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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 Cubes Numerical Reasoning

4 Boats and Streams Time and Distance (Trains) Data Sufficiency Profit and Loss Calendar Clocks Data Interpretation 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)] Geometrical Progression: T n = ar n – 1. S n = a(r n – 1)/(r-1); 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 ) 9. (a 3 +b 3 +c 3 -3abc) = (a+b+c) (a 2 +b 2 +c 2 -ab-bc-ca) If a+b+c = 0, then (a 3 +b 3 +c 3 ) =3abc

8 Problem on Numbers A monkey starts climbing up a tree 20 feet tall. Each hour it claims 3 feet and slips back 2 feet. How much time would it take the monkey to reach the top? (Satyam)

9 Problem on Numbers Solution: = = 18 hours

10 Problem on Numbers The length of the side of a square is represented by x + 2. the length of the side of an equilateral triangle is 2x, if the square and the equilateral triangle have equal perimeter the find the value of x.

11 Problem on Numbers Solution: Side of the square is x + 2 Side of the triangle 2x P = 4(x + 2) = 4x +8 Perimeter = 3*2x = 6x 4x + 8 = 6x x = 4

12 Problem on Numbers On sports day if 30 children were made to stand in a column, 16 column could be formed if 24 children were made to stand in a column. How many columns could be formed? (Satyam)

13 Problem on Numbers Solution: Total no of children = 30 * 16 = 480 No. of column of 24 children each = 480 / 24 = 20

14 Problem on Numbers 5/9 part of the population in a village are males if 30% of the males are married what is the percentage of unmarried females in the total population?

15 Problem on Numbers Solution: Let population be = x Males = 5 x/9 Married Men =30%(5x/9)= x/6 (Married females) Total females = x – 5x/9 = 4x/9 Unmarried females = 4x/9 – x/6 = 5x/18 Percentage = (5x / 18)*100% x = 250 / 9%

16 Problems on Numbers How many terms of the A.P. 1, 4, 7…. are needed to give the sum 715 ?

17 Solution: Sum of n terms of A.P S n = n/2 *[2a+(n-1)d)] S n = 715, n =? a = 1 d = = n/2 *[2*1+(n-1)3] 715 =n/2 *[2+(n-1) = n[2+3n -3] 1430 = n[3n – 1] 1430 = 3n 2 -n 3n 2 – n – 1430 = 0 Solving the Quadratic Equation, n =22 Number of terms needed is 22 Problems on Numbers

18 The sum of the digits of a two-digits number is 8. if the digits are reversed the number is increased by 54. Find the number?

19 Problems on Numbers Solution: x+ y = y+x = 10x+y+54 10y + x - 10x – y = 54 -x + y = Solve equation 1&2 x= 1, y = 7 Required number = 10*1 +7 =17

20 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

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

22 The present ages of three persons are in the proportion of 4:7:9. 8 years ago, the sum of their ages was 56. Find their present ages ? Problems on Ages

23 Solution: Three person’s age ratio = 4:7:9 Sum of their age = 56, after 8 years their sum of their ages = 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

24 Father’s age is three times the sum of the ages of his two children, but twenty years hence his age will be equal to sum of their ages, find the age of Father. Problems on Ages

25 Solution: Father age = 3(x+y) F+20 = x+20+y+20 3x+3y+20 = x+y+40 3x +3y = x+y x-x+3y – y =20 2x+2y = 20 x+y = 10 F = 3*10 =30 The father’s age is 30. Problems on Ages

26 Jalia is twice older than Qurban. When Jalia was 4 years younger, Qurban was 3 years older the difference between their ages is 12 what is the sum of their ages?

27 Problems on Ages Solution: J = 2Q 1 (J – 4) – (Q +3) = 12 2 Solve 1 &2 Sum of their ages = = 57

28 Problems on Ages Ten years ago, Chandrawathi’s mother was 4 times older than her daughter. After 10 years the mother will be twice older than daughter. What is the present age of Chandrawathi?

29 Problems on Ages Solution: Let Chandrawathi’s age 10 years ago x Her mother’s age 10 years ago 4x (4x ) = 2(x+10+10) x = 10 Chandrawathi is x +10 = 20 years

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

31 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

32 Ratio and Proportion Concentration of three wines A, B, and C are 10, 20, and 30 percent respectively. They are mixed in the ratio 2 : 3 : x resulting in a 23% concentration solution. Find x. (Caritor Question)

33 Ratio and Proportion Answer : 10 : 20 : 30 2 : 3 : x Multiplying 20 : 60 : 30x x = 230 x = 5

34 Ratio and Proportion Ajay, Aman, Suman and Geetha rented a house and agreed to share the rent as follow Ajay : Aman = 8:15 Aman : Suman = 5:8 Suman: geetha = 4:5 The part of rent paid by Suman will be ?

35 Ratio and Proportion Solution: Ajay : Aman = 8 : 15 In next ratio Aman is 5 to make that 15 multiply by 3 that is ratio of Aman : Suman = 15:24 Suman and Geetha contribution is 4:5 but Suman previous contribution is 24 to make Suman contribution in 24 multiply by 6 Suman: Geetha = 4*6:5*6 = 24:30 Ratio of Ajay:Aman:Suman:Geetha = 8:15:24:30 Suman ratio = 24/77

36 Ratio and Proportion 60 kg of an alloy A is mixed with 100 kg of alloy B. If alloy A has lead and tin in the ratio 3 : 2 and alloy B has tin and copper in the ratio 1 : 4, then what would be the amount of tin in the new alloy?

37 Ratio and Proportion Solution : The amount of tin in both alloys = (60*2/5) + (100*1/5) = 44 kg

38 Ratio and Proportion In a class composed of x girls and y boys what part of the class is composed of girls? (Satyam Question)

39 Ratio and Proportion Answer: Ratio = x : (x+y) = x / x+y

40 Ratio and Proportion In a factory,the ratio of male workers to female workers was 5:3. If the number of female workers was less by 40.What was the total number of workers in the factory.

41 Ratio and Proportion Solution: Let the number of males is 5x and females is 3x and Total is 8x 5x – 3x = 40 x = 20 Total number of workers in the factory = 8x =8*20 = 160

42 Ratio and Proportion A Mixture contains milk and water in the ratio 5:1.On adding 5 liters of water, the ratio of milk to water becomes 5:2. What is the quantity of milk in the original mixture ?

43 Ratio and Proportion Solution: Let quantity of milk be 5x, and water be x then, 5x = 5 x Hence x=5 Quantity of milk = 5x5 = 25 liters.

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

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

46 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

47 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

48 A certain type of mixture is prepared by mixing brand A at Rs. 9/kg with brand B at Rs. 4/kg. If the mixture is worth Rs. 7/kg how many kg of brand A are needed to make 40 kgs of the mixture?Satyam Question Alligation or Mixture

49 Solution: BA Brand A required = 3*40 / 5 = 24kg Alligation or Mixture

50 A man buys two cows for Rs and sells one so as to lose 6% and the other so as to gain 7.5% and on the whole he neither gains nor loses. How much does each cow cost? Alligation or Mixture

51 Solution: -6%7.5% Ratio is 5 : 4 The cost of first cow = 5*1350/9 =Rs. 750 The cost of second cow = Rs. 600 Alligation or Mixture

52 How much water be added to 14 liters of milk worth Rs. 5.4 per liter so that the value of the mixture may be Rs per liter? Alligation or Mixture

53 Solution: The cost of water is 0. WM Ratio is 4 : 14 The amount of water added for 14 liters is 4 liters. Alligation or Mixture

54 There are 65 students in a class, 39 rupees are distributed among them so that each boy gets 80 paise and girl gets 30 paise. Find the number of boys and girls in that class.

55 Alligation or Mixture Solution: “Money per boy or girl” is considered. Per student = 3900/65 = 60 paise. GirlsBoys Girls : Boys = 2 : 3 Number of boys = 39 Number of girls = 26

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

57 Chain Rule A man completes 5/8 of a job in 10 days. At this ratio how many more days will it take for him to finish the job?

58 Chain Rule Answer: JobsDays 5/810 3/8 x 5x / 8 = (10*3) / 8 (Direct ) The number of days to complete = 6

59 Chain Rule A certain number of men can finish a piece of work in 100 days. If however there were 10 men less it will take 10 days more for the work to be finished. How many men were there originally? (Satyam Question)

60 Chain Rule Solution: MenDays x100 (x-10)110 (indirect ) 100x = 110(x – 10 ) The number of men present originally = 110

61 15 men take 21 days of 8 hours each to do a piece of work. How many days of 6 hours each would it take for 21 women if 3 women do as much work as 2 men? (Satyam Question) Chain Rule

62 Solution: 3 women = 2 men Men DaysHours x 6 x = 15* *6S x = 30 The number of days required is 30. Chain Rule

63 6 cats kill 6 rats in 6 minutes. How many cats will be needed to kill 100 rats in 50 minutes? (Satyam question)

64 Chain Rule Solution: CatsRatsMinutes 666 x10050 x/6 = (100*6)/(6*50) The number of cats needed would be 12.

65 Chain Rule 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons working 6 hours a day, complete the same work?

66 Chain Rule Solution : - Mendayshours x 6 More hours less days ( inverse proportion ) Less men more days ( inverse proportion ) x * 39 * 12  = ---- * ----  x = * 30  x = 13 days.

67 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

68 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. 560, what is the sum? (TCS Question) Partnership

69 Solution : A :B:C 100:65:35 Total = = 200 C’s share = 560 (35*x)/200 = 560 X = 3200 The sum invested is Rs Partnership

70 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

71 Solution: A and B = 3 : 2 Let Profit be x x – 5% x- 5x/100 = 95x/100 A’s share is = 3/5 * 95x/100 = x/100 =285 19x = x = 26500/19 = 1500 Total profit is Rs.1500 Partnership

72 A,B,C subscribe Rs. 50,000 for a business. A subscribes Rs more than B and B subscribes Rs more than C. Out of a total profit of Rs. 35,000 how much does A receive? (TCS Question) Partnership

73 Answer : Ratio A B C x : x : x x = On solving we get x = Ratio AB C 21 : 17 :12 Total = 50 A receives = (21* 35000)/50 = Rs Partnership

74 A began a business with Rs. 450 and B jointed with Rs When did B join if the profit at the end of the year was divided between them in the ratio 2 : 1? (Caritor Question) Partnership

75 Answer : Ratio A : B 450x12 : 300 (12 – x ) 5400 / ( 300( 12 – x )) = 2 / 1 On solving x = 3 B invested money after 3 months Partnership

76 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 at the end of the year? (In the ratio)

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

78 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 Ratio of times taken by A and B= 1 : 3 Time and Work

79 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

80 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

81 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

82 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 = 6 2/3 The work will be completed in 6 2/3 days. Time and Work

83 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 do the work ? Time and Work

84 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 = /120 = 9/120 = 9/240 = 3/80 = 26 2/3 A, B and C can finish the work in 26 2/3 days Time and Work

85 A can do a piece of work in 30 days, and B in 50 days and C in 40 days. If A is assisted by B on one day and by C on the next day. In how many days alternatively work will be completed? Time and Work

86 Solution: A = 1/30, B = 1/50, C = 1/40 A+B = 1/30+1/50 = 8/150 A+C = 1/30+1/40 = 7/120 Work done by A & B and A & C = ( 8/ /120) = 67/600 For 16 days, work done = 67x8 / 600 = 536/600 =67/75 Work left = 1-67/75 = 8/75 17th day A & B are working = 8 / 75 – 8 / 150 = 4 /75 18th day A&C are working = 120 / 7 * 4 / 75 = 32/35 They will finish the work in 17 32/35 days Time and Work

87 A can do a piece of work in 12 days. B is 60% more efficient than A. Find the number of days B takes to do the same piece of work.

88 Time and Work Answer : Work Days x On solving, x =( 100*12) / 160 = 7 ½ Days B will take 7½ days.

89 Time and Work Seven men can complete a work in 12 days. They started the work and after 5 days, two men left. In how many days will the work be complete the remaining work.

90 Time and Work Solution: (7*12) men can complete the work in 1 day 1 man’s 1 day’s work = 1/84 7 men’s 5 day’s work = 1/12 *5 = 5/12 Remaining work = (1-5/12) = 7/12 5 men’s 1 day’s work =5*( 1/84 )= 5/84 5/84 work is done by them in 1 day 7/12 work is done by them in 84/5 *7/12 = 49 / 5 = 9 4/5 days

91 Pipe A can fill a cistern in 20 minutes and pipe B in 30 minutes and pipe C can empty the same in 40 minutes. If all of them work together, find the time taken to fill the tank. (Satyam Question) Time and Work (Pipes)

92 Answer: The time taken is 17 1/7 minutes. Time and Work (Pipes)

93 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)

94 Solution: 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)

95 A cistern is provided by two taps A and B. A can fill it in 20 minutes and B in 25 minutes. Both taps are kept open for 5 minutes and then the second is turned off. How many minutes will be taken to fill the tank ? Time and Work (Pipes)

96 Solution: In one day (A +B) can do = 1/20 + 1/25 = 9/100 In 5 days (A +B ) can do = 5x9/100 = 9/20 Work left = 1 – 9 /20 =11/20 A alone can do the remaining = ( 11/20) x 20 = 11 The cistern will be filled in 11 minutes. Time and Work (Pipes)

97 Two pipes A and B can fill a tank in 20 min. and 40 min. respectively. If both the pipes are opened simultaneously, after how much time A should be closed so that the tank is full in 10 minutes?

98 Time and Work (Pipes) Solution: Let B be closed after x minutes. Then, Part filled by (A+B) in x min. + part by A in ( 10 – x min) =1 x (1/20 +1/40) + (10 – x) * 1/20 = 1 x (3/40) + (10 – x) /20 = 1 3x/40 + (10 –x) / 20 = 1 3x + 20 – 2x = 40 3x – 2x = x = 20 min. A must be closed after 20 minutes.

99 Time and Work (Pipes) Two taps A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full?

100 Time and Work (Pipes) Solution: A’s work in hour = 1/6, B’s work in 1 hour =1/4 (A + B) ’s 2 hr work = 1/6+1/4 = 5/12 (A + B) ’s 4 hr work = 10/12 = 5/6 Remaining Part = 1- 5/6 = 1/6 Now, it is A’s turn and 1/6 part is filled by A in 1 hour. Total time = 4+1 =5

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 = πr l 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 = ½ * 3.2 * 2.8 m2 = 4.48 m2 Cost = Rs.100 * 4.48 = Rs 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 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 area in percentage?

114 Area and Volume Solution: x = - 10, y = - 10 = x + y +xy/100 % = – (10*10/100) % = - 19% 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 If the length of a rectangle is increased by 30% decreased by 20%, then what percent of its area would increase? (TCS Question)

118 Area and Volume Solution: x = 30 y = - 20 (Decreased ) Using formula = x + y + (xy/100)% The increase would be 4%

119 Area and Volume The sides of the right triangular field containing the right angle are x and x + 10 and its area is 5500 m 2. Find the equation to determine its area. (Caritor Question)

120 Area and Volume Answer: The equation is 1 / 2*x(x+10) = 5500 x x – = 0

121 Area and Volume The length and breadth of a rectangular plot are in the ratio 7 : 5. If the length is reduced by 5 m and breadth is increased by 2 m then the area is reduced by 65 m 2. Find the length and breadth of the rectangular plot. (Caritor Question)

122 Area and Volume Answer: Difference in areas = 65 (7x *5x) – (7x – 5) (5x + 2) = 65 x = 5 The length of the rectangle = 7x = 7*5 = 35 m The breadth of the rectangle = 5x = 5*5= 25 m

123 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

124 There are 19 red balls and 1 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)

125 Probability Answer: Probability is 1/2.

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 ? Probability

127 Solution: Probability of selecting Sam and Jessica = 7C1 * 7C1 = 49 n(s) = 49 n (є) = 1 ( selecting Sam and Jessica) P (є) = n (є) / n(s) = 1/49 Not selecting Sam and Jessica = 1- 1/49 The probability = 48/49 Probability

128 There are 5 distinct pairs of white socks and 5 pairs of black socks in a cupboard. In the dark, how many socks do I have to pull out to ensure that I have at least one correct pair of white socks? TCS Question Probability

129 Answer: 12 times Probability

130 In a simultaneous throw of two coins, what is the probability of getting at least one head? TCS Question Probability

131 Answer : Probability = 3/4 Probability

132 Three unbiased coins are tossed. What is the probability of getting at least 2 heads? Probability

133 Answer : The probability is 4/8 = 1/2 Probability

134 A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book?

135 Probability Answer: A does not solve the problem = 1- 90/100 = 1/10 B does not solve the problem = 1- 70/100 = 3/10 Either A or B to solve = 1 – (1/10)*(3/10) = 1- 3/100 = 97/100

136 Probability Two unbiased Dice are thrown. Find the probability that neither a doublet nor a total of 10 will appear ?

137 Probability Solution: A= Doublet (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) B =(6,4),(4,6),(5,5) P(A)= 6 / 36, P(B) = 3 / 36, P(A n B) =1/36 Probability = 1 – P( Au B) = 1 – ( P(A) + P(B) - P(AnB) ) = 7 / 9

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

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

140 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 }

141 The simple interest on a certain sum is 16 / 25 of the sum. Find the rate percent and time if both are numerically equal. Sathyam Question Simple / Compound Interest

142 Solution: R = N and I = 16/25 P R = I * 100 / P*N R*N = 64 where R = N R = 8 The rate of interest is 8% Simple / Compound Interest

143 A sum of S.I. at 13½% per annum amounts to Rs after 4 years. Find the sum. Sathyam Question Simple / Compound Interest

144 Answer: A = P + PNR / 100 A = P (1 +NR/100) The principal is Rs Simple / Compound Interest

145 Simple/compound interest The difference between the compound and simple interest on a certain sum for 2 years at the rate of 8% per annum is Rs.80,What is the sum? Wipro Question

146 Simple/Compound interest Answer : C.I – S.I = P(R/100) 2 80 = P*(8*8/100*100) The sum is Rs.12,500

147 Simple/Compound interest If a sum of money compounded annually amounts of thrice itself in 3 years, in how many years will it become 9 times itself? CTS Question

148 Simple/Compound interest Solution: A = 3P and find n when A = 9P The number of years required = 6 years

149 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?

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

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

152 Simple/Compound interest Solution: A = P(1+R/100) n A = ((1+4/100) 2 (1+4*1/2/100)) = CI = A –P = Compound interest = Rs. 1613

153 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)

154 The average weight of 5 persons sitting in boat is 38 Kg.If the average weight of the boat and the persons sitting in the boat is 52Kg,What is the weight of the boat? Average

155 Soluton: Average weight of 5 persons = 38 Kg Total weight of the 5 persons =38*5=190 Total weight including the boat weight is =52*6=312 Weight of the Boat = = 122 Kg 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 (660 –570)

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: 30 students total age = 30*9=270 Including the teacher’s age = 31*10=310 Diff= =40 years

160 Average What is the average of even numbers from 1 to 81?

161 Average Solution: Average = (last even + 2) / 2 = ( ) / 2 = 41

162 Average The Average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, What is the average of the remaining numbers?

163 Average Solution: Total of 50 numbers = 50*38=1900 Total of 48 numbers = 1900-(45+55)=1800 Average =1800/ 48 = 37.5

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

165 The number of Combinations of ‘n’ things taken ‘r’ at a time in which ‘p’ particular things will always occur is n-p C r-p The number of Combinations of ‘n’ things taken ‘r’ at a time in which ‘p’ particular things never occur is n-p C r Permutations and Combinations

166 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

167 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

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

169 Solution: It is selection so use combination formula Programmers and managers = 6C3 * 4C2 = 20 * 6 = 120 Total number of ways = 120 ways. Permutations and Combinations

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

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

172 Permutations and Combinations There are 5 gentlemen and four ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are together?

173 Permutations and Combinations Solution: 5! = 120 ways

174 Permutations and Combinations In a chess board there are 9 vertical and 9 horizontal lines. Find the number of rectangles formed in the chess board.

175 Permutations and Combinations Solution: 9C 2 * 9C 2 = 1296

176 Permutations and Combinations In how many ways can a cricket team of 11 players be selected out of 16 players, if one particular player is to be excluded?

177 Permutations and Combinations Solution: If one particular player is to be excluded, then selection is to be made of 11 players out of C 11 = 15!/( 11!*4!)=1365 ways

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

179 If Length is increased by X% and Breadth is decreased by Y% What is the Percentage Finding out of Hundred. Increase or Decrease in Area of the rectangle? Formula: X+Y+ XY/100 % Decrease 20% means -20 Percentage

180 Two numbers are respectively 30% & 40% less than a third number. What is second number as a percentage of first?

181 Percentage Solution: Let 3rd number be x. 1 st number = x – 30% of x = x – 30x/100 = 70x/ 100 = 7x/10 2 nd 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 85 5/7%

182 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. The total amount of money spent on machinery and the raw material is?

183 Percentage Solution: Let total salary =100% Salary = 100 Spending: Machinery + Raw material + staff = = 85 Remaining = 100 – 85 = = = ? ( By Chain rule) = 4, 00,000 In this 4, 00,000 75% for machinery and raw material = 4, 00,000* 75/100 = 3, 00,000 Rs. 3, 00,000

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

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

186 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?

187 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 = 37.5%

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

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

190 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?

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

192 Percentage Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight. What is the weight of dry grapes available from 20 kg of fresh grapes?

193 Percentage Solution: Fresh grapes contain 10% pulp 20 kg fresh grapes contain 2 kg pulp Dry grapes contain 80% pulp. 2 kg pulp would contain = 2/0.8 =20/8 = 2.5kg Hence 2.5 kg Dry grapes.

194 Percentage If A is 20% of C and B is 25% of C then what percentage is A of B?

195 Percentage Answer: Given A is 20% of C and B is 25% of C A of B is = (20*100%)/25 = 80% A is 80% of B

196 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

197 Boats and Streams A boat is rowed down a river 40 km in 5 hours and up a river 21 km in 7 hours. Find the speed of the boat and the river.

198 Boats and Streams Solution: Downstream speed (a) = 40/5 = 8 km/h Upstream speed (b)= 21/7 = 3 km/h Speed of the boat =1/2(a+b)= 5.5 km/h Speed of the river=1/2(a-b) = 2.5 km/h

199 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?

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

201 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

202 Solution: Down Stream = Speed of the boat + Speed of the stream = 5 +3 =8 Up Stream = Speed of the boat – Speed 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

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

204 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 = ½ Speed of Downstream = 8,Speed of upstream = 6 Rate of the stream = ½ (8-6) km/hr = 1 km/hr

205 Boats and Streams A Swimmer can swim a certain distance in the direction of current in 5 hr and return the same distance in 7 hr. If the stream flows at the rate of 1 kmph, find the speed of the swimmer in still water?

206 Boats and Streams Answer: Let the Speed of the Swimmer in still water be x Distance covered in going = 5(x+1) km Distance covered in returning = 7(x-1) km 5(x+1) = 7(x-1) x = 6 Speed of the swimmer in still water = 6 kmph.

207 Time and Distance Speed:- Distance covered per unit time is called Speed. Speed = Distance / Time Distance = Speed * Time Time = Distance / Speed

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

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

210 If a man walks at the rate of 5 kmph he misses a train by only 7 minutes. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station. Sathyam Question Time and Distance

211 Solution: x/5-x/6 = 12/60 ( Difference 7+5=12) Solving the equation, x = 6 The distance covered is 6 km. Time and Distance

212 The J and K express from Delhi to Srinagar was delayed by snowfall for 16 minutes and made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed. Find the original speed of the express? Time and Distance

213 Answer: The train saves 16 minutes by travelling faster over a section of 80 km 80/x – 80/(x+10) = 16/60 By solving we get x = 50 Time and Distance

214 If the total distance of a journey is 120 km. If one goes by 60 kmph and comes back at 40 kmph what is the average speed during the journey? Sathyam Question Time and Distance

215 Solution: Average Speed = 2xy / (x + y) The average speed is 48 kmph. Time and Distance

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

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

218 Time and Distance If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. Find his actual distance traveled?

219 Time and Distance Solution: Let the distance be x km. Then, x/10 = (x+20) / 14 14x = 10x x = 200 x = 50 km.

220 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 ?

221 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.

222 Time and Distance (Trains) Two trains 200 m and 150 m long are running on the parallel rails at the rate of 40 km/hr and 45 km/hr. In how much time will they cross each other if they are running in the same direction? Sathyam Question

223 Time and Distance (Trains) Answer: Relative speed = x – y = 45 – 40 = 5 km/h =5*5/18= 25/18 m/s For 25/18 m it takes 1 sec For 350 m it takes = 18 * 350 /25 =252 Time taken = 252 seconds

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

225 Time and Distance (Trains) Solution: To reach 11 poles it takes 24 sec For 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 sec

226 Gain =S.P-C.P Loss =C.P-S.P Loss or gain is always reckoned on C.P. 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

227 A man buys an article for Rs and sells it for Rs Find his gain percentage. Sathyam Question

228 Profit and Loss Answer: Gain 4%

229 Mr. Ravi buys a cooler for Rs For how much should he sell so that there is a gain of 8% Profit and Loss

230 Answer: C.P = 4500 Profit = 8/100*4500 = 360 S.P = C.P + Profit = = 4860 He should sell it for Rs Profit and Loss

231 Sundeep buys two CDs for Rs.380 and sells one at a loss of 22% and the other at a gain of 12%. If both the CDs are sold at the same price, then the cost price of two CDs is ? Profit and Loss

232 Solution: C.P of two CDs =Rs.380 S.P of 1 st CD = x – 22 x /100 S.P of 2 nd CD = y + 12y /100 SP1 = SP2 x – 22x/100 = y – 12y/100 x = 56/39 y x + y = /39y + y = 380 y = 156 x = 224 Cost of the two CDs are Rs. 224 and Rs.156 Profit and Loss

233 A tradesman fixed his selling price of goods at 30% above the cost price. He sells half the stock at this price, one quarter of his stock at a discount of 15% on the original selling price and rest at a discount of 30% on the original selling price. Find the Gain Percent altogether? Profit and Loss

234 Solution: Let us take C.P of goods = 100 The Marked price = 130 =1/2* /4*0.85* /4 *0.7*130 =15.375% Profit and Loss

235 Rajesh buys an article with 25% discount on its Marked Price. He makes a profit of 10% by selling it at Rs 660. Find the marked price?

236 Profit and Loss Solution: S.P = Rs.660 Profit = 10 % C.P = 110 / 100*660 = Rs.600 Let x be M.P 75% of x = 600 x = 800 Marked Price = 800

237 Profit and Loss A shopkeeper gains the cost of 8 meters of thread by selling 40 meters of thread. Find his gain percentage?

238 Profit and Loss Answer: Gain = 20%

239 Profit and Loss Manoj sells a shirt to Yogesh at a profit of 15% and Yogesh sells it to Suresh at a loss of 10%. Find the resultant profit or loss.

240 Profit and Loss Solution: Resultant profit = (x + y + xy/100) The resultant profit is 3.5%

241 Profit and Loss Aditya purchases toffees at Rs. 10 per dozen and sells them at Rs. 12 for every 10 toffees. Find the gain or loss? TCS Question

242 Profit and Loss Answer: Profit = 44

243 Profit and Loss Anirudh bought 8 lemons for a rupee but sells only 6 lemons for a rupee. Find his profit percentage. CTS Question

244 Profit and Loss Answer: Profit 33 1/3%

245 Profit and Loss A radio when sold at a certain price gives a gain of 20%. What will be the gain if it sold for thrice the price? Sathyam Question

246 Profit and Loss Solution: Let C.P be Rs Hence S.P = Rs. 120 Gain = S.P – C.P = 360 – 120 = 260 Gain =Rs. 260

247 Profit and Loss A grocer purchased 80 kg of rice at Rs per kg and mixed it with 120 kg rice Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%? Sathyam Question

248 Profit and Loss Solution: C.P of 200 kg = (80* *16) =3000 S.P of 200 kg = 116% of C.P = 3480 The selling price of 1 kg is Rs

249 Profit and Loss A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks get 12% more than the passing marks. Find the maximum marks. Sathyam Question

250 Profit and Loss Answer: The maximum marks = 100

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

252 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.

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

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

255 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

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

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

258 On what dates of April 2005 did Sunday fall?

259 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 = ( ) -2 = = 12 / 7 = 5 = Friday. 7 1st is Friday means Sunday falls on 3,10,17,24

260 Calendar What was the day on 5 th January 1986

261 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

262 Clock: In every minute, 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

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

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

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

266 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.

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

268 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.

269 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?

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

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

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

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

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

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

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

277 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

278 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:

279 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

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

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

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

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

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

285 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

286 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)

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

288 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

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

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

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

292 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

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

294 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.

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

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

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

298 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?

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

300 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

301 Cubes Answer: = 8

302 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?

303 Cubes Answer:

304 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..

305 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?

306 Cubes Answer:

307 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 ?

308 Cubes Answer:

309 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?

310 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

311 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?

312 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

313 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?

314 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

315 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?

316 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

317 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.

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

319 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?

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


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