Designing IA for AI - Information Architecture Conference 2024
Aptitude test papers
1. APTITUDE TEST PAPERS - FREQUENTLY ASKED QUESTIONS by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india FOR – CSE & PGPSE STUDENTS (CSE & PGPSE are free online programmes open for all, free for all) mobile : 91+9414430763
2. My words..... My purpose here is to give a few questions, which are often asked in aptitude tests and competitive examinations. Please prepare well for your examinations. Please pass this presentation to all those who might need it. Let us spread knowledge as widely as possible. I welcome your suggestions. I also request you to help me in spreading social entrepreneurship across the globe – for which I need support of you people – not of any VIP. With your help, I can spread the ideas – for which we stand....
3. Data sufficiency question ... Are two triangles congruent? a. Both triangles are right angled. b. Both are of same perimeter. Answer : this information is indequate and we cant get the answer on the basis of such limited information
4. Data sufficiency question ? In what time the cylinder completes one rotation? a. Radius of cylinder is given. b. Speed (angular) is given. Answer : BOTH THE STATEMENTS ARE REQUIRED (we get anwer using both)
5. Data sufficiency question ... PQR is an isosceles triangle. What's the length of PQ? a. The length of QR is 7. b. The angle of PQR is 90 degree. ANSWER : BOTH THE STATEMENTS ARE REQUIRED
6. What's the ratio of length of the line segment PQ to the length of arc PQ, when PQ forms the arc of a circle? a. Area of circle is168. b. The segment PQ forms the diameter of circle. Answer : both the statements are required
7. In a 1 km race,A beats B by 28 m or 7sec.Find A's time over the course? Speed of B = 28/7 = 4 m per second total time by B = 1000/4 = 250 seconds total time by A = 250-7 = 243 seconds answer
8. A runs 1 3/4 times as fast as B.If A gives B a start of 84 m,how far must be winning post be so that A and B might reach it at the same time ? Options : 220, 196, 190, 144 solve from options : start with 196 196 – 84 = 112 196/112 = 1.75 this is given in question : 1 3/4 ratio 1 is equal to 1.75 answer
9. A can run 1km in 3 min ,10sec and B can cover same distance in 3 min 20 sec. By what distance A beat B? 1 km = 1000 meters distance covered by B when A finishes : 3.16 / 3.33 * 1000 = 948.9 or 949 meters thus A defeats B by 51 meters. Answer
10. In a 100m race,A runs at 8km per hour.If A gives b a start of 4 m and still beats him by 15 sec,what is the speed of B? Speed of A = 8 * 5/18 = 2.22 meters per second time taken by A = 100/2.22 =45.05 time taken by B = 45.05 +15 = 60.05 seconds distance covered by B = 100-4 = 96 meters speed of B = 96/60.05 =1.6 meters per second speed of B in km / h = 1.6 * 18/5 = 5.76 km per hour answer
11. A and B take part in 100m race .A runs at 5km per hour. A gives B a start of 8 m and still beats him by 8 sec. What is the speed of B? Speed of A = 5 * 5/18 = 1.39 time taken by A = 100/1.39 =71.94 seconds time taken by B 71.94 + 8 = 79.94 sec. Distance covered by B = 100 -8 = 92 meters speed of B -92/79.94 =1.15 speed of B 1.15 * 18/5 = 4.14 km per hour. Answer
12. If length and breadth of a rectangle are increased by 20% each, what is the change in area ? Let us assume L = 20 and B = 10 area = 200 increase : L =20+4 = 24 and B = 12 new area = 288 increase in area is 44% shortcut = X +X +(X*X%) =20+20+4 = 44 %
13. If length and breadth of a rectangle are increased by 30% each, what is the change in area ? Let us assume L = 20 and B = 10 area = 200 increase : L =20+6 = 26 and B = 13 new area = 338 increase in area is 69% shortcut = X +X +(X*X%) =30+30+9 = 69 %
14. If out of length and breadth of a rectangle length is increased by 30% and the other is decreased by 30%, what is the change in area ? Let us assume L = 20 and B = 10 area = 200 increase : L =20+6 = 26 and B = 7 new area = 182 decrease in area is 9% shortcut = decrease of (X*X%) =30*30% = 9% decrease note : it doesnt matter whether length is increased or width is increased, the answer is same in such questions.
15. The ratio of the flow of water in pipes varies inversely as the square of the radius of the pipe, what is the ratio of the rates of flow in two pipes of diameter 2 &4 ? (1/2)^2 and (1/4)^2 ¼ : 1/16 or 4:1 or take radius : 1^2 : (1/2)^2 =1:1/4 or 4:1 answer
16. 4 friends A,B,C,D are studing together in class 12. A and B are good in Hindi but poor in English. A and C are good in Sanskrit, but poor in geography. D and B are good in Maths as well as in Sanskrit. Who is not good in maths but good in Hindi?: A g In H & S p in e, g B G in H & M & S p in e C g in S p in g D g in M,S ANSWER = A
17. A man buys 12 lts of liquid which contain 20% of the liquid and the rest is water. He then mixes it with 10lts of another mixture with 30% of liquid. What is the % of water in the new mixture? Total quantity of mixture = 12+10 =22 total water : 9.6+7 = 16.6 =16.6/22 =75% answer
18. SERIES : b3c, c4e,e6h,h9l,?? L13q answer in the first case, one is added to b, in 2 nd case we add 2 to e , in 3 rd we add 3 to e ... so we will add 5 to l numbers also increase : 3 to 4 (addition of 1), 4 to 6 (addition of 2), 6 to 9 (addition of 3) and 9 to 13 (addition of 4) answer
19. In a class of 50 students 23 speak English 15 speak Hindi and 18 speak Punjabi 3 speak only english and hindi, 6 speak english and punjabi 9 can speak only english, how many speak all the 3 languages? 23 - ( 9+6+3) = 5 answer
20. In which of the system, decimal number 194 is equal to 1234? Options : 3,5,8,16 try with options : try for 5 1*(5^3)+(2*5^2)+(3*5^1)+(4*5^0) =125+50+15+4 =194 so answer is 5 answer
21. If TAFJHH is coded as RBEKGI then RBDJK can be coded as --------- ANS: PCCKJ
22. Find the result of the following expression if, M denotes modulus operation, R denotes round-off, T denotes truncation: M(373,5)+R(3.4)+T(7.7)+R(5.8) 373 / 5 = remainder is 3 3.4 rounded off to 3 7.7 to 7 5.8 to 6 total : = 19 ANS:19
23. A certain number of men can finish a piece of work in 10 days. If however there were 10 men less it will take 10days more for the work to be finished. How many men were there originally? Number of men =x total work =10X 10X = (X-10)(20) 10X=20X-200 X=20 answer
24. In simple interest what sum amount of Rs. 1120/- in 4 years and Rs.1200/- in 5 years? Interest = 1200 – 1120 = 80 the total interest for 5 years = 400 principal = 1200-400 = 800 answer
25. There are total 15 people. 7 speaks french and 8 speaks spanish. 3 do not speak any language. Which part of total people speaks both languages. 7+8 = 15 BUT IT SHOULD BE 12 (because 3 dont speak any language). So there are 3 persons who speak both the languages. Answer
26. A jogger wants to save 1/4 th of his jogging time. He should increase his speed by how much Let us assume that the jogger takes 4 ours for 12 KM. Now he will save 1/4 th so he will take 3 hours for 12 KM. His earlier speed was 12/4 = 3, but his new speed is 12/3 = 4. he increased his speed from 3 to 4 or 33% increase in his speed. Answer (you may assume other figures also but the answer will be the same).
27. A is an integer. Dividing 89 & 125 by A gives remainders 4 & 6 respectively. Find a ? Find HCF of (89-4) and (125-6) HCF of 85 and 119 is 17 answer = 17 answer
28. In a office work is distributed between p persons. If 1/8 members are absent then work increased for each person by what %. Suppose there are 560 units of work among 8 persons. Now there are only 7 persons so each person is doing 80 units of work (instead of 70 units of work). The work increased by : 10 on 70 10/70*100 = 14.29% answer
29. 120, 315, 300, 345, ? We can write these numbers as multiples of 15 : 8, 21,20,23 the next digit should be :
30.
31. RULE : a^0 = 1 IF ZERO IS PUT AS POWER OF ANY NUMBER, IT WILL BECOME 1. THUS IF YOU PUT POWER OF ZERO ON 100000, IT WILL BE EQUAL TO 1.
32. RULE : (a/b)^n = a^n / b^n IF YOU TAKE UP POWER OVER A FRACTION, YOU CAN PUT IT SEPARATELY ALSO. EXAMPLE : (2/3) ^3 = 8/27 ANSWER
33. (27)^2/3 =? 1. FIRST WE PUT 27 AS 3^3 BECAUSE WE KNOW 3*3*3 = 27 = (3^3)2/3 WE CAN CANCEL POWR OF 3 3^2 = 9. ANSWER
34. (1024)^-4/5 =? 1. FIRST PUT 1024 AS 2^X, WE KNOW THAT 2*2*2*2*2..... (TILL 10) = 1024 SO WE CAN WRITE : 1024 = 2^10 = 4^5 (4^5)-4/5 (NOW CANCEL 5) = (4)-4= 1/(4)4 (ANY POWER IN NEGATIVE WILL PUT THE NUMERATOR INTO DENOMINATOR) = 1/256.
35. If 2^(x-1)+ 2^(x+1) = 256 then find the value of x .(X-1) * (X+1) = (X^2 – 1) So we can write it as : =2^(x^2-1) = 256 now factorise 256 = 2*2*2*2*2*2*2*2=2^8 2^(x^2 -1) = 2^8 or X^2 = 9 X = 3 answer
36. Simplify : 125^(2/3) * 25^(1/2)*5*5^(1/2) 125^(2/3) = (5^3)^(2/3) 3 gets cancelled so : 5^2 = 25 =5*5 we can write it as : 5*5*5*5*5^(1/2) =5^4 * 5^ (½) (add the two powers, when there is multiplication of their base) =5^(9/2) answer
37. What is the compounded ratio of : 4:9, duplicate ratio of 3:4, triplicate ratio of 2:3 and 9:7? Duplicate ratio of 3:4 =9:16 (square of 3 and 4 respectively) triplicate ratio of 2:3 = 8:27 (take cube of 2 & 3 respectively). Now let us compound (multiply all the X and Y individuallY ) them : X =4*9*8*9 = 2592 Y = 9*16*27*7=27216 their ratio is 1: 10.5 or 2:21 answer
38. A train 140 m long running at 72 kmph.In how much time will it pass a platform 260m long. Total distance to cover = 140+260 = 400 time = distance / speed speed in meters per second = 72 *5/18 = 20 meters per second time = 400/20 = 20 seconds answer
39. A man is standing on a railway bridge which is 180 m.He finds that a train crosses the bridge in 20 seconds but himself in 8 sec. Find the length of the train and its sppeed Let us assume the length of train = X the speed of train = x meters/8 180 meters are covered by train in (20-8) seconds 180/12 = 15 meters per second. Thus the length of train is : 15*8 = 120 meters. Answer
40. A train 150m long is running with a speed of 54 Km per hour. In what time will it pass a man who is running at a speed of 9 km ph in the same direction in which the train is going Speed of train = 54 * 5/18 = 15 meters per second speed of mann = 9 * 5/18 = 2.5 meters per second distance to cover = 150 time = 150/(15-2.5) =12 seconds answer
41. A train 220m long is running with a speed of 59 k m ph ..In what time will it pass a man who is running at 4 kmph in the direction opposite to that in which train is going. Distance to cover = 220 meters speed = 59+4 = 63 km per hour. In meters per second = 63*5/18 = 17.5 meters per second. Time required : 220/17.5 =12.57 seconds. Answer
42. Two trains 137m and 163m in length are running towards each other on parallel lines,one at the rate of 42kmph & another at 48 mph.In wht time will they be clear of each other from the moment they meet. Distance to cover 137+163 = 300 meters speed = 42+48 = 90 km per hour speed in meters per second = 90 * 5/18 = 25 meters per second time required = 300/25 = 12 seconds answer
43. A train running at 54 kmph takes 20 sec to pass a platform. Next it takes 12 sec to pass a man walking at 6kmph in the same direction in which the train is going.Find length of the train and length of platform Solution : Train v/s man speed = 54 -6 = 48 km per hour speed in m/s =48 * 5/18 = 13.33 m / s length of train = 12*13.33 = 159.6 meters speed for platform =54*5/18 = 15 m / s length of platform+ train = 20*15 = 300 length of platform = 300 – 159 = 140 meters approx.
44. A man sitting in a train which is travelling at 50mph observes that a goods train travelling in opposite direction takes 9 sec to pass him .If the goods train is 150m long find its speed Solution : - Distance travelled = 150 speed =150/9 = 16.66 meters per second or 16.66 * 18/5 = 60 km per hour approx. Thus the speed of goods train is 60-50 = 10 km per hour. Answer
45. Two trains are moving in the same direction at 65kmph and 47kmph. The faster train crosses a man in slower train in18sec.the length of the faster train is Solution : = When the trains are going in same direction, we take difference of their speed. 65-47 =18 km per hour or 5 meters per second distance travelled =(time 18 seconds * speed 5 meters per second) = 18 * 5 = 90 meters. Thus the length of faster train is 90 meters. Answer
46. A train overtakes two persons who are walking in the same direction in which the train is going at the rate of 2kmph and 4kmph and passes them completely in 9 sec and 10 sec respectively. The length of train is Solution : - let us assume the speed of train to be X. (X-2) * 9/3600 = (X-4) *10/3600 9X – 18 = 10X – 40 X=22 km per hour. thus distance travelled = (22-4) * 5/18 = 5 m/s time=10 seconds so length of train = 5*10 = 50 meters. Answer
47. Two stations A & B are 110 km apart on a straight line. One train starts from A at 7am and travels towards B at 20kmph. Another train starts from B at 8am an travels toward A at a speed of 25kmph.At what time will they meet From 7 am to 8 am only A is travelling. It would travel 20 km. Now 90 km is to be covered. 90 / (20+25) =2 hours so at 10 am they will meet.
48. A train travelling at 48kmph completely crosses another train having half its length an travelling inopposite direction at 42kmph in12 sec.It also passes a railway platform in 45sec.the length of platform is Distance by 2 trains = (48+42) = 90 or 25 m/s 25 * 12=300 meters. So the length of train is 200 meters. Platform : 48 * 5/18 =13.33 m/ s 45*13.33 = 600 so length of platform = 600-200=400 meters. Answer
49. Find the time taken by a train 180m long,running at 72kmph in crossing an electric pole Speed of train = 72 * 5/18 = 20 m / s time required = 180/20 = 6 seconds. Answer
50. Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring. Solution: Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 352/7 radius = 8 Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 528/7 radius = 12 thus width of ring = 12-8 = 4 answer
51. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is Solution : area of one circle = (22/7) * 7 * 7 =154 square of on one circle = 14*14 = 196 difference of area : 42 one side 42/4 =10.5 we have 4 cirlces, each has 10.5 cm of space enclosed, so total space enclosed is 42 sq. cm. Answer
52. A semicircular shaped window has diameter of 63cm. Its perimeter equals Circumference of circle = diameter * 22/7 = 63 * 22/7 = 198 it is semicircle so divide by 2 = 99 add diameter also – to denote one side : 99+63 = 162 cm answer
53. The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter. Total area = 5.5 * 3.75 =20.63 multiply it by 800 =Rs.16500
54. The no of revolutions a wheel of diameter 49 cm makes in traveling a distance of 176m is Solution : circumference = 22/7 * 49 = 154 17600 / 154 = 114.29 thus the wheel will make 115 revolutions. Answer
55. .A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field? Solution Area 22/7 * 14 * 14 = 616 time required = 616/100 = 6.16 so cow will take little over 6 days to completely graze the whole field. Answer
56. A man runs round a circular field of radius 49m at the speed of 120 m/hr. What is the time taken by the man to take twenty rounds of the field? Solution : circumference = 2* 22/7 * 49 =308 total distance to be travelled = 308 * 20 = 6160 time required = 6160/120 =51.3 hours.
57. .The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec. What is the speed of motorcycle in km/hr? Speed covered in 1 seconds : 4 * 22/7 *70 =880 cm or 8.8 meters. Speed in km per hour : 8.8 * 18/5 = 31.68 km per hour answer
58. A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be Solution : its circumference is : 2 * 22/7 *56 = 352 when you make a circle out of it, one side will be : 352/4 = 88 thus its area will be : 88*88 =7744 answer
59. The area of the largest triangle that can be inscribed in a semicircle of radius 2 is? Area of triangle = Formula = 1/2 * base * height = ½ * (2+2) *2 =4 answer
60. A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed? The total boundary = 2(90+50)= 280 280/5 = 56 so we will need 56 poles
61. The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is the length of the plot in meter? Total fencing = 5300/26.5 = 200 2(L+B) = 200 2(B+20+B) = 200 2B+20=100 or B=40 and L=60 so length is 60 meters answer
62. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required? Length = X bredth = 20 20X =680, or X = 34 boundary =2(34+20)=108 but we will not cover one side =108-20=88 feet. Answer
63. A rectangular paper when folded into two congruent parts had a perimeter of 34cm for each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper? When we fold from mid of length = L+2B=34 When we fold from mid of width = 2L+B=38 add them = 3L+3B=72 L+B=24, where L =14, B=10 so area of paper =14*10 =140 answer
64. A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is? Diagonal = 52*15/60=13 ½ of perimeter=68*15/60=17 X+Y=17 -- 1 st X^2+Y^2 =169 -- 2 nd square of 1 st equation:X^2+Y^2+2XY=289 2XY=120 or XY =60 thus area of field is 60 sq.meter.
65. The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m Boundary = 10.08/.20 = 50.4 one side is 12.6 area : 158.76 area without pavement : (12.6-6)^2= 43.56 pavement = 115.2 cost = 115.2*.5 = Rs. 57.6 answer
66. Aman walked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges? Let us assume that the side of square is 1. had he walked along edges, he would have travelled 2. he walked on diagonal so he walked sqrt(2) = 1.41 thus he has saved (2-1.41) =.59 or we can say that he has saved 29.5% answer
67. .A man walking at the speed of 4 km p.h. crosses a square field diagonally in 30 minutes. The area of the field is The diagonal is 2 km or 2000 meters side of square is : 2000/sqrt(2) =1414 area of field =2000000 sq. m. Or 200 hectares
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75. Download links for material on English http://www.scribd.com/doc/6583315/English-Improvement-Afterschoool http://www.scribd.com/doc/6583518/English-20-May-Afterschoool http://www.scribd.com/doc/28531795/Mock-Paper-Cat-Rmat-Mat-Sbi-Bank-Po-Aptitude-Tests
81. Free download useful material ... http://www.scribd.com/doc/23393316/general-knowledge http://www.scribd.com/doc/23609752/Group-Discussion-Afterschoool http://www.scribd.com/doc/6583547/General-Knowledge-24-May
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