A horse started to chase a dog that started running two ...



Person travels from A to B with speed of 4kmph and returns back from B with speed of 5 km/h, what is his average speed of journey? 2*xy/(x+y)=2*4*5/(4+5)=4.4A horse started to chase a dog that started running two hrs ago. The horse started to ran with an average speed 22km/hr, horse crossed 10 meters wide road and two small pounds with a depth of 3m, and it crossed two small streets each having a length of 200 meters. After traveling 6 hrs, 1hr after sunset it could catch-up with the dog. Compute the speed of dog?total dis cover by horse=22*6=132total time take by dog is (6+2).so speed of the dog is=132/8=16.52 hours after a freight train leaves Delhi, a passenger train leaves the same station travelling in the same direction at an average speed of 16 km/hr. After travelling 4 hrs the passenger train overtakes the freight train.What was the average speed of the freight train ?total time taken by passenger train is 4 hours with 16 km/hr to cross the freight train which is 4 * 16 = 64 kms ...the same distance the frieght train covers in 2 + 4 hours because it has started 2 hrs early ..which is 6 hrs in total now speed can be calculated by 64/16 = 10.67..A person drove at an average rate of 50 miles per hour for two hours and then increased average rate by 50% for the next 3 hours.If the average rate of speed for the 5 hours was t miles per hour. What is the value of t?dist traveled in 2 hrs=50*2=100and in next ((50*50)/100)=25so speed is 75mile per hrsdist traveled =75*3=225total dist=225+100=325speed =325/5=6565mphans: 65 miles/hrORGiven speed1= 50 miles/hrTime= 2 hrstherefore,distance= 100 milesspeed2=(speed1+50% of speed1)= 75 miles/hrtime= 3 hrstherefore distance=225 milestotal distance=100+225=325 milestotal time=2+3=5 hrsAverage speed=(325/5)=65 miles/hr.An express train takes 1 hour less than a passenger train to travel 132 km between two stations. If the average speed of the express train is 11 km/hr more than that of the passenger train. Find the average speed of the two trains ?if we take avg speed of passanger train as 'x'kmphthen the avg speed of express train is '(x+11)'kmphit's given that the time difference between 2 trains is 1hri.e., 132/x - 132/(x+11) = 1solving above eqn we get x=33so,speed of the passenger train will be 33 kmphand, speed of the express train will be 33 + 11 = 44 kmtherefore, average speed of trains are,[33+44]/2=38.5kmphTwo cars travel in the same direction at 40km/hr at a regular distance. A car comes in the opposite direction at 60km/hr. It meets each car in a gap of 8 seconds. What is the distance between the two cars?222.22 mtrsas car is moving opposite dirn so 60+40=100km/hrso d=100*5/18*8=222.22mtrsA car is traveling at a uniform speed. The driversees a milestone showing a 2-digit number. Aftertraveling for an hour the driver sees anothermilestone with the same digits in reverse order. After another hour the driver sees another milestonecontaining the same two digits with a zero inbetween(0). What is the average speed of the driver.LET THE 1ST MILESTONE BE 10X + YNEXT MILESTONE = 10Y + XLAST MILESTONE = 100X + YSPEED OF THE CAR = 2 - 1 = 9Y - 9X MILES/HRIN 2 HRS THE CAR TRAVELS 18Y - 18X MILESTHEREFORE, 10X + Y + 18Y - 18X = 100X + YSOLVING WE GET Y = 6XBUT X,Y SHOULD BE INTEGERS BETWEEN 0- 9THIS IMPLIES X = 1 , Y = 6SPEED OF THE CAR = 45 MILES/HRA man wishes to cross a river perpendicularly. In still water, he takes 4 minutes to cross the river,but in flowing river he takes 5 minutes. If the river is 100 m wide, the velocity of the flowing water of the river isspeed in still water , s= 100/4 = 25 mtr/minif w is speed of flowing water, thenspeed in flowing water = s-w = 100/5 = 20 mtr/minspeed of flowing water = 25-20= 5 mtr/minRashmi walks to her bus stop at 5 kmph and reaches there late by 10 minutes. On the next day,she increases her speed to 6 kmph and reaches the bus stop 10 minutes early.How far is the bus stop ?let distance x km and time to reach bus stand t hrdistance = speed*timefirst day x= 5*(t+10/60)second day x= 6*(t-10/60)on equating t=11/6 hrthen x= 5*(11/6+10/60) =10 kmA person runs from A to B. He took ?? of the time less to reach B when compare to run at normal Speed. Then how many percentage he has increased his speed?33.3333% he has increased his speed.?Let, Distance(d) = Speed(s)*Time(t)d =s*tWhen t1 =(3/4)t?Then,d = s1*t1Apply t1 value,s*t = s1*(3/4)tThens1/s =4/3So, increase in speed = (s1/s - 1)=4/3 - 1?=1/3=33.33%In a race, Mohan beats Sohan by 40m and Sohan beats Rohan by 80m. Mohan beats Rohan by 104m. Find the length of the race (in m).direct formula is (X=X1*X2/X1+X2-X3) where X1=40, X2=80, X3=104.hence, answer is X (length of the race) = 200 m.length of race = 40*80/(40+80-104) = 40*80/16 = 200 mA man wishes to cross a river perpendicularly. In still water, he takes 4 minutes to cross the river,but in flowing river he takes 5 minutes. If the river is 100 m wide, the velocity of the flowing water of the river isspeed in still water , s= 100/4 = 25 mtr/minif w is speed of flowing water, thenspeed in flowing water = s-w = 100/5 = 20 mtr/minspeed of flowing water = 25-20= 5 mtr/minA locomotive driver travelling at 72 km/hr finds a signal 210 metres ahead of him indicating he should stop. He instantly applies brakes to stop the train. The train retards uniformly and stops 10 metres before the signal post. What time did he take to stop the train?v^2-u^2 = 2ashere v=0u= 72 kmph = 72*5/18 = 20m/secs= 200mhence a= -(20^2)/2*200 = -1 m/sec^2a=(v-u)/t?t= (v-u)/a= -20/-1= 20 secsA man wishes to cross a river perpendicularly. In still water, he takes 4 minutes to cross the river,but in flowing river he takes 5 minutes. If the river is 100 m wide, the velocity of the flowing water of the river isspeed in still water , s= 100/4 = 25 mtr/minif w is speed of flowing water, thenspeed in flowing water = s-w = 100/5 = 20 mtr/minspeed of flowing water = 25-20= 5 mtr/minTim and Elan are 90 km from each other, they start to move each other simultaneously, Tim at speed 10 and Elan 5kmph, if every hour they double their speed what is the distance that Tim will pass until he meet Elan?tim elan10 5 in 1st hour=15km20 10 in second hour=30km40 20 in 3rd hour=60kmbt total distnce in 3rd hour= 90-(15+30)=45kmtotal time= 2+45/60 hourtotal distance covered by tim= 10+20+3/4*40=60km10. A dog taken four leaps for every five leaps of hare but three leaps of the dog is equal to four leaps of the hare. Compare speed?Let the distance covered by dog in 1 leap is x and hare covered 1 leap is y.then, 3x = 4y?=> x =(4/3) y?=> 4x =(16/3) yThen, The ratio of speeds of dog and hare = Ratio of distances covered by them in the same time= 4x : 5y?= (16/3)y : 5y?=(16/3) : 5?= 16:15 Ans...=16:15Apple costs L rupees per kilogram for first 30kgs and Q rupees per kilogram for each additional kilogram. If the price of 33 kilograms is 11.67and for 36kgs of Apples is 12.48 then the cost of first 10 kegs of Apples isa) 3.50 b) 10.53 c) 1.17 d)2.830L+3Q=11.6730L+6Q=12.48------------3Q=.81 Q= .27from that L=0.362 cost of 10 kg apple is 10*.362=3.6A is twice efficient than B. A and B can both work together to complete a work in 7 days. Then find in how many days A alone can complete the work?assume B works in a day = wso A works in a day= 2was given both works for 7 days so total work done is =7(w+2w)=21wNow if A alone want to complete work of 21wso time taken by A is = 21w/2w => 10 and 1/2 daysA work is done by two people in 24 minutes. One of them alone can?do it in 40 minutes. How much time will the other person wiil take to? complete it ??assume A's work is =w1 and B's work is=w2sow/w1=40 and w/(w1+w2)=24 now we have to calculate w/w2=?so by solving both equations we get w/w2 as=60 min so?B's time = 60 mintwo cyclist begin training on oval racecourse at same timethe professional cyclist complete each lap in 4 sec noves take 6 mintue how many mintue after start will both cyclist pass at exactly same spot where they begin to cycle(a)10(b)8(C)14(D)12ans is d.simply LCM of 4 and 6.A hare and a tortoise have a race along a circle of 100 yard diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race?a)37.80b)5 c)40d)837.8Hare starts after tortoise covers 1/5th distance .When hare covers 1/8th of distance tortoise meets hareSo distance covered by tortoise = 1-(1/5 + 1/8)= 27/40Time taken by both is sametime = dist/speedSo (dist/ speed) of tortoise = (dist / speed) of harelet speed of tortoise be t and of hare be h27/40t = 1/8hh = (40/(8*27) )* t = 5/27 * t?Now for the next part hare has to cover 7/8 th distance when tortoise covers 1/8 th distanceso we get1/8t = 7/8hh = 7tso the factor by which h's speed increases = 7t/ (5/27 * t)= 37.8Adam sat with his friends in the Chinnaswamy stadium at Madurai to watch the 100 metres running race organized by the Asian Athletics Association. Five rounds were run. After every round half the teams were eliminated. Finally, one team wins the game. How many teams participated in the race?(a) 30 (b) 32 (c) 41 (d) 54ans b)32final round 2 teams (then only decide one winner)previous round 4 teams(include final round teams plus other two eliminated teams)previous round 8 teams(include next round team plus equal no.of members that is 4+4)like wise ,,in general 2^5=32Two trains move in the same direction at speeds 50 kmph and 32 kmph respectively. A man in the slower train observes that 15 seconds elapse before the faster train completely passes by him. What is the length of faster train?(a) 100m (b) 75m (c) 120m (d) 50mAnswer : (b)Since both trains are moving in the same direction their relative speed is 50-32 kmph = 18kmph = 5m/sNow speed*time = distance which will give length of faster train?So, 15s*5m/s = 75m?In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:(a) 1 hour (b) 2 hours (c) 3 hours (d) 4 hoursAns: 1 hrLet the duration of the flight be x hours600/x - 600 (x+1/2)= 200600/x - 1200/(2x+1) = 200?x(2x + 1) = 32x2 + x - 3 = 0(2x + 3)(x - 1) = 0x can't be negative hence x=1?x = 1 hr.Duration =1 hrWith four fifths of then tank full, a vehicle travels 12 miles. How much distance will the vehicle travel with one thirdtank full?(a) 8.05 km (b) 6.05 km (c) 12km (d) 5 kmif 4/5 full then 12 miles then we have to find 1/3 full =? kmso 4/5 x 5/12=5/12 x 12--->>1/3 full=5 milesbut answer has to be in kmso 1 mile=1.609 kms0 5 miles=5 x 1.609 km=8.045 =8.05 kmA person was fined for exceeding the speed limit by 10mph. Another person was also fined for exceeding thesame speed limit by twice the same. If the second person was traveling at a speed of 35 mph, find the speed limit.(a) 35mph (b) 15 mph (c) 20 mph (d) 30 mph1 person exceeded by 10. Another person twice 10 which is 20. Given he is going at 35mph. So speed limit is 35-20=15mphA 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?speed = (120 + 80)km/h (because direction is opposite hence relative velocity is added)= 500/9 m/stime= 9seclet the lenght of second train is xtotal distance covered= 270 + xtherefore, D= speed*timethus 270 + x = 500/9 * 9x= 500-270=230 mA person run from A to B.He took 1/4 of the time less to reach B when compare to run at normal Speed.Then how many percentage he has increased his speed?Let the distance be constant = D (A to B)consider Speed is x.Thus time = TSecondly, Distance = D, Speed = x + y , Time= (3/4)TThus, D= xT= (x+y)* (3/4)Tor x=3yimplies that previous his speed was 3y and later it becomes 4y.thus % increase =(4-3)/3 *100% =33.33 %?A software engineer starts from home at 3pm for every walk on a route which has level road for some distance and a then a hillock.he walks speed of 4kmph on level ground and then at a speed of 3kmph on the uphill and then down the hill at a speed of6kmph to the level ground and then at a speed of 4kmph to the home if he reaches home at 9pm.what is the distance one way of his route?(A)12km (B)15km (C)18km (D)24km(E)Data Inadequatetime=distance/speedso,let the distance to the level road from home be x kmand the distance to the hillock from the level road be y kmso, (x/4+y/3)+(y/6+x/4)=6x+y=12 km.The price of precious stone is directly proportional to square of its weight. what will be the loss incurred if a stone weighing 28gm and costing 28,000 breaks into two pieces whose weights are in the ratio 15:13?(A)No loss (B)10,000 (C)28,000 (D)9,500 (E)14,000Ans (E) 14,000let the price be P and weight be W.thus P=kW*W where k= proportionality constantk= 28000/28*28= 35.7now, P1= (35.7)*13*13=6033P2= (35.7)*15*15= 8032thus, Loss= (P - (P1+P2))= 13934.5= aprox 14,000Pace length is distance between rear of consecutive footprints. For a old man walking Berghian's formula is n/p=144, where n is number of steps per minute and p is pace length which equals 164 cm.Find the sped of walking in kmphno of steps per minute=144*1.64length of one step=1.64speed=144*1.64*1.64 =387.30m/sspeed=387.30*60/1000=23.23 ansFrancois Pachet, a researcher at Sony Computer Science laboratories is also a jazz musician. He decided to build a robot able to improvise like a pro. Named Continuator, the robot can duet with a live musician in real- time. It listens to a musical phrase and then computes a complementary phrase with the same playing style. If the cost of making the robot is divided between and then computes a complementary phrase with the same playing style. If the cost of making the robot is divided between materials, labour and overheads in the ratio of 4:6:2.If the materials cost $108. the cost of the robot isa) $270?b) $324?c) $216?d) $ 648material cost is 108 then =(108/4)=27total share=4+6+2=12total share cost=(12*27)=324Vinay wlked 2 km to reach school.one day,he was sick,he walked with speed 20% less than before while going and returning.what was the percentage increase i n time,in the entire journey??if 1kmph was his daily speed, then 2km/1kmph=2 hours i.e;1kmph=2 hours then 20% less 0.8kmph=?1/0.8 * 2 = 2.5 hours. then % increase 0.5/2 * 100 =25%TWO STATIONS A & B ARE 110 KM APART. ONE TRAIN STARTS FROM A AT 7 AM, AND TRAVELS TOWARDS B AT 20KMPH. ANOTHER TRAIN STARTS FROM B AT 8 AM AND TRAVELS TOWARDS A AT 25KMPH. AT WHAT TIME WILL THEY MEET?A. 9 AMB. 10 AMC. 11 AMD. 10.30 AMlet T=xhrsA-->Distance=20xKmB-->Distance=25(x-1)km[because diff(7,8) is 1]20x+25(x-1)=110km then x=3hrso 7am+3hr=10am.An athlete decides to run the same distance in 1/4th less time that she usually took. By how percent will she have to increase her average speedlet original speed be s1 and time be t1then s1=d/t1 ---eqn 1and according to quesnew speed be s2 and time given is 3t1/4therefore s2=d/(3t1/4) -----eqn 2dividing eqn 2 by eqn 1s2=4s1/3increased speed = 4s1/3-s1=1s1/3percent increase=[(1s1/3)/s1]*100=33.33%A & B travelling from X to Y. A starts at 12 pm at a speed of 63m/hr. B at 1:30 pm at a speed of 84m/hr.At what time will B be 34m ahead of A?at 7:37pm B will be 34m ahead of A5 . A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel, A, the train catches the cat exactly at the entrance. If the cat moves to the exit, B, the train catches the cat at exactly the exit. The speed of the train is greater than the speed of the cat by what order?(3 : 1)?(4 : 1)(5 :1)(2:1)D / v1 = (3/8) x / v2 ... ( 1 )( D + x ) / v1 = (5/8) x / v2=> D / v1 = (5/8) x / v2 - x / v1 ... ( 2 )From equns. ( 1 ) and ( 2 ),(3/8) x /v2 = (5/8) x / v2 - x / v1=> 3 / (8v2) = 5 (8v2) - 1/ v1=> 1 / v1 = ( 5/8 - 3/8) / v2=> 4v2 = v1hence, answer will be 4:1Two trains st art from Chennai and villupuram spaced 150 k m apart at the same timeand speed as the tr ains start, a bird flies fro m one train towards the other and onreac hing the s econd t rain; it flies back to the first tr ain. T his is repeated till the trainsc ollide. If the speed of the trains is 75 kmph and that of the bird is 100 k mph. How muchdid the bird travel till collision?relative velocity of trains (100+50)kmphtime require to travel 150km=150/150 = 1 hr.distance travel by the bird in this 1 hr. = 100*1= 100 kmNavjivan Express fro m Ahmadabad to Chennai leav es Ahmadabad at 06.30 am andtravels at 50 k m/hr towards B aro da situated 100 k m away. At 7 am Howrah toAhmadabad express leaves Baroda tow ards Ahmadabad and trav els at 40 km/hr . At7.30 Mr.John , the t raffic controller at Baro da r ealises that bot h t he trains are r unningon the s ame t rac k. How much time do es he have to avert a head- o n collision betweenthe two t rains?(a) 15 minut es (b) 20 minut es (c) 12 minutes (d) 18 minutesdistance travel by nav. exp. in 1 hr.= 50*1 = 50 kmdistance travel by hw. ahem. exp. in 1/2 hr = 40*1/2= 20 kmdistance will remain before the collision = 30 kmrelative speeds of trains (50+40)kmphso time require before collision= 30/90= 1/3 hr. = 20 min.?Two c ars A and B start simultaneously fro m two points P and Q with c ertain speedstowar ds eac h other. After reaching a point R, speed of A dec reases by 1/3. It thenmeets B at a po int S where SQ = 2PR. If he speed of A had beco me 1/3 less at the midpoint of RS the c ars would hav e met at T w here ST = PR/4. Find RS: PR.(a) 5: 1 (b) 6: 1 (c) 8:1 (d) 10:1condition (1)(PR/Sa)+(3SR/2Sa)=2PR/Sb-------(1)condition (2)(PR+RS/2)/Sa+3/2Sa=QT/Sb(8PR+4RS+6RS+3PR)/8Sa=QT/Sb=2PR/Sb-----(2)dividing (1) by (2)(11PR+10RS)/(4PR+6RS)=QT/PRON SIMPLIFICATION,(put,PQ=PR+RS+ST+TQ)(11PR+10RS)/(4PR+6RS)=7/4SOLVING FURTHER,RS/PR=8:1so,option (C)A and B are travelling from a distance X and Y. A starts at 12p.m at a speed of 6km/hr and B starts at 1.30p.m at a speed of 8km/hr. At what time will be the 3 km ahead of A.a) 3p.mb) 4.30p.mc) 6p.md) 7.30p.mgo by options...at 7.30 a will go 6*7.5=45kmat 7.30 b will go 8*6=48kmso b ewill be 3 km ahead of aa turtle crossing the field.what is total distance passed by a turtle?consider two following statements X average speed of turtle 2m/min.y had turtlewalked 1 m/min faster than his average it would have been finished 40 mintue earliar.(1)A statement alone is sufficient(2)both statement x and y are needed(3)y is sufficint(4)both are not sufficintwe need both time and speed for getting distance. we get the speed from X. bt we also need the time which we can calculate from Y. so, ans is (2)In a stream running at 2 kmph, a motorboat goes 6 km upstream and back again to the starting?point in 33 minutes. Find the speed of the motorboat in still water.Let the speed of the motorboat in still water be x kmph.?Then, speed downstream = (x + 2) kmph.?Speed upstream = (x - 2) kmph.:. 6/(x + 2) + 6/(x - 2) = 33/60 0r 11X2 – 240x – 44=011X2 – 242X + 2X –44 =0 or 11x(x – 22) + 2(x – 22) = 0or (x-22)(11x+2)=0 or x=22.:. Speed of motorboat in still water = 22 kmph.INITIAL PRICE OF SCOOTER IS 40,000 AND IT IS REDUCED TO 3/4TH OF IT'S PREVIOUS PRICE EVERY YEAR.WHAT WILL BE THE PRICE AFTER 3 YEARS?Original Price = 40000After 1 yr = 30000After 2 yr = 30000*3/4 = 22500After 3 yr = 22500 * 3/4 = 16875 .... Answer ................
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