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Blue Lotus A ptitude Numerical Reasoning. Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn.

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Presentation on theme: "Blue Lotus A ptitude Numerical Reasoning. Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn."— Presentation transcript:

1 Blue Lotus A ptitude Numerical Reasoning

2 Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn Diagram

3 Area and Volume Probability Time and Work (Pipes) SI and CI Average Permutation and Combination Percentage Numerical Reasoning

4 Boats and Streams Time and Distance (Trains) Data Sufficiency Profit and Loss Calendar Clocks Data Interpretation Cubes Numerical Reasoning

5 Problems on Numbers Division Algorithm: Dividend = the number to be divided. Divisor = the number by which it is divided. Dividend / Divisor = Quotient. Quotient * Divisor = Dividend. Quotient * Divisor + Remainder = Dividend.

6 Arithmetic Progression: The nth term of A.P. is given by T n = a + (n – 1)d; Sum of n terms of A.P S n = n/2 *(a + L) or n/2 *[2a+(n-1)d)] a = 1 st term, n = number of term, d= difference, T n = n th term Geometrical Progression: T n = ar n – 1. S n = a(r n – 1)/(r-1); Where a = 1 st term, r = 1 st term / 2 nd term Problems on Numbers

7 Basic Formulae 1. ( a+b) 2 = a 2 + b 2 + 2ab 2. (a-b) 2 = a 2 +b 2 -2ab 3. ( a+b) 2 - (a – b) 2 = 4ab 4. (a+b) 2 + (a – b) 2 = 2 (a 2 +b 2 ) 5. (a 2 – b 2 ) = (a+b) (a-b) 6. (a+b+c) 2 =a 2 +b 2 +c 2 + 2(ab +bc+ca) 7. (a 3 +b 3 ) = ( a+b) (a 2 +ab +b 2 ) 8. (a 3 –b 3 ) = (a-b) (a 2 - ab + b 2 )

8 Problems on Numbers Three numbers are in the ratio 3:4:5. the sum of the largest and the smallest equal to the sum of the third and 52. Find the smallest number ?

9 Problems on Numbers Solution: Let the numbers be 3x, 4x and 5x Then 5x+3x = 4x +52 8x – 4x = 52 4x = 52 x = 52/4 x = 13 The smallest number = 3x = 3*13 = 39.

10 Problems on Numbers What is one half of two third of three fourths of four fifths of five sixth of six sevenths of seven eights of eight ninth of nine tenths of thirty?

11 Problems on Numbers Solution: = ½ * 2/3 *3/4 * 4/5 * 5/6*6/7*7/8*8/9*9/10 *30 = 3

12 Problem on Numbers If the operation ^ is defined by the equator x ^ y = 2x + y what is the value of a in 2 ^ a = a ^ 3? (Sathyam)

13 Problem on Numbers Solution: 2(2) = a ^ a = 2a + 3 a = 1

14 Problem on Numbers There are 150 weight some are 1 kg weight and some are 2 kg weights. The sum of the weights is 260. what is the number of 1 kg weight. (TCS)

15 Problem on Numbers Solution: X + 2Y = 260 X + Y = 150 On Solving Two Equations Y = 110 X + Y = 150 X = 150 – 110 = 40 Kg

16 Problem on Numbers The cost of 1 pencil, 2 pens and 4 erasers is Rs. 22, while the cost of five pencils, four pens and two eraser is 32. how much will 3 pencils, 3 pens and 3 eraser? (TCS)

17 Problem on Numbers Solution: Let Pencil be x, Pens be y, Erasers be z x + 2y + 4z = 22 5x + 4y + 2z = 32 Adding we get 6x+6y+6z = 54 3x + 3y + 3z = 27 3 Pencil, 3 Pens and 3 Eraser is Rs. 27.

18 Problem on Numbers If the numerator of a fraction is increased by 25% and denominator decrease by 20%, the new value is 5/4. what is the original value? (TCS)

19 Problem on Numbers Solution: ( x + 25x/100) / (y – 20y/100) = 5/4 125x / 80y = 5/4 x/y = 5/4 * 80/ 125 = 4/5

20 Problem on Numbers The difference between two numbers is 1/7 of the sum of these two numbers. What is the ratio of the two numbers? (Wipro)

21 Problem on Numbers (x- y ) = 1/7 (x+y) 7( x- y) =( x + y) 7x – 7y = x + y 6x = 8y x/y = 3 / 4

22 Problem on Numbers A fraction has a denominator greater than its numerator by 4. but if you add 10 to the denominator, the value of the fraction would then become 1/8, what is the fraction? (Caritor)

23 Problem on Numbers Solution: x/(x+4+10) = 1/8 x/(x+14) = 1/8 8x =( x + 14) 7x = 14; hence x = 2 x/(x+4 )= 2/(2+4 )= 2/6

24 The ages of two persons differ by 10 years. If 5 years ago, the elder one be 2 times as old as the younger one, find their present ages. Problems on Ages

25 Solution: x - y = 10; x = 10 + y x - 5 = 2(y-5) y = 2y -10 y+5 = 2y -10 2y- y = 15 y=15 and x = 25 Their present ages are 15 years and 25 years. Problems on Ages

26 The present ages of three persons are in the proportion of 4:7:9. Eight years ago, the sum of their ages was 56. Find their present ages ? Problems on Ages

27 Solution: Three person’s ratio = 4:7:9 Total = = 20 Sum of their age = 56, after 8 years their sum = = 80 A’s age = 4/20 *80 = 16 B’s age = 7/20 *80 = 28 C’s age = 9/20 *80 = 36 Their present ages are 16, 28 and 36. Problems on Ages

28 Father’s age is 5 times his son's age.4 years back the father was 9 times older than his son. Find the father's present age? (TCS)

29 Problems on Ages Solution: F = 5S F – 4 = 9(S-4) F – 5s = 0 F – 9S = = -32 4S = 32 S = 8 Father age = 40 years

30 Problems on Ages One year ago Pandit was three times his sister’s age. Next year he will be only twice her age. How old will Pandit be after five years? (TCS)

31 Problems on Ages Solution: (P-1) = 3(S -1) P + 1 = 2( s+1) P – 3S = = -2 P – 2S = 2-1 = 1 S = 3 P – 3(3) = -2 P – 9 = -2 P = = 7 After 5 years = 12

32 Problems on Ages A father is 30 years older than his son, however he will be only thrice as old as his son after 5 years what is father’s present age?

33 Problems on Ages Solution: F = S + 30 F + 5 = 3(S+5) S = 3S S = 20 S= 10 F = = 40

34 Problems on Ages A father is three times as old as his son after 15 years the father will be twice as old as his son’s age at that time. What is the father’s present age ? (TCS)

35 Problems on Ages Solution: F = 3S F + 15 = 2(S +15) Father’s age = 45, Son’s age = 15

36 Ratio and Proportion Ratio: The Relationship between two variables is ratio. Proportion: The relationship between two ratios is proportion.

37 Ratio and Proportion The two ratios are a : b and the sum nos. is x ax bx and a + b a + b Similarly for 3 numbers a : b : c

38 Ratio and Proportion If Rs is divided among A, B, C in the ratio 2 : 3 : 4 what is C’s share?

39 Ratio and Proportion Solution: C’s Share = 4/9*1260 C’s share = Rs. 560

40 Ratio and Proportion To 15 liters of water containing 20% alcohol, we add 5 liters of pure water. What is the % of alcohol?

41 Ratio and Proportion Solution: 15 lit 20 % 20 lit (15+5) x by solving we get = 15% 15% alcohol

42 Ratio and Proportion What number should be added or subtracted from each term of the ratio 17 : 24 so that it becomes equal to 1 : 2

43 Ratio and Proportion Solution: Let the number be x x/24 + x = 1/2 Solving the above equation, The number to be subtracted is 10.

44 Ratio and Proportion The ratio of white balls and black balls is 1:2. If 9 gray balls are added it becomes 2:4:3. Then what is the number of black balls ?

45 Ratio and Proportion Solution: Ratio of all the three balls = 2:4:3 Ratio of two balls before adding gray = 1:2 9 gray ratio =3 3 parts = 9 balls 1 part = 9/3 4 parts =? = 9*4/3 =12 Number of black balls is 12

46 Ratio and Proportion Rs. 770 was divided among A, B and C such that A receives 2/ 9 th of what B and C together receive. Find A’s share?

47 Ratio and Proportion Solution: A = 2/9 (B+C) B+C =9A/2 A+B+C = 770 A + 9A/2 = A = 770*2 A = 140

48 (Quantity of cheaper / Quantity of costlier) (C.P. of costlier) – (Mean price) = (Mean price) – (C.P. of cheaper) Alligation or Mixture

49 Cost of Cheaper Cost of costlier c d Cost of Mixture m d-m m-c (Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)

50 A merchant has 100 kg of salt, part of which he sells at 7% profit and the rest at 17% profit. He gains 10% on the whole. Find the quantity sold at 17% profit? Alligation or Mixture

51 Solution: (17-10) (10-7) 7 : 3 The ratio is 7:3 The quantity of 2nd kind = 3/10 of 100kg = 30kg Alligation or Mixture

52 A 3-gallon mixture contains one part of S and two parts of R. In order to change it to mixture containing 25% S how much R should be added? Alligation or Mixture

53 Answer: R:S 2:1 75%:25% 3:1 1 gallon of R should be added. Alligation or Mixture

54 In an examination out of 480 students 85% of the girls and 70% of the boys passed. How many boys appeared in the examination if total pass percentage was 75% Alligation or Mixture

55 Solution: Number of Boys = 480 * 10/15 Number of Boys = 320 Alligation or Mixture

56 In two varieties of tea, one costing Rs. 25/kg. and the other costing RS. 30/kg are blended to produce blended variety of tea in ratio 2:3. find the cost price of the mixture ? Alligation or Mixture

57 Solution: x (30- 28) (28-25) 2 : 3 Let mixed price be x If you subtract 28 from 30 you will get 2 and if you subtract 25 from 28 you will get 3. Alligation or Mixture

58 A person has Rs He invests a part of it at 3% per annum and the remainder at 8% per annum simple interest. His total income in 3 years is Rs Find the sum invested at different rates of interest.

59 Alligation or Mixture Solution: Average rate of interest = 5% per annum 3%8% 5% 3%2% Investment at 3% per annum = 3x5000/5= 3000 Investment at 8% per annum = 2x5000/5=2000

60 Chain Rule Direct Proportion : A B Indirect Proportion: A B

61 Chain Rule A Garrison of 500 men had provision for 27 days. After 3 days, a reinforcement of 300 men arrived. The remaining food will now last for how many days?

62 Chain Rule Solution: Mendays x 800X = 500x24 X =(500x24)/800 =15 days

63 Chain Rule If 20 men take 15 days to complete a job. In how many days will 25 men finish the work? CTS Question

64 Chain Rule Solution: Men Days x x/15=20/25 x = (20*15)/25 =12 They will take 12 days

65 If 11.25m of a uniform iron rod weighs kg, what will be weight of 6m of the same rod? Chain Rule

66 Solution: length ( m )weight ( kg ) x Since it is a direct proportion, x 6 6 x =  x = The weight of rods x = 22.8 kg Chain Rule

67 A stationary engine has enough fuel to run 12 hours when its tank is 4/5 full. How long will it run when the tank is 1/3 full? TCS Question

68 Chain Rule Answer: Tank hours 4/5 12 1/3 x 4/5 x = 12 * 1/3 It will run for 5 hours

69 Chain Rule 20 men complete one - third of a piece of work in 20 days. How many more men should be employed to finish the rest of the work in 25 more days?

70 Chain Rule Solution: Men days work /3 work done = 1/3 x 25 2/3 remaining = 1-1/3=2/3 More work, more men (direct proportion) More days, less men (indirect Proportion) 1/3 *X = (2/3 )*20*(20 /25) X = 800/25 = 32 More men to be employed = (32-20) 12 More people needed to finish the job

71 Chain Rule 15 men take 21 days of 8 hrs each to do a piece of work. How many days of 6 hrs each would 21 women take, if 3 women do as much work as two men?

72 Chain Rule Solution: 3 women = 2 men 21 women = 14 men MenDaysHrs x6 x/21=(15/14)/(8/6) x= 30 days

73 Types: A invested Rs.x and B invested Rs.y then A:B = x : y A invested Rs. x and after 3 months B invested Rs. y then the share is A:B = x * 12 : y * 9 Partnership

74 Sanjiv started a business by investing Rs After 3 months Rajiv joined him by investing Rs Out of an annual profit of Rs find the share of each. Satyam Question Partnership

75 Solution : * 12: * 9 4 : 3 Sanjiv’s share of profit = (4*37100)/7 = profit = Rajiv’s share of profit = Partnership

76 A sum of money is divided among A, B, C such that for each rupee A gets, B gets 65 paise and c gets 35 paise if c’s share is Rs what is the sum? Partnership

77 Solution A : B : C 100 : 65 : : 13 : 7 Total = = 40 C’ share = 560 7/40 *X =560 X= 3200 Partnership

78 A starts business with Rs.3500 and 5 months after B joins A as his partner. After a year the profits are divided in the ratio of 2:3. How much did B contribute ? Partnership

79 Solution: A :B =3500*12 : 7X : 7X = 2: 3 7X * 2 = *3 X = * 3/14 X = 9000 B’s contribution is Rs.9000 Partnership

80 A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share is Rs. 855 what is the total profit? Partnership

81 Solution: A : B = 3:2 Let Profit be X X – 5% of X X- 5X/100 = 95X/100 3/5 * 95X/100 = X= X = 26500/19 = 1500 Total profit is Rs.1500 Partnership

82 A and B enter into partnership for a year. A contributes Rs.1500 and B Rs After 4 months, they admit C who contributes Rs If B withdraws his contribution after 9 months, find their profit share ratio at the end of the year?

83 Partnership Solution: A: B: C = 1500*12: 2000*9: 2250*8 = 18000: 18000: = 1: 1: 1 Profit share at the end of the year, 1: 1: 1

84 If A can do a piece of work in n days, then A’s 1 day’s work = 1/n If A is thrice as B, then: Ratio of work done by A and B = 3:1 Time and Work

85 Pipes and Cisterns P 1 fills in x hrs. Then part filled in 1 hr is 1/x P 2 empties in y hrs. Then part emptied in 1 hr is 1/y

86 P 1 and P 2 both working simultaneously which fills in x hrs and empties in y hrs resp ( y>x) then net part filled is 1/x – 1/y P 1 can fill a tank in X hours and P2 can empty the full tank in y hours( where x>y), then on opening both pipes, the net part empties in hour 1/y -1/x Pipes and Cisterns

87 One fast typist types same matter in 2 hours and another slow typist types the same matter in 3 hours. If both do combine in how much time will they finish? TCS Question Time and Work

88 Solution: Fast typist = 1/2 ; slow typist = 1/3 ; Together: = 1/2 + 1/3 = 5/6 so 6/5 hrs The work will be completed in 6/5 Hrs. Time and Work

89 A and B can finish a piece of work in 30 days, B and C in 40 days, while C and A in 60 days.In how many days A, B and C together can finish the work ? Time and Work

90 Solution: A + B = 30 days = 1/30 B + C = 40 days = 1/40 C +A = 60 days = 1/60 All work together A+B+C+B+C+A = 1/30 +1/40 +1/60 2(A+B+C) = 1/30+1/40+1/60 = (4+3+2) /120 = 9/120*2 = 9/240 = 3/80 = 26 2/3 A, B and C can finish the work in 26 2/3 days Time and Work

91 10 men can complete a piece of work in 15 days and 15 women can complete the same work in 12 days. If all the 10 men and 15 women work together, In how many days will the work get completed ? Time and Work

92 Solution: 10 men = 15 days means 1day work = 1/15 15 men = 12 days means 1 day work = 1/12 10 men + 15 women = 1/15 + 1/12 = 4+5/60 = 9/60 = 3/20 20/3 days = 6 2/3 days The work will be completed in 6 2/3 days. Time and Work

93 A work done by two people in 24 minutes. One of them can do this work alone in 40 minutes. How much time is required to do the same work by the second person? TCS Question

94 Time and Work Solution : A and B together = 1/24; A = 1/40; B = ? = 1/24 – 1/40 = 2/120 = 1/60 The second person will complete in 60 minutes.

95 A cistern has two taps which fill it in 12 minutes and 15 minutes respectively. There is also a waste pipe in the cistern. When all the pipes are opened, the empty cistern is filled in 20 minutes. How long will a waste pipe take to empty a full cistern ? Time and Work (Pipes)

96 Solution: This problem is based on 2nd method. All the tap work together = 1/12 + 1/15 - 1/20 = 5/60 + 4/60 – 3/60 = 6/60 = 1/10 The waste pipe can empty the cistern in 10 minutes. Time and Work (Pipes)

97 A tap can fill a cistern in 8 hours and another can empty it in 16 hours. If both the taps are opened simultaneously, Find the time ( in hours) to fill the cistern Time and Work (Pipes)

98 Solution: Tap 1 = 1/8 (fill); Tap 2 = 1/16 (empty) = 1/8 – 1/16 = 1 / 16 Total time taken to fill the cistern = 16 hours Time and Work (Pipes)

99 A water tank is 2/5 th full. Pipe A can fill the tank in 10 minutes and the pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?

100 Time and Work (Pipes) Answer : A = 1/10; B = 1/6 = 1/10 -1/6 = - 1/15 Empty in 15 minutes To empty 2/5 of the tank 2/5 * 15 = 6 Time taken (empty)= 6 minutes

101 Area and Volume Cube: Let each edge of the cube be of length a. then, Volume = a 3 cubic units Surface area= 6a 2 sq.units. Diagonal = √3 a units.

102 Cylinder: Let each of base = r and height ( or length) = h. Volume = πr 2 h Surface area = 2 πr h sq. units Total Surface Area = 2 πr ( h+ r) units. Area and Volume

103 Cone: Let radius of base = r and height=h, then Slant height, l = √h 2 +r 2 units Volume = 1/3 πr 2 h cubic units Curved surface area = πrl sq.units Total surface area = πr (l +r) Area and Volume

104 Sphere: Let the radius of the sphere be r. then, Volume = 4/3 πr 3 Surface area = 4 π r 2 sq.units Area and Volume

105 Circle: A= π r 2 Circumference = 2 π r Square: A= a 2 Perimeter = 4a Rectangle: A= l x b Perimeter= 2( l + b) Area and Volume

106 Triangle: A = ½*base*height Equilateral = √3/4*(side) 2 Area of the Scalene Triangle S = (a+b+c)/ 2 A = √ s*(s-a) * (s-b)* (s-c) Area and Volume

107 What is the cost of planting the field in the form of the triangle whose base is 2.8 m and height 3.2 m at the rate of Rs.100 / m 2 Area and Volume

108 Solution: Area of triangular field = 1/2 * 3.2 * 2.8 m 2 = 4.48 m 2 Cost = Rs.100 * 4.48 = Rs.448. Area and Volume

109 Area of a rhombus is 850 cm 2. If one of its diagonal is 34 cm. Find the length of the other diagonal? Area and Volume

110 Solution: 850 = ½ * d1 * d2 = ½ * 34 * d2 = 17 d2 d2 = 850 / 17 = 50 cm Second diagonal = 50cm Area and Volume

111 A grocer is storing small cereal boxes in large cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each small cereal box is 7 inches by 6 inches by 5 inches then what is maximum number of small cereal boxes that can be placed in each large carton ? Area and Volume

112 Solution: No. of small boxes = (25*42*60 ) / ( 7*6*5 ) = boxes of cereal box can be placed. Area and Volume

113 If the radius of a circle is diminished by 10%, what is the change in the area in percentage?

114 Area and Volume Solution: = D1 + D2 – D1*D2 /100 = – 10*10/100 = = 19% New area changed = 19%.

115 Area and Volume A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in the ratio of 6:5. Find the smaller side of the rectangle?

116 Area and Volume Solution: length of wire = 2 πr = (22/7*14*14)cm = 264cm Perimeter of Rectangle = 2(6x+5x) cm = 22xcm 22x =264 x = 12 cm Smaller side = (5*12) cm = 60 cm

117 Area and Volume A man is running around a rectangle. It takes time, 2 times in traveling length than traveling width and the total perimeter is 300 m. Find the Area?.

118 Area and Volume Solution: Breadth = X Length = 2X Area = 2X*X = 2X 2 Perimeter 6X = 300m X = 50m Area = 2x 2 =2*50*50 Area = 5000 sq.m.

119 Probability: P(E) = n(E) / n(S) Addition theorem on probability: n(AUB) = n(A) + n(B) - n(A  B) Mutually Exclusive: P(AUB) = P(A) + P(B) Independent Events: P(A  B) = P(A) * P(B) Probability

120 There are 19 red balls and One black ball. Ten balls are placed in one jar and remaining in one jar. What is probability of getting black ball in right jar ? (Infosys -2008) Probability

121 Answer: Probability is 1/2. Probability

122 There are 5 red shoes 4 green shoes. If one draws randomly a shoe what is the probability of getting a red shoe? CTS Question Probability

123 Answer: The probability is 5/9 Probability

124 A bag contains 2 red, 3 green and 2 blue balls are to be drawn randomly. Two balls are drawn at random. What is the probability that the balls drawn contain only blue balls ? Probability

125 Answer : The probability is 1/21 Probability

126 Sam and Jessica are invited to a dance party. If there are 7 men and 7 women in total at the dance and 1 woman and 1 man are chosen to lead the dance, what is the probability that Sam and Jessica will not chosen to lead the dance ?

127 Probability Answer: The Probability of Selecting = 1/7*7 = 1/49 The Probability of not Selecting = 1-1/49 = 48/49

128 Probability The letters of the word SOCIETY are placed in a row. What is the probability that the three vowels come together?

129 Probability Answer: Required Probability = (5!*3! )/7! = 1/7

130 Simple Interest = PNR / 100 Amount A = P + PNR / 100 When Interest is Compound annually: Amount = P (1 + R / 100) n Compound Interest = A - P Simple / Compound Interest

131 Half-yearly C.I.: Amount = P (1+(R/2)/100) 2n Quarterly C.I. : Amount = P (1+(R/4)/100) 4n Simple / Compound Interest

132 Simple/compound interest Difference between C.I and S.I for 2 years = P*(R/100) 2. Difference between C.I and S.I for 3 years = P{(R/100) 3 +3 (R/100) 2 }

133 What is the S.I. on Rs at 18% per annum for the period from 4 th Feb 1995 to 18 th April 1995 Sathyam Question Simple / Compound Interest

134 Answer: Time = = 73 days = 73/365 = 1/5 P = 3000; R = 18%; = PNR/100 = 3000*1*18/100*5 The simple interest is Rs. 108 Simple / Compound Interest

135 A sum of money doubles itself at C.I. in 15 years. In how many years will it become eight times? Satyam Question Simple / Compound Interest

136 Solution: A = P(1+R/100) 15 2P =P(1+R/100) 15 ; 2 = (1+R/100) 15 If A = 8P 8P = P(1+R/100) n 2 3 = ( 1+R/100) [(1+R/100) 15 ] 3 = (1+R/100) n n = 3*15 It will take a period of 45 years. Simple / Compound Interest

137 Raja borrowed a certain money at a certain rate of S.I. After 5 years, he had to pay back twice the amount that he had borrowed. What was the rate of interest? TCS Question Simple / Compound Interest

138 Solution: A = 2P A = P + PNR/100 2P = P(1+NR/100) 2 = (1+5*R/100) 1 = R/20 The rate of interest is 20% Simple / Compound Interest

139 Simple/compound interest In simple interest what sum amounts to Rs in 4 years and Rs in 5 years? CTS Question

140 Simple/Compound interest Answer : Interest for 1 year = 1200 – 1120 = 80 Interest for 4 year = 80*4 = 320 A = 1120 P = A – P = 1120 – 320 The Principal is Rs. 800

141 Simple/Compound interest A simple interest amount for Rs for 6 months is Rs What is the annual rate of interest? CTS Question

142 Simple/Compound interest Solution: P = 5000; N = 6/12 = ½ I = 200 R = I *100 / P*N =200*100*2/5000*1 = 40/5 = 8% The annual rate of interest is 8%

143 Simple/Compound interest A man earns Rs. 450 as an interest in 2 years on a certain sum invested with a company at the rate of 12% per annum. Find the sum invested.

144 Simple/Compound interest Solution: P = I*100/R*N = 450*100/12*2 Principal = Rs. 1875

145 Simple/Compound interest If Rs. 85 amounts to Rs. 95 in 3 years, what Rs. 102 will amount in 5 years at the same rate percent?

146 Simple/Compound interest Solution: Let P = Rs. 85; A = Rs. 95; I = 10/3 in 1 year Rate = I*100/P*N =4% ( app) Amount = P+PNR/100 = =122 Hence the amount in 5 years = Rs. 122

147 Simple/Compound interest What will be the difference between S.I and C.I on a sum of Rs put for 2 years at 5% per annum?

148 Simple/Compound interest Solution: C.I – S.I = P (R/100) 2 = 4500(5/100) 2 = Difference = Rs

149 Simple/Compound interest What will be the C.I on Rs for 2½ years at 4% per annum?

150 Simple/Compound interest Solution: A = P(1+R/100) n A = [(1+4/100) 2 ( 1+(4*1/2) / 100)] = C.I = A – P C.I = Compound interest = Rs. 1613

151 Average Average is a simple way of representing an entire group in a single value. “Average” of a group is defined as: X = (Sum of items) / (No of items)

152 Total temperature for the month of September is 840C. If the average temperature of that month is 28C. Find out the days in the month of September? Average

153 Solution Number of days= 840/28=30 days Average

154 The painter is paid x rupees for painting every 10m of a wall and y rupees for painting every extra meter. During one week, he painted 10m on Monday, 13m on Tuesday, 12m on Wednesday, 11m on Thursday and 12m on Friday. What is his average daily earning in rupees for the five day week? Average

155 Solution: Day 1 = x, Day 2 = x+3y, Day 3 = x+2y, Day 4 = x+y, Day 5 = x+2y Average = (x+ x+3y+ x+2y+ x+y+ x+2y) / 5 = 5x+8y / 5 = 5x/5 + 8y / 5 Average for 5 days is x+ (8y/5) Average

156 The average of 11 observations is 60. If the average of 1st five observations is 58 and that of last five is 56, find sixth observation?

157 Average Solution: 5 observations average = 58 Sum = 58*5 = 290 Last 5 observation average = 56 Sum = 56*5 = 280 Total sum of 10 numbers = 570 ( ) Total sum of 11 numbers = 660 (11*60) 6th number = 90 ( )

158 Average The average of age of 30 students is 9 years. If the age of their teacher included, it becomes 10 years. Find the age of the teacher?

159 Average Solution: Total age of 30 students = 30*9 = 270 Including the teacher’s age = 31*10 = 310 Difference is = = 40 years

160 Permutations and Combinations Factorial Notation: n! = n(n-1)(n-2)… Number of Permutations: n!/(n-r)! Combinations: n!/r!(n –r)!

161 A foot race will be held on Saturday. How many different arrangements of medal winners are possible if medals will be for first, second and third place, if there are 10 runners in the race … Permutations and Combinations

162 Solution: n = 10 r = 3 n P r = n!/(n-r)! = 10! / (10-3)! = 10! / 7! = 8*9*10 = 720 Number of ways is 720. Permutations and Combinations

163 To fill a number of vacancies, an Employer must hire 3 programmers from 6 applicants, and two managers from 4 applicants. What is total number of ways in which she can make her selection ? Permutations and Combinations

164 Solution: It is selection so use combination formula Programmers and managers = 6C 3 * 4C 2 = 20 * 6 = 120 Total number of ways = 120 ways. Permutations and Combinations

165 In how many ways can the letters of the word BALLOON be arranged so that two Ls do not come together? Permutations and Combinations

166 Solution: Total arrangement = 7! / 2!*2! (L and O occurred twice) =1260 Ls come together (BAOON) (LL) = 6! / 2! = 3* 4* 5*6* = 360 Ls not come together 1260 – 360 = 900 Number of ways = 900. Permutations and Combinations

167 A man has 7 friends. In how many ways can he invite one or more of them to a party?

168 Permutations and Combinations Solution: In this problem, the person is going to select his friends for party, he can select one or more person, so = 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7 = 127 Number of ways is 127

169 Percentage By a certain Percent, we mean that many hundredths. Thus, x Percent means x hundredths, written as x%

170 Percentage Two numbers are respectively 30% and 40% less than a third number. What is the second number as a percentage of first?

171 Percentage Solution: Let 3rd number be x. 1st number = x – 30% of x = x – 30x/100 = 70x/ 100 = 7x/10 2nd number = x – 40% of x = x – 40x/100 = 60x/ 100 = 6x/10 Suppose 2 nd number = y% of 1 st number 6x / 10 = (y/100 )* 7x /10 y = 600 / 7 y = 85 5/7 Hence it is 85 5/7%

172 Percentage After having spent 35% of the money on machinery, 40% on raw material, 10% on staff, a person is left with Rs.60,000. What is the total amount of money spent on machinery and the raw materials?

173 Percentage Solution: Let total salary =100% Spending: Machinery + Raw material + staff = 35%+40%+10% = 85% Remaining percentage = 100 %– 85% = 15% 15 % of X = X = 4, 00,000 In this 4, 00,000 75% for machinery and raw material = 4, 00,000* 75/100 = 3, 00,000

174 Percentage If the number is 20% more than the other, how much percent is the second number less than the first?

175 Percentage Solution: Let X =20 = X / (100+X) *100% = 20 /120 *100% =16 2/3% The percentage is 16 2/3%

176 Percentage An empty fuel tank of a car was filled with A type petrol. When the tank was half empty, it was filled with B type petrol. Again when the tank was half empty, it was filled with A type petrol. When the tank was half – empty again, it was filled with B type petrol. What is the percentage of A type petrol at present in the tank?

177 Percentage Solution: Let capacity of the tank be 100 liters. Then, Initially: A type petrol = 100 liters After 1 st operation: A = 100/2 = 50 liters, B = 50 liters After 2 nd operation: A = 50 / 2+50 = 75 liters, B = 50/2 = 25 liters After 3 rd operation: A = 75/2= 37.5 liters, B = 25/2 +50 = 62.5 liters Required Percentage of Type A is = 37.5%

178 Percentage Find the percentage increase in the area of a Rectangle whose length is increased by 20% and breadth is increased by 10%

179 Percentage Answer: Percentage of Area Change=( X +Y+ XY/100)% = *10/100 =32%

180 Percentage If A’s income is 40% less than B’s income, then how much percent is B’s income more than A’s income?

181 Percentage Answer: Percentage = R*100%/(100-R) = (40*100)/ (100-40) =66 2/3%

182 Percentage One side of a square is increased by 30%. To maintain the same area by how much percentage the other side will have to be decreased?

183 Percentage Answer: Percentage = r*100%/(100+r) = (30*100) / 130 = 23 1/3%

184 Boats and streams Up stream – against the stream Down stream – along the stream u = speed of the boat in still water v = speed of stream Down stream speed (a)= u+v km / hr Up stream speed (b)=u-v km / hr u = ½(a+b) km/hr V = ½(a-b) km / hr

185 A man can row Upstream at 12 kmph and Downstream at 16 kmph. Find the man’s rate in still water and rate of the current? Boats and streams

186 Solution: Rate in still water = 1/2 ( ) = 14 Kmph Rate of Current = 1/2 (16 – 12 ) = 2 Kmph Boats and streams

187 A Boat is rowed down a river 40 km in 5 hr and up a river 21 km in 7 hr. Find the speed of the boat and the river? Boats and streams

188 Solution: Speed of the Boat Downstream = 40/7 = 8 (a) Speed of the Boat Upstream = 21/7 = 3 (b) Speed of the Boat = 1/2 ( a + b ) = 1/2 ( 8+3 ) = 5.5 Kmph Speed of the River = 1/2 ( a – b ) = 1/2 (8 – 3) = 2.5 kmph Boats and streams

189 A boat’s crew rowed down a stream from A to B and up again in 7 ½ hours. If the stream flows at 3km/hr and speed of boat in still water is 5 km/hr. find the distance from A to B ? Boats and streams

190 Solution: Down Stream = Sp. of the boat + Sp. of the stream = 5 +3 =8 Up Stream = Sp. of the boat – Sp. of the stream = 5-3 = 2 Let distance be X Distance/Speed = Time X/8 + X/2 = 7 ½ X/8 +4X/8 = 15/2 5X / 8 = 15/2 5X = 15/2 * 8 5X = 60 X =12 Boats and streams

191 Boats and Streams A boat goes 40 km upstream in 8 hours and 36 km downstream in 6 hours. Find the speed of the boat in still water in km/hr?

192 Boats and Streams Solution: Speed of the boat in upstream = 40/8 = 5 km/hr Speed of the boat in downstream= 36/6 =6 km/hr Speed of the boat in still water = (5+6 ) / 2 = 5.5 km/hr

193 Boats and Streams A man rows to place 48 km distant and back in 14 hours. He finds that he can row 4 kmph with the stream in the same time as 3 Kmph against the stream. Find the rate of the stream?

194 Boats and Streams Solution: Down stream 4 km in x hours. Then, Speed Downstream = 4/x km/hr, Speed Upstream = 3/x km/hr 48/ (4/x) + 48/(3/x) = 14 x = 1/2 Speed of Downstream = 8, Speed of upstream = 6 Rate of the stream =1/2 (8-6) km/hr = 1 km/hr

195 Time and Distance Speed:- Distance covered per unit time is called speed. Speed = distance/time (or) Distance = speed*time (or) Time = distance/speed

196 Distance covered α Time (direct variation). Distance covered α speed (direct variation). Time α 1/speed (inverse variation). Time and Distance

197 Speed from km/hr to m/sec - (Multi by 5/18). Speed from m/sec to km/h, - (Multi by 18/5). Average Speed:- Average speed = Total distance traveled Total time taken Time and Distance

198 From height of 8 m a ball fell down and each time it bounces half the distance back. What will be the distance traveled? Sathyam Question Time and Distance

199 Solution: = ….etc. The total distance traveled is 24 m Time and Distance

200 Two cars are 15 km apart. One is running at a speed of 50 kmph and the other at 40 kmph. How much time will it take for the two cars to meet? Sathyam Question Time and Distance

201 Solution: Time taken =Distance / (S1 – S2) = 15 / (50 – 40) = 15 / 10 = 1.5 It will take 1½ hours. Time and Distance

202 The center of a storm shifts 22.5 miles in 1 hour. At the same rate what time will it take to move 60 miles? TCS Question Time and Distance

203 Answer: For 22.5 miles it takes 1 hour It means for 60 miles T = D / S Time taken = 60 / 22.5 It will take 2 2/3 hours. Time and Distance

204 By walking at ¾ of his usual speed, a man reaches office 20 minutes later than usual. Find his usual time?

205 Time and Distance Solution: Usual time = Numerator * late time = 3*20 = 60

206 Time and Distance A man on motorcycle rides 110 miles in 330 minutes. What is his average speed in miles per hour? TCS Question

207 Time and Distance Answer: Speed = D / T =110*60 /330 The average speed = 20 miles/hour

208 Time and Distance (Trains) A train starts from Delhi to Madurai and at the same time another train starts from Madurai to Delhi after passing each other they complete their journeys in 9 and 16 hours, respectively. At what speed does second train travels if first train travels at 160 km/hr ?

209 Time and Distance (Trains) Solution: Let x be the speed of the second train S1 / S2 = √T2/T1 160/x = √16/9 160/x = 4/3 x = 120 The speed of second train is 120km/hr.

210 Time and Distance (Trains) Two hours after a freight train leaves Delhi a passenger train leaves the same station traveling in the same direction at an average speed of 16 km/hr. After traveling 4 hours the passenger train overtakes the freight train. What was the average speed of the freight train?Wipro Question

211 Time and Distance (Trains) Solution : Speed of Passenger train = 16 kmph Distance = 16*4 = 64 Speed of freight train = Distance / ( S1 + S2 ) = 64 / (4+2) = 64/6 = 10.6 km/hr The average speed = 10.6 km/hr

212 Time and Distance (Trains) There are 20 poles with a constant distance between each pole. A train takes 24 sec to reach the 12 pole. How much time will it take to reach the last pole ?

213 Time and Distance (Trains) Solution: To cross 11 poles it is taking 24 sec To cross 19 poles it will take x time Poles time x 11x = 19 * 24 x = 19* 24 /11 x = sec It reaches the last pole in 41.45sec

214 Time and Distance (Trains) 120 m long train crosses the pole after 2½ sec. Find how much time it takes to cross 140 m long platform? Caritor Question

215 Time and Distance (Trains) Solution: To cross 120 m it is taking 2 ½ sec. (5/2sec) To cross ( )=260 m it will take x sec 120x = 260*5/2 (apply chain rule) = 5 5/12 It takes 5 5/12 seconds.

216 Time and Distance (Trains) A train X speeding with 120 kmph crossed another train Y, running in the same direction, in 2 minutes. If the lengths of the trains X and Y be 100 m and 200m respectively, what is the speed of train Y?

217 Time and Distance (Trains) Solution: Let the speed of train Y be x km/hr Relative Speed of X to Y = (120 –x) km/hr = [(120 –x)*5/18] m/sec =( 600 – 5x) / 18 m/sec T = D / Rel. Speed 300 / (600 – 5x /18) = 120 ( 2 Minutes ) 5400 = 120 (600 -5x) x = 111 m/sec.

218 Profit and Loss Cost Price. - CP Selling Price. - SP Profit or Gain. - P = SP – CP Loss. - L = CP - SP

219 . Gain% = [(Gain*100)/C.P.] Loss% = [(Loss*100)/C.P.] S.P. = ((100 + Gain%)/100)C.P. S.P. = ((100 – Loss%)/100)C.P. Profit and Loss

220 Anu bought a necklace for Rs. 750 and sold it for Rs Find her Loss percentage?. Profit and Loss

221 Solution: CP = 750, SP = 675 L = 750 – 675 = 75 Loss% = loss /CP *100 = 75*100/750 = 10% Loss 10% Profit and Loss

222 A shopkeeper bought a watch for Rs. 400 and sold it for Rs What is his profit percentage? TCS Question Profit and Loss

223 Solution: CP = 400; SP = 500 P = 500 – 400 = 100 Profit % = Profit /CP *100 = 100*100/400 = 25% Profit 25% Profit and Loss

224 By Selling 15 Mangoes, a Fruit vendor recovers the cost price of 20 Mangoes. Find the profit percentage? Profit and Loss

225 Solution: The expenditure and the revenue are equated, Percentage of profit =Goods left*100 Goods sold = ( 5*100 ) /15 = 33.3% Profit and Loss

226 A shopkeeper loses 7% by selling a cricket ball for Rs. 31. for how much should he sell the ball so as to gain 5%

227 Profit and Loss Solution: First case S.P = Rs. 31 and Loss% = 7% C.P = [100/(100 – loss%)]*S.P = (100*31) / (100-7) = 100 / 3 Second case, C.P = Rs 100 / 3 and gain% = 5% S.P = [(100+gain 5%) / 100]*C.P = [ (100+5 ) / 100] * 100/3 = Rs. 35

228 Profit and Loss What is the selling price of a Toy car if the cost of the car is Rs. 60 and a profit of 10% over selling price is earned? CTS Question

229 Profit and Loss Solution: Profit = 60*10/100 = 6 Selling Price = C.P + Profit = Selling Price = Rs. 66

230 Profit and Loss Find the single discount to a series discount 20%, 10% and 5%.

231 Profit and Loss Answer: SP = [( 80*90*95 )/ 100*100*100 ]* CP = CP Discount = (1 – 0.684) * 100% = *100 % Discount = 31.6%

232 Calendar Odd days: 0 = Sunday 1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday

233 Calendar Month code: Ordinary year J = 0 F = 3 M = 3 A = 6 M = 1 J = 4 J = 6 A = 2 S = 5 O = 0 N = 3 D = 5 Month code for leap year after Feb. add 1.

234 Calendar Ordinary year = (A + B + C + D ) take remainder 7 Leap year = (A + B + C + D) – take remainder 7

235 What is the day of the week on 30/09/2007? Calendar

236 Solution: A = 2007 / 7 = 5 B = 2007 / 4 = 501 / 7 = 4 C = 30 / 7 = 2 D = 5 ( A + B + C + D )-2 = = ( ) = 14/7 = 0 = Sunday 7 Calendar

237 What was the day of the week on 13 th May, 1984? Calendar

238 Solution: A = 1984 / 7 = 3 B = 1984 / 4 = 496 / 7 = 6 C = 13 / 7 = 6 D = 2 ( A + B + C + D) -3 = = 14/7= 0, Sunday. Calendar

239 On what dates of April 2005 did Sunday fall?

240 Calendar Solution: You should find for 1st April 2005 and then you find the Sundays date. A = 2005 / 7 = 3 B = 2005 / 4 = 501 / 7 = 4 C = 1 / 7 = 1 D = 6 (A + B + C + D) -2 = = 12 / 7 = 5 = Friday. 7 1st is Friday means Sunday falls on 3, 10, 17, 24

241 Calendar What was the day on 5 th January 1986?

242 Calendar Solution: A = 1986 / 7 = 5 B = 1986 / 4 = 496/7 = 6 C = 5 / 7 = 5 D = 0 (A + B + C + D) -2 = = = 14 / 7 = Sunday 7

243 Clock: In every Hour, the minute hand gains 55 minutes on the hour hand In every hour both the hands coincide once.  = (11m/2) – 30h (hour hand to min hand)  = 30h – (11m/2) (min hand to hour hand) If you get answer in minus, you have to subtract your answer with 360 o Clocks

244 Find the angle between the minute hand and hour hand of a clock when the time is 7:20.

245 Solution:  = 30h – (11m/2) = 30 (7) – 11 20/2 = 210 – 110 = 100 Angle between 7: 20 is 100 o Clocks

246 How many times in a day, the hands of a clock are straight?

247 Clocks Solution: In 12 hours, the hands coincide or are in opposite direction 22 times a day. In 24 hours, the hands coincide or are in opposite direction 44 times a day.

248 Clocks How many times do the hands of a clock coincide in a day?

249 Clocks Solution: In 12 hours, the hands coincide or are in opposite direction 11 times a day. In 24 hours, the hands coincide or are in opposite direction 22 times a day.

250 Clocks At what time between 7 and 8 o’clock will the hands of a clock be in the same straight line but, not together?

251 Clocks Solution:h = 7  = 30h – 11m/2 180 = 30 * 7 – 11 m/2 On simplifying we get, 5 5/11 min past 7

252 Clocks At what time between 5 and 6 o’clock will the hands of a clock be at right angles?

253 Clocks Solution:h = 5 90 = 30 * 5 – 11m/2 Solving 10 10/11 minutes past 5

254 Clocks Find the angle between the two hands of a clock at 15 minutes past 4 o’clock

255 Clocks Solution: Angle  = 30h – 11m/2 = 30*4 – 11*15 / 2 The angle is 37.5 o

256 Clocks At what time between 5 and 6 o’clock are the hands of a clock together?

257 Clocks Solution: h = 5 O = 30 * 5 – 11m/2 m = 27 3/11 Solving 27 3/11 minutes past 5

258 In interpretation of data, a chart or a graph is given. Some questions are given below this chart or graph with some probable answers. The candidate has to choose the correct answer from the given probable answers. Data Interpretation

259 1. The following table gives the distribution of students according to professional courses: __________________________________________________________________ Courses Faculty ___________________________________ Commerce Science Total Boys girls Boys girls ___________________________________________________________ Part time management C. A. only Costing only C. A. and Costing __________________________________________________________________ On the basis of the above table, answer the following questions:

260 The percentage of all science students over Commerce students in all courses is approximately: (a) 20.5 (b) 49.4 (c) 61.3 (d) 35.1 Data Interpretation

261 Answer: Percentage of science students over commerce students in all courses = 35.1% Data Interpretation

262 What is the average number of girls in all courses ? (a) 15 (b) 12.5 (c) 16 (d) 11 Data Interpretation

263 Answer: Average number of girls in all courses = 50 / 4 = 12.5 Data Interpretation

264 What is the percentage of boys in all courses over the total students? (a) 90 (b) 80 (c) 70 (d) 76 Data Interpretation

265 Answer: Percentage of boys over all students = (450 x 100) / 500 = 90% Data Interpretation

266 Data Sufficiency Find given data is sufficient to solve the problem or not. A.If statement I alone is sufficient but statement II alone is not sufficient B.If statement II alone is sufficient but statement I alone is not sufficient C.If both statements together are sufficient but neither of statement alone is sufficient. D.If both together are not sufficient

267 Data Sufficiency What is John’s age? I.In 15 years will be twice as old as Dias would be II.Dias was born 5 years ago. (Wipro)

268 Data Sufficiency Answer: c) If both statements together are sufficient but neither of statement alone is sufficient.

269 Data Sufficiency What is the distance from city A to city C in kms? I.City A is 90 kms from city B. II.City B is 30 kms from city C

270 Data Sufficiency Answer: d) If both together are not sufficient

271 Data Sufficiency If A, B, C are real numbers, Is A = C? I.A – B = B – C II.A – 2C = C – 2B

272 Data Sufficiency Answer: D. If both together are not sufficient

273 Data Sufficiency What is the 30 th term of a given sequence? I.The first two term of the sequence are 1, ½ II.The common difference is -1/2

274 Data Sufficiency Answer: A.If statement I alone is sufficient but statement II alone is not sufficient

275 Data Sufficiency Was Avinash early, on time or late for work? I.He thought his watch was 10 minute fast. II.Actually his watch was 5 minutes slow.

276 Data Sufficiency Answer: D. If both together are not sufficient

277 Data Sufficiency What is the value of A if A is an integer? I.A 4 = 1 II.A = 0

278 Data Sufficiency Answer: B. If statement II alone is sufficient but statement I alone is not sufficient

279 Cubes A cube object 3”*3”*3” is painted with green in all the outer surfaces. If the cube is cut into cubes of 1”*1”*1”, how many 1” cubes will have at least one surface painted?

280 Cubes Answer: 3*3*3 = 27 All the outer surface are painted with colour. 26 One inch cubes are painted at least one surface.

281 Cubes A cube of 12 mm is painted on all its sides. If it is made up of small cubes of size 3mm, and if the big cube is split into those small cubes, the number of cubes that remain unpainted is

282 Cubes Answer: = 8

283 Cubes A cube of side 5 cm is divided into 125 cubes of equal size. It is painted on all 6 sides. 1.How many cubes are coloured on only one side? 2.How many cubes are coloured on only two side? 3.How many cubes are coloured on only three side? 4.How many cubes are not coloured?

284 Cubes Answer:

285 Cubes A cube of 4 cm is divided into 64 cubes of equal size. One side and its opposite side is coloured painted with orange. A side adjacent to this and opposite side is coloured red. A side adjacent to this and opposite side is coloured green? Cont..

286 Cubes 1.How many cubes are coloured with Red alone? 2.How many cubes are coloured orange and Red alone? 3.How many cubes are coloured with three different colours? 4.How many cubes are not coloured? 5.How many cubes are coloured green and Red alone?

287 Cubes Answer:

288 Cubes A 10*10*10 cube is split into small cubes of equal size 2*2*2 each. A side and adjacent to it is coloured Pink. A side adjacent to Pink and opposite side is coloured Blue. The remaining sides are coloured yellow. 1.Find the no. of cubes not coloured? 2.Find the no. of cubes coloured blue alone? 3.Find the no. of cubes coloured blue & pink & yellow? 4.Find the no. of cubes coloured blue & pink ? 5. Find the no. of cubes coloured yellow & pink ?

289 Cubes Answer:

290 Venn Diagram If X and Y are two sets such that X u Y has 18 elements, X has 8 elements, and Y has 15 elements, how many element does X n Y have?

291 Venn Diagram Solution: We are given n (X uY) = 18, n (X) = 8, n (Y) =15. using the formula. n( X n Y) = n (X) + n (Y) - n ( X u Y) n( X n Y) = – 18 n( X n Y) = 5

292 Venn Diagram If S and T are two sets such that S has 21elemnets, T has 32 elements, and S n T has 11 elements, how many element elements does S u T have?

293 Venn Diagram Answer: n (s) = 21, n (T) = 32, n ( S n T) = 11, n (S u T) = ? n (S u T) = n (S) + n( T) – n (S n T) = – 11 = 42

294 Venn Diagram If A and B are two sets such that A has 40 elements, A u B has 60 elements and A n B has 10 elements, how many element elements does B have?

295 Venn Diagram Answer: n ( A) = 40, n ( n B) = 60 and n ( A n B) = 10, n ( A u B) = n ( A) + n (B) – n ( A n B) 60 = 40 + n (B) – 10 n (B) = 30

296 Venn Diagram In a group of 1000 people, there are 750 people who can speak Hindi and 400 who can speak English. How many can Speak Hindi only?

297 Answer: n( H u E) = 1000, n (H) = 750, n (E) = 400, n( H u E) = n (H) + n (E) – n( H n E) 1000 = – n ( H n E) n ( H n E) = 1150 – 100 = 150 No. of people can speak Hindi only _ = n ( H n E) = n ( H) – n( H n E) = 750 – 150 = 600

298 Venn Diagram In a class of 100 students, the number of students passed in English only is 46, in maths only is 46, in commerce only is 58. the number who passes in English and Maths is 16, Maths and commerce is 24 and English and commerce is 26, and the number who passed in all the subject is 7. find the number of the students who failed in all the subjects.

299 Venn Diagram Solution: No. of students who passed in one or more subjects = = 91 No of students who failed in all the subjects = = 9

300 Venn Diagram In a group of 15, 7 have studied Latin, 8 have studied Greek, and 3 have not studied either. How many of these studied both Latin and Greek?

301 Venn Diagram Answer: 3 of them studied both Latin and Greek.


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