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# Blue Lotus

Blue Lotus. Aptitude Numerical Reasoning. Numerical Reasoning. Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn Diagram. Numerical Reasoning. Area and Volume Probability Time and Work (Pipes) SI and CI Average

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## Blue Lotus

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1. Blue Lotus Aptitude Numerical Reasoning

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

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

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

5. Problems on Numbers Arithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d; Sum of n terms of A.P Sn = n/2 *(a + L) or n/2 *[2a+(n-1)d)] a = 1st term, n = number of term, d= difference, Tn = nth term Geometrical Progression: Tn = arn – 1. Sn = a(rn – 1)/(r-1); a = 1st term , r = 1st term / 2nd term

6. Basic Formulae 1. ( a+b)2 = a2 + b2 + 2ab 2. (a-b)2 = a2 +b2 -2ab 3. ( a+b)2 -(a – b)2 = 4ab 4. (a+b)2 + (a – b)2 = 2 (a2 +b2) 5. (a2 – b2) = (a+b) (a-b) 6. (a+b+c)2 =a2 +b2 +c2 + 2(ab +bc+ca) 7. (a3 +b3) = ( a+b) (a2 –ab +b2) 8. (a3 –b3) = (a-b) (a2 +ab + b2) 9. (a3+b3+c3 -3abc) = (a+b+c) (a2+b2+c2-ab-bc-ca) If a+b+c = 0, then (a3+b3+c3) =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. 5000. 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 + 100 (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/100 DR = 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 78633. 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 89601. It’s reverse is 10968 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. Problems on Ages 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 ?

26. Problems on Ages 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.

27. Problems on Ages 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 x2 years” and also said “ We are talking about 20th century” What is the year of birth of grand father?

32. Problem on Ages Solution: 20th century means 1900 – 2000 year Perfect square 44 * 44 = 1936 (present) Year of birth = 1936 - 44 = 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 = 40 + 2N 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 = 50 -42 =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 = 6x Saraswathi’s age = 7x Four years hence [(6x + 1) + 4] / [(7x+1) + 4] = 7 / 8 Simplifying x =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/r12 R2 α 1/r22 R1: R2 = 1/r12: 1/r22 = 1/12: 1/22 = 1/1: 1 / 4 = 4:1 Ratio of rates of flow is 4:1

47. Ratio and Proportion I participated in a race. 1/5th of the participants were before me and 5/6th 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

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