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# Numerical reasoning I

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A detailed presentation about the various in numerical reasoning.

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• Suppose x is for pen and y is for pencils then Cost of 36 pens and 42 pencils is Rs 460/- =&gt; 36x + 42y =460 =&gt;2(18x + 21y) =460 =&gt; 18x + 21 y =460/2 =230 Now the cost of 18 pens and 21 pencils? 18x+21y=? From (1) Ans is Rs 230/-

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• @Abhijeet Dwivedi his income is 18000.....x-0.8x-0.1x=1800 solving income(x)=18000

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• @Tirupati Rao 0.75x-(4/7)x=125 on solving x=700

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• In an examination, out of 80 students 85% of the girls and 70%

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• number of student studying in colleges Aand B are in the ratio of 3:4 respectively. if 50 more students join college A and there is no change in the number of student in college BY,the respective ratio became 5:6, what is the number in college b?

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### Numerical reasoning I

1. 1. Numerical Reasoning
2. 2. Problem on Numbers
3. 3. Problems on NumbersArithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d Sum of n terms of A.P S = n/2 *[2a+(n-1)d)]Geometrical Progression: Tn = arn – 1. Sn = a(rn – 1) / (r-1)
4. 4. Basic Formulae1. ( a+b)2 = a2 + b2 + 2ab2. (a-b)2 = a2 +b2 -2ab3. ( a+b)2 - (a – b)2 = 4ab4. (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)
5. 5. Problem - 1A 2 digit number is 3 times the sum of its digitsif 45 is added to the number. Its digits areinterchanged. The sum of digits of the numberis?
6. 6. SolutionThe number is 3 times the sum of its digits45 is added = 4 +5 = 9So, common numbers in 3 and 9th table.9, 18, 27, 36, 45….27 + 45 = 722 + 7 = 9 or 4 + 5 = 9
7. 7. Problem - 2A number when divided by 119 leaves aremainder of 19. If it is divided by 17. It willleave a remainder of?
8. 8. Solution= 19/17 = 2 remainder
9. 9. Problem - 3A boy was asked to find the value of 3/8 of sum ofmoney instead of multiplying the sum by 3/8 hedivided it by 8/3 and then his answer by Rs.55.Find the correct answer?
10. 10. Solution8/3 – 3/8 = 55/24 = 55/55/24 = 24
11. 11. Problem - 4A man spends 2/5rd of his earning. 1/4th of theexpenditure goes to food, 1/5th on rent, 2/5th ontravel and rest on donations. If his total earningis Rs.5000, find his expenditure on donations?
12. 12. Solution5000*2/5 = 2000Remaining amount has given as donation2000* (1/5 + 2/5 + ¼)Total amount = 200*17/20 = 17002000 – 1700 = 300
13. 13. Problem - 5From a group of boys and girls 15 girls leave.There are then left, 2 boys for each girl. Afterthis 45 boys leave, there are then left 5 girls foreach boy, find the number of girls in thebeginning?
14. 14. Solution15 girls leave = 2 boys for each girl45 boys leave = 5 girls fro 1 boyLet the boys be x; Girls = x/2 +15After the boys have left,No.of boys = x – 45 and girls = 5(x-45)x/2 = 5(x-45)X = 2(5x-22)X = 10x – 450X =5050/2 +15 =40
15. 15. Problem - 6An organization purchased 80 chairs froRs.9700. For chairs of better quality they paidRs.140 each and for each of the lower gradechair they paid Rs.50 less. How many betterquality chairs did the organization buy?
16. 16. SolutionBetter quality chairs = x;Lower quality = 80 –xPrice of better quality = Rs.140, Lower quality = 140-50 = 90140*x + 90(80-x) = 9700140x + 7200 – 90x = 970050x = 9700 – 7200;50x = 2500X = 50
17. 17. Problem - 7A labour is engaged for 30 days, on the conditionthat Rs.50 will be paid for everyday he worksand Rs.15 will be deducted from his wages foreveryday he is absent from work. At the end of30 days he received Rs.850 in all. For how manydays did he wanted?
18. 18. SolutionTotal wages = 30*50 = 1500 (without Absent)Wages received in 30 days = 850 (with Absent)Let the labourer work for x daysAbsent = 30 – x50x – (30-x)15 = 85050x -450 +15x = 85065x = 1300X = 1300/65 = 20 days
19. 19. Problem - 8The rent is charged at Rs.50 per day for first 3days Rs.100 per day next 5 days, and 300 perday thereafter. Registration fee is 50 at thebeginning. If a person had paid Rs.1300 for hisstay how many days did he stay?
20. 20. Solution3 days = 150 + 50 = 2005 days = 100*5 = 500 = 200 + 500 = 7001300 – 700 = 6002 days = 300*2 = 600= 5 + 3 + 2 = 10 days
21. 21. Problem - 9In a school 20% of students are under the age of8 years. The number of girls above the age of 8years is 2/3 of the number of boys above the ageof 8 years and amount to 48. What is the totalnumber of students in the school?
22. 22. SolutionGirls above 8 yrs = 48Boys above 8 yrs = 48 / 2/380% of students above 8 yrs = 48 + 72 = 120 80 120 20 x80x = 120*20X = 120*20/80 = 30Total No.of students = 120+30 = 150
23. 23. Ratio and Proportion
24. 24. Ratio and ProportionRatio : Relationship between two variables. =a:bProportion : Relationship between two ratios. =a:b::c:dProportion Calculation = a*d : b*c
25. 25. Problem - 1The ratio of number of boys to that of girls in aschool is 3:2. If 20% boys and 25% of girls arescholarship holders, find the percentage of theschool students who are not scholarship holders?
26. 26. SolutionLet the total number of students be 100Boys = 100*3/5 = 60Girls = 100*2/5 = 40S. holders = 60*20/100 = 12, non S. holders = 60 -12 = 48Girls s. holders = 40*25/100 = 10,Non s. holders = 40 – 10 = 30Students who do not have scholarship = 48 + 30 = 7878/100*100 = 78%
27. 27. Problem - 2The cost of diamond varies as the square of itsweight. A diamond weighing 10 decigrams costsRs. 32000. Find the loss incurred when it breaksinto two pieces whose weights are in the ratio2:3?
28. 28. Solution1st piece = 10*2/5 = 42nd piece = 10*3/5 = 6Cost of the diamond varies as square of its weight42 : 62 102 = 100k16k : 36k100k – 52 k = 48k(loss)100k = 32000; k = 32048*320 = 15360
29. 29. Problem - 3The ratio of the first and second class faresbetween two railway stations 4 : 1 and the ratioof the number of passengers traveling by firstand second class is 1:40. If the total of Rs.1100is collected as fare from passengers of bothclasses what was the amount collected from firstclass passengers?
30. 30. SolutionFare = 4 : 1Passengers traveling = 1 : 40Amount = No. pas * fare = 4*1 :10*1 = 4 : 40= 1:10Total amount = 1100.First class passengers‟ amount = 1*1100/11= 100
31. 31. Problem - 4A vessel contains a mixture of water and milk inthe ratio 1:2 and another vessel contains themixture in the ratio 3:4. Taking 1 kg each fromboth mixtures a new mixture is prepared. Whatwill be the ratio of water and milk in the newmixture?
32. 32. Solution1st vessel = water = 1/3 , milk = 2/32nd vessel = water = 3/7, milk = 4/7Water = 1/3 + 3/7 = 16/21Milk = 2/3 + 4/7 = 26/2116 : 26 = 8:13
33. 33. Problem - 5Ratio of the income of A, B, C last year 3 : 4 : 5.The ratio of their individual incomes of last yearand this year are 4:5, 2:3 and 3:4 respectively. Ifthe sum of their present income is Rs.78,800.Find the present individual income of A, B andC.
34. 34. SolutionA‟s Present Income = 5/4*3x = 15x/4B‟s Present Income = 3/2*4x = 12x/2C‟s Present Income = 4/7*5x = 20x/715x/4 + 6x+20x/3 = 78,800197x/12 = 78,800X = 945600/197X = 4,800A‟s Present income = 15x/4 = 15*4800/4 = 18,000B‟s Present income = 6*x = 6*4,800 = 28,800C‟s Present income = 20x/3 = 20*4800/3 = 32,000
35. 35. Problem - 6Of the three numbers, the ratio of the first andthe second is 8:9 and that of the second and thirdis 3:4. If the product of the first and thirdnumbers is 2,400, then find the second number?
36. 36. Solutiona:b=8:9b:c=3:4b : c = 3*3 : 4*3 = 9 : 12a : b : c = 8 : 9 : 12Product of first and third = 8k * 12k = 240096k2 = 2400; k2 = 2400/96 = 25k=5Second number = 9 * 5 = 45
37. 37. Problem - 7Annual income of A and B are in the ratio of 4 : 3and their annual expenses are in the ratio 3 : 2. Ifeach of them saves Rs.600 at the end of the year,what is the annual income of A?
38. 38. SolutionIncome = 4 : 3, Expenses = 3 : 2Savings 600 eachA‟s income = 4x, expenses = 3x,savings = x i.e 600Income = 4*600 : 3*600A : B = 2400 : 1800A income = 2400
39. 39. Problem - 8The property of a man was divided among hiswife, son and daughter according to his will asfollows. Wife‟s hare is equal to 6/7th of son‟sshare and daughter share is equal of 4/7th ofSon‟s. If the son and daughter together receivesRs.1,02,300. How much does his wife get?
40. 40. SolutionLet the Son‟s share be x.Daughter‟s share = x*4/7 = 4x/7Wife‟s share = x* 6/7 = 6x/7X + 4x/7 = 1,02,3007x + 4x = 1,02,300X = 1,02,300 /11 = 65,100Wife Share = 65,100 *6/7 = Rs. 55, 800
41. 41. Problem - 9A pot containing 81 litres of pure milk of themilk 1/3 is replaced by the same amount ofwater. Again 1/3 of the mixture is replaced bythe same amount of water. Find the ratio of milkto water in the new mixture?
42. 42. Solution Milk : WaterInitial = 81 : 01/3 removed = 54 : 271/3 mixture = 36 : 45Ratio of Milk and Water = 4 : 5
43. 43. Problem - 10729 ml of mixture contains milk and water are inthe ratio 7 : 2. How much more water is to beadded to get a new mixture containing milk andwater in the ratio of 7 : 3.
44. 44. SolutionWater = 729 * 2/9 = 162 Ratio Water 2 162 3 x2x = 3*162/2 = 243243 – 161 = 81 ml water is to be added
45. 45. Problem - 11Price of a scooter and a television set are in theratio 3 : 2. If the scooter costs Rs.600 more thanthe television set, then find the price oftelevision?
46. 46. SolutionDiff. in ratio = 3 – 2 = 11 ratio is 600 means, the television cost is 2 ratio so, cost of television = 1200
47. 47. Problem - 12The annual income and expenditure of man andhis wife are in the ratio of 5:3 and 3:1respectively, if they decide to save equally andfind their balance is 4000. Find their income atthe end of the year?
48. 48. SolutionMan and Wife income = 5 : 3 = 2 (diff)Man and Wife Expenses = 3 : 1 = 2 (diff)so, both of them are saving ratio of 2Total saving of Man and Women = 4000, individual saving 2000So, Man income = 5000 and Women income = 3000
49. 49. Problem - 13In a class room, ¾ of the boys are above 160 cmin height and they are in 18 number. Also out ofthe total strength, the boys are only 2/3 and therest are girls. Find the total number of girls in aclass?
50. 50. Solution¾ of the boys in 18 numbers means, ¼ of the boys = 6Total number of boys = 18+6 = 24Ratio Number2/3 241/3 x2/3*x = 24*1/3x = 24/2 = 12 Girls
51. 51. Problem - 14Rs. 770 was divided among A, B and C such thatA receives 2/ 9th of what B and C togetherreceive. Find A‟s share?
52. 52. SolutionA = 2/9 (B+C)B+C =9A/2A+B+C = 770A + 9A/2 = 77011A = 770*2A = 140
53. 53. Problem - 15A sporting goods store ordered an equalnumber of white and yellow balls. The tennisball company delivered 45 extra white ballsmaking the ratio of white balls to yellow balls1/5 : 1/6. How many white tennis balls did thestore originally order for?
54. 54. SolutionLet the number of yellow balls be x (x + 45) : x = 1/5 : 1/6Solving the above equation,The number of white balls originally orderedwould be = 225 balls
55. 55. Alligation and Mixture
56. 56. Alligation and MixtureAlligation : It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.(Quantity of cheaper / Quantity of costlier) (C.P. of costlier) – (Mean price)= -------------------------------------- (Mean price) – (C.P. of cheaper)
57. 57. Alligation or Mixture Cost of Cheaper Cost of costlier c d Cost of Mixture md-m m-c(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)
58. 58. Problem -1Three glasses of size 3 lit, 4 lit and 5 lit containmixture of milk and water in the ratio of 2:3, 3:7and 4:11 respectively. The content of all thethree glasses are poured into a single vessel.Find the ratio of milk and water in the resultingmixture.
59. 59. Solution1st Vessel = Milk = 3*2/5 = 6/5 = Water = 3*3/5 = 9/52nd Vessel: = Milk = 4*3/10 = 12/10 = Water = 4*7/10 = 28/103rd Vessel: = Milk = 5*4/15 = 20/15 = Water = 5*11/15 = 55/15Milk : Water = 6/5 +12/10 + 20/15 : 9/5 + 28/10 + 55/15 = 18/15 + 18/15 +20/15 : 27/15 + 42/15+55/15 = 56 : 124 (or) 14:31
60. 60. Problem - 2How many kg of tea worth Rs. 25 per kg must beblended with 30 kg tea worth Rs. 30 per kg, sothat by selling the blended variety at Rs.30 perkg there should be a gain of 10%?
61. 61. Solution30*100/110 = 300/1125 30 300/1130/11 25/1130 : 256 : 536 : 30kg
62. 62. Problem - 3A man buys cows for Rs. 1350 and sells one soas to lose 6% and the other so as to gain 7.5%and on the whole he neither gains nor loses. Howmuch does each cow cost?
63. 63. Solution 6 7.5 0 7.5 6 15 12 5 : 41350*5/9 = 7501350 *4/9 = 600
64. 64. Problem - 4There are 65 students in a class, 39 rupees aredistributed among them so that each boy gets80p and each girl gets 30p. Find the number ofboys and girls in a class.
65. 65. SolutionGirls Boys30 80 6020 302 : 365*2/5 = 2665*3/5 = 39
66. 66. Problem - 5A person covers a distance 100 kms in 10 hrPartly by walking at 7 km per hour and rest byrunning at 12 km per hour. Find the distancecovered in each part.
67. 67. SolutionSpeed = Distance / Time = 100 / 10 = 10 7 12 10 2 : 3Time taken in 7 km/hr = 10 * 2/5 = 44*7 = 28 kmTime taken in 12 km/hours = 10*3/5 = 612*6 = 72 km
68. 68. Problem - 6A merchant has 100 kg of salt, part of whichhe sells at 7% profit and the rest at 17% profit.He gains 10% on the whole. Find the quantitysold at 17% profit?
69. 69. Solution 7 17 10 (17-10) (10-7) 7 : 3The quantity of 2nd kind = 3/10 of 100kg = 30kg
70. 70. Problem - 7In what ratio two varieties of tea one costingRs. 27 per kg and the other costing Rs. 32 perkg should be blended to produce a blendedvariety of tea worth Rs. 30 per kg. How muchshould be the quantity of second variety of tea,if the first variety is 60 kg?
71. 71. Solution 27 32 30 2 3Quantity of cheaper tea = 2Quantity of superior tea 3Quantity of cheaper tea =2*x/5 = 60 , x=150Quantity of superior tea = 3 * 150/5 = 90 kg
72. 72. Problem - 8A 3-gallon mixture contains one part of S andtwo parts of R. In order to change it to mixturecontaining 25% S how much R should beadded?
73. 73. Solution R : S 2 : 1 75% : 25% 3 : 11 gallon of R should be added.
74. 74. Problem - 9Three types of tea A,B,C costs Rs. 95/kg, Rs.100/kg. and Rs 70/kg respectively. How manykg of each should be blended to produce 100 kgof mixture worth Rs.90/kg given that thequantities of B and C are equal?
75. 75. Solution B+C/2 A 85 95 90 5 5Ratio is 1:1 so A = 50 , B + C = 50The quantity would be 50 : 25 : 25
76. 76. Problem - 10In what proportion water must be added tospirit to gain 20% by selling it at the cost price?
77. 77. SolutionProfit%=20%Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3 0 10 25/3 5/3 25/3The ratio is 1: 5
78. 78. Problem - 11In an examination out of 480 students 85% ofthe girls and 70% of the boys passed. Howmany boys appeared in the examination if totalpass percentage was 75%
79. 79. SolutionSolution:70 85 7510 5Number of Boys = 480 * 10/15Number of Boys = 320
80. 80. Problem - 12A painter mixes blue paint with white paint sothat the mixture contains 10% blue paint. In amixture of 40 litre paint how many litre of bluepaint should be added, so that the mixturecontains 20% of blue paint?
81. 81. SolutionQuantity of blue paint in the mixture = 10% of 4040*10/100 = 440 – 4 = 36 litreLet x litre blur paint can be mixed4+x/30 = 20/80 = 4+x = 9x=5
82. 82. Problem - 13From a 100 litre mixture containing water andmilk equal proportion, 10 litres of mixture isreplaced by 10 litres of water in successiontwice. At the end, what is the ratio of milk andwater?
83. 83. Solution Milk Water10 lit(1st) 50 : 50 45 : 45 45 : 552nd 10 lit 40.5 : 49.5Add Water 40.5 : 59.5 81 : 119
84. 84. Problem - 14In a mixture of 400 gms, 80% is copper, sliver is20%. How much copper is to be added, so thatthe new mixture has 84% copper?
85. 85. Solution 400*80/100 = 320 Copper 400*20/100 = 80 SliverPercen Mixture 80 320 84 x= 320*84/80 = 336(320+x) = (400+x) 84/100320+x = 400+84/100 + 84x/10016x/100 = 336 – 320; 16x/100 = 16; x = 100
86. 86. Problem - 15A jar full of whisky contains 50% alcohol. A partof this whisky is replaced by another containing30% alcohol and now the percentage of alcoholwas found to be 35%. Find the quantity ofwhisky replaced?
87. 87. Solution 50 30 35 5 : 155 : 15 = 1 : 3Replaced = 3/4
88. 88. Partnership
89. 89. Type - 1A invest = 10000B invest = 15000Profit = 5000Find their Individual Share ?A : B = 10000 : 15000 = 2 : 3A‟s Share = 5000*2/5 = 2000B‟s Share = 5000*3/5 = 3000This is a first and basic step for any Partnership Problem.
90. 90. Type - 2A invest = 5000,After 3 months B joined A, with an investment of 3000Profit at the end of the year = 3500Find their Share ?Any thing happen after a month, like a person joining a business, or withdraw from business or withdraw some amount means given amount is for month. Cont…
91. 91. Type - 2A : B = 5000 : 3000 = 5*12 : 3*9 = 60 : 27 = 20 : 9A‟s share = 3500*20/29 = 2413.7B‟s Share = 3500*9/29 = 1086.3
92. 92. Type - 3A invest 5000B invest 6000After 3 months A withdraw amount 1000, after 5 months a withdraw amount 1000 again.Profit at the end of the Year = 5000Find their Share ?A = 5*3 + 4*5 + 3*4 = 15 +20 + 12 = 47B = 6*12 = 72
93. 93. Type - 3A‟s share = 5000* 47/119 = 1974.8B‟s share = 5000*72/119 = 3025.2
94. 94. Type - 4A invest twice as much as B, B invest 1/3rd of C. At the end of the year their Profit is 6000. Find their Share?A = 2BB = 1/3CC=xA : B : C = 2x/3 : x/3 : xA : B : C = 2x/3 : x/3 : 3x /3A:B:C=3:2:6A‟s Share = 6000*3/11 = 1636B‟s Share = 6000*2/11 = 1091C‟s Share = 6000*6/11 = 3273
95. 95. Problem - 1A, B and C started a business in partnership byinvesting Rs.12000 each. After 6 months, C leftand after 4 months D joined with his capital ofRs.24,000. At the end of a year, a profit ofRs.8,500 shared among all the partners. Find B‟sshare?
96. 96. These are all the basic types remaining we will see when we solve problems.
97. 97. Solution A : B : C : D 12000 : 12000 : 12000 : 24000 1 : 1 : 1 : 2 1*12 : 1*12 : 1*6 : 2*2 12 : 12: 6 : 4 6 : 6 : 3 : 2B‟s share = 6/17*8500 = 3000
98. 98. Problem - 2A, B and C enter into partnership. A contributesone third of the capital while B contributes asmuch as A and C together contributed. If theprofit at the end of the year amounted to Rs.840.What would be B‟s share?
99. 99. SolutionA‟s share = 1/3 of the capitalA‟s share = 1/3*840 = 280B‟s share = A + C = 280 + xA + B + C = 840280 + 280 + x + x = 840560 + 2x = 8402x = 840 – 560X = 140B‟s share = 280+140 = 420
100. 100. Problem - 3Akilesh and Jaga enter into a partnership.Akilesh contributing Rs.8000 and Jagacontributing Rs.10000. At the end of 6 monthsthey introduce Prakash, who contributesRs.6000. After the lapse of 3 years, they find thathe firm has made a profit of Rs.9660. FindPrakash‟s share?
101. 101. SolutionAkilesh : Jaga : Prakash 8 : 10 : 6 4 : 5 : 3 4*36 : 5836 : 3*30 144 : 180 : 90 8 : 10 : 5Prakash‟s share = 9660*5/23 = 2100
102. 102. Problem - 4Priya and Vijay enter into partnership. Priyasupplies whole of the capital amounting toRs.45,000 with the conditions that the profit areto be equally divided and that Vijay pays Priyainterest on half of the capital of 10% p.a. butreceives Rs.120 per month for carrying on theconcern. Find their total yearly profit. WhenVijay‟s income is one half of Priya‟s income?
103. 103. Solution45,000 *1/2 = 22,50022,500 *10//100 = 2250 (interest p.a)Vijay receives Rs.120 per month = 120*12 = 1440Total profit be xRatio of Profit sharing = 1 : 1Priya‟s income = x/2+2250Vijay‟s income = x/2 – 2250 +14401 Priya = ½ VijayPriya Income = Twice of Vijay incomex/2 + 2250 = 2(x/2 – 2250 +1440)X+4500/2 = 2(x/2 – 810)X+4500/2 = x – 1620 = x +4500 = 2x – 3240X = 7740Total Profit of the year = 7740+1440 = 9,180
104. 104. Problem - 5Revathy and Shiva are partners sharing profits inthe ratio of 2:1. They admit Pooja intopartnership giving her 1/5th share in profitswhich she acquires from Revathy and Shiva inthe ratio of 1:2. Calculate the new profit sharingratio?
105. 105. SolutionPooja gets her share of 1/5th of total share of Profit from Revathy and Shiva in the ratio 1 : 2From Revathy = 1/3*1/5 = 1/15From Shiva = 2/3*1/5 = 2/15Total Pooja share = 1/15+2/15 = 3/15 = 1/15Revathy share = 2/3 – 1/15 = 9/15Shiva share = 1/3 – 2/15 = 3/15Shares = Revathy : Shiva : Pooja = 3 : 1 : 1
106. 106. Problem - 6A and B started a partnership business investingsome amount in the ratio of 3 : 5. C joined themafter six months with an amount equal to that ofB. In what proportion should the profit at the endof 1 year be distributed among A, B and C?
107. 107. SolutionLet the investment, 3 : 5 : 5 3*12: 5*12 : 5*6 36 : 60 : 30 6 : 10 : 5
108. 108. Problem - 7If 4(A‟s capital) = 6(B‟s capital) = 10 (C‟scapital) then out of a profit of rs.4650. Find C‟sshare?
109. 109. SolutionLet the unknown value be xx/4 : x/6 : x/1015x/60 : 10x/60 :6x/6015 : 10 : 6C‟s share = 6/31*4650 = Rs. 900
110. 110. Problem - 8A, B, C subscribe Rs.50,000 fro business. Asubscribes Rs.4000 more than B and B Rs.5000more than C. Out of total profit of Rs.35,000.Find A‟s share?
111. 111. SolutionC = x,B = x + 5000A = x+5000+4000 = x + 9000x +x+5000 +x+9000 = 500003x+14000 = 500003x = 50000 – 140003x = 36000,x = 12000C : B : A12000 : 17000 : 21000A = 35000*21/50 = 14,700
112. 112. Problem - 9A and B are partners in a business, A contributes¼ of he capital for 15 months and B received 2/3of the profit. For how long B‟s money was used?
113. 113. SolutionB = 2/3A = 1/3A : B = 1/3 : 2/3 = 1 : 2Investment1/4x+15 : 3/4x*y15x/4 : 3xy/415x/4 : 3xy/4 : : 1 : 230x/4 = 3xy/4Y = 30x/4 * 4/3x = 10 months
114. 114. Problem - 10A, B and C invests Rs.4,000, Rs.5,000 andRs.6,000 respectively in a business and A gets25% of profit for managing the business and therest of the profit is divided by A, B and C inproportion to their investment. If in a year, Agets Rs.200 less than B and C together, what wasthe total profit for the year?
115. 115. SolutionTotal Profit = 10025% for managing the business = 100 – 25 = 75%A : B : C 4000 : 5000 : 6000 4 : 5 : 64x : 5x : 6x = 25x100*15x/75 = 20xA gets 4x + 25% of 20x = 4x + 20x *25/100 = 9xB = 5x, C = 6x(5x + 6x) – 9x = 20011x – 9x = 2002x = 200; x = 100Total Profit 20x = 20*100 = 2000
116. 116. Problem - 11A and B entered into partnership with capitals inthe ratio of 4 : 5. After 3 months, A withdraw ¼of his capital and B withdraw 1/5 of his capital.The gain at the end of 10 months was Rs.760.Find the share of B?
117. 117. SolutionA : B4 : 54000 : 5000A‟s share = 4000*1/4 = 4000 – 1000 = 3000B‟s share = 5000*1/5 = 5000 – 1000 = 4000A : B 3*4+3*7 : 5*3 +4*7 12 + 21 : 15+28 33 : 4360*43/76 = 430
118. 118. Problem - 12Rs. 1290 is divided between A, B and C. So, thatA‟s share is 1 ½ times B‟s and B‟s share is 1 ¾times C. What is C‟s share?
119. 119. SolutionA : B = 1 ½ : 1 = 3/2 : 1 = 3 : 2B : C = 1 ¾ : 1 = 7/4 : 1 = 7 : 4A :B :C =3*7(A) : 2*7(B) : 7*2(B) : 4*2(C) = 21 : 14 : 8B = 1290*8/43 = Rs.240
120. 120. Problem - 13A man starts a business with a capital ofRs.90000 and employs an assistant. From theyearly profit he keeps an amount equal to 4 ½ ofhis capital and pay 35% of the remainder of theprofits. Find how much the assistant receives in ayear, in which profit is Rs.30,000.
121. 121. SolutionInvestment = 90,0004 ½ of investment = 9/2/100*90000 = Rs.4050Profit = 30,000 – 4050 = 25,95035/100*25,950 =9082.50
122. 122. Problem - 14A and B invest in a business in the ratio 3 : 2. If5% of the total profit goes to charity and A‟sshare is Rs. 855, what is the total profit %?
123. 123. SolutionLet the total profit be Rs. 100After paying charity A‟s share = 3/5 *95 = 57If A‟s share is Rs. 57, the total profit is 100If A‟s share is Rs. 855, the total profit is 100 * 855/57 = Rs. 1500The total profit = Rs. 1500
124. 124. Problem - 15A,B,C entered into a partnership by making aninvestment in the ratio of 3 : 5 : 7. After a yearC invested another Rs. 337600 while A withdrewRs. 45600. The ratio of investments thenchanged into 24: 59 : 167. How much did Ainvest initially?
125. 125. SolutionSolution:Let the investments of A, B, and C be 3x, 5x, 7x(3x – 45600) : 5x : (7x + 337600) = 24 : 59 : 167 (3x – 45600)/5x = 24/59 x = 47200Initial investment of A = 47200 * 3 = Rs. 141600
126. 126. Problems on Age
127. 127. Problem - 1The age of the Father is 4 times the age of hisSon. If 5 years ago, Father‟s age was 7 times theage of his Son, what is the Father‟s present age?
128. 128. SolutionF = 4SF - 5 = 7(S - 5)4S – 5 = 7S – 353S = 30S = 10Father‟s age = 4* 10 = 40 years
129. 129. Problem - 2The age of Mr. Gupta is four times the age of hisSon. After Ten years, the age of Mr. Gupta willbe only twice the age of his Son. Find the presentage of Mr. Gupta‟s Son.
130. 130. SolutionG = 4SG + 10 = 2 ( S + 10)4S + 10 = 2S + 202S = 10S=5Son‟s Age = 5 years
131. 131. Problem - 310 years ago Anu‟s mother was 4 times olderthan her daughter. After 10 years, the motherwill be twice as old as her daughter. Find thepresent age of Anu.
132. 132. SolutionTen years before:M – 10 = 4(A – 10 )M – 10 = 4A – 40M = 4A – 40 + 10M = 4A – 30Ten Years After:M + 10 = 2(A + 10)M + 10 = 2A + 20M = 2A + 20 – 10M = 2A + 104A – 30 = 2A + 102A = 10 + 302A = 40: Anu‟s Age = 20
133. 133. Problem – 4The sum of the ages of A and B is 42 years. 3years back, the ages of A was 5 times the age ofB. Find the difference between the present agesof A and B?
134. 134. SolutionA + B = 42A = 42 – BA – 3 = 5 ( B – 3)A – 3 = 5B – 1542 – B – 3 = 5B – 1542 – 3 + 15 = 5B + B54 = 6BB = 54 /6 = 9A = 42 – B; A = 42 – 9 = 33Difference in their ages = 33 – 9 = 24 Years
135. 135. Problem - 5The sum of the ages of a son and father is 56years. After 4 years, the age of the father will be3 times that of the son. Find their respectiveages?
136. 136. SolutionF + S= 56S = 56 – FF + 4 = 3 (S + 4)F + 4 = 3 (56 – F + 4)F + 4 = 168 – 3F + 124F = 168 + 12 – 44F = 176 ; F = 44S = 56 – F ; S = 56 – 44 = 12Father Age = 44; Son Age = 12
137. 137. Problem – 6The ratio of the ages of father and son at presentis 6:1. After 5 years, the ratio will become 7:2.Find the Present age of the son.
138. 138. Solution6x + 5/x + 5 = 7/212x + 10 = 7x + 3512x – 7x = 35 – 105x = 25x = 25 / 5x = 5 yearsSon age = 1* 5 = 5 years
139. 139. Problem - 7The ages of Ram and Shyam differ by 16 years.Six years ago, Shyam‟s age was thrice as that ofRam‟s. Find their present ages?
140. 140. SolutionS = R + 16S – 6 = 3(R – 6)S – 6 = 3R – 18R + 16 – 6 = 3R – 18R + 10 = 3R – 182R = 28 ; R = 14Shyam‟s Age = 14 + 16 = 30.
141. 141. Problem - 8A man‟s age is 125% of what it was 10 yearsago, 83 1/3% of what it will be after 10 years.What is his present age?
142. 142. SolutionLet the age be x125% of (x – 10) = 83 1/3 % of (x +10)125/100 * x – 10 = 250/ 300 * x +105/4 x – 10 = 5/6 x – 105x / 4 – 5x / 6 = 50/6 + 50/45x /12 = 250/125x = 250 ; x = 50 years
143. 143. Problem - 93 years ago, the average age of a family of 5members was 17. A baby having born, theaverage age of the family is the same today.What is the age of the child?
144. 144. SolutionAverage age of 5 members = 17Total age of 5 members = 17*5 = 853 years later, the age of 5 members will be= 85 + 15 = 100100 + x / 6 = 17100 + x = 17*6100 + x = 102x = 102 – 100 = 2 years
145. 145. Problem - 10The sum of the age of father and his son is 100years now. 5 years ago their ages were in theratio of 2 : 1. The ratio of the ages of father andhis son after 10 years will be?
146. 146. SolutionF + S = 1005 years ago 2 : 15 years agoF + S = 100 – 10 = 9090*2/3 = 60 : 30Present age = 65 : 3510 years ago = 75 : 45=5:3
147. 147. Problem - 11Six years ago, Sushil‟s age was triple the age ofSnehal. Six years later, Sushil‟s age will be 5/3of the age of Snehal. What is the present age ofSnehal?
148. 148. SolutionSix years ago,Snehal = x; Sushil = 3xSix years later,3x + 6+6 = 5/3(x+6+6)9x +36 = 5x+604x = 60 – 36X=6Present Age of Snehal = 6+6 = 12 years
149. 149. Problem - 12Susan got married 6 years ago. Today her age is1¼ times that at the time of her marriage. Herson is 1/6 as old as she today. What is the age ofher son?
150. 150. Solution6 years ago Susan got married.So her son‟s age will be less than 6 years.Let as consider, her son‟s age is 5 years.Susan‟s Age is 5*6 = 30 yrs, since the son is 1/6th of Susan‟s age.6 years ago her age must have been 24 yrs24*1 ¼ = 24*5/4 = 30 yrsAs it satisfies the conditions her son‟s age is 5 years
151. 151. Problem - 13My brother is 3 years elder to me. My father was28 years of age when my sister was born, whilemy mother was 26 years of age, when I wasborn. If my sister was 4 years of age when mybrother was born, then, what was the age of myfather and mother respectively when my brotherwas born?
152. 152. SolutionMy brother was born 3 years before I was born and 4 years after my sister was bornFather‟s age when brother was born = 28 + 4 = 32 yearsMother‟s age when brother was born = 26 – 3 = 23 years
153. 153. Problem - 14If 6 years are subtracted from the present age ofGagan and the reminder is divided by 18, thenthe present age of his grandson Aunp is obtained.If Anup is 2 years younger to Madan whose ageis 5 years, then what is Gagan‟s present age?
154. 154. SolutionAnup‟s age = 5 – 2 = 3 yearsLet Gagan‟s age be x= x – 6 / 18 = 3x – 6 = 3*18 ; x – 6 = 54x = 54 + 6Gagan‟s age = 60
155. 155. Problem - 15Ramu‟s grandfather says, “ Ram, I am now 30years older than your father. 15 years ago, I was2½ times as old as your father”. How old is thegrandfather now?
156. 156. SolutionLet the father‟s age be x.Grandfather‟s age will be 30 + x15 years ago,X + 30 – 15 = 5/2 (x – 15)X + 15 = 5/2 (x – 15)2x + 30 = 5x – 75105 = 3xX = 105 / 3 = 35Grandfather‟s age = 35 + 30 = 65
157. 157. Average
158. 158. AverageAverage = Sum of Quantities Number of QuantitiesSum of quantities = Average*Number of Quantities.Number of quantities = Sum of Quantities Average
159. 159. Problem - 1The average age of a class of 22 students is 21years. The average increases by 1 when theteacher‟s age is also included. What is the age ofthe teacher?
160. 160. SolutionTotal age of the students be xx/22 = 21; x = 21*22= 462Teacher‟s age is also includedx/23 = 22; x = 22*23 = 506Total age of 23 people – Total age of 22 peoplewill be the age of teacher506 – 462 = 44 yearsThe age of teacher = 44
161. 161. Problem - 2The average of 7 numbers is 25. The average offirst three of them is 20 while the last three is 28.What must be the remaining number?
162. 162. SolutionAverage of 7 numbers = 25,Sum of 7 numbers = 25* 7 = 175Avg. of first three numbers = 20, 20* 3 = 60Avg. of last three numbers = 28, 28*3 = 84The 4th number = 175 – (60+84) = 175 – 144= 31
163. 163. Problem - 3The average age of a team of 10 people remainsthe same as it was 3 years ago, when a youngperson replaces one of the member. How muchyounger was he than the person whose place hetook?
164. 164. SolutionLet Average be x10 members‟ Average = 10xAverage of 10 members (including new one) issame as it was 3 yrs ago.Now 10*3 = 30 years have increased, so a personof 30 years should have replaced to keep theaverage as same.
165. 165. Problem - 4The average age of a couple was 26 years at thattime of their marriage. After 11 years of marriagethe average age of the family with 3 childrenbecome 19 years. What is the average age of theChildren?
166. 166. SolutionAverage of parents ages is 26, sum= 26*2 = 52Parents age after 11 years = 52 +22 = 74Average age of Family = 19, Sum = 19*5 = 95Sum of family‟s age – Sum of parents‟ age= 95 – 74 = 21Sum of the ages of 3 children = 21,Average Age = 21/3 = 7 yrs
167. 167. Problem - 59 members went to a hotel for taking meals.Eight of them spent Rs. 12 each on their mealsand the ninth person spent Rs. 8 more than theaverage expenditure of all the nine. What wasthe total money spent by them?
168. 168. SolutionAverage = x/9Amount Spent by 8 members = 12 * 8 = 9696 + x/9 + 8 = x104 = x – x/9104 = 8x/98x = 104 *9 = 936x = 936/8 = 117
169. 169. Problem - 7A batsman makes a score of 87 runs in the 17thinning and thus increases his average by 3. Findhis average after 17th innings?
170. 170. Solution17th innings avg. = x, Runs = 17x16th innings avg. = x -3, Runs = 16 (x -3)16 (x-3) + 87 = 17x16x – 48 +87 = 17xX = 39
171. 171. Problem - 7There are 24 students in a class. One of them,who was 18 yrs old left the class and his placewas filled up by the newcomer. If the average ofthe class thereby was lowered by one month,what is the age of the newcomer?
172. 172. SolutionAverage reduced by 1 month,24 * 1 = 2 yearsSo, the newcomer‟s age is 18 -2 = 16 years
173. 173. Problem - 8The average of marks in mathematics for 5students was found to be 50. Later, it wasdiscovered that in the case of one student themark 48 was misread as 84. What is the correctaverage?
174. 174. SolutionDifference = 84 – 48 = 3636 /5 = 7.2 (Increased)The corrected average = 50 – 7.2 = 42.8
175. 175. Problem - 9The average salary of all the workers in a factoryis Rs. 8000. The average salary of 7 techniciansis Rs. 12000 and the average salary of the rest isRs. 6000. What is the total number of workers inthe factory?
176. 176. SolutionMembers Avg. 7 12000 X 6000 6x = 7*12 X = 7812/6 = 14 Total no. of workers = 7 + 14 = 21
177. 177. Problem - 10Average salary of all the 50 employees including5 officers of the company is Rs. 850. If theaverage salary of the officers is 2500, find theaverage salary of the remaining staff of thecompany.
178. 178. Solutionx/50 = 850; x = 42,5005 officers‟ salary = 2500*5 = 1250050 – 5 members = 42500 – 1250045 members = 30000Avg. salary of 45 members = 30000/45 = 667(App)
179. 179. Problem - 11Find the average of 8 consecutive odd numbers21,23,25,27,29,31,33,35
180. 180. Solution1st number + last Number /2= 21 + 35 /2 = 28
181. 181. Problem - 12A train covers 50% of the journey at 30 km/hr,25% of the journey at 25 km/hr, and theremaining at 20 km/hr. Find the average speed ofthe train during entire journey.
182. 182. SolutionTotal Journey = 100 kmS = Distance / Time = 100 / 5/3 + 1/1 + 5/4= 100 * 12 /20+12+15= 1200/47 = 25 25/47 km/hr
183. 183. Problem - 13The average of 10 numbers is 7. What will be thenew average if each number is multiplied by 8?
184. 184. SolutionIf numbers are multiplied by 8,Average also to be multiplied by 8= 7*8 = 56{or}x/10 = 7x = 10*7 = 70= 70* 5 = 560 /10 = 56
185. 185. Problem - 14The mean marks of 10 boys in a class is 70%whereas the mean marks of 15 girls is 60%.What is the mean marks of all 25 students?
186. 186. SolutionBoys = x/10 = 70 = 700Girls = x/15 = 60 = 90010 + 15 = 700 + 90025 = 16001600/25 = 64%
187. 187. Problem - 15Of the three numbers the first is twice the secondand the second is thrice the third. If the averageof the three numbers is 10, what are thenumbers?
188. 188. SolutionA = 2xB=xC = x/32x + x + x/3/3 = 106x + 3x + x /9 = 106x + 3x + x = 9010x = 90 ; x = 9.A = 18, B = 9, C = 3
189. 189. Percentage
190. 190. Percentage• By a certain Percent, we mean that manyhundredths.• Thus, x Percent means x hundredths, writtenas x%
191. 191. Percentage•Finding out of Hundred.If Length is increased by X% and Breadth isdecreased by Y% What is the percentageIncrease or Decrease in Area of the rectangle? Formula: X+Y+ XY/100 % Decrease 20% means -20
192. 192. Problem -1When 75% of the Number is added to 75%, theresult is the same number. What is the number?
193. 193. SolutionPercentage Number 75 x+75 100 x100x + 7500 = 75x25x = 7500x = 300
194. 194. Problem - 2A tank is full of milk. Half of the milk is soldand the tank is filled with water. Again half ofthe mixture is sold and the tank is filled withwater. This operation is repeated thrice. Find thepercentage of milk in the tank after the thirdoperation?
195. 195. SolutionMilk Water100 050 50(1st) 25 75 (2nd) 12.5 87.5 (3rd)After 3 operation Milk 12.5%
196. 196. Problem 3A large water-melon weighs 20kg with 96% ofits weight being water. It is allowed to stand inthe sun and some of the water evaporates so thatnow, only 95% of its weight is water. What willbe its reduced weight?
197. 197. Solution20 *96/100=19.2kg of waterLet the evaporated water be x19.2-x=95%(20-x)19.2-x=95(20-x)/1001920-100x=1900-95x5x=20 ;x=420-4=16kg.
198. 198. Problem 4The population of a city is 155625. Forevery1000 men, there are 1075 women. If 40%of men and 24% of women be literate, then whatis the percentage of literate people in the city?
199. 199. SolutionRatio of men and women=1000:1075=40:43Number of men=40*155625/83=75000Number of women=155625-7500=80625Number of literate men=75000*40/100=3000Number of literate women=80625*24/100=19350Literate people =30000+19350=49350Percentage of literate people=49350/155625*100=2632/83=31 59/83%
200. 200. Problem 5300 grams of sugar solution has 40% sugar in it.How much sugar should be added to make it50% in the solution?
201. 201. SolutionGrams Sugar300 40%X 50%50x = 40*300x = 40*300/50 = 240300 – 240 =60 Kg
202. 202. Problem - 6A man lost 12½% of his money and afterspending 70% of the remainder, he has Rs. 210left. How much did the man have at first?
203. 203. SolutionLet the amount be 100Then, 100.00 – 12.50 = 87.5070% of 87.50 = 87.50 *70/100 =61.25The remaining amount will be Rs. 26.25 Initial Final 100 26.25 X 21026.25x = 21000; x = 21000/26.25 = 800
204. 204. Problem - 7During one year the population of a townincreases by 10% and during next year itdiminished by 10%. If at the end of the secondyear, the population was 89,100, what was thePopulation at the beginning of first year?
205. 205. SolutionLet the population be 1001st Year = 100 + 10 = 1102nd Year = 110 * 10/100 = 110 -11 = 99Percentage Population99 89100100 x99x = 89100*100;x = 8910000/99 = 90000
206. 206. Problem - 8When a number is first increased by 20% andthen again 20% by what percent should theincrease number be reduced to get back theoriginal number?
207. 207. SolutionLet the number be 10020% increase = 100*20/100 = 20New Value = 120Again increase by 20% = 120*20/100 = 24New value = 144Increased amount = 44/144*100 = 30 5/9%
208. 208. Problem - 9The number of students studying Arts,Commerce and Science in an institute were inthe ratio 6 : 5 : 3 respectively. If the number ofstudents in Arts, Commerce and science wereincreased by 10%, 30% and 15% respectively,what was the new ratio between number ofstudents in the three streams?
209. 209. SolutionA:C:S6:5:36x : 5x : 3x6x*110/100 : 5x*130/100 : 3x*115/1006x*110 : 5x*130 : 3x*115660 : 650 : 345132 : 130 : 69
210. 210. Problem - 10In measuring the sides of rectangle errors of 5%and 3% in excess are made. What is the errorpercent in the calculated area?
211. 211. SolutionArea = xyX = 5% Excess = 100* 5/100 = 105Y = 3% Excess = 100*3/100 = 103103*105/100 = 10815/100 = 108.15Error – Actual = 108.15 – 100 = 8.15% Excess
212. 212. Problem - 11In a certain examination there were 2500candidates. Of them 20% of them were girls andrest were boys. If 5% of boys and 40% of girlsfailed, what was the Percentage of candidatespassed?
213. 213. SolutionGirls = 2500*20/100 = 500Boys = 2500*80/100 = 2000Students who failed wereBoys = 2000*5/100 = 100Girls = 500*40/100 = 200Total Failed Students = 300Total Pass students = 2500 – 300 = 2200Pass Percentage = 2200/2500*100 = 88%
214. 214. Problem - 12A person saves every year 20% of his income. Ifhis income increases every year by 10% then hissaving increases by?
215. 215. SolutionEvery year saving, if the income is Rs. 100 = 100 *20/100 =Rs. 20Salary increases = 110*20/100 = 22Percentage increase (Savings) = 2/20*100 = 10%
216. 216. Problem - 13On a test containing 150 questions carrying 1mark each, meena answered 80% of the firstanswers correctly. What percent of the other 75questions does she need to answer correctly toscore 60% on the entire exam?
217. 217. SolutionRequired correct answer = 150*60/100 = 90 Questions need to be correct.80% of 75 questions = 60 q answered correctly.Remaining 30 questions need be correct out of 75= 30/75*100 = 40
218. 218. Problem - 14A boy after giving away 80% of his pocketmoney to one companion and 6% of theremainder to another has 47 paise left with him.How much pocket money did the boy have in thebeginning?
219. 219. SolutionLet the amount be 100To the first companion = 100*80/100 = 80Remaining = 100 – 80 = 20To the 2nd person = 20*6/100 = 1.20The remaining = Rs.18.80 or 1880 paiseInitial Final100 1880X 471880x = 47*100x = 4700/1880 = 2.5
220. 220. Problem - 15The length of a rectangle is increased by 10%and breath decreased by 10%. Then the area ofthe new rectangle?
221. 221. SolutionI – D – I*D /10010 -10 – 10*10/1000 – 1 = -1Decrease by 1%
222. 222. Profit and Loss
223. 223. Profit and Loss• 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.
224. 224. Problem - 1A trade man allows two successive discount of20% and 10%. If he gets Rs.108 for an article.What was its marked price?
225. 225. SolutionI1 + I2 – I1*I2/10020 + 10 – 20*10 /100= 28%Discount = 28%, 72 Percent Cost is 108Then 100percent cost = 72 108 100 x100*108/72 = 150
226. 226. Problem - 2A trade man bought 500 metres of electric wireat 75 paise per metre. He sold 60% of it at profitof 8%. At what gain percent should he sell theremainder so tas to gain 12% on the whole
227. 227. Solution500* 60/100 = 3008 X 12300 200300 : 200 = 6 : 48 18 126 4Remainder at 18% Profit
228. 228. Problem - 3A man purchased a box full of pencils at the rateof 7 for Rs. 9 and sold all of them at the rate of 8for Rs. 11. in this bargains he gains Rs. 10. Howmany pencils did the box contains.
229. 229. SolutionLCM = 7 and 8 = 5656 pencil cost price = 8*9 = 7256 Pencil selling price = 7*11Profit = 77 – 72 = Rs. 5 for 56 pencilRs. 5 for 56 pencil means , for Rs. 10 the pencils are 112
230. 230. Problem - 4A cloth merchant decides to sell his material atthe cost price, but measures 80cm for a metre.His gain % is?
231. 231. Solution100 – 80 = 20 cm differenceActual = 8020/80*100 = 25% Gain
232. 232. Problem - 5Sales of a book decrease by 2.5% when its priceis hiked by 5%. What is the effect on sales?
233. 233. SolutionLet the sales be 100 – 2.5 = 97.5Profit = 100+5 = 105Sales Profit97.5 105100 X100x = 97.5*105x = 97.5*105/100 = 102.375100 – 102.375 = 2.375 = 2.4 profit (app)
234. 234. Problem - 6A dealer buys a table listed at Rs.1500 and getssuccessive discount of 20% and 10%. He spendsRs. 20 on transportation and sells it at a profit of10%. Find the selling price of the table.
235. 235. SolutionDiscount = 20+10 – 20*10/100 = 28%Actual price = 100 – 28 = 72100 1500 72 x72*1500/100 = 1080Transport = 1080 +20 = 1100100 1100 110 x1100*110/100 = 1210
236. 236. Problem - 7A fridge is listed at Rs. 4000. due to the offseason, a shopkeeper announces a discount of5%. What is the S.P?
237. 237. Solution= 4000*95/100 = 3800
238. 238. Problem - 8If the cost price of 9 pens is equal to the S.P of11 pens. What is the gain or loss?
239. 239. Solution= 11 – 9 = 2= 2/11*100 = 18 2/11% loss
240. 240. Problem - 9A machine is sold for Rs.5060 at a gain of 10%what would have been the gain or loss percent ifit had been sold Rs.4370?
241. 241. SolutionS.P = Rs.5060 = Gain = 10%C.P = 100/110*5060 = 4600IF S.P = Rs.4370 and C.P = Rs.4600Loss = 230Loss % = 230/4600 * 100 = 5% loss
242. 242. Problem - 10A person purchased two washing machines eachfor Rs.9000. he sold one at a loss of 10% andother at a gain of 10%. What is his gain or loss?
243. 243. SolutionEach Rs.9000. one is 10% profit and other is10% loss. So No profit and No loss
244. 244. Problem - 11Four percent more is gained by selling an articlefor Rs.180, then by selling if for Rs.175. then itsC.P is?
245. 245. SolutionLet the cost price = Rs. X4% of x = 180 – 175 = 4x/100 = 54x = 500; x = 500/4 = 125
246. 246. Problem - 12An article is sold at a profit of 20%. If it hadbeen sold at a profit of 25%. It would havefetched Rs.35% more. The Cost Price of thearticle is?
247. 247. SolutionLet C.P = Rs. X125% of x – 120% of x = 355% of x =Rs.35 = x = 35*100/5 = 700C.P = Rs. 700
248. 248. Problem - 13A reduction of 20% in the price of orangeenables a man to buy 5 oranges more for Rs. 10.The price of an orange before reduction was,
249. 249. Solution20% Rs. 10 = Rs.2Reduced price of 5 oranges = Rs. 2Reduced price of 1 oranges = 40 pOriginal price = 40/ 1- 0.20 = 400/8 = 50 Paise
250. 250. Problem - 14A man sells two horses for Rs.1475. The costprice of the first is equal to the S.P of the second.If the first is sold at 20% loss and the second at25% gain. What is his total gain or loss? ( inrupees)
251. 251. SolutionLet cost price of 1st horse = S.P of 2nd = xC.P of 2nd = S.P of 2nd * 100/125 = x*100/125 = 4x/5S.P of 1st = C.P of 1st *80/100 = x*80/100 = 4x/5Neither loss nor gain
252. 252. Problem - 15Rekha sold a watch at a profit of 15%. Had hebought it at 10% less and sold it for Rs. 28 less,he would have gained 20%. Find the C.P of theWatch.
253. 253. SolutionC.P be Rs. XFirst S.P = 115% of x = 23x/20 and second C.P = 90% x = 9x/10Second S.P = 120% of 9x/10 = 120/100 * 9x/10 = 27x/25Given 23x/20 – 27x/25 = 28 = 115x – 108x/100 = 287x/100 = 28 = x = 28*100/7 = 400C.P = Rs.400
254. 254. Probability
255. 255. Probability• Probability: P(є) = n(є) / 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)
256. 256. Problem - 1Four cards are drawn at random from a pack of52 playing cards. Find the probability of gettingall face cards?
257. 257. Solutionn(E) = 52C4n(S) = 12C4 = 12C4/52C4
258. 258. Problem - 2Four persons are to be chosen at random from agroup of 3 men, 2 women and 4 children. Findthe probability of selecting 1 man, 1 woman or 2children?
259. 259. SolutionTotal 3 M + 2 W + 4 C = 9 C 4 = 126n (E) = 3C1 * 2C1 * 4C2 = 3636/126 = 2/7
260. 260. Problem - 3A word consists of 9 letters, 5 consonants and 4vowels. Three letters are chosen at random.What is the probability that more than onevowels will be selected?
261. 261. Solutionn(E) = 9C3 = 84More than one Vowels. So,2V +1C or 3 V4C2 *5C1 + 4C3 = 34= 34/84 = 17/42
262. 262. Problem - 4A bag contains 10 mangoes out of which 4 arerotten. Two mangoes are taken out together. Ifone of them was found to be good, then what isthe probability that the other one is also good?
263. 263. Solution10 mangoes – 4 are rotten = 6 good mangoesGetting good mangoes = 6C1/10C1 = 6/10Getting second mango to be good = 5/91st and 2nd mangoes6/10 *5/9 = 1/3
264. 264. Problem - 5Out of 13 applicants for a job there are 5 womenand 8 men. It is desired to select 2 persons forthe job. What is the probability that at least oneof the selected person will be a woman?
265. 265. Solutionn(E) = 13C2 = 78n(S) = 1m and 1 w or 2 w = 8C1*5C1 + 5C2 = 50 = 50/78 = 25/39
266. 266. Problem - 6Two cards are drawn at random from a pack of52 cards. What is the probability that either bothare black or both are queen?
267. 267. SolutionP(A) = Both are BlackP(B) = Both are QueenP(AnB) = Both are queen and BlackP(A) = 26C2/52C2 = 325/1326P(B) = 4C2 /52C2 = 6/1326P(AnB) = 2C2 /52C2 = 1/1326325/1326 + 6/1326 - 1/ 1326 = 55/221
268. 268. Problem -7A man and his wife appear in an interview fortwo vacancies in the same post. The probabilityof husband‟s selection is 1/7 and the probabilityof wife‟s selection is 1/5. Find the probabilitythat only one of them is selected?
269. 269. SolutionHusband‟s Selection = 1/7;Not getting selected = 1 – 1/7 = 6/7Wife‟s selection = 1/5;Not getting selected = 1 – 1/5 = 4/5Only one of them is selected =(Husband‟s Selection + Wife Not getting selected) or(Wife‟s selection + Husband‟s Not getting selected)= (1/7*4/5) + 1/5*6/7) = 2/7
270. 270. Problem - 8Four persons are chosen at random from a groupof 3 men, 2 women and 4 children. What is thechance that exactly 2 of them are children?
271. 271. Solution3 + 2 + 4 = 9C4 = 1264 members 2(M and W) + 2(boy)5C2 + 4C2 = 60= 60 / 126 = 10/21
272. 272. Problem - 9Prakash can hit a target 3 times in 6 shots, Priyacan hit the target 2 times in 6 shots and Akhileshcan hit the target 4 times in 4 shots. What is theprobability that at least 2 shots hit the target?
273. 273. SolutionPrakash hitting = 3/6; not hitting = 3/6Priya hitting = 2/6; not hitting = 4/6Akilesh = 4/4 = 1At least 2 shots hit target = 3/6*4/6 + 3/6*2/6 = ½
274. 274. Problem - 10There are two boxes A and B. A contains 3 whiteballs and 5 black balls and Box B contains 4white balls and 6 black balls. One box is taken atrandom and what is the probability that the ballpicked up may be a white one?
275. 275. Solution(Box A is selected and a ball is picked up ) or(Box B is selected and a ball is picked up)½*3/8 + ½*4/10 = 31/80
276. 276. Problem - 11A bag contains 6 white balls and 4 black balls.Four balls are successively drawn withoutreplacement. What is the probability that they arealternately of different colour?
277. 277. SolutionSuppose the balls drawn are in the order white,black, white, black…= 6/10 *4/9*5/8*3/7 = 360/5040Suppose the balls drawn are in the order black,white, black, white…= 4/10*6/9*3/8*5/7 = 360/5040360/5040 +360/5040 = 1/7
278. 278. Problem - 12A problem in statistics is given to four studentsA, B, C and D. Their chances of solving it are1/3, ¼, 1/5 and 1/6 respectively. What is theprobability that the problem will be solved?
279. 279. SolutionA is not solving problem = 2/3,B is not solving problem = ¾C not solving problem = 4/5D not solving problem = 5/62/3*3/4*4/5*5/6 = 1/3All together the probability of solving theproblem = 1 -1 /3 = 2/3
280. 280. Problem - 13There are 8 questions in an examination eachhaving only 2 answers choices „Yes‟ or „No‟. Allthe questions carry equal marks. If a studentmarks his answer randomly, what is theprobability of scoring exacting 50%?
281. 281. SolutionEach questions having 2 ways of answering,1 question = 2!........ 8 question = 2! = 2!*2!*2!*2!*2!*2!*2!*2! = 256To get 50%, 4 questions need to be correct,8c4 = 8*7*6*5/1*2*3*4 = 70 = 70/256 = 35/128
282. 282. Problem - 14A group consists of equal number of men andwomen. Of them 10% of men and 45% ofwomen are unemployed. If a person is randomlyselected from the group find the probability forthe selected person to be an employee.
283. 283. SolutionLet the number of men is 100 and women be 100Employed men and women = (100-10)+(100-45) = 145 Probability = 145 / 200 = 29 / 40
284. 284. Problem - 15The probability of an event A occurring is 0.5and that of B is 0.3. If A and B are mutuallyexclusive events. Find the probability thatneither A nor B occurs?
285. 285. SolutionIt is Mutually exclusive events P(A n B)=0Probability = 1 – ( P(A) + P (B) – P(A n B) ) = 1 – (0.5 + 0.3 – 0) = 0.2
286. 286. Permutation and Combination
287. 287. Permutation and CombinationPermutation means ArrangementCombination means Selection
288. 288. Permutation and Combination• Permutations: Each of the arrangements which can be made by taking some (or) all of a number of items is called permutations.np = n(n-1)(n-2)…(n-r+1)=n!/(n-r)! r• Combinations: Each of the groups or selections which can be made by taking some or all of a number of items is called a combination.nC = n!/(r!)(n-r)! r
289. 289. Types1. How many ways of Arrangement possible by using word SOFTWARE? SOFTWARE = 8!2. How many ways of arrangement Possible by using word SOFTWARE, vowels should come together. SFTWR (OAE) = 6! * 3!
290. 290. Types3. How many ways of Arrangement Possible by using word SOFTWARE, vowels should not come together? SFTWR ( ARE) Not together = Total arrangement – Vowels together = 8! – (5! * 3!)
291. 291. Types4. How many ways of arrangement possible by using word MACHINE, so that vowels occupy only ODD places. - - - - - - - (7 places) MCHN (AIE) 4 Consonant and 3 vowels. 7 places = 4 ODD places, 3 EVEN placesVowels = 4P3 = 4!Consonant = 4P4 = 4!Total Number of arrangement = 4!*4!
292. 292. Types5. How many ways of arrangement possible by using word ARRANGEMENT Letter‟s Repetition = 2(A) 2(R) 2 (E) 2 (N) = 11!/2!*2!*2!*2!In a given problem, any letter is repeated more than once that should be divided with total number.
293. 293. Problem - 1A committee of 5 is to be formed out of 6 gentsand 4 ladies. In how many ways this can bedone, when at least 2 ladies are included?
294. 294. Solutiona. 2 ladies * 3 Gents4C2 * 6 C3 = 120b. 3 ladies * 2 Gents4C3 * 6C2 = 60c. 4 ladies * 1 Gent4C4 *6C1 = 1*6 = 6Total ways = 120 +60 +6 = 186
295. 295. Problem - 2It is required to seat 5 men and 4 women in arow so that the women occupy the even places.How many such arrangements are possible?
296. 296. SolutionTotal places = 9Odd places = 5Even places = 44 even places occupied by 4 women= 4P4 = 4! = 245 odd places occupied by 5 men= 5P4 = 5! = 120Total ways = 120*24 = 2880 ways
297. 297. Problem - 3A set of 7 parallel lines is intersected by anotherset of 5 parallel lines. How many parallelogramsare formed by this process?
298. 298. SolutionTwo parallel lines from the first set and any twofrom the second set will from a parallelogram.7C2 *5C2 = 21 * 10 = 210
299. 299. Problem - 4There are n teams participating in a footballchampionship. Every two teams played onematch with each other. There were 171 matcheson the whole. What is the value of n?
300. 300. SolutionTotal number of matches played = nC2nC2 = 171n(n-1)/2= 171n2 – n – 342 = 0(n+18) (n-19) = 0 n = 19
301. 301. Problem - 5In an examination, a candidate has to pass ineach of the 6 subjects. In how many ways can hefail?
302. 302. Solution6C1 + 6C2 + 6C3 + 6C4+6C5+6C61 + 6 + 15 + 20 + 15 + 6 = 63 ways
303. 303. Problem - 6In how many ways can a pack of 52 cards bedistributed to 4 players, 17 cards to each of 3 andone card to the fourth player?
304. 304. Solution17 cards can be given to 1st player = 52 C172nd player = 35C173rd player = 18C174th player = 1= 52C17*35C17*18C17= 52!/17!35! * 35!/17!*18! * 18!/17!*1!= 52!/(17!)3
305. 305. Problem - 7A foot race will be held on Saturday. How manydifferent arrangements of medal winners arepossible if medals will be for first, second andthird place, if there are 10 runners in the race …
306. 306. Solutionn = 10r=3n P r = n!/(n-r)! = 10! / (10-3)! = 10! / 7! = 8*9*10 = 720Number of ways is 720.
307. 307. Problem - 8To fill a number of vacancies, an employer musthire 3 programmers from among 6applicants, and two managers from 4 applicants.What is total number of ways in which she canmake her selection ?
308. 308. SolutionIt is selection so use combination formulaProgrammers and managers = 6C3 * 4C2 = 20 * 6 = 120 Total number of ways = 120 ways.
309. 309. Problem - 9A man has 7 friends. In how many ways canhe invite one or more of them to a party?
310. 310. SolutionIn this problem, the person is going to select hisfriends for party, he can select one or moreperson, so addition = 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7 = 127 Number of ways is 127
311. 311. Problem - 9Find the number of different 8 letter wordsformed from the letters of the word EQUATIONif each word is to start with a vowel
312. 312. SolutionFor the words beginning with a vowel, the firstletter can be any one of the 5 vowels, theremaining 7 places can be filled by 7P7 = 5040The number of words = 5 * 5040 = 25200
313. 313. Problem - 10In how many different ways can the letters of theword TRAINER be arranged so that the vowelsalways come together?
314. 314. SolutionA,I,E can be arranged in 3! Ways (5! * 3!) / 2! = 360 ways
315. 315. Problem - 11In how many different ways can the letters of theWord DETAIL be arranged so that the vowelsmay occupy only the odd positions?
316. 316. Solution___ ___ ___ ___ ___ ___3P3 = 3! = 1*2*3 = 63P3 = 3! = 1*2*3 = 6 = 6*6 = 36
317. 317. Problem - 12There are 5 red, 4 white and 3 blue marbles in abag. They are taken out one by one and arrangedin a row. Assuming that all the 12 marbles aredrawn, find the number of differentarrangements?
318. 318. SolutionTotal number of balls = 12Of these 5 balls are of 1st type (red), 4 balls are the 2nd type and 3 balls are the 3rd type.Required number of arrangements = 12!/5!*4!*3! = 27720
319. 319. Problem - 135 men and 5 women sit around a circular table,the en and women alternatively. In how manydifferent ways can the seating arrangements bemade?
320. 320. Solution5 men can be arranged in a circular table in 4 ways = 24 waysThere are 5 seats available for 5 women they can be arranged in 5 waysNo. of ways = 5!*4! = 2880 ways
321. 321. Problem - 14In a chess board there are 9 vertical and 9horizontal lines. Find the number of rectanglesformed in the chess board.
322. 322. SolutionSolution: 9C2 * 9C2 = 1296
323. 323. Problem - 15In how many ways can a cricket team of 11players be selected out of 16 players, If oneparticular player is to be excluded?
324. 324. SolutionSolution: If one particular player is to be excluded, then selection is to be made of 11 players out of 15. 15C11= 15!/( 11!*4!)=1365 ways
325. 325. Area and Volume
326. 326. Area and VolumeCube:• Let each edge of the cube be of length a. then,• Volume = a3cubic units• Surface area= 6a2 sq.units.• Diagonal = √3 a units.
327. 327. Area and VolumeCylinder:• Let each of base = r and height ( or length) = h.• Volume = πr2h• Surface area = 2 πr h sq. units• Total Surface Area = 2 πr ( h+ r) units.
328. 328. Area and VolumeCone:• Let radius of base = r and height=h, then• Slant height, l = √h2 +r2 units• Volume = 1/3 πr2h cubic units• Curved surface area = πr l sq.units• Total surface area = πr (l +r)
329. 329. Area and VolumeSphere:• Let the radius of the sphere be r. then,• Volume = 4/3 πr3• Surface area = 4 π r2sq.units
330. 330. Area and VolumeCircle: A= π r 2Circumference = 2 π rSquare: A= a 2Perimeter = 4aRectangle: A= l x bPerimeter= 2( l + b)
331. 331. Area and VolumeTriangle:A = 1/2*base*heightEquilateral = √3/4*(side)2Area of the Scalene Triangle S = (a+b+c)/ 2A = √ s*(s-a) * (s-b)* (s-c)
332. 332. Problem - 1A rectangular sheet of size 88 cm * 35 cm is bentto form a cylindrical shape with height 35 cm.What is the area of the base of the cylindricalshape?
333. 333. SolutionThe circumference of the circular region = 88 cm2 r = 88r = 88*7/22*2 = 14 cmArea of the base = r2 = 22/7*14*14 v= 616 cm2
334. 334. Problem - 2The radius of the base of a conical tent is 7metres. If the slant height of the tent is 15metres, what is the area of the canvas required tomake the tent?
335. 335. SolutionR=7mL = 15 mArea of Canvas required = Curved Surface Areaof cone rl = 22/7*7*15 = 330 sq.m
336. 336. Problem - 3Three spherical balls of radius 1 cm, 2 cm and 3cm are melted to form a single spherical ball. Inthe process, the material loss was 25%. Whatwould be the radius of the new ball?
337. 337. SolutionVol. of sphere = 4/3 r3Vol. of 3 small spherical balls = 4/3 ( 13+23+33)= 4/3 (1+8+27) = 4 /3 (36) = 48Material loss = 25%Vol. of the single spherical ball = 48 *75/100 = 48 * ¾ = 36 V = 4/3 r3 = 36 r3 = 36*3/4 = 27 r = 3 cm
338. 338. Problem - 4A rectangular room of size 5m(l)*4m(w)*3m(h)is to be painted. If the unit of painting is Rs. 10per sq.m, what is the total cost of painting?
339. 339. SolutionArea of 4 walls = 2h(l+b)The area to be painted includes the 4 walls andthe top ceiling.Area to be painted = 2h (l+b) +lb = 2*3 (5+4) + 5*4 = 54+20 = 74 sq.m.Total cost of painting = 74*10 = Rs.740
340. 340. Problem - 5The radius of a sphere is r units. Each of theradius of the base and the height of a rightcircular cylinder is also r units. What is the ratioof the volume of the sphere to that of thecylinder?
341. 341. SolutionVol. of sphere = 4/3 r3 and Vol. of Cylinder = r2h = r3Required Ratio = 4/3 r3: r3 = 4/3 : 1 =4:3
342. 342. Problem - 6A measuring jar of internal diameter 10 cm ispartially filled with water. Four equal sphericalballs of diameter 2 cm each, are dropped in itand they sink down in the water completely.What will be the increase in the level of water inthe jar?
343. 343. SolutionRadius of each ball = 1 cmVol. of 4 balls = 4* 4/3 (r)3 = 16/3 cm3Vol. of water raised in the Jar = Vol. of 4 ballsLet h be the rise in water level, thenArea of the base *h = 16/3 *5*5*h = 16/3H = 16/3*25 = 16/75 cm
344. 344. Problem - 7What is the cost of planting the field in the formof the triangle whose base is 2.8 m and height3.2 m at the rate of Rs.100 / m2
345. 345. SolutionArea of triangular field = ½ * 3.2 * 2.8 m2 = 4.48 m2 Cost = Rs.100 * 4.48 = Rs.448..
346. 346. Problem - 8Find the length of the longest pole that can beplaced in a room 14 m long, 12 m broad, and 8 mhigh.
347. 347. SolutionLength of the longest pole = Length of the diagonal of the room = √(142 + 122 + 82) = √ 404 = 20.09 m
348. 348. Problem - 9 Area of a rhombus is 850 cm2. If one of itsdiagonal is 34 cm. Find the length of the otherdiagonal.
349. 349. Solution850 = ½ * d1 * d2 = ½ * 34 * d2 = 17 d2 d2 = 850 / 17 = 50 cm Second diagonal = 50cm
350. 350. Problem - 10A grocer is storing small cereal boxes in largecartons that measure 25 inches by 42 inches by 60inches. If the measurement of each small cerealbox is 7 inches by 6 inches by 5 inches then what ismaximum number of small cereal boxes that can beplaced in each large carton ?
351. 351. SolutionNo. of Boxes = 25*42*60 / 7*6*5 = 300300 boxes of cereal box can be placed.
352. 352. Problem - 11If the radius of a circle is diminished by 10%,what is the change in area in percentage?
353. 353. Solution= x + y + xy/100= -10 - 10 + 10*10/100= -19% Diminished area = 19%.
354. 354. Problem - 12A circular wire of radius 42 cm is bent in theform of a rectangle whose sides are in the ratioof 6:5. Find the smaller side of the rectangle?
355. 355. Solutionlength of wire = 2 πr = (22/7*14*14)cm = 264cmPerimeter of Rectangle = 2(6x+5x) cm = 22xcm 22x =264 x = 12 cm Smaller side = (5*12) cm = 60 cm
356. 356. Problem - 13A beam 9m long, 40cm wide and 20cm deep ismade up of iron which weights 50 kg per cubicmetre. Find the weight of the Beam.
357. 357. SolutionVol. of the Beam = lbh = 9*40/100*10/100 = 72 m3Weight of the iron beam is given as lm3 = 50 kg72/100 m3 = 72/100*50 = 36 kg
358. 358. Problem - 14If the length of a rectangle is reduced by 20%and breadth is increased by 20%. What is thepercentage change in the area?
359. 359. Solution x + y + (xy/100)% = - 20 + 20 – 400/100 = -4The area would decrease by 4%
360. 360. Problem - 15Find the number of bricks measuring 25 cm inlength, 5 cm is breadth and 10 cm in height for awall 40 m long, 75 cm broad and 5 metres inheight?
361. 361. SolutionVol. of the wall = 40*72/100*5 = 150 m3Vol. of 1 bricks = 25/100*5/100*10/100 = 1/80 m3Number of bricks required = 150/1/800 = 150*800 = 120000
362. 362. Calendar
363. 363. CalendarOdd days:0 = Sunday1 = Monday2 = Tuesday3 = Wednesday4 = Thursday5 = Friday6 = Saturday
364. 364. CalendarMonth code: Ordinary yearJ=0 F=3M=3 A=6M=1 J=4J=6 A=2S=5 O=0N=3 D=5 Month code for leap year after Feb. add 1.
365. 365. CalendarOrdinary year = (A + B + C + D )-2 -----------------------take remainder 7Leap year = (A + B + C + D) – 3 ------------------------- take remainder 7
366. 366. Problem - 111th January 1997 was a Sunday. What day ofthe week on 7th January 2000?
367. 367. Solution11th Jan 1997 = Sunday11th Jan 1998 = Monday11th Jan 1999 = Tuesday11th Jan 2000 = Wednesday7th Jan 2000 is on Saturday
368. 368. Problem - 2What day of the week was on 5th June 1999?
369. 369. SolutionA+B+C+D – 2 / 7A = 1999/7 = 4B = 1999/4 = 499/7 = 2C = June = 4D = 5/7 = 5= 4+2+4+5 – 2/7 = 13/7 = 5 = Saturday
370. 370. Problem - 3On what dates of August 1988 did Friday fall?
371. 371. SolutionA = 1988 / 7 = 0B = 1988/4 = 497/7 = 0C=3D=x0+0+3+x+3/7 = x/7 = 5(Friday)Friday falls on = 5,12,19,26
372. 372. Problem - 4India got independence on 15 August 1947.What was the day of the week?
373. 373. SolutionA = 1947/7 = 1B = 1947/4 = 486/7 = 3C = 15/7 = 1D=21+3+1+2 – 2 /7 = 5/7 = Friday
374. 374. Problem - 57th January 1992 was Tuesday. Find the day ofthe week on the same date after 5 years. i.e on7th January 1997.
375. 375. Solution7th January 1992 = Tuesday7th January 1993 = Thursday (Leap)7th January 1994 = Friday7th January 1995 = Saturday7th January 1996 = Monday ( Leap)7th January 1997 = Tuesday
376. 376. Problem - 6The first Republic day of India was celebrated on26th January 1950. What was the day of theweek on that date?
377. 377. SolutionA = 1950/7 = 4B = 1950/4 = 487/7 = 4C=0D = 26/7 = 54+4+0+5 – 2/7 = 11/7 = 4 = Thursday
378. 378. Problem - 7Find the Number of times 29th day of the monthoccurs in 400 consecutive year?
379. 379. Solution1 year = 1 (Ordinary Year)1 year = 12 (Leap Year)400 years = 97 leap year97 * 12 = 1164303*11 = 3333= 1164+3333 = 4497 times
380. 380. Problem - 8If 2nd March 1994 was on Wednesday, 25 Jan1994 was on,
381. 381. SolutionA = 1994/7 = 6B = 1994/4 = 498/7 = 1C=0D = 25/7 = 4= 6 + 1 + 0 + 4 – 2 / 7 = 3 = Tuesday
382. 382. Problem - 9Calendar for 2000 will serve also?
383. 383. Solution= 2000 + 2001 + 2002 + 2003 + 2004= 2 + 1 + 1 + 1 + 2 = 7 (Complete Week)2005
384. 384. Problem - 10If Pinky‟s 1st birthday fell in Jan 1988 on one ofthe Monday‟s, the day on which are was born is,
385. 385. SolutionJan = 1988 = MondayJan = 1987 = Sunday
386. 386. Problem - 11Akshaya celebrated her 60th birthday on Feb 24,2000. What was the day?
387. 387. SolutionA = 2000 /7 = 7B = 2000/4 = 500/7 = 3C=3D = 24/7 = 0= 7+3+3+0-3/7 = 10/7 = 3 = Wednesday
388. 388. Problem - 12On what dates of April 2008 did Sunday Fall?
389. 389. SolutionCalculate for 1st April 2008A = 2008/7 = 6B = 2008/4 = 502/7 = 5C = 1/7 = 1D=0= 6+5+1+0 – 3/ 7 = 2 = Tuesday1st April on Tuesday, then 1st Sunday fall on 6.Sunday falls on 6, 13, 20, 27.
390. 390. Problem - 13Today is Friday. After 62 days it will be,
391. 391. Solution62 / 7 = 6 days after Friday then it will be Tuesday
392. 392. Problem - 14What will be the day of the week on 1st Jan2010?
393. 393. SolutionA=1B=5C=0D=1= 1+5+0+1 – 2/ 7 = 5/7 = 5 = Friday
394. 394. Problem - 15What is the day of the week on 30/09/2007?
395. 395. CalendarSolution:A = 2007 / 7 = 5B = 2007 / 4 = 501 / 7 = 4C = 30 / 7 = 2D=5 ( A + B + C + D )-2 = ----------------------- 7 = ( 5 + 4 + 2 + 5) -2 ----------------------- = 14/7 = 0 = Sunday 7
396. 396. Clock
397. 397. ClocksClock:Angle between hour hand and minute hand = (11m/2) – 30hAngle between minute hand and hour hand =30h – (11m/2)
398. 398. Problem - 1What is the angle between the minute hand andhour hand when the time is 2.15?
399. 399. Solution= 11 m/2 – 30(h) = 11 15/2 – 30(2) = 11(7.5) – 60 = 82.5 – 60 = 22 1/2
400. 400. Problem - 2At what time between 5 and 6 o‟clock the handsof a clock coincide?
401. 401. SolutionCoinciding Angle = 0Min. hand to hour hand = 25 min apart60/55*25 = 12/11 * 25 = 300/11 = 27 3/11min past 5
402. 402. Problem - 3At what time between 12 and 1 o‟clock both thehands will be at right angles?
403. 403. SolutionRight angle = 90 degrees = 30(h) – 11 m/290 = 30(12) – 11 m/2180 = 360 – 11m11m = 360 – 180M = 180/1116 4/11 past 12
404. 404. Problem - 4Find at what time between 7 and 8 o‟clock willthe hands of a clock be in the same straight linebut not together?
405. 405. SolutionMinute hand to hour hand = 35 min apartStraight line not together = 30 min apartDifference = 35 – 30 = 5 min= 60/55*5 = 12/11*5 = 60/11= 55 5 / 11 past 7
406. 406. Problem - 5At what time between 5 and 6 are the hands ofthe clock 7 minutes apart?
407. 407. Solution7 min space behind the hour hand:25 min – 7 min = 18 min60/55 *18 = 216/11 = 19 7/11 min past 57 Min space ahead the hour hand25 min + 7 min = 32 min60/55*32 = 12/11*32 = 384/11= 34 10/11 min past 5
408. 408. Problem - 6A clock strikes 4 and takes 9 seconds. In order tostrike 12 at the same rate what will be the timetaken?
409. 409. Solution Strike Sec 3 (interval) 9 11 x3x = 11*9X = 11*9/3 = 33 Sec