The document provides a collection of aptitude test questions and their solutions. Some key questions covered include: calculating percentages in mixtures, work problems, interest rate problems, sets and Venn diagrams, time and work problems related to trains, and geometry problems involving circles, spheres, cylinders. The questions are meant to help students prepare for competitive exams.

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3. 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

4. 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

5. 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

6. 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

7. 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).

8. 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

9. 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

10. 120, 315, 300, 345, ? We can write these numbers as multiples of 15 : 8, 21,20,23 the next digit should be :

12. 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.

13. 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

14. (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

15. (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.

16. 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

17. 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

18. 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

20. WHAT IS A FUNCTION It is a rule. It says that given a value of X what will be the value of f(x). here F is not multiplied to X, but it only shows function of X. Itis also written as Y = f(x)

21. What are the types of functions ? There are many classifications : 1. even and odd functions 2.

22. What is inverse function ? Inverse function puts the relation of Y to X (function puts the relation of X to Y). Thus if function = convert celsius to feirenheit F(c) = 9/5 (C+32) now inveserse function will be to convert feirehheit into celcius : f(f) = 5/9 (f-32)

23. What is even function ? When the outcome of f(x) = f(-x), it is called even function. Example: F(x) = X^2+x^4 suppose X is 2, f(x) = 20 we get the same outcome in the case of f(-2)

24. What is odd function ? When the outcome of f(x) = -f(-x), it is called even function. Example: F(x) = X^3+x^5 suppose X is 2, f(x) = 40, but f(-2) = -40 thus we can say that f(x)=-f(-x) answer

25. What is composite function ? If Y = f(x) and X = g(u), then we have function of function. This is called composite function. Suppose first function says that Y = X^2 and it is given than X = Z^2, and Z = 2 then Y = 16 thus this is composite function. Answer

26. What is limit? As X approaches certain value it is called limit example : x = 1.7 and it approaches 1.8 thus f(x)=2x = 3.6 (it approaches 3.6 answer

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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.

33. 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

34. 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

35. 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

36. 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.

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

44. .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

45. 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.

46. .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

47. 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

48. 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

49. 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

50. 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

51. 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

52. 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

53. 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.

54. 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

55. 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

56. .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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