Topic - Permutation & Combination

Solved ? - Yes

Source - 2IIM ]]>

Topic - Permutation & Combination

Solved ? - Yes

Source - 2IIM ]]>

a) 3

b) 4

c) 7

d) 5 ]]>

Time taken to meet for the first time = T/relativespeed = T/(6−b) or T/(b−6)

Time taken for a lap for A = T/6

Time taken for a lap for B = T/b

So, time taken to meet for the first time at the starting point = LCM (T/6, T/b)=T/HCF(6,b)

Number of meeting points on the track = Time taken to meet at starting point/Time taken for first meeting

= Relative speed / HCF (6, b).

So, in essence we have to find values for b such that (6−b)/HCF(6, b) = 2 or (b−6)/HCF(6, b) = 2

b = 2, 10, 18 satisfy this equation. So, there are three different values that b can take.

]]>a) 7.5 × √2 times the side of the square ABCD

b) 7.5 × √2 times the side of the square EFGH

c) 7.5 times the side of the square ABCD

d) 7.5 times the side of the square EFGH ]]>

Lakshman and Kanika meet for the second time at H where Jagadeesh also meets them for the first time.

Let us assume that the side of the big square ABCD to be ‘2a’; Half of this length, given by AE would be a, and distance EF would be √2a.

Lakshman and Kanika start from points B and D respectively at speeds ‘l’ kmph and ‘k’ kmph respectively and travel towards each other along the sides of the square ABCD. They are at a distance of 4a from each other. Since they are at diametrically opposite points, the relative distance would be 4a irrespective of the directions they choose to travel in. So, to meet for the first time, they would have travelled a distance of 4a together.

To meet for the second time, they would have travelled a further 8a together. Essentially, between them, they would have to cover the entire perimeter of the square to meet again.

So by the time they meet for the second time, they would have covered a distance of 12a together. Their speeds are in the ratio 1: 3. So, Lakshman would have travelled 3a and Kanika would have travelled 9a.

Or, Lakshman travels in the direction BADC, while Kanika would have travelled in the direction DABC. They meet for the first time at E and the second time at H.

In the same time, Jagadeesh travels along the square EFGH in the anti-clockwise direction at ‘j’ kmph and meets Lakshman and Kanika. While Jagadeesh meets the other two for the first time, we do not know how many laps he has completed by then.

The ratio of Lakshman’s speed to that of Jagadeesh is 1: 5√2 so, they would have travelled distances in the same ratio as well. So, if Lakshman has travelled a distance of 3a, Jagadeesh should have travelled a distance of 3a x 5√2 to reach H.

Or, Jagadeesh travels 52a to reach H. Answer choice (a)

]]>a) 60 kmph

b) 80 kmph

c) 20 kmph

d) 40 kmph ]]>

AB/V1 − AB/V2 = AB/V2 − AB/V3

240/V1 − 240/V2 = 1

v3 = 2v1

Let v1, v2 and v3 be the speeds of the cars.

Condition I states that the cars leave in equal intervals of time and arrive at the same time. Or, the difference in the time taken between cars 1 and 2 should be equal to the time taken between cars 2 and 3.

We get AB/V1−AB/V2 = AB/V2−AB/V3

As the second car arrived at C an hour earlier than the first, we get a second equation 240/V1−240/V2 = 1

The third car covered 240 + 80 kms when the first one covered 240 – 80 kms. Therefore, 320/V3=160/V1

This gives us v3 = 2v1

From condition 1, we have AB/V1 − AB/V2 = AB/V2 − AB/V3

Substituting v3 = 2v1, this gives us AB/V1−AB/V2=AB/V2−AB/2V1

or 3xAB/2xV1=2xAB/V2

or v2 = 4xv1/3

Solving 240/v1−240/v2 = 1, we get 60/v1 =1 or, v1 = 60 kmph

=> v2 = 80 kmph and v3 = 120 kmph

Answer choice (a)

a) 5 : 4 : 2

b) 4 : 3 : 2

c) 5 : 4 : 3

d) 3 : 2 : 1 ]]>

After three laps C would have traveled a distance of 3 * (t - 2d) = 3t - 6d.

After 3 laps C is in the same position as B was at the end of one lap. So, the position after 3t - 6d should be the same as t - d. Or, C should be at a distance of d from the end of the lap. C will have completed less than 3 laps (as he is slower than A), so he could have traveled a distance of either t - d or 2t - d.

=> 3t - 6d = t - d

=> 2t = 5d

=> d = 0.4t

The distances covered by A, B and C when A completes a lap will be t, 0.6t and 0.2t respectively. Or, the ratio of their speeds is 5 : 3 : 1.

In the second scenario, 3t - 6d = 2t - d => t = 5d => d = 0.2t.

The distances covered by A, B and C when A completes a lap will be t, 0.8t and 0.6t respectively. Or, the ratio of their speeds is 5 : 4 : 3.

The ratio of the speeds of A, B and C is either 5 : 3 : 1 or 5 : 4 : 3.

]]>a) 90 km

b) 120 km

c) 20 km

d) 40 km ]]>

Let the distance from Delhi to Gurgaon be ‘d’ km. The first 30 km he travels at his usual speed. However, the remaining ‘d-30’ km he travels at a reduced speed.

To travel ‘d’ km he usually takes T hr. Therefore, to travel ‘d - 30’ km he should ideally take (d−30)xT/d hr. However, this is only if he travels at his usual speed. It is given that he traveled only at 4/5th of his usual speed. Because of this he would have taken 5/4th of the time to travel the remaining distance, i.e., he takes 1/4th of the time extra. This is given to be 45 minutes (or 3/4th hr)

x/4 * (d−30)xT/d = 3/4 ........ (1)

On the other hand, had the same thing happened after he travelled 48 km, he would have reached only 36 minutes or 3/4 hrs late. Hence,

x/4 * (d−48)xT/d = 3/5 ........ (2)

Dividing (1) by (2) and solving for d, we get d = 120 km.

Answer choice (b)

a) 60

b) 36

c) 24

d) 48 ]]>

Distance traveled by car A before meeting car B = a * t. Likewise distance traveled by car B before meeting car A = b * t.

Distance traveled by car A after meeting car B = a * 54. Distance traveled by car B after meeting car A = 24 * b.

Distance traveled by car A after crossing car B = distance traveled by car B before crossing car A (and vice versa).

=> at = 54b ---------- (1)

and bt = 24a -------- (2)

Multiplying equations 1 and 2

we have ab * t^2 = 54 * 24 * ab

=> t^2 = 54 * 24

=> t = 36

So, both cars would have traveled 36 minutes prior to crossing each other. Or, B would have taken 36 + 24 = 60 minutes to travel the whole distance. Choice (A).

]]>a) Twice

b) Thrice

c) Once

d) 5 times ]]>

A will take 50/3 = 16.67 seconds to reach Q, so between 10s and 16.67 seconds they meet once.

First, let us have a table where we know times when A, B would be at the end points

A | B | ||
---|---|---|---|

Time (in sec) | Position | Time (in sec) | Position |

16.67 | Q | 10 | Q |

33.33 | P | 20 | P |

50 | Q | 30 | Q |

66.67 | P | 40 | P |

83.33 | Q | 50 | Q |

100 | P | 60 | P |

We see that the critical time slots are multiples of 10 seconds and multiples of 16.67 seconds.

So, let us break the time from 0 to 60 seconds into these slots and see in which direction A and B travel.

Time | Position A | Position B | Do they cross? |
---|---|---|---|

0 to 10 | P to Q | P to Q | |

10 to 16.67 | P to Q | Q to P | Yes |

16.67 to 20 | Q to P | Q to P | |

20 to 30 | Q to P | P to Q | Yes |

30 to 33.33 | Q to P | Q to P | |

33.33 to 40 | P to Q | Q to P | Yes |

40 to 50 | P to Q | P to Q | |

50 to 60 | Q to P | Q to P |

So, they cross each other thrice totally. Between 10 and 16.67 seconds for the first time; between 20 and 30 seconds for the second time; and between 33.33 to 40 seconds for the third time.

Answer Choice (B)

a) 36 km/hr

b) 45 km/hr

c) 67.5 km/hr

d) 18 km/hr ]]>

Their relative speed = Sum of the two speeds = 45 + 54 kmph. = 99kmph. = 99 * 5/18m/s = 55/2m/s = 27.5m/s.

The relative distance = 220m. So, time they will take to cross each other = 220/27.5 = 8 seconds.

Now, car A has to overtake car B within 8 seconds. The relative distance = 50 m

=> Relative speed should be at least 50/8 m/s = 6.25 m/s

=> 6.25 * 18/5 kmph = 22.5 kmph

Car B travels at 45kmph, so car A should travel at at least 45 + 22.5 = 67.5 kmph

]]>a) 3 m/s

b) 4 m/s

c) 5 m/s

d) 8 m/s ]]>

Laplength/RelativeSpeed = 24; Or relative speed = 5 m/s.

The question says – Now, if B had started in the opposite direction to the one he had originally started, they would have met for the first time after 40 seconds.

Now, B would have crossed each other after 40 seconds if B had reversed direction. This is higher than the 24 seconds it takes them to cover the relative distance of a lap in the first instance.

Or, in the first instance they were travelling towards each other.

Or, a + b = 5 m/s.

They meet for the first time 16 seconds after they start the race.

a) 7.5 hours and 15 hours

b) 7.5 hours and 18 hours

c) 8 hours and 18 hours

d) 7.5 hours and 20 hours ]]>

The second part is more interesting, because the speed of the boats change when they change direction. Boat P is quicker, so it will reach the destination sooner. Boat P will reach City B in 10 hours 300/30. When boat P reaches city B, boat Q will be at a point 100kms from city B.

After 10 hours, both P and Q will be travelling upstream,

P's speed = 20 km/hr

Q's speed = 10 km/hr

Relative speed = 10km/hr

Q is ahead of P by 100 kms

P will catch up with Q after 10 more hours Distance/RelativeSpeed − 100/10.

So, P and Q will meet after 20 hours at a point 200 kms from city B.

Correct Answer: 7.5 hours and 20 hours

]]>a) 300 kms from A

b) 280 kms from A

c) 305 kms from A

d) 295 kms from A ]]>

Travels 200km @ 40kmph

the next 200km @ 50kmph and

the final 200km @ 60kmph

So, Bus A will be at a distance of 200km from city M after 5 hours, and at a distance of 400km after 9 hours, and reach N after 12 hours and 20 mins

Bus B

Travels at an overall average speed of 50kmph, so will take 12 hours for the entire trip.

So, Bus B will travel

160 kms in the first 4 hours

200 kms in the next 4 hours

and 240 in the final 4 hours

So, both buses cross each other when they are in their middle legs.

After 5 hours, bus A will be at a position 200kms from city M. At the same time, bus B will be at a distance 210kms from city N (4 * 40 + 50).

The distance between them will be 190kms (600 - 200 - 210). Relative speed = Sum of the the two speeds = 50 + 50 = 100 kmph.

Time taken = 190/100 = 1.9 hours. = 1 hour and 54 minutes. So, the two buses will meet after 6 hours and 54 minutes. Bus B will have travelled 210 + 95 = 305 kms. So, the two buses will meet at a point that is 305 kms from City N and 295 kms from city A.

]]>a) 12 m/s

b) 14.8 m/s

c) 12.4 m/s

d) 7.6 m/s ]]>

car of length = 4 m

length of truck = 20 m

Speed of truck = 36 km/hr

Approach:

Total distance to be covered by the car 'd' = 20 + 4 = 24 m

Let speed of car be S1 m/s

Speed of truck S2 = 36 km/hr = 10 m/s

Both the car and the truck are travelling in the same direction and we know that to overtake the truck, the speed of the ca should be more than that of the truck.

Hence, relative speed = |S1 - S2|

Time to overtake = 10s

S = d/t

|S1 - 10| = 24/10

S1 - 10 = 2.4

S1 = 10 + 2.4 = 12.4 m/s

a) 400 m

b) 810 m

c) 500 m

d) 310 m ]]>