#2
 
 
Re: Solved paper for TCS Aptitude questions
Here I am giving you solved question paper for placement aptitude examination of TCS in a file attached with it so you can download it easily.. Some questions are given below : TCS Aptitude solved paper: 1) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ? Sol) log 0.317=0.3332 and log 0.318=0.3364, then log 0.319=log0.318+(log0.318log0.317) = 0.3396 2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ? Sol) x= 2 kg Packs y= 1 kg packs x + y = 150 * * .......... Eqn 1 2x + y = 264 * .......... Eqn 2 Solve the Simultaneous equation; x = 114 so, y = 36 ANS : *Number of 2 kg Packs = 114. 3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed? 6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00 Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360). When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours). Hence, the time at the destination place is 12 noon  5:20 hours = 6: 40 AM 4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ? Sol) Since it is moving from east to west longitide we need to add both ie,40+40=80 multiply the ans by 4 =>80*4=320min convert this min to hours ie, 5hrs 33min It takes 8hrs totally . So 85hr 30 min=2hr 30min So the ans is 10am+2hr 30 min =>ans is 12:30 it will reach 5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation. Very urgent. Sol) see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple.... 6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes. (n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure 7) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week Ans:4 good, 1 fair n 2 bad days Sol) Go to river catch fish 4*9=36 7*1=7 2*5=10 36+7+10=53... take what is given 53 good days means  9 fishes so 53/9=4(remainder=17) *if u assume 5 then there is no chance for bad days. fair days means  7 fishes so remaining 17  17/7=1(remainder=10) if u assume 2 then there is no chance for bad days. bad days means 5 fishes so remaining 1010/5=2days. * Ans: 4 good, 1 fair, 2bad. ==== total 7 days. x+y+z=7 eq1 9*x+7*y+5*z=53 eq2 multiply eq 1 by 9, 9*x+9*y+9*z=35 eq3 from eq2 and eq3 2*y+4*z=10eq4 since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4 for first *y=1,z=2 then from eq1 x= 4 so 9*4+1*7+2*5=53.... satisfied now for second y=3 z=1 then from eq1 x=3 so 9*3+3*7+1*5=53 ......satisfied so finally there are two solution of this question (x,y,z)=(4,1,2) and (3,3,1)... 8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X? Sol) Let no. of fish x catches=p no. caught by y =r r=5p. r+p=42 then p=7,r=35 9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings? suppose total income is 100 so amount x is getting is 80 y is 70 z =60 total=210 but total money is 300 300210=90 so they are getting 90 rs less 90 is 30% of 300 so they r getting 30% discount 10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income? Sol) incomes:3:4 expenditures:4:5 3x4y=1/4(3x) 12x16y=3x 9x=16y y=9x/16 (3x4(9x/16))/((4x5(9x/16))) ans:12/19 11) If G(0) = 1 G(1)= 1 and G(N)=G(N1)  G(N2) then what is the value of G(6)? ans: 1 bcoz g(2)=g(1)g(0)=1+1=2 g(3)=1 g(4)=1 g(5)=2 g(6)=1 12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages? Sol) A can copy 50 pages in 10 hrs. A can copy 5 pages in 1hr.(50/10) now A & B can copy 70 pages in 10hrs. thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x> no. of pages B can copy in 10 hrs.] so, B can copy 9 pages in 1hr. therefore, to copy 26 pages B will need almost 3hrs. since in 3hrs B can copy 27 pages. 13) what's the answer for that : A, B and C are 8 bit no's. They are as follows: A > 1 1 0 0 0 1 0 1 B > 0 0 1 1 0 0 1 1 C > 0 0 1 1 1 0 1 0 (  =minus, u=union) Find ((A  C) u B) =? To find AC, We will find 2's compliment of C and them add it with A, That will give us (AC) 2's compliment of C=1's compliment of C+1 =11000101+1=11000110 AC=11000101+11000110 =10001001 Now (AC) U B is .OR. logic operation on (AC) and B 10001001 .OR . 00110011 The answer is = 10111011, Whose decimal equivalent is 187. 14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........ Sol) it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000 so starting add of 2x3 will be 3012. 15) In a twodimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5). Sol) initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 1) * 4 = 212 ( 1 is to deduct the 1 element ) 3000 + 212 = 3212 16) Which of the following is power of 3 a) 2345 b) 9875 c) 6504 d) 9833 17) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ? Sol) M=sqrt(100N) N is increased by 1% therefore new value of N=N + (N/100) * * * * * * * *=101N/100 M=sqrt(100 * (101N/100) ) Hence, we get M=sqrt(101 * *N) 18) 1)SCOOTER  AUTOMOBILE A. PART OF 2.OXYGEN WATER  B. A Type of 3.SHOP STAFF FITTERS C. NOT A TYPE OF 4. BUG REPTILE D. A SUPERSET OF 1)B * *2)A * * * 3)D * * 4)C 19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand this is the step by step solution: a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min. after this speed of return increase by 50% so 50%of 18 mph=9mph Total speed of returnig=18+9=27 Then in return it take 27/27=1 hour then total time in joureny=1+1:30+00:30 =3 hour so it will *come at 8+3 hour=11 a.m. So Ans==11 a.m 20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ? Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338. 21) The temperature at Mumbai is given by the function: t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM? 22) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. *Sum of the low temperature and highest temp. is 100 degrees. Then what is the low temp? Sol) Let highest temp be x so low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2 total temp=x+x/6+x/2=100 therefore, x=60 Lowest temp is 40 23) In Madras, temperature at noon varies according to t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. Ans. At 9pm 7.5 more Sol) In equestion *first put t=9, we will get 34.5...........................(1) now put t=4, we will get 27..............................(2) so ans=34.527 * * * * *=7.5 24) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product? a) 780 * * b) 1040 * c) 1590 * d) 1720 Sol) x*53x*35=540=> x=30 therefore, 53*30=1590 Ans 25) How many positive integer solutions does the equation 2x+3y = 100 have? a) 50 * b) 33 * c) 16 * *d) 35 Sol) There is a simple way to answer this kind of Q's given 2x+3y=100, take l.c.m of 'x' coeff and 'y' coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula constant / (l.cm of x coeff and y coeff) 26) The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders. a) 65 * *b) 55 * c) 50 * d) 45 Sol) Let a = fixed cost and k = variable cost and n = number of boarders total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k solving above 2 eqns, 30001750 = 25k i.e. 1250 = 25k i.e. k = 50 therefore, substituting this value of k in either of above 2 eqns we get a = 500 (a = 300050*50 = 500 or a = 1750  25*50 = 500) so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500 so cost per head = 5500/100 = 55 27) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens? a) 37.5% * * b) 62.5% * c) 50% * d) None of these Sol) Let, 5 pens + 7 pencils + 4 *erasers = x *rupees so 10 pens *+ 14 pencils + 8 erasers = 2*x rupees also mentioned, 6 pens + 14 pencils + 8 erarsers = 1.5*x rupees so (106) = 4 pens = (21.5)x rupees so 4 pens = 0.5x rupees => 8 pens = x rupees so 5 pens = 5x/8 rupees *= 5/8 of total (note x rupees is total amt paid byamal) i.e 5/8 = 500/8% = 62.5% is the answer 28) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race? Sol) x + x+6 = rs 68 2x + 6 = 68 2x = 686 2x = 62 x=31 x is the amt lost in I race x+ 6 = 31+6=37 is lost in second race then my friend lost 37 + 4 = 41 Rs 29) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms. i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10 iii) Same question with n=9 to me the chances are when n=3 (i) nC1= 3C1 =3 for 10 boxes .. 10*3=30 (ii) 10C1=10 for 10 boxes ....10*10=100 (iii)9C1=9 for 10 boxes .....10*9=90 30) (11/6) (11/7).... (1 (1/ (n+4))) (1(1/ (n+5))) = ? leaving the first numerater and last denominater, all the numerater and denominater will cancelled out one another. Ans. 5/(n+5) 31) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part? Sol) the clock normally has 12 hr three parts x,y,z x+y+z=12 x=y+z 2x=12 x=6 so the largest part is 6 hrs 32) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels Sol) 4/5 full tank= 12 mile 1 full tank= 12/(4/5) 1/3 full tank= 12/(4/5)*(1/3)= 5 miles 33) wind blows 160 miles in 330min.for 80 miles how much time required Sol) 160 miles= 330 min 1 mile = 330/160 80 miles=(330*80)/160=165 min. 34) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was travelling at a speed of 35 mph. find the speed limit Sol) (x+10)=(x+35)/2 solving the eqn we get x=15 35) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3. what is the answer he actually has to get. Sol) Assume 1 1* 3 = 3 1*1/3=1/3 so he has to got 1/3 this is the exact answer 36) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot. 37) Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp. ans is 40 c. Sol) let x be the highest temp. then, x+x/2+x/6=100. therefore, x=60 which is the highest temp and 100x=40 which is the lowest temp. 38) car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up. Sol) let feul consumed in coming up is x. thus equation is: x+1.25x=4.5ans:2gallons 39) A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person Sol) Two people work together in 24 mins. So, their one day work is (1/A)+(1+B)=(1/24) One man can complete the work in 40mins one man's one day work (1/B)= (1/40) Now, (1/A)=(1/24)(1/40) (1/A)=(1/60) So, A can complete the work in 60 mins. 40) In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability? That a randomly chosen employee is a male or female? Sol) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is 0.264 41) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin Sol) the minimum number of wieghtings needed is just 5.as shown below * * * * * *(1) * *80>3030 * * * * * *(2) * * *1515 * * * * * *(3) * * * 77 * * * * * * *(4) * * *33 * * * * * * * * * * * * *(5) * * *11 42) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples? 2x+3y+4z=15 3x+2y+z=10 adding 5x+5y+5z=25 x+y+z=5 that is for 1 orange, 1 bannana and 1 apple requires 5Rs. so for 3 orange, 3 bannana and 3 apple requires 15Rs. i.e. 3x+3y+3z=15 43) In 8*8 chess board what is the total number of squares refers Sol) odele discovered that there are 204 squares on the board We found that you would add the different squares  1 + 4 + 9 + 16+ 25 + 36 + 49 + 64. Also in 3*3 tic tac toe board what is the total no of squares Ans 14 ie * 9+4(bigger ones)+1 (biggest one) If you ger 100*100 board just use the formula the formula for the sum of the first n perfect squares is * * * * * * * * * * * *n x (n + 1) x (2n + 1) * * * * * * * * * * * *______________________ * * * * * * * * * * * * * * * * *6 if in this formula if you put n=8 you get your answer 204 44) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish. Sol) Faster one can do 1/2 of work in one hourslower one can do 1/3 of work in one hourboth they do (1/2+1/3=5/6) th work in one hour.so work will b finished in 6/5=1.2 hour i e 1 hour 12 min. 45) If Rs20/ is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive? Here is the answer Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42) ==> (42 * 20 )/70 ==> 12 Rs 46) An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she cannt type.? 47) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held? the answer is 138 matches Sol) since one player gets a bye in each round,he will reach the finals of the tournament without playing a match. therefore 137 matches should be played to detemine the second finalist from the remaining 138 players(excluding the 1st player) therefore to determine the winner 138 matches shd be played. 48) One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate? Sol) Vol. of rect. plate= 8*11*2=176 area of rod=(22/7)*(8/2)*(8/2)=(352/7) vol. of rod=area*length=vol. of plate so length of rod= vol of plate/area=176/(352/7)=3.5 49) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank? 3 ft 7.5 inches 50) A man has to get airmail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horsecart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive? ans:10min:::assume *he started at 1:00,so at 1:30 he met cart. He returned home at 2:00.so it took him 1 hour for the total jorney.by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.so the actual time of plane is at 1:40.The cart travelled a time of 10 min before it met him.so the plane is 10 min early. 51) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked? 52) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years. Sol) THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE *GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4 ans: 1.5 mt 4 (x+(3/5)x)=88x/5=2x=5/4 after 2 years x=(3/5)*(5/4)*2 =1.5 53) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code? sol) Assume the digit xx xx xx (six digits) First Two digit * * * xx * 3=111 * * * * * * * * * * *xx=111/3=37 ( first two digits of 1 is not divisible by 3 * so we can use 111) Second Two digit xx*6=222 * * * * * * * * xx=222/6=37 ( first two digits of 2 is not divisible by 6 * so we can use 222) Thrid Two digit * *xx*9=333 * * * * * * * * * xx=333/9=37 ( first two digits of 3 is not divisible by 9 * so we can use 333) 54) There are 4 balls and 4 boxes of colours yellow, pink, red and green. Red ball is in a box whose colour is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball? ans is green... Sol) Yellow box can have either of pink/yellow balls. if we put a yellow ball in "yellow" box then it wud imply that "yellow" is also the colour of the box which has the red ball(becoz acordin 2 d question,d box of the red ball n the ball in the yellow box have same colour) thus this possibility is ruled out... therefore the ball in yellow box must be pink,hence the colour of box containin red ball is also pink.... =>the box colour left out is "green",,,which is alloted to the only box left,,,the one which has yellow ball.. 55) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color. Ans:he should pick 6 ball totally. Sol) Suppose he picks 5 balls of all different colours then *when he picks up the sixth one, it must match *any on of the previously drawn ball colour. thus he must pick 6 balls 56) What is the number of zeros at the end of the product of the numbers from 1 to 100 Sol) For every 5 in unit palce one zero is added so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place. Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 *11 zeros are added for 25,50,75 3 extra zeros are added so total no of zeros are 10+11+3=24 57) 10 Digit number has its first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4's..till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?.(6marks) ans:2100010006 2shows that two 1's in the ans 1shows that one 2 in ans 0shows no 3 in the ans 0shows no 4 in the ans 0shows no 5 in the ans 1shows one 6 in the ans 0shows no 7 in the ans 0shows no 8 in the ans 0shows no 9 in the ans 6shows six 0's in the ans 58) There are two numbers in the ratio 8:9. if the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9. Find the larger number? sol) 8x:9x initialy 8x+ 12 : 9x  19 = 5x:9x 8x+12 = 5x > x = 4 9x = 36 not sure about the answer .. 59) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time AB = 3 Bc = 5 ac = 8 so sum of diff = 8+3+5 = 16 kgs 60) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark? ans is 11 coz each had atleast 1 shot done so 10 + 1 = 11 n 9 + 2 = 11 so d ans is 11 61) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time? sol) A shows 11:57, B shows 12:03, C shows 12:02, and D shows 11:06 therefore, max time is for B 62) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall. Sol) The falling height is proportional to the squere of the time. Now, the falling height is 64cm at 2sec so, the proportional constant is=64/(2*2)=16; so, at 6sec the object fall maximum (16*6*6)cm=576cm; Now, the object may be situated at any where. if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec. 63) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings Ans) first 50 ings. run= 50*50=2500 51st ings. avg 51. so total run =51*51=2601. so run scored in that ings=26012500=101 runs. 64) Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over? a) 3rd * b) 6th * c) 9th * d) 7th Ans is d) 7th day Sol) In d 1st day Anand does 1/7th of total work similarly, Bithu does 1/8th work in d 2nd day hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168 remaining work = (16873)/168 = 95/168 again after 6 days of work, remaining work is = (9573)/168 = 22/168 and hence Anand completes the work on 7th day.(hope u understood.) 65) A man, a women and a child can do a piece of work in 6 days,man can do it in 14 days, women can do it 16 days, and in how many days child can do the same work? The child does it in 24 days 66) A: 1 1 0 1 1 0 1 1 B: 0 1 1 1 1 0 1 0 C: 0 1 1 0 1 1 0 1 Find ( (AB) u C )==? Hint : 109 AB is {A}  {A n B} A: 1 1 0 1 1 0 1 1 B: 0 1 1 1 1 0 1 0 by binary sub. ab = 01100001 (10=1, 11=0,00=0, n for the 1st 3 digits 110011=011) now (ab)uc= 01100001 * * * * * or * * *01101101 gives 1101101... convert to decimal equals 109
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#3
 
 
Re: Solved paper for TCS Aptitude questions
I am doing preparation of Placement Test of TCS. I need some solved question papers of Aptitude Test. So is there any one who will provide source from where I can download solved question papers of Aptitude for TCS Placement?

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Re: Solved paper for TCS Aptitude questions
Here you want solved question papers of Aptitude for TCS Placement Test, so I am giving following papers: TCS Sample Solved Paper Aptitude 1 1. Ray writes a two digit number. He sees that the number exceeds 4 times the sum of its digits by 3. If the number is increased by 18, the result is the same as the number formed by reversing the digits. Find the number. a) 35 b) 42 c) 49 d) 57 Solution: Let the two digit number be xy. 4(x + y) +3 = 10x + y .......(1) 10x + y + 18 = 10 y + x ....(2) Solving 1st equation we get 2x  y = 1 .....(3) Solving 2nd equation we get y  x = 2 .....(4) Solving 3 and 4, we get x = 3 and y = 5 2. a, b, c are non negitive integers such that 28a+30b+31c = 365. a + b + c = ? a) Greater than 14 b) less than or equal to 11 c) 13 d) 12 In a calender, Number of months having 28 days = 1 Number of months having 30 days = 4 Number of months having 31 days = 7 28 x 1 + 30 x 4 + 31 x 7 = 365 Here, a = 1, b = 4, c = 7. a+b+c = 12 3. George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours. George, paul and hari start the same work at 9 am, while george stops at 11 am, the remaining two complete the work. What time will the work complete? a) 11.30 am b) 12 noon c) 12.30 pm d) 1 pm Let the total work = 120 units. As George completes this entire work in 8 hours, his capacity is 15 units /hour Similarly, the capacity of paul is 12 units / hour the capacity of Hari is 10 units / hour All 3 started at 9 am and worked upto 11 am. So total work done upto 11 am = 2 x (15 + 12 + 10) = 74 Remaining work = 120  74 = 46 Now this work is to be done by paul and hari. 46 / (12 + 10) = 2 hours (approx) So work gets completed at 1 pm 4. If x^y denotes x raised to the power y, Find last two digits of (1141^3843) + (1961^4181) a) 02 b) 82 c) 42 d) 22 Remember 1 raised to any power will give 1 as unit digit. To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power. So the Last two digits of the given expression = 21 + 61 = 82 5. J can dig a well in 16 days. P can dig a well in 24 days. J, P, H dig in 8 days. H alone can dig the well in How many days? a) 32 b) 48 c) 96 d) 24 Assume the total work = 48 units. Capacity fo J = 48 / 16 = 3 units / day Capacity of P = 48 / 24 = 2 units / day Capacity of J, P, H = 48 / 8 = 6 units / day From the above capacity of H = 6  2  3 = 1 So H takes 48 / 1 days = 48 days to dig the well 6. If a lemon and apple together costs Rs.12, tomato and a lemon cost Rs.4 and an apple costs Rs.8 more than a lemon. What is the cost of lemon? L + A = 12 ...(1) T + L = 4 .....(2) L + 8 = A Taking 1 and 3, we get A = 10 and L = 2 7. 3 mangoes and 4 apples costs Rs.85. 5 apples and 6 peaches costs 122. 6 mangoes and 2 peaches costs Rs.144. What is the combined price of 1 apple, 1 peach, and 1 mango. a) 37 b) 39 c) 35 d) 36 Sol: Note: It is 114 not 144. 3m + 4a = 85 ..(1) 5a + 6p = 122 ..(2) 6m + 2p = 114 ..(3) (1) x 2 => 6m + 8a = 170 (3) => 6m + 2p = 114 Solving we get 8a  2p = 56 ..(4) (2) => 5a + 6p = 122 3 x (4) = 24a  6p = 168 Solving we get a = 10, p = 12, m = 15 So a + p + m = 37 8. An organisation has 3 committees, only 2 persons are members of all 3 committee but every pair of committee has 3 members in common. what is the least possible number of members on any one committee? a) 4 b) 5 c) 6 d) 1 Sol: Total 4 members minimum required to serve only on one committee. 9. There are 5 sweets  Jammun, kaju, Peda, Ladu, Jilebi which can be consumed in 5 consecutive days. Monday to Friday. A person eats one sweet a day, based on the following constraints. (i) Ladu not eaten on monday (ii) If Jamun is eaten on Monday, Ladu should be eaten on friday. (iii) Peda is eaten the day following the day of eating Jilebi (iv) If Ladu eaten on tuesday, kaju should be eaten on monday based on above, peda can be eaten on any day except a) tuesday b) monday c) wednesday d) friday From the (iii) clue, peda must be eaten after jilebi. so Peda should not be eaten on monday. 10. If YWVSQ is 25  23  21  19  17, Then MKIGF a) 13  11  8  7  6 b) 1  2357 c) 9  8  7  6  5 d) 7  8  5  3 MKIGF = 13  11  9  7  6 Note: this is a dummy question. Dont answer these questions 11. Addition of 641 + 852 + 973 = 2456 is incorrect. What is the largest digit that can be changed to make the addition correct? a) 5 b) 6 c) 4 d) 7 Sol: 641 852 963  2466 Largest among tens place is 7, so 7 should be replaced by 6 to get 2456 TCS Sample Solved Paper Aptitude 2 1. The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked x? a. c b. a c. d d. b Ans: C Explanation: If you fold the above picture at the dotted lines, X and C are opposite to each other. 2. In how many ways a team of 11 must be selected from 5 men and 11 women such that the team must comprise of not more than 3 men? a. 1565 b. 1243 c. 2256 d. 2456 Ans: C 3. Given that 0 < a < b < c < d, which of the following the largest ? a.(c+d) / (a+b) b.(a+d) / (b+c) c.(b+c) / (a+d) d.(b+d) / (a+c) Sol: A Explanation: Take a = 1, b = 2, c = 3, d = 4. option A is clearly true. 4. Eesha bought 18 sharpeners for Rs.100. She paid 1 rupee more for each white sharpener than for each brown sharpener. What is the price of a white sharpener and how many white sharpener did she buy ? a. Rs.5, 10 b. Rs.6, 10 c. Rs.5, 8 d. Rs.6, 8 Sol: B Explanation: Just check the options. If she bought 10 white sharpeners at Rs.6 per piece, She has spent Rs.60 already. And with the remaining Rs.40, she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener. 5. The fourteen digits of a credit card are to be written in the boxes shown above. If the sum of every three consecutive digits is 18, then the value of x is : a. 3 b. cannot be determined from the given information. c. 2 d. 1 Sol : A Explanation: Let us assume right most two squares are a , b Then Sum of all the squares = 18 x 4 + a + b .......... (1) Also Sum of the squares before 7 = 18 Sum of the squares between 7, x = 18 and sum of the squares between x , 8 = 18 So Sum of the 14 squares = 18 + 7 + 18 + x + 18 + 8 + a + b (2) Equating 1 and 2 we get x = 3 6. Four people each roll a four die once. Find the probability that at least two people will roll the same number ? a. 5/18 b. 13/18 c. None of the given choices d. 1295/1296 Sol: B Explanation: The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 x 5 x 4 x 3 Now total possibilities of rolling a dice = 647. Jake can dig a well in 16 days. Paul can dig the same well in 24 days. Jake, Paul and Hari together dig the well in 8 days. Hari alone can dig the well in a. 96 days b. 48 days c. 32 days d. 24 days Sol: Explanation: Simple one. Let the total work to be done is 48 meters. Now Jake can dig 3 mts, Paul can dig 2 mts a day. Now all of them combined dug in 8 days so per day they dug 48/8 = 6 mts. So Of these 8 mts, Hari capacity is 1 mt. So he takes 48 /1 = 48 days to complete the digging job. 8. Eesha bought 18 sharpeners for Rs.100. She paid 1 rupee more for each white sharpener than for each brown sharpener. What is the price of a white sharpener and how many white sharpener did she buy ? a. Rs.5, 10 b. Rs.6, 10 c. Rs.5, 8 d. Rs.6, 8 Ans: Explanation: This question can be solved easily by going through options. A. White sharpener total cost: Rs.5 x 10 = Rs.50. Brown sharpeners cost = Rs.4 x 8 = 32. Total cost is only Rs.82. Wrong option. B. White sharpener total cost: Rs.6 x 10 = Rs.60. Brown sharpeners cost = Rs.5 x 8 = 40. Total cost is Rs.100. Correct option. 9. The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is 109. If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order. Find the number. a. 773 b. 683 c. 944 d. 863 Ans: D Explanation: Check options. Sum of the squares should be equal to 109. Only Options B and D satisfying. When we subtract 495, only 863 becomes 368. 10. Mark told John "If you give me half your money I will have Rs.75. John said, "if you give me one third of your money, I will have Rs.75/ How much money did John have ? a. 45 b. 60 c. 48 d. 37.5 Ans: B Explanation: Let the money with Mark and John are M and J respectively. Now M + J/2 = 75 M/3 + J = 75 Solving we get M = 45, and J = 60. 11. Eesha has a wheat business. She purchases wheat from a local wholesaler of a particular cost per pound. The price of the wheat of her stores is $3 per kg. Her faulty spring balance reads 0.9 kg for a KG. Also in the festival season, she gives a 10% discount on the wheat. She found that she made neither a profit nor a loss in the festival season. At what price did Eesha purchase the wheat from the wholesaler ? a. 3 b. 2.5 c. 2.43 d. 2.7 Ans: C Explanation: Faulty spring balance reads 0.9 kg for a kg" means that she sells 1 kg for the price of 0.9 kgs, so she looses 10% of the price because of the faulty spring balance. She looses another 10% because of the discount. So, she actually sells 1 kg for $3×0.9×0.9=$2.43 and since at that price she made neither a profit nor a loss, then Eesha purchase the wheat from the wholesaler for $2.43. 12. Raj goes to market to buy oranges. If he can bargain and reduce the price per orange by Rs.2, he can buy 30 oranges instead of 20 oranges with the money he has. How much money does he have ? a. Rs.100 b. Rs.50 c. Rs.150 d. Rs.120 Ans: D Explanation: Let the money with Raj is M. So M20−M30=2. Check options. Option D satisfies. 13. A city in the US has a basketball league with three basketball teams, the Aziecs, the Braves and the Celtics. A sports writer notices that the tallest player of the Aziecs is shorter than the shortest player of the Braves. The shortest of the Celtics is shorter than the shortest of the Aziecs, while the tallest of the Braves is shorter than the tallest of the Celtics. The tallest of the Braves is taller than the tallest of the Aziecs. Which of the following can be judged with certainty ? X) Paul, a Brave is taller than David, an Aziec Y) David, a Celtic, is shorter than Edward, an Aziec a. Both X and Y b. X only c. Y only d. Neither X nor Y Ans: B Sol: We solve this problem by taking numbers. Let the shortest of Braves is 4 feet. Then tallest of Aziecs is less than 4. So let it be 3 feet. A > 2  3 B > 4  6 C > 1  7 From the above we can safely conclude X is correct. but Y cannot be determined. 14. There are 3 classes having 20, 24 and 30 students respectively having average marks in an examination as 20,25 and 30 respectively. The three classes are represented by A, B and C and you have the following information about the three classes. a. In class A highest score is 22 and lowest score is 18 b. In class B highest score is 31 and lowest score is 23 c. In class C highest score is 33 and lowest score is 26. If five students are transferred from A to B, what can be said about the average score of A; and what will happen to the average score of C in a transfer of 5 students from B to C ? a. definite decrease in both cases b. can't be determined in both cases c. definite increase in both cases d. will remain constant in both cases Ans: B Explanation: Class A average is 20. And their range is 18 to 22 Class B average is 25. And their range is 23 to 31 Class A average is 30. And their range is 26 to 33 If 5 students transferred from A to B, A's average cannot be determined but B's average comes down as the highest score of A is less than lowest score of B. If 5 students transferred from B to C, C's average cannot be determined the B's range fo marks and C's range of marks are overlapping.
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