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**Instructions**

Directions for the following two questions: Cities A and B are in different time zones. A is located 3000 km east of B. The table below describes the schedule of an airline operating non-stop flights between A and B. All the times indicated are local and on the same day.

Assume that planes cruise at the same speed in both directions. However, the effective speed is influenced by a steady wind blowing from east to west. It reduces or increases the speed of plane byÂ 50 km per hour depending on direction of flight.

**Question 1:Â **What is the time difference between A and B?

a)Â 1 hour and 30 minutes

b)Â 2 hours

c)Â 2 hours and 30 minutes

d)Â 1 hour

e)Â Cannot be determined

**1)Â AnswerÂ (D)**

**Solution:**

Let the speed of the plane be p Kmph.

So the speed of plane from A to B will be ‘p+50’ and the speed from B to A will be ‘p-50’.

We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.

So we have $\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$.

We can clearly see that speed of the plane is 550 which satisfies the above equation.

So for the journey of B to A,Â the plane takes $\frac{3000}{550-50} = 6$ hrs.

So time at B when plane reaches at A is 2 pm .

Hence the time difference between A and B is 1 hr.

Alternatively,

Let speed of flight be s,

Since A is to the east of B, A is ahead of be in time

Let A be ahead of B in time by a hours

Departure from A = 4PM, Arrival at B = 8PM

Travel time = 8 – 4 +a = 4 + a

Since City B is behind city A by ‘a’ hours, the actual travel time is ‘a’ hours more than the difference of local times.

Similarly when one travels from B to A, since B is ahead of A by ‘a’ hrs, actual travel time is ‘a’ hours less than total

i.e. B->A Travel time = (3PM – 8AM) – a = 7 -a

Total distance travelled = Speed $\times$ Time taken ….(1)

From A to B, the wind is favourable / in same direction as flight

Hence from (1), we have

A->B >>> $3000 = (s+50)(4+a) => 3000(7-a) = (s+50)(4+a)(7-a)$ …(2)

B->A >>> $3000 = (s-50)(7-a) => 3000(4+a) = (s-50)(7-a)(4+a)$ …(3)

(2) – (3) => $3000(3-2a) = 100(7-a)(4+a) => a^2-63a+62=0 => a=1/62$

Hence the time difference between A and B is 1 hr.

**Question 2:Â **What is the planeâ€™s cruising speed in km per hour?

a)Â 700

b)Â 550

c)Â 600

d)Â 500

e)Â Cannot be determined.

**2)Â AnswerÂ (B)**

**Solution:**

Let the speed of the plane be p Kmph.

So the speed of plane from A to B will be ‘p+50’ and the speed from B to A will be ‘p-50’.

We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.

So we have $\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$.

On substituting the options, we can clearly see that speed of the plane is 550 which satisfies the above equation.

**Question 3:Â **Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60Â° to AB. Also, C is located between south and southwest of A with AC at 30Â° to AB. The latest time by which Rahim must leave A and still catch the train is closest to

a)Â 6 : 15 am

b)Â 6 : 30 am

c)Â 6 :45 am

d)Â 7 : 00 am

e)Â 7 : 15 am

**3)Â AnswerÂ (B)**

**Solution:**

According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle.

So, total time taken by train is 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.

Time taken by Rahim to travel byÂ car is around 6.2 hrs. So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.

**Instructions**

DIRECTIONS for the following three questions: Answer the questions on the basis of the information given below.

A city has two perfectly circular and concentric ring roads, the outer ring road (OR) being twice as long as the inner ring road (IR). There are also four (straight line) chord roads from E1, the east end point of OR to N2, the north end point of IR; from N1, the north end point of OR to W2, the west end point of IR; from W1, the west end point of OR, to S2, the south end point of IR; and from S1 the south end point of OR to E2, the east end point of IR. Traffic moves at a constant speed of $30\pi$ km/hr on the OR road, 20$\pi$ km/hr on the IR road, and 15$\sqrt5$ km/hr on all the chord roads.

**Question 4:Â **The ratio of the sum of the lengths of all chord roads to the length of the outer ring road is

a)Â $\sqrt5 : 2$

b)Â $\sqrt5 : 2\pi$

c)Â $\sqrt5 : \pi$

d)Â None of the above.

**4)Â AnswerÂ (C)**

**Solution:**

Let the radius of outer circle be 2R and the centre of both the circles be O.

Triangle $ON_2E_1$and all the other 3 similar trianglesÂ form a right angle at the centre.

Let the radius of the inner ring road be R

The radius of outer will be 2R as theÂ circumferenceÂ of the outer ring road is double that of the inner ring road.

So, in triangleÂ $ON_2E_1$Â using Pythagoras theorem the value of chords come out to be $\sqrt5$ * R so the total length of the chords 4 * $\sqrt5$ * R and circumference is equal to 2 *Pi*2R. The ratio gives option C.

**Question 5:Â **Amit wants to reach N2 from S1. It would take him 90 minutes if he goes on minor arc S1 – E1 on OR, and then on the chord road E1 – N2. What is the radius of the outer ring road in kms?

a)Â 60

b)Â 40

c)Â 30

d)Â 20

**5)Â AnswerÂ (C)**

**Solution:**

We know that the total time taken is 1.5 hrs. Calculating the individual time taken and the adding and then equating to 1.5.

Let R be the radius of the outer-ring road.

$\frac{\pi*R}{2*30*\pi} + \frac{\sqrt{5}*R}{2*15*\sqrt{5}}$ = 1.5 solving we get R=30.

**Question 6:Â **Amit wants to reach E2 from N1 using first the chord N1 – W2 and then the inner ring road. What will be his travel time in minutes on the basis of information given in the above question?

a)Â 60

b)Â 45.

c)Â 90

d)Â 105

**6)Â AnswerÂ (D)**

**Solution:**

Let the radii of 2 circles be R and r respectively such that R=2*r. Triangle O$N_2$$E_1$and all the other 3 similar triangles form a right angle at the centre . So using pythagoras theorem the value of chords come out to be $\sqrt5$ * R/2 . Hence the total distance traveled is $\sqrt5$ * R/2 + 0.5*R*pi. Total time required can be calculated by distance / speed which comes out to be 3.5*R. Among options only 105 is integral multiple of 3.5.

**Question 7:Â **In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

a)Â 20 min

b)Â 15 min

c)Â 10 min

d)Â 5 min

**7)Â AnswerÂ (C)**

**Solution:**

Let A , B and f,s be the distance traveled and speed of the fastest and the slowest person respectively. Also f=2s so in the given time A=2B. Since the ration of the speeds is 2:1, they will meet at 2-1 points = 1 pont.

Both meet each other for first time at starting point . let b travel distance equal to 1 circumference i.e. 1000m so A=2000m . Both meet after 5 min so speed of slowest is 1000/5=200m/min . So speed of the fastest is 400m/min. So time taken by A to complete race 4000/400 = 10 min

**Question 8:Â **Two boats, traveling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far apart are they (in kms) one minute before they collide.

a)Â 1/12

b)Â 1/6

c)Â 1/4

d)Â 1/3

**8)Â AnswerÂ (C)**

**Solution:**

The relative speed is 15 km/hr = 15 km/60 min = 0.25 km/min = 250 m/min.

Therefore, one minute before they collide, they are at a distance of 250m.

**Question 9:Â **If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what sped must he cycle to get there at noon?

a)Â 11 km/hr

b)Â 12 km/hr

c)Â 13 km/hr

d)Â 14 km/hr

**9)Â AnswerÂ (B)**

**Solution:**

Let the distance to be travelled be D.

In the first case, D/10 = t

In the second case, D/15 = t-2

=> D/15 = D/10 – 2

=> 2D = 3D – 60

=> D = 60 km and T = 6 hours

Therefore, to get there at noon, he has to travel at 60/5 = 12 km/hr

**Question 10:Â **A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is $\pi$ rÂ during the first 30 seconds, $\pi$ r/2Â during next one minute, $\pi$ r/4Â during next 2 minutes, $\pi$ r/8Â during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?

a)Â 4

b)Â 8

c)Â 16

d)Â 32

**10)Â AnswerÂ (C)**

**Solution:**

Let radius be 1 units and p = 3.14 or $\pi$ . So circumference is $2*\pi$.

According to given condition distance covered in first 1/2 mins = $\pi$/2 km, distance covered in next 1 min = $\pi$/2 km, distance covered in next 2 mins = $\pi/2$ km and finally distance covered in next 4 minutes =Â $\pi/2$ km.

Time taken to cover first round = 1/2 + 1 + 2 + 4 = 7.5 minutes.

Now time taken to cover $\pi/2$ is in GP.

For the second round the time taken is = 8+16+32+64 = 120

Ratio = 120/7.5 = 16

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**Question 11:Â **Karan and Arjun run a 100-meter race, where Karan beats Arjun 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?

a)Â Karan and Arjun reach the finishing line simultaneously.

b)Â Arjun beats Karan by 1 metre

c)Â Arjun beats Karan by 11 metres.

d)Â Karan beats Arjun by 1 metre.

**11)Â AnswerÂ (D)**

**Solution:**

The speeds of Karan and Arjun are in the ratio 10:9. Let the speeds be 10s and 9s.

Time taken by Karan to cover 110 m = 110/10s = 11/s

Time taken by Arjun to cover 100 m = 100/9s = 11.11/s

Therefore, Karan reaches the finish line before Arjun. From the options, the only possible answer is d).

**Question 12:Â **Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?

a)Â 3

b)Â 3.5

c)Â 4

d)Â 4.5

e)Â 5

**12)Â AnswerÂ (C)**

**Solution:**

Let the distance be D.

Time taken by Arun = D/30

Time taken by Barun = D/40

Now, D/40 = D/30 – 2

=> 3D = 4D – 240

=> D = 240

Therefore time taken by Arun to cover 240 km = 240/30 = 8 hr

Time Kiranmala takes to cover 240 km = 240/60 = 4 hr

So, Kiranmala has to start 4 hours after Arun.

**Instructions**

Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.

Checkout: **CAT Free Practice Questions and Videos**

**Question 13:Â **At what time do Ram and Shyam first meet each other?

a)Â 10 a.m.

b)Â 10:10 a.m.

c)Â 10:20 a.m.

d)Â 10:30 a.m.

**13)Â AnswerÂ (B)**

**Solution:**

Let the time at which they meet be t minutes past 10.

So, distance run by Ram + distance run by Shyam = 10 km

=> (60+t)5/60 [t+60 because he would have traveled for 9 am to 10 am and t minutes more before meeting Shyam]+ (15+t)*10/60 [15+t because he would have traveled from 9:45 to 10:00 and t minutes more]= 10

=> 300+5t+150+10t = 600 => t = 10

So, they meet at 10.10 am