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Patni Computer Solutions Placement Papers
I need the Patni Computer Solutions Placement Papers, will you please provide here?
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Re: Patni Computer Solutions Placement Papers
You are looking for the Patni Computer Solutions Placement Papers i am giving here: 1. IKBUTA 1) $8@721 2) ?8@72? 3) %8@72% 4) ?8@72% 5) None of these 2. EMPRJH 1) 9653#4 2) ?9653# 3) %653#% 4) ?653#? 5) None of these 3. IPAUHM 1) ?5174? 2) %5174% 3) $51746 4) $51476 5) None of these 4. RFHKJA 1) 3©48#1 2) 483©#1 3) ?©48#? 4) %©48% 5) None of these 5. TMRBFJ 1) ?63@©? 2) %63@©% 3) 236@©# 4) 263@©# 5) None of these 6.KTJUFA 1) 82#7©1 2) ?2#7©? 3) %2#7©% 4) $2#7©1 5) None of these 7. UMBKPE 1) ?6@85? 2) 76@85© 3) ?6@85© 4) 76@85% 5) None of these 8. Find the approximate value of the following equation. 8149.72 + 6935.25 – 1756.79 – 3567.42 = ? 1) 9750 2) 9760 3) 9765 4) 9755 5) 9770 9. A milkman buys some milk. If he sells it at Rs 5 per litre, he loses Rs 200. But when he sells it at Rs 6 per litre, he gains Rs 150. How much milk did he purchase? 1) 50 litres 2) 31 litres 3) 350 litres 4) 200 litres 5) None of these For detailed paper here is attachment; |
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Re: Patni Computer Solutions Placement Papers
As you wanted to get Job in patni Computer Solution let me tell you that there placement drive consist of three Rounds they are : Apptitude Test as first Round Technical Round Interview the selection will be based on the performance of the both Three rounds and the interview . As you aski8ng for the Aptitude question paper here I am providng you the question paper the questions are as follow: 1)A train covers a distance in 50 min ,if it runs at a speed of 48kmph on an average.The speed at which the train must run to reduce the time of journey to 40min will be. 1. Solution:: Time=50/60 hr=5/6hr Speed=48mph distance=S*T=48*5/6=40km time=40/60hr=2/3hr New speed = 40* 3/2 kmph= 60kmph 2)Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is? 2. Solution:: Let total distance be S total time=1hr24min A to T :: speed=4kmph diistance=2/3S T to S :: speed=5km distance=1-2/3S=1/3S 21/15 hr=2/3 S/4 + 1/3s /5 84=14/3S*3 S=84*3/14*3 = 6km 3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr. the usual time is. 3. Solution:: Usual speed = S Usual time = T Distance = D New Speed is ¾ S New time is 4/3 T 4/3 T – T = 5/2 T=15/2 = 7 ½ 4)A man covers a distance on scooter .had he moved 3kmph faster he would have taken 40 min less. If he had moved 2kmph slower he would have taken 40min more.the distance is. 4.Solution:: Let distance = x m Usual rate = y kmph x/y – x/y+3 = 40/60 hr 2y(y+3) = 9x ————–1 x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2 divide 1 & 2 equations by solving we get x = 40 5)Excluding stoppages,the speed of the bus is 54kmph and including stoppages,it is 45kmph.for how many min does the bus stop per hr. Solution:: Due to stoppages,it covers 9km less. time taken to cover 9 km is [9/54 *60] min = 10min 6)Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively.wht time will they take to be 8.5km apart, if they walk in the same direction 6.Solution:: The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph Distance between them is 8.5 km Time= 8.5km / 0.5 kmph = 17 hrs 7)2 trains starting at the same time from 2 stations 200km apart and going in opposite direction cross each other ata distance of 110km from one of the stations.what is the ratio of their speeds. 7. Solution:: In same time ,they cover 110km & 90 km respectively so ratio of their speed =110:90 = 11:9 8)Two trains start from A & B and travel towards each other at speed of 50kmph and 60kmph resp. At the time of the meeting the second train has traveled 120km more than the first.the distance between them Solution:: Let the distance traveled by the first train be x km then distance covered by the second train is x + 120km x/50 = x+120 / 60 x= 600 so the distance between A & B is x + x + 120 = 1320 km 9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the theft is discovered at 3pm and the owner sets off in another car at 75kmph when will he overtake the thief Solution:: Let the thief is overtaken x hrs after 2.30pm distance covered by the thief in x hrs = distance covered by the owner in x-1/2 hr 60x = 75 ( x- ½) x= 5/2 hr thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm Rest of the questions you may found in the file given below:
__________________ Answered By StudyChaCha Member |