track length for John be 'a' and for Mary be 'b'

So, Distance travelled by John = 9a

Distance travelled by Mary = 5b

Now, Time taken by John = 9a/6 hours

Time taken by Mary = 5b/7.5 hours

We know that Time taken by John = Time taken by Mary

⇒ 9a/6 = 5b/7.5

⇒ a = 4b/9

Total track length = 325 meters

So, 4b/9 + b = 325 meters

⇒  13b/9 = 325 meters

⇒ b = 225 meters

⇒ a = 100 meters

Mary jogs at 7.5 Kmph = 7.5 × 5/18

So, time taken = $\frac{100}{(7.5\times {\displaystyle \frac{5}{18}})}$ = 48 seconds.

Hence, 48.

Answer: Option A

Explanation :

We are told that, by the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.

So, the speeds of A and B are in the ratio 60:40 or 3:2

Hence, the time they take to cover a particular distance will be in the ratio 2:3

We know that A covers 60% of the distance at 10:00 AM and covers 100% of the distance at 10:12 AM.

That means A takes 12 minutes to cover 40% of the track. So to cover the entire track he must have taken 12 + 12 + 6 = 30 minutes. (because 40% + 40% + 20% = 100%)

Since the time taken by A and B to complete the track are in the ratio 2:3, the time taken by B to complete the track will be 45 minutes.

At 10:00 AM, B has covered 40% of the track. If we can find out what time does B take to complete the remaining 60% of the track, we can find the finish time of B.

Time required to complete 60% of the track = 60% of 45 = 27 minutes.

Hence, B complete one single round at 10:27 AM.

Hence, option (a).