Raashid Shah

Q.1 I and J starts running on the circular track in opposite directions.They start from a point M and meet for the first,second and third time at m & n respectively.What is the ratio of speeds of I and J ?A) 2 : 1B) 1 : 2C) 2 : 3D) 3 : 2Ans. Athey meet 3 times in a circle so there ratio should be 2:1 Q.2 There are two points P and Q diametrically opposite on a circular road of circumference 20km. A car starts at P and makes 4 rounds. It starts of with a speed of 5km/hour and then increases its speed by 5km every time it reached Q. What is the interval between the 2nd time it passes through Q and the 4th time it passes through Q?A) 8/3 hrsB) 7 hrsC) 7/3 hrsD) 3 hrsAns. Cstarting from 0 time to reach Q for the 1st time – 2hrs (at 5 kmph it would take 2hrs to cover 10 km)2nd time – (2+2)hrs (at (5+5)kmph, it would take it further 2 hrs to cover a distance of 20km)3rd time – (2+2+1+(1/3))hrs (at (10+5)kmph,it would take it futher 1+(1/3)hrs to come to Q)4th time – (2+2+1+(1/3)+1)hrs (at (15+5)kmph,it would take it futher 1 hr to again come to Q) thus the difference between the 4th and 2nd time is 1+1+(1/3) = 7/3 hrs Q.3 A Lizard has to climb 40 feet on a tree. The Lizard climbs 1 foot in 1st day, 2 foot on second day, and so on. But every night it slides down half of the distance, which it climbed on the corresponding day. On which day will the Lizard climb on the tree ?A) 1stB) 12th C) 13thD) 14thAns. CThe distance covered by the Lizard forms a series with differences in AP with a common difference of 0.5.1st day the distance travelled is 0.5, next day total distance from starting point is 1.5,3rd day total distance from starting point is 3,4th-5 and so on.so differences are 1,1.5,2,2.5,3,3.5…….having a common difference of 0.5.for n =12 distance covered from starting point is 39 feet from start.Hence on 13th day it will climb on the tree. Q.4 Two buses F and G are operated between cities I and J. One day F and G started from I and J respectively towards J and I respectively. After the two buses cross each other, the bus F takes 5 hours to reach city J and the bus G takes 3 hours 12 minutes to reach city I. After how much time from the start did the buses F and G cross each other.A) 8 hours 12 minutesB) 4 hours 24 minutesC) 4 hours 6 minutesD) 4 hoursAns. DTake t to be the required time. Same distance will be covered by F in 5 hours as that of G in t hours. Similarly, G in 16/5 hrs and F in t hours. Then compare the ratio of speeds, t/(16/5) = 5/t Q.5 In a survey, it is found that in the year 2012 approximately1/3 of the airline passengers traveling to or from the Bangalore used airport A. If the number of such passengers that used airport B was 1/ 2 of the number that used airport A and 4 times the number that used airport C. If the survey consists of 37.3 unique passengers that used only one of these three airports, then the approximate number passengers (in millions) used airport C in that year is:A) 18.6B) 9.3C) 6.2D) 2.0Ans. DKA = 37.3/3 = 12.43MA = KA/2 = 6.215MA = 4 LALA = MA/4 = 6.215/4 = 1.552.0 is the closest approximation Q.6 A stairway is 8 ft high such that each step accounts for half a foot upward & one-foot forward. An ant begins from ground level, What distance will it travel to reach the top of the stairway ?A. 24B.23C.22D.25Ans. BAnt will travel 16 times upwards half a foot,So upward distance travelled=16*0.5= 8ftAlso, the ant will travel 15 times a foot steps, so forward distance travelled= 15*1= 15ftTotal distance travelled 8+15 = 23 ft Q.7 One-third of a journey was covered at a rate of 25 km per hour, one-fourth at the rate of 30 km per hour and the rest at the rate of 50 km per hour. Find the average speed of the whole journey. A) 33 1/3 km/hB) 66 1/3 km/hC) 36 1/6 km/hD) 63 1/3 km/h Ans. A Let the total distance of the journey be 12 km Distance covered at the speed of 25 km/h = 4 kmDistance covered at the speed of 30 km/h = 3 kmDistance covered at the speed of 50 km/h = 5 kmAverage speed = = = 33 km/h Q.8 A Formula One driver covers some distance at the speed of 60 mph. With what speed should he return so that his average speed of the total journey comes out to be 120 mph? A) 180 mphB) 120 mphC)150 mphD) Not possibleAns. D Using the funda that the average speed of a body can never be the double or more than any of the speed. So the average speed cannot be 120 kmph. Hence, can’t possible.        

Q.1 Vikas has an appointment near the train station in the city at 10:30 am.To get to the city, Vikas needs to catch a bus & then a train. The bus leaves from Vikas’s stop every 10 min. from 9:00 am.The bus trip to the train station takes 25 min. From 9.30 am the train leaves for the city every 20 min. the train takes 15 min. to get to the city. If Vikas wants to arrive as close as possible to the appointment time , what is the latest bus he would catch ? A) 9.10 am B) 9.20 am C) 9.30 am D) 9.40 am Ans. DGiven, The train leaves for the city every 20 minutes from 9:30 AM. So,The timing of train = 9:30 AM, 9:50 AM, 10:10 AM, 10:30 AM and so on.Train takes 15 minutes to get to the city but Vikas has an appointment at 10:30 AM. So, he will take train at 10:10 AM.Now, the bus trip to the train station takes 25 minute. So,10:10 AM – 25 min = 9:45 AMGiven, bus leaves from Vikas’s stop every 10 minutes from 9:00 AM. So,The bus timing are = 9:00 AM, 9:10 AM, 9:20 AM, 9:30 AM, 9:40 AM, 9:50 AM and so on.Thus, the latest bus which Vikas can catch is 9:40 AM. Q.2 A bus left point for point y. Two hours later a car left x for y and arrived at y at the same time as the bus. If the car and the bus left simultaneously from the opposite ends x and y towards each other, they would meet 1.33 hours after the start. How much time for the bus to travel from x to y. A) 3 B) 4 C) 5 D) 6 Ans. B take d as distance btwn x-y, car Speed=c bus speed=b by prob: d/c+b=80/60=4/3——(1) Now, as both car and bus reach y at same time let time taken by car is t Thus d/c=t and d/b=t+2 From (1)-> c+b/d=3/4 or c/d+b/d=3/4 Thus 1/t +1/t+2 =3/4 Forming eqn: 3t^2-2t-8=0 Solving: t=2 So bus time= t+2=4 Q.3 Rohan and sohan travel with uniform speed from P to Q and Q to P respectively, via the same route. Rohan starts at 5:00 a.m. while Sohan starts four hours later. They meet, on the way, at 1:10 p.m. At what time do they reach their destinations, if both of them reach at the same time? A) 7 B) 8 C) 6:30 D) 7:30 Ans. A Speed of Rohan=xSpeed of Sohan=y Rohan begins from 5 am and continues for 4 hrs alone till 9 am. So, distance covered is 4x.At 9 am, Sohan begins with a speed of y.Now, after 25/6 hrs (4 hrs 10 mins), Rohan meets Sohan . So, both have covered 25/6x and 25/6y distances respectively.Thus, total distance becomes 4x+(25/6)x+(25/6)y. Now, it is given both reach end point at the same time. Thus, total time taken by Rohan will be 4 hours more than SohanThis gives us the equation{4x+(25/6)x+(25/6)y}/x = [{4x+(25/6)x+(25/6)y}/y] +4Solving it, 5y=7x=>y/x= 7/5 Now we need to find out time taken by Rohan to travel (25/6)y distance as we know the time taken to traverse previous distances. So, it will be 25y/6x=(25/6)*(7/5)= 35/6= 5 hrs 50 mins So, total time taken by Rohan= 4+ 4 hrs 10 mins + 5 hrs 50 mins= 14 hrs i.e. 7 pm Q.4 Tripti leaves office at 6.00 p.m. and catches a 6.30 p.m. train that arrives in her town at 7.00 p.m. Her father leaves home to pick her up at 7.00 p.m. from the station as she gets off the train. Yesterday, Tripti left her office early and took a 6.00 p.m. train and arrived at 6.30 p.m. As her father was not there to pick her up, she started walking towards home. Her father left home at the usual time, saw his daughter walking, turned around, picked her up and drove home arriving there 10 min earlier than usual. For how long did Tripti walk before her father picked her up?A) 10B) 15C) 20D) 25Ans. DAs her father reached home 10 minutes earlier than the usual time, he saw his daughter 5 minutes before the scheduled time and he turned back (saving 5 minutes each side).So, he met her at 6.55 pmi.e. she walked for 25 minutes from the station towards her home before her father picked her up Q.5 A leaves Mumbai at 6am and reaches Bangalore at 10am. B leaves Bangalore at 8am and reaches Mumbai at 11:30am. At what time do they cross each other ?A) 10am B) 8:32 am C) 8:56am D) 9 :20 am Ans. C Let us assume that the distance between Mumbai and Banglore is 28kms (LCM of their time travelling so that we can make calculations easy)Now :speed of A is 28/4 = 7 kmph speed of B is 28/3.5 = 8 kmph Since A to hours earlier than B and the total distance travelled by both we be equal to the total distance between the cities.Let us assume that they met after t hours from the time A started.Hence,7t+8(t-2) = 28t= 44/15hence t = 2 hrs and 56 mins which is 8:56 AM      

Q.1 There are 2 cars moving in opposite directions. the distance between them is 300cm. They move forward for 100cm with a speed of 50 cm/s, and comes back 50 cms at 25cm/s. After how many seconds will the cars collide? A) 10 seconds B) 11 seconds C) 12 seconds D) 13 seconds Ans. C 12 seconds — first 50cms 4 sec then 50 cms 4 sec and so…on Q.2 There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position. In the mean time the whole platoon has moved ahead by 50m.The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed. A) 120.71 meters B) 120.61 meters C) 121.71 meters D) 121.61 meters Ans. A It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered – while person moving forward and backword – are equal.Let’s assume that when the last person reached the first person, the platoon moved X meters forward.Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.Now, as the ratios are equal,(50+X)/X = X/(50-X)(50+X)*(50-X) = X*XSolving, X=35.355 metersThus, total distance covered by the last person= (50+X) + X= 2*X + 50= 2*(35.355) + 50= 120.71 metersNote that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total length of the platoon (50 meters) twice. true, but that’s the relative distance covered by the last person i.e. assuming that the platoon is stationary. Q.3 A car travels from B at a speed of 20 km/hr. The bus travel starts from A at a time of 6 A.M. There is a bus for every half an hour interval. The car starts at 12 noon. Each bus travels at a speed of 25 km/hr. Distance between A and B is 100 km. During its journey , The number of buses that the car encounter is A) 14 B) 15 C) 16 D) 17 Ans. D A——————- BBus Car25km/hr 20 km/hr6:00 AM 12:00 noonTherefore, Speed of bus = 25/6 & Speed of car = 20/12 = 5/3 Thus, S bus + SCar = 25/6 + 5/3 = 35/6 Therefore, Distance is given i.e. = 100 km Therefore, distance/Speed i.e. 100 * 6 / 3 5 = 17 Q.4 One-third of a journey was covered at a rate of 25 km per hour, one-fourth at the rate of 30 km per hour and the rest at the rate of 50 km per hour. Find the average speed of the whole journey.A) 33 1/3 km/hB) 66 1/3 km/hC) 36 1/6 km/hD) 63 1/3 km/hAns. A Let the total distance of the journey be 12 km Distance covered at the speed of 25 km/h = 4 kmDistance covered at the speed of 30 km/h = 3 kmDistance covered at the speed of 50 km/h = 5 kmAverage speed = = = 33 km/h Q.5 A Formula One driver covers some distance at the speed of 60 mph. With what speed should he return so that his average speed of the total journey comes out to be 120 mph?A) 180 mphB) 120 mphC) 150 mphD) Not possibleAns. D        

Q.1 A monkey climbed up greased pole ascends 10 metres & slips 2 metres in an alternate min. If the pole is 56 metres high, how long will it take him to reach the top ?A) 12 min.B) 12.8 min.C) 8 min.D) Can’t be determinedAns. B10 m and – 2m in total 2 minutes => Net height climbed = 10-2 = 8 m in 2 minutes = 4m/ min.In 14 minutes he will cover 14*4 = 56 m.Now, in the next minute he will cover 10m in ‘1’ min.=> Speed = 10/60 m/sec.He has to cover 64-56 = 8 metres more.Time taken to cover these 8m = 8/(10/60) = 48 seconds. Q.2 Hemant rides at the rate of the 10 km/hr but stops for 10 min.to take rest at the end of every 15 km. How long will he take to go a distance of 100 km ?A) 10 hrsB) 12 hrsC) 11 hrsD) 14 hrsAns. CHe will take rest = 100/15 = 6 timesTime taking in taking rest = 6*10 = 60 min. = 1 hrTime taken to cover the distance = 100/10 = 10 hrsTotal time = 1+10 = 11 hrs Q.3 A car travels 25 km an hour faster than a bus & takes 10 hrs less for a journey of 500km. the speed of the car & bus are :A) 60 & 35B) 50 & 55C) 70 & 45D) 80 & 55Ans. BLet speed of car be x km/hrTherefore speed of bus = x-25 km/hr500/x-25 – 500/x = 10500 (x – x + 25)/ x (x-25) = 10500 * 25 = 10x (x – 25)(x – 25)x = 50 * 25Therefore, x=50 km/hr & x – 25 =25 km/hr Q.4 A goods train starts from the Bombay to Nandgaon & runs at a speed of 32 km/hr. Three hours later, passenger train also starts from the Bombay & reaches Nandgaon at the same time as the goods train. If the speed of the passanger train exceeds the speed of the goods train by 16km/hr. What is the distance between Bombay & Nandgaon ?A) 330 kmB) 250 kmC) 412 kmD) 288 kmAns. DDistance covered by the goods train in 3 hrs = 32*3 = 96 kmRelative speed of passanger train = 16 km/hr& speed = distance /Time16 = 96/TimeTime = 6hrsSpeed of the passenger train = 32 + 16 = 48km/hrTherefore, Distance between Bombay & Nandgaon = 48 * 6= 288 km Q.5 Two trains leave Delhi at the same time . One travels North at 60 Km/hr & the other travels South at 40 km/hr. After how many hours will the trains be 150 km apart ?A) 4/3B) 3/4C) 15/2D) 3/2Ans. DRelative speed = 60 + 40 = 100Time = Distance / Speed= 150/100 = 3/2 hr    

Q.1 Two men A & B walk simultaneously from Point Q, a distance of 210m at the speed 30km/hr & 40km/hr respect to B reaches Q, returns immediately & meets A at R. Find the distance from P to R ?A) 18 kmB) 19 kmC) 20 kmD) 21 kmAns. ADistance travelled by A= 2*distance between the two points * ( a/a+b)Distance travelled by B = 2 * distance between the two points * (a / a+b)Therefore, distance travelled by A = 2 * 210 / 30 / 10= 18 km Q.2 A man covers a certain distance by the car driving at 70km/hr & he returns back to the starting point riding on a scooter at 55km/hr. Find the speed ?A) 61.6B) 62.5C) 63.6D) 64.5Ans. A2 * s1 * s2 / s1 + s22 * 70 * 55 / 70 + 55= 61.6 km/hr Q.3 A person covers 10km. At 4km/hr & then travels further 21km at 6 km/hr. Find his average speed for the total journey ?A) 31 / 6B) 32 / 6C) 33 /6D) 34 / 6Ans. AT1 = 10 / 4 = 5 / 2T2 = 21 / 6 = 7 / 2Total time = 5/2 + 7/2 = 12 / 2= 6 hrsNow, Average speed = 21 + 10 / 6= 31 / 6 Q.4 A person travelled a distance of 50 km in 8 hr. He covered a part of the distance on the foot at the rate of 4 km/hr & a part on a bicycle at the rate of 10 km/hr. How much distance did he travel on foot? A) 10 km B) 20 km C) 30 km D) 40 km Ans. B Let x hour he travel by foot4x + 10(8-x) = 5080 – 6x = 506x = 30, x=5 hrNow putting in place of x, we get distance Q.5 Two cars X & Y start from two places A & B which are 700 km apart at 9:00 am. Both the cars runs at an average speed of 60 km/hr. Car X stops at 10:00 am while the other car Y continues to run without stopping. When the two cars cross each other ? A) 4:20 pm B) 3:20 pm C) 4:10 pm D) 2:40 pm Ans. A Since speed of X & Y are 60km/hrDistance b/w A & B is 700 kmDistance travelled by X upto 10 am is 60 km since x stops at 10:00 am & distance travelled by Y upto 11 am is 120 kmNow the distance between them = 700-180 = 520 kmNow, let they will meet at a distance of x from X’s positionTherefore, distance travelled by x = x km& distance travelled by y = 520-x kmTherefore, x/60 = 520-x/602x = 520X = 260Or t = 260/60 = 4 1/3Thus they will cross each other after 4 1/3hrs after 11:00 am3:20 pm    

Q.1 A bus left point X for point Y. two hours later a car left point X for Y and arrived at the same time as the bus. If the car and bus left simultaneously from the opposite ends X and Y towards each other they would meet 1.33 hours after the start . How much time did the bus take to travel from X to Y A) 2hr B) 6hr C) 4hr D) 8hr Ans. C Q.2 Sardar khan went out wasseypur at 10:00 am at a speed of 50km/hr. Ramadhir finds out about it at 11:00 am & sends his man to kill sardar khan. Ramadhir’s men travels at a speed of 60km/hr. What will be the distance between them 1hr before they meet & at what time they will meet ?A) Distance- 10 km , time-4 pmB) Distance- 11 km , time-5 pmC) Distance- 11.5 km , time-5.30 pmD) Distance-12 km , time- 6 pmAns. A11 am sardar khan would be 50km12pm he would be 100km1pm he would be 150kmNow,11am ramadhir men (gap=50km)12pm Ramadhir men would be 60km (gap=40km)1pm Ramadhir men would be 120km (gap=30km)Therefore,Relative speed =60-50 = 10km/hrOr 50/10 = 5hrNow calculate 5hr from 11am; i.e. 4pmDistance between them = 10km Q.3 A man sets out to cycle from Delhi to Rohtak & at the same time another man starts from Rohtak to delhi on cycle. After passing each other they complete their journey in 3 1/3 & 4 4/5hrs. At what rate does the second man cycle if the first cycles at 8km/hr ?A) 6 2/3B) 7 2/3C) 8 2/3D) 9 2/3Ans. AIf two persons start from same time in opposite direction from two points, thenA’s speed : B’s speed = vb : va8 /3 = v4 4/5 / v3 9/3= 6/5Or B’s speed = 8*5/6 = 6 2/3 Q.4 When a person covers a distance between his house & office at 50km/hr he is late by 20 min. but when he travels at 60km/hr he reaches 10min. early. What is the distance between his house & office ?A) 150B) 160C) 170D) 180Ans. AProduct of speed / difference of speed * difference between arrival time60*50 / 60-50 * 60 / 60150 Q.5 Running at 5/4 of his usual speed an athlete improves his timing by 5 min.How much time he usually takes to run the same distance ?A) 25 min. B) 26 min. C) 27 min. D) 28 min. Ans. AUsual time = Improvement in time / 1- old speed/new speed5 / 1-1/ 5/45/ 1-4/5=5/ 1/5=25min.      

Q.1 On a railway route between the two places A & B, there are 20 stations on the way. If 4 new stations are to be added how, many types of new tickets will be required if each ticket is issued for a one way journey ?A) 48B) 14C) 108D) 96Ans. CNumber of stations=10+2+4 = 16Thus for each stations number of stations for which tickets are to be issued = 15Number of tickets for each new stations = 30Total number of tickets = 30*4 – 4*3 = 108 Q.2 In a road network covering 10 cities, city C can be reached only from City A or B. The distance from A to C is 50km & the same from B to C is 25 km.The shortest distance from A to B is 40 km. The shortest distance from city P to A is 200 km & the shortest distance from City P to B is 230 km. Then the shortest distance from city P to C is:A) 255 kmB) 250 kmC) 225 kmD) 265 kmAns. BSince city C can be reached from A or B only, P is not directly connected to city C.From P we will have to go either to A or B, before going to city C.i.e. PA + AC = 200 + 50 = 250 KM Q.3 Arun, Barun & Kiranmala start from the same place & travel in the same direction at speed of 30,40 & 60 km/hr. Barun starts two hours after Arun . If Barun & Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start ?A) 3B) 4C) 3.5D) 5Ans. BArun covers 60 km in 2 hrsBarun takes 60/10 = 6hrs to meet Arun In 2+6 = 8 hrsArun travels 8 * 30 = 240 kmKiranmala also overtakes both at this pointShe takes 240/60 = 4hr to overtake ArunHence, Kiranmala starts 8-4 = 4 hrs after Arun Q.4 I walk a certain distance & ride back & take 6 ½ hours altogether. I could walk both the ways in 7¾ Hours. How long would it take me to ride both the ways ?A) 5 hrs 15 min.B) 4 hrs 25 minC) 7 hrs 8 minD) 3 hrs 12 m inAns. AWalking + riding = 13/2 hrs–iWalking + walking = 7 ¾ hrsOr 2 walking = 31/4 hrs—ii2 * equation 1 – equation 2 2 riding = 13-31/4 = 21/4 5hrs 15min Q.5 A thief is spotted by a policeman from a distance of 200 meters. When the policeman starts the chase thief also starts running. Assuming the speed of the thief 10 km an hour & that of the policeman 12 kilometers an hour , how far will have the thief run before he is overtaken ?A) 2.5 kmB) 1 kmC) 1.5 kmD) 5 kmAns. BRelative speed = 12-10 = 2km/hrTherefore the thief will be caught after = 0.2/2 = 1/10 hrThus distance covered by the thief before he gets caught = 10 * 1/10 = 1km