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

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

1 Blue Lotus Aptitude 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 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); a = 1 st term, r = 1 st term / 2 nd term Problems on Numbers

6 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

7 Problems on Numbers A man ate 100 Bananas in 5 days, each day eating 6 more than the previous day. How many bananas did he eat on the first day?

8 Problems on Numbers Solution: First day be x Then x+6, x+12, x+18, x+24 5x + 60 = 100 x=8 In first day he ate 8 Bananas.

9 Problems on Numbers The traffic light at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively. If they all change simultaneously at 10:20:00 hours then at what time they again change simultaneously?

10 Problems on Numbers Solution: LCM (48,72,108) = 432 seconds = 7 minutes 12 seconds The next change will be at 10:27:12 hours

11 Problems on Numbers Raja had 85 currency notes in all, some of which were of Rs. 100 denomination and the remaining of Rs. 50 denomination. The total amount of all these currency notes was Rs How much amount did she have in the denomination of Rs. 50?

12 Problems on Numbers Solution: Let x be the number of Rs. 50 notes 50x (85-x) = 5000 Solving the above equation, Amount = Rs. 3500

13 Problems on Numbers Mr. Raja is on tour and he has Rs. 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by Rs. 3. For how many days is Mr. Raja on tour?

14 Problems on Numbers Solution: Let x be the number of days 360 / x – 360 / (x+4) = 3 Solving the above equation, The number of days of the tour x = 20 days

15 Problems on Numbers If the numerator of a fraction is increased by 25% and the denominator is decreased by 20% the new value is 5/4. What is the original value?

16 Problems on Numbers Solution: NR = x + 25x/100DR = y -20y/100 (5x/4)/(4y/5) =5/4 Solving the equation, The fraction is 4/5

17 Problem on Numbers There is a circular pizza with negligible thickness that is cut into X pieces by 4 straight line cuts. What is the maximum and minimum value of x respectively?

18 Problem on Numbers Answer: Maximum = 11. Minimum = 5.

19 Problem on Numbers A man was engaged on a job for 30 days on condition that he would get a wage of Rs. 10 for the day he works but he has to pay a fine of Rs. 2 for each day of his absence. If he gets Rs. 216 at the end. Find the number of days he was absent?

20 Problem on Numbers Solution: 10x – 2(30 –x) = 216 x = 23 Absent for = (30 – 23) = 7 days

21 Problems on Numbers I bought a car with a peculiar 5-digit number license plate which on reversing could still be read. On reversing the value is increased by What is the original number if all the digits were different from 0 - 9?

22 Problems on Numbers Solution: Only 0,1,6,8,9 can be reversed. Hence the number is It’s reverse is and difference is 78633

23 Problems on Numbers Naval collected 8, spiders and beetles in a little box. When he counted the legs he found that they were altogether 54. How many beetles and how many spiders did he collect?

24 Problems on Numbers Solution: Spiders have 8 legs and beetles have 6 legs. S + B = 8,8S + 6B = 54. Solving the above equations The number of spiders = S = 3 The number of beetles = B = 5

25 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 ? Problems on Ages

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

27 A man’s age is 125% of what it was 10 years ago, but 83 1/3% of what it will be after ten years. What is his present age?

28 Problems on Ages Solution: Let the present age be x years, Then, 125% of (x-10) = x 83 1/3% of (x+10) = x 125% of (x-10) = 83 1/3 % of (x+10) 5/4 (x-10) = 5/6 (x+10) 5x/12 = 250 / 12 x = 50 years

29 Problem on Ages My age three years hence multiplied by 3 and from that subtracted three times my age three years ago will give you my exact age. Find my age?

30 Problem on Ages Solution: (x+3)3 – 3(x-3) = x x =18

31 Problem on Ages A boy asks his father “ What is the age of grand father? Father replied “ He is x years old in x 2 years” and also said “ We are talking about 20 th century” What is the year of birth of grand father?

32 Problem on Ages Solution: 20 th century means 1900 – 2000 year Perfect square 44 * 44 = 1936 (present) Year of birth = = 1892

33 Problem on Ages A wizard named Nepo says “ I am only three times my son's age. My father is 40 years more than twice my age. Together the three of us are a mere 1240 years old. How old is Nepo?

34 Problem on Ages Solution: N = 3S F = N F + N + S = 1240 Solve 3 equations Age of Nepo = 360 years

35 Problems on Ages Joe’s father will be twice his age 6 years from now. His mother was twice his age 2 years before. If Joe will be 24 two years from now, what is the difference between his father’s & mothers age? (TCS)

36 Problems on Ages Solution: F + 6 = 2(J +6) M-2 = 2(J-2) Joe is now 22, F = 2(22) +12 – 6 = 50 M = 2(22) – 4+2 = 42 Difference = =8

37 Problems on Ages One year ago the ratio of Baskaran’s and Saraswati’s age was 6 : 7 respectively. Four years hence, this ratio would become 7 : 8. How old is Saraswathi?

38 Problems on Ages Solution: One year ago, Baskaran’s age = 6xSaraswathi’s age = 7x Four years hence [(6x + 1) + 4] / [(7x+1) + 4] = 7 / 8 Simplifyingx =5 Saraswathi’s present age is 7x +1 = 36 years

39 Puzzle Can you make 120 by using 5 zeros?

40 Puzzle Solution: Fact (0! + 0! + 0! + 0! + 0!) = Fact (5) = 5! = 120

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

42 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

43 Ratio and Proportion A sporting goods store ordered an equal number of white and yellow balls. The tennis ball company delivered 45 extra white balls making the ratio of white balls to yellow balls 1/5 : 1/6. How many white tennis balls did the store originally order for? (TCS Question)

44 Ratio and Proportion Solution: Let the number of yellow balls be x (x + 45) : x = 1/5 : 1/6 Solving the above equation, The number of white balls originally ordered would be = 225 balls

45 Ratio and Proportion The ratio of the rate of flow of water in pipes varies inversely as the square of the radius of the pipes. What is the ratio of the rates of flow in two pipes of diameters 2 cm and 4 cm respectively?

46 Ratio and Proportion Solution: R1 = 1st Pipe R2 = 2nd Pipe R1 α 1/r1 2 R2 α 1/r2 2 R1: R2 = 1/r1 2 : 1/r2 2 = 1/1 2 : 1/2 2 = 1/1: 1 / 4 = 4:1Ratio of rates of flow is 4:1

47 Ratio and Proportion I participated in a race. 1/5 th of the participants were before me and 5/6 th of them behind me. Find the total number of participants. (Infosys Question)

48 Ratio and Proportion Solution: Let the total number of participants be x. (x – 1)/5 + 5 (x-1)/6 = x Solving the above equation, The total number of participants would be = 31

49 Ratio and Proportion If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D?

50 Ratio and Proportion Solution : (A /B *B/C*C/D) =2/3*4/5*6/7 A : D = 16 : 35

51 Ratio and Proportion A bag contains 50 paise, 25 paise and 10 paise coins in the ratio 5 : 9 : 4, amounting to Rs Find the number of coins of each type.

52 Ratio and Proportion Solution : The actual ratio would be 5 * 50/100 : 9 * 25/100 : 4 * 10/100 = 50 : 45 : 8 Value of 50 paise coins=206*50/103=100 25paise coins =206*45/103=90 10paise coins =206*8/103 =16 Deriving from this ratio, Number of 50 paise coins = 100*2=200 Number of 25 paise coins = 90*4 =360 Number of 10 paise coins = 16*10 =160

53 Ratio and Proportion There are 20 poles with a constant distance between each pole. A car takes 24 seconds to reach the 12 th pole. How much time will it take to reach the last pole?

54 Ratio and Proportion Solution: 11 : 19 = 24 : x By Solving x = seconds

55 Ratio and Proportion Annual incomes of A and B is in the ratio 5 : 4 and their annual expenses bear a ratio of 4 : 3. If each of them saves Rs. 500 at the end of the year then find the annual income.

56 Ratio and Proportion Solution: IncomeA : B = 5x : 4x Expenses A : B = 4y : 3y 5x – 4y = 500 and 4x – 3y = 500 Solving the above equations, A’s annual income = Rs B’s annual income = Rs. 3500

57 Puzzle Can you draw three concentric circles with a line passing through their center without lifting hand?

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

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

60 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

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

62 In what ratio two varieties of tea one costing Rs. 27 per kg and the other costing Rs. 32 per kg should be blended to produce a blended variety of tea worth Rs. 30 per kg. How much should be the quantity of second variety of tea, if the first variety is 60 kg? Alligation or Mixture

63 Solution: Quantity of cheaper tea=2 Quantity of superior tea3 Quantity of cheaper tea =2*x/5 = 60, x=150 Quantity of superior tea = 3 * 150/5 = 90 kg Alligation or Mixture

64 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 be neither gains nor loss. What does each cow cost?

65 Alligation or Mixture Solution: Ratio = 5 : 4 Cost of first cow = 1350*5/9=Rs. 750 Cost of second cow = Rs. 600

66 Three types of tea A,B,C costs Rs. 95/kg, Rs. 100/kg. and Rs 70/kg respectively. How many kg of each should be blended to produce 100 kg of mixture worth Rs.90/kg given that the quantities of B and C are equal? Sathyam Question Alligation or Mixture

67 Answer: B+C/2A Ratio is 1:1 so A = 50, B + C = 50 The quantity would be 50 : 25 : 25 Alligation or Mixture

68 In what proportion water must be added to spirit to gain 20% by selling it at the cost price? Alligation or Mixture

69 Solution: Profit%=20% Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/ /3 5/325/3 The ratio is 1: 5 Alligation or Mixture

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

71 Chain Rule If 36 men can do a piece of work in 25 days then in how many days 15 men will do it?

72 Chain Rule Solution: MenDays x(Indirect) x = 36 * 25/15 x = 60 days

73 If 12 carpenters working 6 hours a day can make 460 chairs in 24 days, how many chairs will 18 carpenters make in 36 days, each working 8 hours a day? Chain Rule

74 Solution: MenDaysHoursChairs x x/460=18*8*36/12*6*24 x=1380 They will make 1380 chairs. Chain Rule

75 If 1 man can load 1 box in a truck in 9 minutes and a truck can hold up to 8 boxes. How many trucks completely are loaded when 18 men work for 1.5 hours? ( CTS )

76 Chain Rule Solution: MenBoxMinTrucks x x = 18 *1* 90/1*8 * 9 = 22.5 The number of trucks completely loaded = 22

77 Chain Rule A garrison of 3300 men has provisions for 32 days when given at a rate of 850 grams per head. At the end of 7 days reinforcement arrives and it was found that now the provisions will last 8 days less when given at the rate of 825 grams per head. How many more men can it feed?(Sathyam question)

78 Chain Rule Solution: MenDaysGrams x25825 x = 3300 * 25 * 850/17 * 825 = – 3300 = 1700 It can feed 1700 more men.

79 Chain Rule If 3 men or 6 boys can do a piece of work in 10 days, working 7 hours a day. How many days will it take to complete a piece of work twice as large with 6 men and 2 boys working together for 8 hours a day?

80 Chain Rule Solution: One man’s work = 2 boys’ work So, 6 men and 2 boys = 6 (2) + 2 = 14 boys BoysHoursDaysWork x2 x = 6 * 7 * 10 * 2/14 * 8 They will finish the work in 7 ½ days.

81 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

82 A and B are two partners in a business. A contributes Rs for 5 months and B Rs. 750 for 4 months. If total profit is Rs. 450, find the respective shares.

83 Partnership Solution: Ratio is1200 * 5 : 750 * 4 = 2 : 1 Profit share of A = 2/3 * 450 = Rs. 300 Profit share of B = Rs. 150

84 Partnership 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 %?

85 Partnership Solution: Let the total profit be Rs. 100 After paying charity A’s share = 3/5 *95 = 57 If A’s share is Rs. 57, the total profit is 100 If A’s share is Rs. 855, the total profit is 100 * 855/57 = Rs The total profit = Rs. 1500

86 Partnership A,B,C entered into a partnership by making an investment in the ratio of 3 : 5 : 7. After a year C invested another Rs while A withdrew Rs The ratio of investments then changed into 24: 59 : 167. How much did A invest initially?

87 Partnership Solution: Let the investments of A, B, and C be 3x, 5x, 7x (3x – 45600) : 5x : (7x ) = 24 : 59 : 167 (3x – 45600)/5x = 24/59 x = Initial investment of A = * 3 = Rs

88 A,B,C started a business each investing Rs After 5 months A withdrew Rs. 5000, B withdrew Rs and C invested Rs more. At the end of the year a total profit of Rs was recorded. Find the share of each. (Satyam Question) Partnership

89 Solution: (20000* *7) : (20000* *7) : (20000* *7) The ratio is 205 : 212: 282 A’s share = 205/699 * = Rs B’s Share = Rs C’s share = Rs Partnership

90 A starts business with a capital of Rs B and C join with some investments after 3 and 6 months respectively. If at the end of a year the profit is divided in the ratio 2 : 3 : 5 respectively what is B’s investment in the business? Partnership

91 Solution: (1200 * 12) : ( x * 9) : (y * 6) = 2 : 3 : 5 (1200 * 12)/2 = 9x/3 x = 2400 B’s investment = Rs Partnership

92 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)

93 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

94 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

95 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

96 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

97 A can do a piece of work in 8 days. A undertook to do it for Rs With the help of B, he finishes the work in 6 days. Find B’s share. Time and Work

98 Solution: (A + B)’s 1 day work = 1/6 A’s 1 day work = 1/8 B’s 1 day work = 1/24 Ratio of their work = 3 : 1 B gets ¼ * 320 = Rs. 80 Time and Work

99 5 men or 8 women do equal amount of work in a day. A job requires 3 men and 5 women to finish the job in 10 days. How many women require to finish the job in 14 days? Satyam Question Time and Work

100 Solution: 5 men = 8 women, so 1 man = 8/5 women 3m + 5w = 3 (8/5) w + 5 w = 49/5 w DaysWomen 1049/5 14x The number of women required = 7 Time and Work

101 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 alternatively work will be completed in ? Time and Work

102 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 2 days work done by A & B = 8/ /120) = 67/ days work done = 67 * 8/600 = 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 4/75 = 120/7* 4/75 = 32/35 They will finish the work in /35 = 17 32/35 days Time and Work

103 Two men undertake to do a piece of work for Rs First man alone can do this work in 7 days while the second man alone can do this work in 8 days. If they working together complete this work in 3 days with the help of a boy, how should money be divided?

104 Time and Work Solution: Wages of the first man for 3 days = work done by him in 3 days * Rs = 3/7 * 1400 = Rs. 600 Wage of second man for 3 days = work done by him in 3 days * Rs = 3 / 8 *1400 = Rs. 525 Wages of the boy for 3 days = Rs – Rs. ( ) = Rs. 275 There shares will be 600, 525, 275.

105 Time and Work 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work? Satyam Question

106 Time and Work Answer: 1 man’s work = x 1 boy’s work = y 2x + 3y = 1/10and 3x + 2y = 1/8 Solving the above equations, The required number of days = 12½ days

107 Pipes A and B running together can fill a cistern in 6 minutes. If B takes 5 minutes more than A to fill the cistern then find the time in which A and B will fill the cistern separately. Time and Work (Pipes)

108 Answer: 1/x + 1/(x+5) = 1/6 Solving the above equation, Pipe A will fill the cistern in 10 minutes. Pipe B will fill the cistern in 15 minutes. Time and Work (Pipes)

109 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)

110 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)

111 One fill pipe A is 3 times faster than second fill pipe B and takes 32 minutes less than the fill pipe B. when will the cistern be full if both pipes are opened together ? Time and Work (Pipes)

112 Solution: B = x A = 3x B = x (1/3) = x/3 x – 32 =x/3 3x – 96 = x 3x – x = 96 2x = 96 x = 96/2 = 48 B = = 16 A and B together = 1/16 + 1/48 = 1/12 The cistern will be filled in 12 minutes when both the pipes are opened together. Time and Work (Pipes)

113 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?

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

115 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?

116 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

117 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.

118 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

119 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

120 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

121 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

122 Triangle: A = 1/2*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

123 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

124 Solution: Area of triangular field = ½ * 3.2 * 2.8 m2 = 4.48 m2 Cost = Rs.100 * 4.48 = Rs Area and Volume

125 Find the length of the longest pole that can be placed in a room 14 m long, 12 m broad, and 8 m high.

126 Area and Volume Solution: Length of the longest pole = Length of the diagonal of the room = √( ) = √ 404= m

127 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

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

129 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

130 Solution: No. of Boxes = 25*42*60 / 7*6*5 = boxes of cereal box can be placed. Area and Volume

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

132 Area and Volume Solution: = x + y + xy/100 = *10/100 = -19% Diminished area = 19%.

133 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?

134 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

135 Area and Volume A farmer owns a square field with a pole in one of the corners to which he tied his cow with a rope whose length is about 10 m. What is the area available for the cow to graze? (Caritor Question)

136 Area and Volume Answer: Area = 78.5 m 2

137 Area and Volume If the length of a rectangle is reduced by 20% and breadth is increased by 20%. What is the percentage change in the area? (Infosys Question)

138 Area and Volume Solution: x + y + (xy/100)% = – 400/100 = -4 The area would decrease by 4%

139 Area and Volume A power unit is there by the bank of the river of 750 m width. A cable is made from power unit to power a plant opposite to that of the river and 1500 m away from the power unit. The cost of the cable below water is Rs. 15 per meter and cost of cable on the bank is Rs. 12 per meter. Find the total cost.(TCS Question)

140 Area and Volume Answer: Cable under water is 750m, on the bank 750m (750 * 15) + (750 * 12) = Total cost = Rs

141 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) Where P(A  B)=0 Independent Events: P(A  B) = P(A) * P(B) Probability

142 There are 6 pairs of gloves of different sizes. In how many ways can you choose one for the left hand and one for the right hand such that they are not of the same pair? (Caritor Question) Probability

143 Solution: The number of possibilities are 6 * 6 = 36 They should not be from the same pair = =30 30 ways. Probability

144 A speaks the truth 80% of the times, B speaks the truth 60% of the times. What is the probability that they tell the truth at the same time? (Wipro Question) Probability

145 Answer : P(Both tell truth) = P(A) * P (B) = 0.8 * 0.6 P = 0.48 Probability

146 A group consists of equal number of men and women. Of them 10% of men and 45% of women are unemployed. If a person is randomly selected from the group find the probability for the selected person to be an employee. (Satyam Question) Probability

147 Solution: Let the number of men is 100 and women be 100 Employed men and women = (100-10)+(100-45) = 145 Probability = 145 / 200 = 29 / 40 Probability

148 The probability of an event A occurring is 0.5 and that of B is 0.3. If A and B are mutually exclusive events. Find the probability that neither A nor B occurs?

149 Probability Solution: It is Mutually exclusive events P(A n B)=0 Probability = 1 – ( P(A) + P (B) – P(A n B) ) = 1 – ( – 0) = 0.2

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

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

152 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 }

153 If the compound interest on a certain sum for two years is Rs and simple interest is Rs. 60, then the rate of interest per annum is ? Simple / Compound Interest

154 Solution: SI for 2 years 60 = P*2*R/100 PR/ 100 = 30 C.I – S.I = P (R/100) 2 P(R/100 * R/100) = * R/100 =0.6 = 2% The rate of interest is 2% Simple / Compound Interest

155 Find in what time a given sum of money will quadruple itself at simple interest at the rate of one paise per rupee per month ? Simple / Compound Interest

156 Solution: 4P after n years 4P = P+ P*n*12/100 4 = 1+ 12n/100 n = 25 years The period is 25 years. Simple / Compound Interest

157 If Rs is invested at 5% simple interest if the interest is added to the principal every ten years, the amount will become Rs after how many years ? Simple / Compound Interest

158 Solution: Amount = Principle + interest for 10 years = ( 1000*10*5/100 ) = after 10 years 2000 after n years 500 = 1500*5*n/100 n = 6 2/3year Total /3 = 16 2/3 Total time is 16 2/3 years Simple / Compound Interest

159 Simple/compound interest A sum of money amounts to Rs after 3 years and to Rs after 6 years on C.I. Find the sum. (Satyam Question)

160 Simple/Compound interest Solution: P(1 + R/100) 3 = P(1 + R/100) 6 = Dividing(1 + R/100) 3 = 10035/6690 = 3/2 Substitute in equation 1 then P(3/2)=6690 P = 6690 * 2/3 = 4460 The sum is Rs.4460

161 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?

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

163 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)

164 Average A batsman makes a score of 90 runs in the 17 th innings and then increases his average by 3. Find his average after 17 th innings. (Infosys Question)

165 Average Solution: Let the average of 17 th innings be x Average after the 16 th innings = x-3 16(x-3) + 90 = 17x x = 42

166 Average 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?

167 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)

168 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?

169 Average Solution: 30 students total age = 30*9=270 Including the teacher’s age = 31*10=310 Difference = 310 – 270 = 40 years

170 Average Of the three numbers second is twice the first and is also thrice the third. If the average of the three numbers is 44, find the largest number.

171 Average Answer: Let the first number be x Average = (x + 2x + 2x/3)/3 = 44 Largest number is 72

172 Average In a coconut groove, (x+2) trees yield 60 nuts per year, x trees yield 120 nuts per year and (x-2) trees yield 180 nuts per year. If the average yield per year per tree be 100 find x.

173 Average Answer: [(x+2)*60 + (x*120) + (x-2) *180] =100 [x+2+x+x-2] By simplifying x = 4

174 Average In a cricket team of 11 boys one player weighing 42 kg is injured and his place is taken by another player. If the average weight of the team is increased by 100 g as a result of this, then what is the weight of the new player?

175 Average Solution: Average weight of 11 boys is increased by 0.1 kg The total increase in weight = 0.1 * 11 = 1.1 kg Weight of the new boy = = 43.1 kg

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

177 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

178 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

179 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

180 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

181 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

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

183 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 addition = 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7 = 127 Number of ways is 127

184 Permutations and Combinations If nP 5 = 20 nP 3 then what is the value of n?

185 Permutations and Combinations Solution: n!/(n-5)! = 20 * n!/(n-3)! n = 8

186 Permutations and Combinations Find the number of different 8 letter words formed from the letters of the word EQUATION if each word is to start with a vowel

187 Permutations and Combinations Solution: For the words beginning with a vowel, the first letter can be any one of the 5 vowels, the remaining 7 places can be filled by 7P 7 = 5040 The number of words = 5 * 5040 = 25200

188 Permutations and Combinations In how many different ways can the letters of the word TRAINER be arranged so that the vowels always come together?

189 Permutations and Combinations Solution: A,I,E can be arranged in 3! Ways (5! * 3!) / 2! = 360 ways

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

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

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

193 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 n umber 6x / 10 = y/100 * 7x /10 y = 600 / 7 y = 85 5/7 85 5/7%

194 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?

195 Percentage Solution: Let total salary =100% Salary = 100 Spending: Machinery + Raw material + staff = = 85 Remaining percentage = 100 – 85 = = = ? By chain rule, we get In this , 75% for machinery & raw material = 4, 00,000* 75/100 =Rs

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

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

198 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/ liters, B = 25/2 +50 = 62.5 liters Required Percentage = 37.5%

199 Percentage Answer: Increase % in the area of the new rectangle =x + y + (xy/100)% = (200/100) =32% Increase 32%

200 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?

201 Percentage Answer: B’s income = 40/(100-40) * 100% = 66 2/3%

202 Percentage Ramesh loses 20% of his pocket money. After spending 25% of the remainder he has Rs. 480 left. What was his pocket money?

203 Percentage Answer: x (1-20/100) (1-25/100) = 480 Solving the equation, x = 800 Pocket money = Rs. 800

204 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

205 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

206 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 x =12 Boats and streams

207 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?

208 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

209 Boats and Streams A man rows to place 48km 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?

210 Boats and Streams Solution: Down stream 4km in x hours. Then, Speed of Downstream = 4/x km/hr, Speed of Upstream = 3/x km/hr 48/ (4/x) + 48/(3/x) = 14, x = 1/2 Speed of Down stream = 8, Speed of upstream = 6 Rate of the stream = ½ (8-6) km/hr = 1 km/hr

211 Boats and Streams In a stream running at 2 km/h a motor boat goes 10 km upstream and back again to the starting point in 55 minutes. Find the speed of the motor boat in still water.

212 Boats and Streams Solution: Let the speed be x km/h 10/(x+2) + 10/(x-2)=55/60 Solving the above equation, x = 22 km/h

213 Boats and Streams A swimmer can swim a certain distance in the direction of current in 5 hours and return the same distance in 7 hours. If the stream flows at the rate of 1 km/h, find the speed of the swimmer in still water.

214 Boats and Streams Solution: Let the speed of the swimmer in still water be x Downstream speed = (x+1) * 5 Upstream speed = (x-1) * 7 5(x+1) = 7(x-1) Solving the equation, x = 6 Speed of the swimmer = 6 km/h

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

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

217 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

218 The jogging track in a sports complex is 726m in circumference. Suresh and his wife start from the same point and walk in opposite direction at 4.5km/hr and 3.75km/hr respectively. They will meet for the first time in how many minutes ? Time and Distance

219 Solution: Suresh speed m/hr = 4.5*5/18 *60*60 = 4500 His wife speed m/hr = 3.75*5/18*60*60 = 3750 Meeting time 4500/60 x +3750/60x = 756 m x = 5.28 m They will meet after 5.28minutes Time and Distance

220 Ram travels from P to Q at 10km/hr and returns at 15km/hr. Shyam travels P to Q and returns at 12.5km/hr. If he takes 12 minutes less than ram then what is the distance between P and Q ? Time and Distance

221 Solution: Average Speed of Ram = 2(10*15)/25 = 12km/hr Average Speed of Shyam = 2(12.5 *12.5)/25 = 12.5 Ram’s speed in m/sec = 12*5/18 = 10/3 Shyam’s speed in m/sec = 12.5*5/18 = 625/36 (x/ 10/3) – (x/625/36) = 12*60 3x/10 – 30x/625 = 720 = 30 Distance between P and Q is 30 km Time and Distance

222 If I walk 30 miles/hr I reach 1 hour before and if I walk 20 miles/hr I reach 1 hour late. Find the distance between two points and if the exact time of reaching destination is 11 a.m. then find the speed with which I walk? Infosys Question Time and Distance

223 Solution: x/30-x/20 = 2 By Solving the equation, The total distance between two points = 120 miles TheAverage speed required = 24 miles/hr Time and Distance

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

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

226 Time and Distance A car travels a certain distance taking 7 hours in forward journey. During the return journey it increased the speed by 12 km/hr and takes the time of 5 hours. What is the distance traveled? (Satyam Question)

227 Time and Distance Solution: Let speed be x, D = S * T =7x 7x = (x+12) 5 Solving the above equation, x = 30 D = 7 * 30 = 210 Actual Distance = 210 km

228 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 ?

229 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.

230 Time and Distance (Trains) How long will a train 100 m long traveling at 72 km/hr take to overtake another train 200 m long traveling at 54 km/h? (Satyam Question)

231 Time and Distance (Trains) Solution: They are traveling in the same direction. Hence the Relative speed is x -y 72 – 54 = 18 km/h 18 * 5/18 = 5 m/sec To travel 5 m it takes 1 second To travel 300m it will take300/5 = 60 seconds It will take 60 seconds or 1 minute.

232 Time and Distance (Trains) Rajee starts her journey from Delhi to Bhopal and simultaneously Sheela starts from Bhopal to Delhi. After crossing each other they finish their remaining journey in 5 4/9 hours and 9 hours respectively. What is Sheela’s speed if Rajee’s speed is 36 km/hr.

233 Time and Distance (Trains) Answer: Rajee’s speed= Square root of (t2/t1) Sheela’s speed The speed of Sheela = 28 km/hr

234 Time and Distance (Trains) Excluding stoppages, the speed of a train is 54 km/hr and including stoppages it is 45 km/hr. For how many minutes does the train stop per hour?

235 Time and Distance (Trains) Solution: x = _____________ x/45 + x/36 Take L.C.M and find the x value x = 12 It stops 12 minutes/hr

236 Time and Distance (Trains) A train 130 m long passes a bridge in 21 seconds moving with a speed of 90 km/hr. Find the length of the bridge?

237 Time and Distance (Trains) Solution: Let the length of the bridge be x Speed of the train = (length of the train + length of the bridge) / Time taken 90*5/18 = (130 +x )/21 Length of the bridge = 395 m

238 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

239 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

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

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

242 Solution: S.P = (100+8%/100)/CP = (108/100) * 4500 = 4860 He should sell it for Rs Profit and Loss

243 If the manufacturer gains 20%, the wholesale dealer 25%, and the retailer 35%, then find the cost of production if the retail price is Rs

244 Profit and Loss Solution:Let C.P be x Gain35% of 25% of 20% of x = /100 * 125/100 * 120/100 * x = 1265 x = Rs.624.7

245 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.

246 Profit and Loss Solution: Resultant profit = (x + y + xy/100) = 15 – 10 – (150/100) = 3½% The resultant profit is 3.5%

247 Profit and Loss Mr. Verma sold his scooter for Rs at a gain of 5%. Find the cost price of the scooter.

248 Profit and Loss Solution: C.P = (100/(100+P%))*S.P = (100/(100+5) )* = Cost price = Rs

249 Profit and Loss A man bought a horse and a cart. If he sold the horse at 10% loss and the cart at 20% gain, he would not lose anything, but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. What is the cost of one horse and one cart? Satyam Question

250 Profit and Loss Solution : C.P of horse = x; C.P of cart = y 10x/100 = 20y/100 5x/100 = 5y/ Solving the above two equations, C.P of horse = x = Rs. 400 C.P of cart = y = Rs. 200

251 Profit and Loss A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs less he would have gained 30%. Find the cost price of the article. Satyam Question

252 Profit and Loss Solution: First C.P = x, S.P = 125% of x = 5x/4 Second C.P = 80x/100 = 4x/5 S.P= 130/100 * 4x/5 = 26x/5 5x/4 - 26x/25 = x = 50 The cost price is Rs. 50

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

254 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.

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

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

257 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

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

259 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

260 On what dates of April 2005 did Sunday fall?

261 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

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

263 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

264 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

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

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

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

268 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.

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

270 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.

271 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?

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

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

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

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

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

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

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

279 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

280 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:

281 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

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

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

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

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

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

287 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

288 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)

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

290 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

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

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

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

294 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

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

296 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.

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

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

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

300 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?

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

302 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

303 Cubes Answer: = 8

304 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?

305 Cubes Answer:

306 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..

307 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?

308 Cubes Answer:

309 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 ?

310 Cubes Answer:

311 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?

312 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

313 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?

314 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

315 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?

316 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

317 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?

318 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

319 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.

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

321 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?

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


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