#2
 
 
Re: Infosys Solved Placement Papers
Here are the Infosys placement exam solved questions: Q15 marks) a wall clock looses 10 mins in 1 hr. 4 one hours of wall clock table clock gains 10mins for 1 hour of table clock alarm clock looses 5 mins for 1hr of aramm clock, wrist watch gains 5min they were set correctly at 4 pm what was the time of wrist watch at 6 pm(of correct time clock) Q2:3 marks Mr. Jonh was a inspector of falt finding in balancing sscal. He found that 1arm was longer than the other one. When 8 pyramids were kept in the longer the cubes required were in the short arm. When 1 pyramid is kept on longer arm the cubes required at shorts arm was 6. if the correct weight of pyramid is 1 ounce.? Is the true weight of one cube. Q3:6marks They were 2 twins one of them say nothing but lies on Monday, Wednesday and Friday and truth on other days. 2nd one says nothing but lies on Tuesday, Thursday, and Saturday. Only on sun both speak truth. Q4: (5 marks) 3 (A, B, C) players along with their wife (D, E, F) play a golf game, who scores less will win. Altogether there were 18 games. C score 100, D  102, E – 106. 94 was the score of B. 2 other man A, B scores 98 and 96. B's wife beats A's wife. It was found that 2 pair score identical then identify the pairs. For detailed paper here is attachment:
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#7
 
 
Re: Infosys Solved Placement Papers
As you asking for the Sample paper of the Aptitude test of the Infosys ,here I am giving you the sample paper of the Aptitude test . the test consist the following questions Solve the following question based on the information provide i. Students A, B, C, D, E, and F participated in a selfevolution test of Quant‟s and Data (D.I) ii. Total marls of A in quant‟s was just above C and in D.I just above F was just above C in D.I but he scored less than D in Quant‟s iii. Got more marks than D and E in D.I but did not perform as well in Quant‟s as in D.I compared to D and E iv. One is in between C and D in Quant‟s and C and A in D.I Got the highest make in D.I ? 1) A 2) B 3) C 4) Data InadequateSolve the following question based on the information provide i. Students A, B, C, D, E, and F participated in a selfevolution test of Quant‟s and Data (D.I) ii. Total marls of A in quant‟s was just above C and in D.I just above F was just above C in D.I but he scored less than D in Quant‟s iii. Got more marks than D and E in D.I but did not perform as well in Quant‟s as in D.I compared to D and E iv. One is in between C and D in Quant‟s and C and A in D.I Got the highest make in D.I ? 1) A 2) B 3) C 4) Data InadequateSolve the following question based on the information provide i. Students A, B, C, D, E, and F participated in a selfevolution test of Quant‟s and Data (D.I) ii. Total marls of A in quant‟s was just above C and in D.I just above F was just above C in D.I but he scored less than D in Quant‟s iii. Got more marks than D and E in D.I but did not perform as well in Quant‟s as in D.I compared to D and E iv. One is in between C and D in Quant‟s and C and A in D.I Got the highest make in D.I ? 1) A 2) B 3) C 4) Data Inadequate Probability questions Question 1 In a bus stand, there are two services namely A and B. Every 10 minutes buses will leave from A and this service works from 6.10 am to 2 pm. The service at B starts at 2.20 pm and for every 20 minutes buses will leave from the bus stand. Find the probability of getting bus from service B between 2.20 pm to 2.50 pm, if service A is late by 1 hour. a) 1/2 b) 1/3 c) 1/4 d) 1/5 Answer : b) 1/3 Solution : Usually service A starts at 6.10 am. Since the service is late by 1 hour, the first and last bus will leave by 7.10 am and 3 pm respectively. Note that, there is a bus for every 10 minutes. Number of buses leaving between 2.20 pm to 2.50 pm is 4 and the timings are 2.20 pm, 2.30pm, 2.40pm and 2.50pm. There is a bus for every 20 minutes from service B. Number of buses leaving between 2.20 pm to 2.50 pm is 2 and they start at 2.20 pm and 2.40 pm. Therefore, 4 buses from A and 2 buses from B are available. The probability of getting bus from B = buses from B / total number of buses from A and B. = 2/6 = 1/3. Question 2 From a railway station, trains leave for every 15 minutes and 25 minutes to city A and city B respectively. First train to city A and city B start at 9 am and 10.15 am respectively. If a man arrives to the station in between 11.25 am and 12.25 pm then the probability of getting train for city A is: a) 1/4 b) 4/7 c) 3/5 d) 2/5 Answer : b) 4/7. Solution : The man wants to go to city A and he arrives station in between 11.25 am and 12.25 pm. First train to city A is at 9 am and there is a train for every 15 minutes. Trains for city A will leave at the following times : 9 am, 9.15 am, 9.30 am,…,11.30 am, 11.45 am, 12 pm, 12.15pm, and so on. Number of trains for city A between 11.25 am and 12.25 pm is 4. First train to city B is at 10.15 am and there is a train for every 25 minutes. Trains for city B will leave at the following times: 10.15 am, 10.40 am, 11.05 am, 11.30 am, 11.55 am, 12.20 pm, and so on. Number of trains for city B between 11.25 am and 12.25 pm is 3. Probability of getting train for city A between 11.25 am and 12.25 pm = Number trains for city A from 11.25 am to 12.25 pm / Total number of trains for city A and B from 11.25 am to 12.25 pm = 4/7. Question 3 There are two bus stands, namely X and Y. Buses leave from X for every 30 minutes and its first bus starts at 8.05 am. Every hour number of buses leaving from Y increases by 1 and its first bus starts at 7 am. From Y there is only 1 bus for the 1st hour. Any bus from either of the bus stations takes 15 minutes to reach a nearby bus stop. Suppose a person reaches the stop in between 12.15 pm and 1.15 pm. The probability that the person will get a bus from Y is: a) 3/4 b) 1/3 c) 1 d) 1/4 Answer : a) 3/4 Combination questions Question 1 In a dance school, there are 14 girls and 15 boys. In how many ways, a group consisting of 3 girls and 7 boys can be formed for a competition? a) 2342340 b) 2232340 c) 2132340 d) 2132840 Answer : a) 2342340 Solution : Now, use the combination method, nCr = n! / r! (nr)! 3 are selected from 14 girls in 14C3 ways = 14! / 3! (14  3)! = 14! / 3!11! = 364 ways. And 7 are selected from 15 boys in 15C7 ways = 15! / 7! (15  7)! = 15! / 7!8! = 6435 ways Hence the required number of possible ways = 364 x 6435 = 2342340 ways. Question 2 The players for district football team are selected from three different colleges and each college has 5 top players. Except captain and goalkeeper, each 3 are selected from each 3 colleges. Find the number of possible ways of such selection. a) 1000 b) 1500 c) 4500 d) none of these Answer : a) 1000 Solution : Total number of players = 11 2 players(captain & goalkeeper)selected in 1 way. Now, use the combination method, nCr = n! / r! (nr)! 3 out of 5 can be selected in 5C3 ways = 5! / 3! (5  3)! = 10 Similarly, other 3 are selected from other two colleges. Hence total number of possible ways = 1 x 10 x 10 x 10 = 1000 Question 3 In a college cricket club of 22 players, captain and wicketkeeper are selected from seniors team. For remaining players they have to select from 8 seniors and 12 juniors. Now, for a new team of 11 players except captain and wicketkeeper, out of 9 they select 4 from juniors and 5 from seniors. Find the number of methods to select players of new team. a) 12220 b) 27720 c) 15870 d) none of these Answer : b) 27720 Puzzele problems Question 1 Find X's age which equals the number of grand children of a man who has 4 sons and 4 daughters. Each daughter of the man's wife have 3 sons and 4 daughters and each son of the man's wife have 4 sons and 3 daughters. a) 40 b) 56 c) 64 d) none of these Answer : b) 56 Solution : We have to find the number of grand children of the man. Given that, he had 4 sons and 4 daughters. Each son has 4 sons and 3 daughters and each daughter has 3 sons and 4 daughters. Therefore total number of grandsons = 4x4 + 4x3 = 16 + 12 = 28 And total number of grand daughters = 4x3 + 4x4 = 28 Total number of grandchildren is 28+28 = 56. Hence the required age is 56. Question 2 A man have many daughters, each daughter have as many sons as her sisters. The product of the number of daughters and grandsons of the man lies between 40 and 50. Find the number of daughters of the man. a) 6 b) 4 c) 3 d) 8 Answer : b) 4 Solution: Let N be the number of daughters of the man. Then each daughter has N1 sisters. Given that, each daughter has as many sons as her sisters. That is, each of them has N1 sons. Number of grandsons of the man = N(N1) And the required product = N(N1) x N = N x N(N1) which lies between 40 and 50. If N = 1 then N x N(N1) = 0. f N = 2 then N x N(N1) = 2 x 2(1) = 4 If N = 3 then N x N(N1) = 3 x 3 x 2 = 18 If N = 4 then N x N(N1) = 4 x 4 x 3 = 48 If N = 5 then N x N(N1) = 5 x 5 x 4 = 100 Therefore the possible value of N is 4. Hence the man has 4 daughters. Question 3 Two men A and B have equal number of daughters and A have 1 more son than B. Each son and daughter of A have 2 sons and 2 daughters and each son and daughter of B have 3 sons and 3 daughters. If A and B have equal number of grandchildren then find the number of sons of B. a) 3 b) 4 c) 5 d) none of these. Answer : d) none of these Solution : Let D be the number of daughters of A and B. Let S be the number of sons of A. Then S1 be the number of sons of B. Now, number of grandsons of A = 2S + 2D. Number of granddaughters of A = 2S + 2D Number of grand children of A = 2S + 2D + 2S + 2D = 4S + 4D. Number of grandsons of B = 3(S1) + 3D = 3S + 3D  3 Number of granddaughters of B = 3(S1) + 3D = 3S + 3D  3 Number of grand children of A = 3S + 3D  3 + 3S + 3D  3 = 6S + 6D  6 And, 4S + 4D = 6S + 6D  6 2S + 2D = 6 S + D = 3 Then the possibilities of (S,D) are (1,2), (2,1), (0,3), (3,0) We have to find the value of S1. S1 cannot be a negative number, so (0,3) is not possible. Therefore, possible S1 values are 0,1,2. Hence the answer is option d.
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#9
 
 
Re: Infosys Solved Placement Papers
I have a sample paper of Infosys Placement exam. So here I am providing you as you want. Question 1 An express A starts at 2.30 pm from Karnal station and travel towards Sirsa station at a speed of 80 km per hour. Another express B starts at 4.30 pm from Karnal to Sirsa at the speed of 100 km per hour. How far away from station Karnal will the two trains meet? a) 400 km b) 100 km c) 800 km d) 450 km Answer : c) 800 km. Solution : Train A starts at 2.30 pm from KARNAL to SIRSA. Train B starts at 4.30 pm from KARNAL to SIRSA. Let they meet H hours after 2.30 pm. We have to find the distance traveled by A in H hours or by B in (H2) hours. Given that, speed of A = 80 km/hr Distance traveled by A in H hours at 80km/hr = (distance = speed x time) = 80H km. … (1) Speed of B = 100 km/hr. Note that, it starts at 4.30 pm, that is 2 hours later than A. Therefore, we would find the distance in (H2) hours. Distance traveled by B in (H2) hours at 100km/hr = 100(H2) km … (2) We know that trains meet after H hours and this means distance covered by train A in H hours equals distance covered by train B in H2 hours. Therefore, from (1) and (2), we get 80H = (H2)100 20H = 200 H = 10 hours. That is, they meet after 10 hours from2.30 pm. Hence, the required distance = 80H = 80 x 10 = 800 km. Question 2 A train P starts from Delhi at 8 pm, reaches its destination UP at 1 am. Another train Q starts from UP at 8 pm and reaches Delhi at 2 am. The two trains P and Q will cross each other at : a) 10.45 pm b) 12.45 pm c) 11.45 pm d) none of these. Answer : a) 10.45 pm. Solution : Let the distance between UP and Delhi be X km. Time taken by P to reach UP (to cover X km) = 8 pm to 1 am = 5 hours. Speed of P = distance / time = X/5 km/hr. Time taken by Q to reach Delhi (to cover X km) = 8 pm to 2 am = 6 hours. Speed of Q = X/6 km/hr. Let them meet Y hours after 8 pm. Distance covered by P at Y hours at X/5 km/hr = XY/5 km Distance covered by Q at Y hours at X/6 km/hr = XY/6 km. Since they travel in opposite directions, during the time when they cross each other, the sum of the distance covered by the first and second trains will be equal to the distance between Delhi and destination at UP. i.e., XY/5 + XY/6 = X XY (1/5 + 1/6 ) = X Y (1/5 + 1/6 ) = 1 Y (11/30) = 1 11Y = 30. Y = 30/11 = 2.72 hours Therefore, they meet after 2.72 hours from 8 pm. That is, 8 pm + 2 hours + 45 minutes (nearly). Hence, the answer is 10.45 pm. Question 3 Chennai express starts at 8 am from A towards B and Mumbai express starts at 8 am from B towards A. After their meet, Chennai express takes 18 hours to reach B and Mumbai express takes 8 hours to reach A. Then which of the following is their speed’s ratio? a) 1 : 2 b) 2 : 3 c) 2 : 5 d) 1 : 3 Answer : b) 2 : 3. Solution : Note: “ If two trains start at the same time from A and B towards each other and after they take a and b hours in reaching B and A respectively, then A’s speed : B’s speed = sqrt(b) : sqrt (a).” Here they start at same time ( 8 am) and travels in opposite direction. (After their meet), Chennai express takes 18 hours to reach B and Mumbai express takes 8 hours to reach A. b = 8 and a = 18. By using the above note, required ratio = sqrt(8) : sqrt(18) = 2 sqrt(2) : 3 sqrt(2) = 2 : 3. Hence, the answer is 2 : 3. Paper 1.One guy has Rs. 100/ in hand. He has to buy 100 balls. One football costs Rs. 15/, One Cricket ball costs Re. 1/ and one table tennis ball costs Rs. 0.25 He spend the whole Rs. 100/ to buy the balls. How many of each balls he bought? 2.The distance between Station Atena and Station Barcena is 90 miles. A train starts from Atena towards Barcena. A bird starts at the same time from Barcena straight towards the moving train. On reaching the train, it instantaneously turns back and returns to Barcena. The bird makes these journeys from Barcena to the train and back to Barcena continuously till the train reaches Barcena. The bird finally returns to Barcena and rests. Calculate the total distance in miles the bird travels in the following two cases: (a) The bird flies at 90 miles per hour and the speed of the train is 60 miles per hour. (b) the bird flies at 60 miles per hour and the speed of the train is 90 miles per hour 3.A tennis championship is played on a knockout basis, i.e., a player is out of the tournament when he loses a match. (a) How many players participate in the tournament if 15 matches are totally played? (b) How many matches are played in the tournament if 50 players totally participate? 4.When I add 4 times my age 4 years from now to 5 times my age 5 years from now, I get 10 times my current age. How old will I be 3 years from now? 5.A rich merchant had collected many gold coins. He did not want anybody to know about them. One day, his wife asked, "How many gold coins do we have?" After pausing a moment, he replied, "Well! If I divide the coins into two unequal numbers, then 37 times the difference between the two numbers equals the difference between the squares of the two numbers." The wife looked puzzled. Can you help the merchant's wife by finding out how many gold R 6.A set of football matches is to be organized in a "roundrobin" fashion, i.e., every participating team plays a match against every other team once and only once. If 21 matches are totally played, how many teams participated? 7.Glenn and Jason each have a collection of cricket balls. Glenn said that if Jason would give him 2 of his balls they would have an equal number; but, if Glenn would give Jason 2 of his balls, Jason would have 2 times as many balls as Glenn. How many balls does Jason have? 8.Suppose 8 monkeys take 8 minutes to eat 8 bananas. (a) How many minutes would it take 3 monkeys to eat 3 bananas? (b) How many monkeys would it take to eat 48 bananas in 48 minutes 9.It was vacation time, and so I decided to visit my cousin's home. What a grand time we had! In the mornings, we both would go for a jog. The evenings were spent on the tennis court. Tiring as these activities were, we could manage only one per day, i.e., either we went for a jog or played tennis each day. There were days when we felt lazy and stayed home all day long. Now, there were 12 mornings when we did nothing, 18 evenings when we stayed at home, and a total of 14 days when we jogged or played tennis. For how many days did I stay at my cousin's place? 10.A 31" x 31" square metal plate needs to be fixed by a carpenter on to a wooden board. The carpenter uses nails all along the edges of the square such that there are 32 nails on each side of the square. Each nail is at the same distance from the neighboring nails. How many nails does the carpenter use? Answer and Explanation. 1. F + C + T = 100eq1 15F + C + 0.25T = 100eq2 eq1=eq2 .solve to get F=3T/56 ;F=3,T=56,C=41 2.a) There is no need to consider their meeting pt at all.the train has been running for 90miles/(60miles/hr)=1.5hrs.bird flies till train reaches destination frm strting pt.so bird flies for1.5hrs at the vel given(90).so dist=1.5*90=135miles b) time of train=1hr.so dist of bird=60*1=60miles 3.(a) u don't need to sum it up.since it's a knock out only 1 person emerges winner finally.so15+1=16is answer.becos after15 matches finally we shud've 15losers and 1winner. (b) 49:its always one less than no of players as per the idea given above.so no need to check okay cos its always true.ans is 49. 4.Let x= current age 4(x+4)+5(x+5)=10x ;so x=R 41 years 5.37(xy)=x^2y^2. u no tht x^2y^2=(xy)(x+y).so (xy) cancels on both sides to give x+y=37.so sum of unequal halves=37 which is the req answer. 6.R ans:7 teams okay.for a match u need 2 teams.suppose there r totally 'n 'teams. Now uve to choose 2 teams out of 'n' teams.so answer =no of such choices=no. of possible combinations. So we've ans = nC2(ncombination2)=21;solve to get n=7. Sol: n(n1)/2=21. so n=7.if u don't understand c the graph below each team plays no. of matches=no of teams ahead of it. One bar '' represents one team.        7 6 5 4 3 2 1 0 21 last team is written as 0 matches becos this team has already played with all other teamshence sum of matches =6+5+4+3+2+1=21 which is correct only if no of teams =7 7.R 14 1. G+2=j2 2. 2(G2)=J+2. solve these 2 to get J=14 8.a). Sol:each mky takes 8 min to eat a banana b).ans:8m=48 m=6 9.Use sets and venn diagram to solve such questions.a,b ,aub,anb etc. 12=tennis+leave 18=jog +leave so jogtennis=6 again jog+tennis=14.so solve and get jog=10,leave=8,tennis=4.so tot=22 10.Ans= 32*2 + 30*2=124
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