TCS Placement Paper | MCQ 8

21 May, 2019
This is a TCS model placement paper for aptitude preparation. This placement paper will cover aptitude questions that are asked in TCS recruitment drives and also strictly follows the pattern of questions asked in TCS interviews. It is recommended to solve each one of the following questions to increase your chances of clearing the TCS interview.
Identify the missing number in the series: 2, 5, __, 19, 37, 75?a) 16b) 12c) 10d) 9Answer: d) 9Solution:2 * 2 + 1 = 55 * 2 – 1 = 99 * 2 + 1 = 1919 * 2 – 1 = 37 and so onA rectangle is divided into four rectangles with area 70, 36, 20, and x. What is the value of ‘x’?a) 350/7b) 350/11c) 350/9d) 350/13
Video Player is loading.Play VideoPlayMuteCurrent Time 0:00/Duration 0:00Loaded: 0%0:00Stream Type LIVESeek to live, currently behind liveLIVERemaining Time -0:00 1xPlayback RateChaptersChaptersDescriptionsdescriptions off, selectedCaptionscaptions settings, opens captions settings dialogcaptions off, selectedEnglishAudio TrackPicture-in-PictureFullscreenThis is a modal window.The media could not be loaded, either because the server or network failed or because the format is not supported.Beginning of dialog window. Escape will cancel and close the window.TextColorWhiteBlackRedGreenBlueYellowMagentaCyanTransparencyOpaqueSemi-TransparentBackgroundColorBlackWhiteRedGreenBlueYellowMagentaCyanTransparencyOpaqueSemi-TransparentTransparentWindowColorBlackWhiteRedGreenBlueYellowMagentaCyanTransparencyTransparentSemi-TransparentOpaqueFont Size50%75%100%125%150%175%200%300%400%Text Edge StyleNoneRaisedDepressedUniformDropshadowFont FamilyProportional Sans-SerifMonospace Sans-SerifProportional SerifMonospace SerifCasualScriptSmall CapsReset restore all settings to the default valuesDoneClose Modal DialogEnd of dialog window.Answer: c) 350/9Solution:Since the areas of the rectangles are in proportion we can say,=> 70/x = 36/20=> x = 350/9If VXUPLVH is written as SURMISE, what is SHDVD written as?a) PEASAb) PBASAc) PEBSBd) None of the aboveAnswer: a) PEASASolution:It is a question of coding-decoding where,V is written as S (V – 3 = S)X is written as U (X – 3 = U)and so on.Similarly, SHDVD will be written as PEASAAman owes Bipul Rs 50. He agrees to pay Bipul over a number of the consecutive day starting on Monday, paying a single note of Rs 10 or Rs 20 on each day. In how many different ways can Aman repay Bipul. (Two ways are said to be different if at least one day, a note of a different denomination is given)a) 5b) 6c) 7d) 8Answer: d) 8Solution:Aman can pay Bipul in all 10 rupees note in 5 days = 5 * 10 = 50 rupees = 1 wayAman can pay Bipul in 3 ten rupee note and 1 twenty rupee note = 4!/(3! * 1!) = 4 waysAman can pay Bipul in 1 ten rupee note and 2 twenty rupee note = 3!/(1! * 2!) = 3 waysSo in all Aman can pay Bipul in 8 ways.Salim bought a certain number of oranges at a rate of 27 oranges for rupees 2 times M, where M is an integer. He divided these oranges into two equal halves, one part of which he sold at the rate of 13 oranges for Rs M and the other at the rate of 14 oranges for Rs M. He spent and received an integral no of rupees, but bought the least number of oranges. How many did he buy?a) 980b) 9828c) 1880d) 102660Answer: b) 9828Solution:Let Salim buy 2x number of oranges.So he buys 27 oranges at a price of 2M.He buys 1 orange at a price of 2M/27or, x oranges cost him Rs. 2Mx/27Now he sells x oranges at the rate of 13 oranges for Rs. MSo he sells 1 orange at Rs. M/13and x oranges at Rs Mx/13Same goes for 14 oranges which are Mx/14,According to the question, 2Mx/27, Mx/13, Mx/14 are integersSo, x oranges must be divisible by 27, 13 and 14The lcm of 27, 13 and 14 = 4914 or 2x = 9828In a football match, 16 teams participate and are divided into 4 groups. Every team from each group will play with each other once. The top 2 winning teams will move to the next round and so on the top two teams will play the final match. So how many minimum matches will be played in that tournament?a) 40b) 14c) 43d) 50Answer: c) 43Solution:Total matches to be played = 4C2 = 6 matches.So total number of matches played in the first round = 6 * 4 = 24 matchesNow top two teams from each group progress to the next round. These 8 teams are to be divided into 2 groups.Total matches played in the second round = 6 × 2 = 12 matchesNow 4 teams progress to the next round. Total matches played in the third round = 6 * 1 = 6matchesFrom this round, 2 teams progress to the next round. And final will be played between them.Total matches = 24 + 12 + 6 + 1 = 43There are 12 letters and exactly 12 envelopes. There is one letter to be inserted randomly into each envelope. What is the probability that exactly 1 letter is inserted in an improper envelope?a) 1b) 0c) 10!d) None of theseAnswer: b) 0Solution:This is a question of very common sense in which,12 letters are to be inserted in 12 envelopes, 1 in each, so if one letter is inserted into a wrong envelope there has to be another letter which is to be inserted into another wrong envelope. So the probability of this is 0.A hollow space on the earth surface is to be filled. The total cost of filling is Rs. 20000. The cost of filling per cubic-meter is Rs 225. How many times is a size of 3 cubic-meter soil required to fill the hollow space?a) 29.62b) 30.32c) 88.88d) 43.64Answer: a) 29.62Solution:The total cost of filling = 20, 000Cost of filling 1 cubic meter = Rs. 225So cubic meters to be filled = 20, 000/225 = 88.89 meter-cubeNow we need to find the three times of 88.89 to be filled = 88.89/3 = 29.63So the closest match is 29.62A 7-digit number is to be formed with all different digits. If the digits at the extreme right and extreme left are fixed to 5 and 6 respectively, find how many such numbers can be formed?a) 120b) 30240c) 6720d) None of theseAnswer: c) 6720Solution:If the digits at extreme left and right are fixed as 5 and 6, then the number of digits left = 8So the in-between 5 places can be filled in 8 * 7 * 6 * 5 * 4 ways= 6720 waysThere are five tires in a sedan (four road tires and one spare) which is to be used equally in a journey to travel 40, 000 km. The number of km of use of each tyre wasa) 32000b) 8000c) 4000d) 10000Answer: a) 32000Solution:The total km travelled by the sedan = 40, 000 kmSince every tire capacity’s = 40, 000/5 = 8000 km eachSo total distance covered by each tire = 8000*4 = 32000 km each will be travelled by each tire after being worn out after every 8000 km.My Personal Notes
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Identify the missing number in the series: 2, 5, __, 19, 37, 75?a) 16b) 12c) 10d) 9Answer: d) 9Solution:2 * 2 + 1 = 55 * 2 – 1 = 99 * 2 + 1 = 1919 * 2 – 1 = 37 and so on
Answer: d) 9
Solution:2 * 2 + 1 = 55 * 2 – 1 = 99 * 2 + 1 = 1919 * 2 – 1 = 37 and so on
A rectangle is divided into four rectangles with area 70, 36, 20, and x. What is the value of ‘x’?a) 350/7b) 350/11c) 350/9d) 350/13
Video Player is loading.Play VideoPlayMuteCurrent Time 0:00/Duration 0:00Loaded: 0%0:00Stream Type LIVESeek to live, currently behind liveLIVERemaining Time -0:00 1xPlayback RateChaptersChaptersDescriptionsdescriptions off, selectedCaptionscaptions settings, opens captions settings dialogcaptions off, selectedEnglishAudio TrackPicture-in-PictureFullscreenThis is a modal window.The media could not be loaded, either because the server or network failed or because the format is not supported.Beginning of dialog window. Escape will cancel and close the window.TextColorWhiteBlackRedGreenBlueYellowMagentaCyanTransparencyOpaqueSemi-TransparentBackgroundColorBlackWhiteRedGreenBlueYellowMagentaCyanTransparencyOpaqueSemi-TransparentTransparentWindowColorBlackWhiteRedGreenBlueYellowMagentaCyanTransparencyTransparentSemi-TransparentOpaqueFont Size50%75%100%125%150%175%200%300%400%Text Edge StyleNoneRaisedDepressedUniformDropshadowFont FamilyProportional Sans-SerifMonospace Sans-SerifProportional SerifMonospace SerifCasualScriptSmall CapsReset restore all settings to the default valuesDoneClose Modal DialogEnd of dialog window.Answer: c) 350/9Solution:Since the areas of the rectangles are in proportion we can say,=> 70/x = 36/20=> x = 350/9
Chapters
descriptions off, selected
captions settings, opens captions settings dialog
captions off, selected
English
This is a modal window.
Beginning of dialog window. Escape will cancel and close the window.
End of dialog window.
Answer: c) 350/9
Solution:Since the areas of the rectangles are in proportion we can say,=> 70/x = 36/20=> x = 350/9
If VXUPLVH is written as SURMISE, what is SHDVD written as?a) PEASAb) PBASAc) PEBSBd) None of the aboveAnswer: a) PEASASolution:It is a question of coding-decoding where,V is written as S (V – 3 = S)X is written as U (X – 3 = U)and so on.Similarly, SHDVD will be written as PEASA
Answer: a) PEASA
Solution:It is a question of coding-decoding where,V is written as S (V – 3 = S)X is written as U (X – 3 = U)and so on.Similarly, SHDVD will be written as PEASA
Aman owes Bipul Rs 50. He agrees to pay Bipul over a number of the consecutive day starting on Monday, paying a single note of Rs 10 or Rs 20 on each day. In how many different ways can Aman repay Bipul. (Two ways are said to be different if at least one day, a note of a different denomination is given)a) 5b) 6c) 7d) 8Answer: d) 8Solution:Aman can pay Bipul in all 10 rupees note in 5 days = 5 * 10 = 50 rupees = 1 wayAman can pay Bipul in 3 ten rupee note and 1 twenty rupee note = 4!/(3! * 1!) = 4 waysAman can pay Bipul in 1 ten rupee note and 2 twenty rupee note = 3!/(1! * 2!) = 3 waysSo in all Aman can pay Bipul in 8 ways.
Answer: d) 8
Solution:Aman can pay Bipul in all 10 rupees note in 5 days = 5 * 10 = 50 rupees = 1 wayAman can pay Bipul in 3 ten rupee note and 1 twenty rupee note = 4!/(3! * 1!) = 4 waysAman can pay Bipul in 1 ten rupee note and 2 twenty rupee note = 3!/(1! * 2!) = 3 waysSo in all Aman can pay Bipul in 8 ways.
Salim bought a certain number of oranges at a rate of 27 oranges for rupees 2 times M, where M is an integer. He divided these oranges into two equal halves, one part of which he sold at the rate of 13 oranges for Rs M and the other at the rate of 14 oranges for Rs M. He spent and received an integral no of rupees, but bought the least number of oranges. How many did he buy?a) 980b) 9828c) 1880d) 102660Answer: b) 9828Solution:Let Salim buy 2x number of oranges.So he buys 27 oranges at a price of 2M.He buys 1 orange at a price of 2M/27or, x oranges cost him Rs. 2Mx/27Now he sells x oranges at the rate of 13 oranges for Rs. MSo he sells 1 orange at Rs. M/13and x oranges at Rs Mx/13Same goes for 14 oranges which are Mx/14,According to the question, 2Mx/27, Mx/13, Mx/14 are integersSo, x oranges must be divisible by 27, 13 and 14The lcm of 27, 13 and 14 = 4914 or 2x = 9828
Answer: b) 9828
Solution:Let Salim buy 2x number of oranges.So he buys 27 oranges at a price of 2M.He buys 1 orange at a price of 2M/27or, x oranges cost him Rs. 2Mx/27Now he sells x oranges at the rate of 13 oranges for Rs. MSo he sells 1 orange at Rs. M/13and x oranges at Rs Mx/13Same goes for 14 oranges which are Mx/14,According to the question, 2Mx/27, Mx/13, Mx/14 are integersSo, x oranges must be divisible by 27, 13 and 14The lcm of 27, 13 and 14 = 4914 or 2x = 9828
In a football match, 16 teams participate and are divided into 4 groups. Every team from each group will play with each other once. The top 2 winning teams will move to the next round and so on the top two teams will play the final match. So how many minimum matches will be played in that tournament?a) 40b) 14c) 43d) 50Answer: c) 43Solution:Total matches to be played = 4C2 = 6 matches.So total number of matches played in the first round = 6 * 4 = 24 matchesNow top two teams from each group progress to the next round. These 8 teams are to be divided into 2 groups.Total matches played in the second round = 6 × 2 = 12 matchesNow 4 teams progress to the next round. Total matches played in the third round = 6 * 1 = 6matchesFrom this round, 2 teams progress to the next round. And final will be played between them.Total matches = 24 + 12 + 6 + 1 = 43
Answer: c) 43
Solution:Total matches to be played = 4C2 = 6 matches.So total number of matches played in the first round = 6 * 4 = 24 matchesNow top two teams from each group progress to the next round. These 8 teams are to be divided into 2 groups.Total matches played in the second round = 6 × 2 = 12 matchesNow 4 teams progress to the next round. Total matches played in the third round = 6 * 1 = 6matchesFrom this round, 2 teams progress to the next round. And final will be played between them.Total matches = 24 + 12 + 6 + 1 = 43
There are 12 letters and exactly 12 envelopes. There is one letter to be inserted randomly into each envelope. What is the probability that exactly 1 letter is inserted in an improper envelope?a) 1b) 0c) 10!d) None of theseAnswer: b) 0Solution:This is a question of very common sense in which,12 letters are to be inserted in 12 envelopes, 1 in each, so if one letter is inserted into a wrong envelope there has to be another letter which is to be inserted into another wrong envelope. So the probability of this is 0.
Answer: b) 0
Solution:This is a question of very common sense in which,12 letters are to be inserted in 12 envelopes, 1 in each, so if one letter is inserted into a wrong envelope there has to be another letter which is to be inserted into another wrong envelope. So the probability of this is 0.
A hollow space on the earth surface is to be filled. The total cost of filling is Rs. 20000. The cost of filling per cubic-meter is Rs 225. How many times is a size of 3 cubic-meter soil required to fill the hollow space?a) 29.62b) 30.32c) 88.88d) 43.64Answer: a) 29.62Solution:The total cost of filling = 20, 000Cost of filling 1 cubic meter = Rs. 225So cubic meters to be filled = 20, 000/225 = 88.89 meter-cubeNow we need to find the three times of 88.89 to be filled = 88.89/3 = 29.63So the closest match is 29.62
Answer: a) 29.62
Solution:The total cost of filling = 20, 000Cost of filling 1 cubic meter = Rs. 225So cubic meters to be filled = 20, 000/225 = 88.89 meter-cubeNow we need to find the three times of 88.89 to be filled = 88.89/3 = 29.63So the closest match is 29.62
A 7-digit number is to be formed with all different digits. If the digits at the extreme right and extreme left are fixed to 5 and 6 respectively, find how many such numbers can be formed?a) 120b) 30240c) 6720d) None of theseAnswer: c) 6720Solution:If the digits at extreme left and right are fixed as 5 and 6, then the number of digits left = 8So the in-between 5 places can be filled in 8 * 7 * 6 * 5 * 4 ways= 6720 ways
Answer: c) 6720
Solution:If the digits at extreme left and right are fixed as 5 and 6, then the number of digits left = 8So the in-between 5 places can be filled in 8 * 7 * 6 * 5 * 4 ways= 6720 ways
There are five tires in a sedan (four road tires and one spare) which is to be used equally in a journey to travel 40, 000 km. The number of km of use of each tyre wasa) 32000b) 8000c) 4000d) 10000Answer: a) 32000Solution:The total km travelled by the sedan = 40, 000 kmSince every tire capacity’s = 40, 000/5 = 8000 km eachSo total distance covered by each tire = 8000*4 = 32000 km each will be travelled by each tire after being worn out after every 8000 km.
Answer: a) 32000
Solution:The total km travelled by the sedan = 40, 000 kmSince every tire capacity’s = 40, 000/5 = 8000 km eachSo total distance covered by each tire = 8000*4 = 32000 km each will be travelled by each tire after being worn out after every 8000 km.
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