This document provides practice questions and tips in business mathematics. It contains multiple choice questions related to topics like ratios, percentages, profit and loss, time and work, averages, simple and compound interest, discounts, and permutations and combinations. The questions are intended to help students prepare for competitive exams in subjects like commerce and management.

1. PRACTICE QUESTIONS AND TIPS IN BUSINESS MATHEMATICS 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. Find the present worth of Rs.930 due 3 years hence at 8% per annum.Aso find the discount? Formula Present worth = : (100* amount) / (100 + rate * time ) =(100*930) / (100 +(8*3)) =93000/124 750 discount is simple interest on present value so : (750*8*3) /100 =180 answer

4. The truediscount on a bill due 9 months hence at 12% per annum is Rs540.Find the amount of the bill and its presentworth? Present worth = (interest *100)/((rate *time) =(540*100)/(12*.75) 6000 thus the amount of the bill = present worth + true discount =6000+540 =6540 answer

5. The TrueDiscount on a certain sum of money due 3 years hence is Rs.250 and SimpeInterest on the same sum for same time and same rate is Rs375 find sum and rate%? Suppose present value is X RTX / 100 = 250 (X+250) *rt / 100 = 375 solving these we get : (X+250)/X =375/250 250X+62500 =375X -125X = -62500 or X = 500 RT5=250 or RT = 50 or r = 16.33 amount = (500+250) = 750 answer

6. What is the true discount on Rs. 1400, rate of interest is 10% and time is 4 years? Present value = (100*1400)/(100+10*4) =140000/140 =1000 true discount is simple interest on present value interest = PRT / 100 =(1000 * 10 * 4 )/ 100 =400 ANSWER

7. WHAT IS BANKERS DISCOUNT ON RS. 1400 AS PER PREVIOUS QUESTION? BANKER'S DISCOUNT IS SIMPLE INTEREST ON AMOUNT FORMULA = PRT / 100 =(1400 * 10 * 4) / 100 =560 BANKER'S GAIN IS THE DIFFERENCE OF BANKER'S DISCOUNT AND TRUE DISCOUNT = 560-400 = 160 BANKER'S GAIN IS SIMPLE INTEREST ON TRUE DISCOUNT =(400 * 10*4) / 100 = 160 ANSWER

8. The difference between SimpeInterest and TrueDiscount on a certain sum of money for 6 months at 25/2% per annum is Rs25. Find the sum. True discount = (25*100)/(1/2 *25/2) =2500*4 /25 10000/25 = 400 Present worth is : (400*100)/(1/2 *25/2) =160000/25 =6400 6400 is the present value, so amount is 6400+400 =6800

9. A bill falls due in 1 year.The creditor agrees to accept immediate payment of the half and to defer the payment of the other half for 2 years .By this arrangement he gains Rs.40.what is the amount of bill,if the money be worth 25/2%? Let us assume that the bill amount is Rs. 200 amount received =100+present value of 100 (after 2 years) PV of 100 = (100 * 100 )/ (100+25) =80 thus payment received = 80 present value of bill amount = (100*200)/(100+25) = 160 his profit is 180-160 = 20. thus bill amount is : 40/20 * 200 = 400 answer

10. Equation -7x+1=5-3x will be satisfied for X equal to ? Options :a. 1, b. -1, c 4, d9 answer -1 here you must use options to get the answer. When you put x=-1, you can see that the equation is balanced.

11. What is X if X/30 = 2/45 ? Options : 5,7,1.3, none of these answer : 1.3

12. Products of two numbers is 3200 and the quotient when the larger number is divided by smaler is 2. what are the numbers? Options : (16,200), (160,20), (60,30),(40,80) answer : (40,80) - both the conditions must be fulfilled.

13. The denominator of a fraction exceeds the numerator by 2. if 5 is added to the numerator, the fraction increases by unity. What is the fraction. Options : 5/7, 1/3, 7/9, 3/5 answer : try with last option 3/5 = .6 and 8/5 = 1.6 thus there is a increase of unity. It also fulfill other condition. Thus this is the answer

14. The 4 th part of a number exceeds the 6 th part by 4, what are the numbers ? Options : 84,44,48, none of these answer : 48 4 th part of 48 = 12 6 th part of 48 = 8 their difference is 4, so this is the answer

15. Diagonal of rectangle is 5, one of the side is 4, what is the area of rectangle ? Options : 10,12,20 none of these when one of the side is 4, the other side is : 5^2 – 4^2 = 9 take the square root of 9 = 3 so area = 4 * 3 = 12 answer

16. Divide 56 in two parts such that 3 times the first part exceeds 1/3 rd of the 2 nd part by 48. what are the parts? Options : 20,36 25,31 24,32 none of these answer : 20 & 36 3 times the first part = 60 48 + 1/3 *36 = 60 answer

17. If a number of which the half is greater than 1/5 th of the number by 15, then the number is ? Options : 40,50,80 none of these try with options : 50 ½ of 50 = 25 1/5 th = 10 their difference is 15 so this is the answer

18. What is the solution of the set : 1. 3x+4Y = 7 and 2. 4X – Y = 3? Options : 1,-1, 2,1 1,1 1,-2 answer :try the options : put 1,1 in the first equation, when X =1, and Y =1, we get 3+4 = 7 and in 2 nd equation : 4-1 =3 so answer is 1,1

19. A number consists of 3 digit number, middle one is zero, sum of the other digits is 9 . The number formed by interchanging the 1 st and 3 rd digits is more than the original number by 297 . what is the number? OPTIONS : 408, 307, 306 , NONE OF THESE ANSWER : 306 603 – 306 = 297 ANSWER

20. FIND ROOTS OF THE FOLLWOING EQUATIONS : X^2-5X+6 = 0 FORMULA TO FIND ROOT : -b+ - sqrt(b^2 – 4ac) / 2a a=1, b=-5, c=6 = 5 + - sqrt (25 -24) / 2 =(5+1)/2 = 3 (5-1)/2 = 2 roots are 3,2 answer

21. The sum of 2 numbers is 8 and the sum of their squares is 34 . taking one of the numbers as X form an equation in X and hence find the numbers. The numbers are ? Options : 7,10 4,4 3,5 2,,6 answer : 3, 5 3 + 5 = 8 and sum of their squares : 9+25 = 34

22. Sum of 2 numbers is 45 and the mean proportional between them is 18, what are the numbers ? Options : 15,30, 32,14, 36,9 25,15 answer : 36,9 their total is 45 and their mean proportion = sqrt (9*36) = 18 so this is the answer

23. How many types of series is there ? 1. arithematic series (example : 1,2,3,4,5,6,7) 2. geometric series (example : 1,2,4,8,16..) 3. harmonic series (example 1/2, 1/4,1/6,1/8 ...)

24. What is AP ? Arithematic progression : here there are two things : A = starting term D = difference between two terms (which is always equal) example : 3,10,13,20 ... here A = 3, D = 7.

25. What is GP ? Geometric progression here ratio (ratio means what you get when you divide the 2 nd term from the 1 st term). Here we have two things : A and R A = starting term (1 st term) R = ratio (2 nd term / 1 st term). Ratio remains same for the whole series

26. What is HP? Harmonic progression It is the reverse of AP example : 2,4,6,8, is AP but 1/2, 1/4, 1/6, 1/8 is HP.

27. What is the sum of series ? AP =there are 2 formula : Sum = n/2 (2a + n-1)d) or sum = n/2 (1 st term + last term)

28. What is the sum of series in GP? There are 3 possibilities : 1. series is infinite and r<1 2. series is not infintie and r<1 3. series is not infinite and r>1

29. Sum of infinite series ? We can add an infinite series by the following formula : sum = a / (1-r) (here r is always less than 1). a= starting term, and r=ratio

30. What is the sum of the infinite series, if the series is : 64,32,16 ....? We can see it is an infinite series a = 64, r = (32/64 = .5 sum = 64/ (1- .5) = 64/ .5 = 128 answer

31. Sum of finite series when r<1? Sum = a(1-r^n) / (1-r)

32. What is the sum of 1 st 5 terms of the series : 64,32,16 ....? A=64, r = .5 n = 5 64 * (1 - (.5)^5) / (1-.5) =64 (1-.03125)/.5 =128*.96785 =124 approx. Answer

33. Sum of fininte series when r>1 ? Formula : sum = a (r^n – 1)/(r-1)

34. What is the sum of this series for 8 terms : 2,4,8,16,32.... a=2,r=4/2 = 2, n=8 r=2 nd term / 1 st term 2(2^(8) – 1) / (2-1) =2(256-1) =2*255 =510 answer

35. What is the sum of this series for 5 terms : .7,.77, .777, .7777? Let us simplify this term. Let us take 7 out of this series : .1,.11,.111,.1111. .... it is still neither GM nor AM now let us multiply this series by 9 .9,.99,.999, .9999.. it is still not a GM now let us put it like this : (1-.1),(1-.01),(1-.001)....this is now a GM so solve it....

36. contd.... Now let us add the series : the series has two components : 1 and .1^n so let us add 1 = for 5 terms = 5 now let us add .1^n = .1(1-(.1)^5) )/(1-.1) =1/9 (.99999) = .11111 7/9 (5 -.11111) = 3.8 answer

37. What is the sum of this series? 5/6, 5/18, 5/54 .... this is infinite series A = 5/6 R = 5/18 divided by 5/6, we get 1/3 solve it : 5/6 (1 – 1/3) = 15/12 or 1.25 answer

38. 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

39. 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)

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

41. 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)

42. 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)

43. 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

44. 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

45. 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

46. 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

47. 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

48. 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

49. 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

50. 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

51. 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.

52. 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

53. 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

54. 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

55. 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.

56. 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

57. 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

58. 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

59. 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

60. 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

61. 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

62. 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

63. .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

64. 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.

65. .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

66. 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

67. 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

68. 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

69. 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

70. 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

71. 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

72. 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.

73. 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

74. 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

75. .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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