Slide 1

Numerical Reasoning Problem 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) Problems on NumbersBasic 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)

Problem - 1A 2 digit number is 3 times the sum of its digits if 45 is added to the number. Its digits are interchanged. The sum of digits of the number is?

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 = 72 2 + 7 = 9 or 4 + 5 = 9Problem - 2A number when divided by 119 leaves a remainder of 19. If it is divided by 17. It will leave a remainder of?Solution = 19/17 = 2 remainderProblem - 3A boy was asked to find the value of 3/8 of sum of money instead of multiplying the sum by 3/8 he divided it by 8/3 and then his answer by Rs.55. Find the correct answer?Solution 8/3 3/8 = 55/24 = 55/55/24 = 24Problem - 4A man spends 2/5rd of his earning. 1/4th of the expenditure goes to food, 1/5th on rent, 2/5th on travel and rest on donations. If his total earning is Rs.5000, find his expenditure on donations?Solution5000*2/5 = 2000Remaining amount has given as donation2000* (1/5 + 2/5 + )Total amount = 200*17/20 = 17002000 1700 = 300

Problem - 5From a group of boys and girls 15 girls leave. There are then left, 2 boys for each girl. After this 45 boys leave, there are then left 5 girls for each boy, find the number of girls in the beginning?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 =40Problem - 6An organization purchased 80 chairs fro Rs.9700. For chairs of better quality they paid Rs.140 each and for each of the lower grade chair they paid Rs.50 less. How many better quality chairs did the organization buy?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 = 50Problem - 7A labour is engaged for 30 days, on the condition that Rs.50 will be paid for everyday he works and Rs.15 will be deducted from his wages for everyday he is absent from work. At the end of 30 days he received Rs.850 in all. For how many days did he wanted?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 daysProblem - 8The rent is charged at Rs.50 per day for first 3 days Rs.100 per day next 5 days, and 300 per day thereafter. Registration fee is 50 at the beginning. If a person had paid Rs.1300 for his stay how many days did he stay?Solution3 days = 150 + 50 = 2005 days = 100*5 = 500 = 200 + 500 = 7001300 700 = 6002 days = 300*2 = 600 = 5 + 3 + 2 = 10 daysProblem - 9In a school 20% of students are under the age of 8 years. The number of girls above the age of 8 years is 2/3 of the number of boys above the age of 8 years and amount to 48. What is the total number of students in the school?SolutionGirls above 8 yrs = 48Boys above 8 yrs = 48 / 2/380% of students above 8 yrs = 48 + 72 = 120 80120 20 x80x = 120*20X = 120*20/80 = 30Total No.of students = 120+30 = 150

Ratio and ProportionRatio and ProportionRatio : Relationship between two variables. = a : bProportion : Relationship between two ratios.= a : b : : c : dProportion Calculation = a*d : b*cProblem - 1The ratio of number of boys to that of girls in a school is 3:2. If 20% boys and 25% of girls are scholarship holders, find the percentage of the school students who are not scholarship holders?

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 = 78 78/100*100 = 78% Problem - 2The cost of diamond varies as the square of its weight. A diamond weighing 10 decigrams costs Rs. 32000. Find the loss incurred when it breaks into two pieces whose weights are in the ratio 2:3?

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 = 15360Problem - 3The ratio of the first and second class fares between two railway stations 4 : 1 and the ratio of the number of passengers traveling by first and second class is 1:40. If the total of Rs.1100 is collected as fare from passengers of both classes what was the amount collected from first class passengers?

SolutionFare = 4 : 1Passengers traveling = 1 : 40 Amount = No. pas * fare = 4*1 :10*1 = 4 : 40 = 1:10Total amount = 1100. First class passengers amount = 1*1100/11 = 100Problem - 4A vessel contains a mixture of water and milk in the ratio 1:2 and another vessel contains the mixture in the ratio 3:4. Taking 1 kg each from both mixtures a new mixture is prepared. What will be the ratio of water and milk in the new mixture?

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 = 8:13Problem - 5Ratio of the income of A, B, C last year 3 : 4 : 5. The ratio of their individual incomes of last year and this year are 4:5, 2:3 and 3:4 respectively. If the sum of their present income is Rs.78,800. Find the present individual income of A, B and C.SolutionAs Present Income = 5/4*3x = 15x/4Bs Present Income = 3/2*4x = 12x/2Cs Present Income = 4/7*5x = 20x/715x/4 + 6x+20x/3 = 78,800197x/12 = 78,800X = 945600/197X = 4,800As Present income = 15x/4 = 15*4800/4 = 18,000Bs Present income = 6*x = 6*4,800 = 28,800Cs Present income = 20x/3 = 20*4800/3 = 32,000Problem - 6Of the three numbers, the ratio of the first and the second is 8:9 and that of the second and third is 3:4. If the product of the first and third numbers is 2,400, then find the second number?Solution a : b = 8 : 9 b : c = 3 : 4

b : c = 3*3 : 4*3 = 9 : 12 a : b : c = 8 : 9 : 12Product of first and third = 8k * 12k = 240096k2 = 2400; k2 = 2400/96 = 25 k = 5Second number = 9 * 5 = 45Problem - 7Annual income of A and B are in the ratio of 4 : 3 and their annual expenses are in the ratio 3 : 2. If each of them saves Rs.600 at the end of the year, what is the annual income of A?SolutionIncome = 4 : 3, Expenses = 3 : 2Savings 600 eachAs income = 4x, expenses = 3x, savings = x i.e 600

Income = 4*600 : 3*600A : B = 2400 : 1800A income = 2400

Problem - 8The property of a man was divided among his wife, son and daughter according to his will as follows. Wifes hare is equal to 6/7th of sons share and daughter share is equal of 4/7th of Sons. If the son and daughter together receives Rs.1,02,300. How much does his wife get? SolutionLet the Sons share be x.Daughters share = x*4/7 = 4x/7Wifes 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, 800Problem - 9A pot containing 81 litres of pure milk of the milk 1/3 is replaced by the same amount of water. Again 1/3 of the mixture is replaced by the same amount of water. Find the ratio of milk to water in the new mixture?SolutionMilk : WaterInitial =81 : 01/3 removed = 54 : 271/3 mixture = 36 : 45

Ratio of Milk and Water = 4 : 5Problem - 10729 ml of mixture contains milk and water are in the ratio 7 : 2. How much more water is to be added to get a new mixture containing milk and water in the ratio of 7 : 3.SolutionWater = 729 * 2/9 = 162

RatioWater 2 162 3 x2x = 3*162/2 = 243243 161 = 81 ml water is to be addedProblem - 11Price of a scooter and a television set are in the ratio 3 : 2. If the scooter costs Rs.600 more than the television set, then find the price of television?SolutionDiff. in ratio = 3 2 = 11 ratio is 600 means, the television cost is 2 ratio so, cost of television = 1200Problem - 12The annual income and expenditure of man and his wife are in the ratio of 5:3 and 3:1 respectively, if they decide to save equally and find their balance is 4000. Find their income at the end of the year?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 2000

So, Man income = 5000 and Women income = 3000Problem - 13In a class room, of the boys are above 160 cm in height and they are in 18 number. Also out of the total strength, the boys are only 2/3 and the rest are girls. Find the total number of girls in a class?Solution of the boys in 18 numbers means, of the boys = 6Total number of boys = 18+6 = 24RatioNumber2/3 241/3 x2/3*x = 24*1/3 x = 24/2 = 12 Girls

Problem - 14Rs. 770 was divided among A, B and C such that A receives 2/ 9th of what B and C together receive. Find As share?

SolutionA = 2/9 (B+C)B+C =9A/2A+B+C = 770A + 9A/2 = 77011A = 770*2A = 140

Problem - 15A 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? SolutionLet 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 Alligation and MixtureAlligation 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)

Alligation or Mixture Cost of Cheaper Cost of costlier c d

Cost of Mixture m

d-m m-c

(Cheaper quantity) : (Costlier quantity) = (d m) : (m c)

Problem -1 Three glasses of size 3 lit, 4 lit and 5 lit contain mixture of milk and water in the ratio of 2:3, 3:7 and 4:11 respectively. The content of all the three glasses are poured into a single vessel. Find the ratio of milk and water in the resulting mixture.Solution1st Vessel = Milk = 3*2/5 = 6/5 = Water = 3*3/5 = 9/5 2nd 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) 14:31Problem - 2How many kg of tea worth Rs. 25 per kg must be blended with 30 kg tea worth Rs. 30 per kg, so that by selling the blended variety at Rs.30 per kg there should be a gain of 10%?Solution 30*100/110 = 300/112530300/1130/1125/1130:256:536:30kgProblem - 3A man buys cows for Rs. 1350 and sells one so as to lose 6% and the other so as to gain 7.5% and on the whole he neither gains nor loses. How much does each cow cost?Solution 67.507.5615125:41350*5/9 = 7501350 *4/9 = 600Problem - 4There are 65 students in a class, 39 rupees are distributed among them so that each boy gets 80p and each girl gets 30p. Find the number of boys and girls in a class.SolutionGirlsBoys30806020302: 365*2/5 = 2665*3/5 = 39Problem - 5A person covers a distance 100 kms in 10 hr Partly by walking at 7 km per hour and rest by running at 12 km per hour. Find the distance covered in each part. Solution Speed = Distance / Time = 100 / 10 = 107 12102: 3Time taken in 7 km/hr = 10 * 2/5 = 4 4*7 = 28 kmTime taken in 12 km/hours = 10*3/5 = 6 12*6 = 72 kmA 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? Problem - 6 7 17 10 (17-10) (10-7) 7 : 3The quantity of 2nd kind = 3/10 of 100kg = 30kg

SolutionIn 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?Problem - 727323023Quantity of cheaper tea=2Quantity of superior tea3

Quantity of cheaper tea =2*x/5 = 60 , x=150Quantity of superior tea = 3 * 150/5 = 90 kgSolutionA 3-gallon mixture contains one part of S and two parts of R. In order to change it to mixture containing 25% S how much R should be added?Problem - 8R:S2:175%:25%3:11 gallon of R should be added.

SolutionThree 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?Problem - 9B+C/2A85959055Ratio is 1:1 so A = 50 , B + C = 50

The quantity would be 50 : 25 : 25

SolutionIn what proportion water must be added to spirit to gain 20% by selling it at the cost price?Problem - 10Profit%=20%Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3

010 25/35/325/3

The ratio is 1: 5 SolutionIn an examination out of 480 students 85% of the girls and 70% of the boys passed. How many boys appeared in the examination if total pass percentage was 75%Problem - 11SolutionSolution:70 85 7510 5

Number of Boys = 480 * 10/15Number of Boys = 320

Problem - 12A painter mixes blue paint with white paint so that the mixture contains 10% blue paint. In a mixture of 40 litre paint how many litre of blue paint should be added, so that the mixture contains 20% of blue paint?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 = 9 x = 5Problem - 13From a 100 litre mixture containing water and milk equal proportion, 10 litres of mixture is replaced by 10 litres of water in succession twice. At the end, what is the ratio of milk and water?SolutionMilkWater10 lit(1st) 50: 50 45: 45 45: 552nd 10 lit 40.5: 49.5Add Water 40.5: 59.5 81: 119 Problem - 14In a mixture of 400 gms, 80% is copper, sliver is 20%. How much copper is to be added, so that the new mixture has 84% copper?Solution400*80/100 = 320 Copper400*20/100 = 80 SliverPercenMixture 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

Problem - 15A jar full of whisky contains 50% alcohol. A part of this whisky is replaced by another containing 30% alcohol and now the percentage of alcohol was found to be 35%. Find the quantity of whisky replaced? Solution5030355:155 : 15 = 1 : 3Replaced = 3/4PartnershipType - 1A invest = 10000B invest = 15000Profit = 5000Find their Individual Share ?A : B = 10000 : 15000 = 2 : 3As Share = 5000*2/5 = 2000Bs Share = 5000*3/5 = 3000This is a first and basic step for any Partnership Problem.Type - 2A invest = 5000,After 3 months B joined A, with an investment of 3000 Profit 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.ContType - 2 A : B = 5000 : 3000 = 5*12 : 3*9 = 60 : 27 = 20 : 9As share = 3500*20/29 = 2413.7Bs Share = 3500*9/29 = 1086.3Type - 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 = 72Type - 3 As share = 5000* 47/119 = 1974.8Bs share = 5000*72/119 = 3025.2Type - 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 : 6As Share = 6000*3/11 = 1636Bs Share = 6000*2/11 = 1091Cs Share = 6000*6/11 = 3273

Problem - 1A, B and C started a business in partnership by investing Rs.12000 each. After 6 months, C left and after 4 months D joined with his capital of Rs.24,000. At the end of a year, a profit of Rs.8,500 shared among all the partners. Find Bs share?These are all the basic types remaining we will see when we solve problems.

SolutionA:B:C:D 12000 : 12000 : 12000 : 240001:1:1:21*12 :1*12 :1*6: 2*212:12:6:46:6:3:2Bs share = 6/17*8500 = 3000Problem - 2A, B and C enter into partnership. A contributes one third of the capital while B contributes as much as A and C together contributed. If the profit at the end of the year amounted to Rs.840. What would be Bs share?

SolutionAs share = 1/3 of the capitalAs share = 1/3*840 = 280Bs share = A + C = 280 + xA + B + C = 840280 + 280 + x + x = 840560 + 2x = 8402x = 840 560 X = 140Bs share = 280+140 = 420

Problem - 3Akilesh and Jaga enter into a partnership. Akilesh contributing Rs.8000 and Jaga contributing Rs.10000. At the end of 6 months they introduce Prakash, who contributes Rs.6000. After the lapse of 3 years, they find that he firm has made a profit of Rs.9660. Find Prakashs share?SolutionAkilesh:Jaga:Prakash 8:10:64:5:34*36:5836:3*30144:180:908:10:5Prakashs share = 9660*5/23 = 2100

Problem - 4Priya and Vijay enter into partnership. Priya supplies whole of the capital amounting to Rs.45,000 with the conditions that the profit are to be equally divided and that Vijay pays Priya interest on half of the capital of 10% p.a. but receives Rs.120 per month for carrying on the concern. Find their total yearly profit. When Vijays income is one half of Priyas income?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 : 1Priyas income = x/2+2250Vijays income = x/2 2250 +14401 Priya = 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

Problem - 5Revathy and Shiva are partners sharing profits in the ratio of 2:1. They admit Pooja into partnership giving her 1/5th share in profits which she acquires from Revathy and Shiva in the ratio of 1:2. Calculate the new profit sharing ratio?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

Problem - 6A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined them after six months with an amount equal to that of B. In what proportion should the profit at the end of 1 year be distributed among A, B and C?SolutionLet the investment,3:5:53*12: 5*12: 5*6 36:60:306:10:5

Problem - 7If 4(As capital) = 6(Bs capital) = 10 (Cs capital) then out of a profit of rs.4650. Find Cs share?SolutionLet the unknown value be xx/4:x/6:x/1015x/60: 10x/60:6x/60:10:6Cs share = 6/31*4650 = Rs. 900Problem - 8A, B, C subscribe Rs.50,000 fro business. A subscribes Rs.4000 more than B and B Rs.5000 more than C. Out of total profit of Rs.35,000. Find As share?SolutionC = x, B = x + 5000A = x+5000+4000 = x + 9000 x +x+5000 +x+9000 = 500003x+14000 = 500003x = 50000 140003x = 36000,x = 12000C:B:A12000 : 17000 : 21000A = 35000*21/50 = 14,700Problem - 9A and B are partners in a business, A contributes of he capital for 15 months and B received 2/3 of the profit. For how long Bs money was used?SolutionB= 2/3A = 1/3A : B = 1/3 : 2/3 = 1 : 2Investment 1/4x+15 : 3/4x*y15x/4 : 3xy/415x/4 : 3xy/4 : : 1 : 230x/4 = 3xy/4Y = 30x/4 * 4/3x = 10 monthsProblem - 10A, B and C invests Rs.4,000, Rs.5,000 and Rs.6,000 respectively in a business and A gets 25% of profit for managing the business and the rest of the profit is divided by A, B and C in proportion to their investment. If in a year, A gets Rs.200 less than B and C together, what was the total profit for the year?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

Problem - 11A and B entered into partnership with capitals in the ratio of 4 : 5. After 3 months, A withdraw of 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?SolutionA:B4:5 : 5000As share = 4000*1/4 = 4000 1000 = 3000Bs 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

Problem - 12Rs. 1290 is divided between A, B and C. So, that As share is 1 times Bs and Bs share is 1 times C. What is Cs share?SolutionA:B = 1 : 1 = 3/2 : 1 = 3 : 2B:C = 1 : 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

Problem - 13A man starts a business with a capital of Rs.90000 and employs an assistant. From the yearly profit he keeps an amount equal to 4 of his capital and pay 35% of the remainder of the profits. Find how much the assistant receives in a year, in which profit is Rs.30,000.SolutionInvestment = 90,0004 of investment = 9/2/100*90000 = Rs.4050Profit = 30,000 4050 = 25,95035/100*25,950 =9082.50Problem - 14A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and As share is Rs. 855, what is the total profit %?SolutionLet the total profit be Rs. 100After paying charity As share = 3/5 *95 = 57If As share is Rs. 57, the total profit is 100If As share is Rs. 855, the total profit is 100 * 855/57 = Rs. 1500The total profit = Rs. 1500Problem - 15A,B,C entered into a partnership by making an investment in the ratio of 3 : 5 : 7. After a year C invested another Rs. 337600 while A withdrew Rs. 45600. The ratio of investments then changed into 24: 59 : 167. How much did A invest initially?SolutionSolution:Let the investments of A, B, and C be 3x, 5x, 7x(3x 45600) : 5x : (7x + 337600) = 24 : 59 : 167(3x 45600)/5x = 24/59x = 47200Initial investment of A = 47200 * 3 = Rs. 141600

Problems on AgeProblem - 1The age of the Father is 4 times the age of his Son. If 5 years ago, Fathers age was 7 times the age of his Son, what is the Fathers present age?

SolutionF = 4S F - 5 = 7(S - 5)4S 5 = 7S 353S = 30S = 10Fathers age = 4* 10 = 40 years

Problem - 2The age of Mr. Gupta is four times the age of his Son. After Ten years, the age of Mr. Gupta will be only twice the age of his Son. Find the present age of Mr. Guptas Son.SolutionG = 4SG + 10 = 2 ( S + 10)4S + 10 = 2S + 202S = 10 S = 5Sons Age = 5 years

Problem - 310 years ago Anus mother was 4 times older than her daughter. After 10 years, the mother will be twice as old as her daughter. Find the present age of Anu.SolutionTen years before:M 10 = 4(A 10 )M 10 = 4A 40M = 4A 40 + 10M = 4A 30 Ten Years After:M + 10 = 2(A + 10)M + 10 = 2A + 20M = 2A + 20 10 M = 2A + 104A 30 = 2A + 102A = 10 + 302A = 40: Anus Age = 20Problem 4 The sum of the ages of A and B is 42 years. 3 years back, the ages of A was 5 times the age of B. Find the difference between the present ages of A and B?SolutionA + B = 42 A = 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

Problem - 5The sum of the ages of a son and father is 56 years. After 4 years, the age of the father will be 3 times that of the son. Find their respective ages?

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

Problem 6The ratio of the ages of father and son at present is 6:1. After 5 years, the ratio will become 7:2. Find the Present age of the son.Solution6x + 5/x + 5 = 7/212x + 10 = 7x + 3512x 7x = 35 105x = 25x = 25 / 5 x = 5 yearsSon age = 1* 5 = 5 years

Problem - 7The ages of Ram and Shyam differ by 16 years. Six years ago, Shyams age was thrice as that of Rams. Find their present ages?SolutionS = R + 16S 6 = 3(R 6)S 6 = 3R 18R + 16 6 = 3R 18R + 10 = 3R 182R = 28 ; R = 14 Shyams Age = 14 + 16 = 30.Problem - 8A mans age is 125% of what it was 10 years ago, 83 1/3% of what it will be after 10 years. What is his present age?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 yearsProblem - 93 years ago, the average age of a family of 5 members was 17. A baby having born, the average age of the family is the same today. What is the age of the child?SolutionAverage age of 5 members = 17Total age of 5 members = 17*5 = 85 3 years later, the age of 5 members will be = 85 + 15 = 100100 + x / 6 = 17100 + x = 17*6100 + x = 102x = 102 100 = 2 years Problem - 10The sum of the age of father and his son is 100 years now. 5 years ago their ages were in the ratio of 2 : 1. The ratio of the ages of father and his son after 10 years will be?SolutionF + S = 1005 years ago 2 : 15 years agoF + S = 100 10 = 9090*2/3 = 60 : 30Present age = 65 : 35 10 years ago = 75 : 45 = 5 : 3

Problem - 11Six years ago, Sushils age was triple the age of Snehal. Six years later, Sushils age will be 5/3 of the age of Snehal. What is the present age of Snehal?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 yearsProblem - 12Susan got married 6 years ago. Today her age is 1 times that at the time of her marriage. Her son is 1/6 as old as she today. What is the age of her son?Solution6 years ago Susan got married.So her sons age will be less than 6 years.Let as consider, her sons age is 5 years.Susans Age is 5*6 = 30 yrs, since the son is 1/6th of Susans age.6 years ago her age must have been 24 yrs24*1 = 24*5/4 = 30 yrsAs it satisfies the conditions her sons age is 5 yearsProblem - 13My brother is 3 years elder to me. My father was 28 years of age when my sister was born, while my mother was 26 years of age, when I was born. If my sister was 4 years of age when my brother was born, then, what was the age of my father and mother respectively when my brother was born?SolutionMy brother was born 3 years before I was born and 4 years after my sister was bornFathers age when brother was born = 28 + 4 = 32 yearsMothers age when brother was born = 26 3 = 23 yearsProblem - 14If 6 years are subtracted from the present age of Gagan and the reminder is divided by 18, then the present age of his grandson Aunp is obtained. If Anup is 2 years younger to Madan whose age is 5 years, then what is Gagans present age? SolutionAnups age = 5 2 = 3 yearsLet Gagans age be x = x 6 / 18 = 3x 6 = 3*18 ; x 6 = 54x = 54 + 6Gagans age = 60Problem - 15Ramus grandfather says, Ram, I am now 30 years older than your father. 15 years ago, I was 2 times as old as your father. How old is the grandfather now? SolutionLet the fathers age be x.Grandfathers 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 = 35Grandfathers age = 35 + 30 = 65AverageAverageAverage = Sum of Quantities Number of QuantitiesSum of quantities= Average*Number of Quantities.

Number of quantities = Sum of Quantities AverageProblem - 1The average age of a class of 22 students is 21 years. The average increases by 1 when the teachers age is also included. What is the age of the teacher?SolutionTotal age of the students be xx/22 = 21; x = 21*22= 462Teachers 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 = 44Problem - 2The average of 7 numbers is 25. The average of first three of them is 20 while the last three is 28. What must be the remaining number?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

Problem - 3The average age of a team of 10 people remains the same as it was 3 years ago, when a young person replaces one of the member. How much younger was he than the person whose place he took?SolutionLet Average be x10 members Average = 10xAverage of 10 members (including new one) is same as it was 3 yrs ago.Now 10*3 = 30 years have increased, so a person of 30 years should have replaced to keep the average as same.Problem - 4The average age of a couple was 26 years at that time of their marriage. After 11 years of marriage the average age of the family with 3 children become 19 years. What is the average age of the Children?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 familys age Sum of parents age= 95 74 = 21Sum of the ages of 3 children = 21,Average Age = 21/3 = 7 yrsProblem - 59 members went to a hotel for taking meals. Eight of them spent Rs. 12 each on their meals and the ninth person spent Rs. 8 more than the average expenditure of all the nine. What was the total money spent by them?SolutionAverage = x/9Amount Spent by 8 members = 12 * 8 = 9696 + x/9 + 8 = x104 = x x/9104 = 8x/98x = 104 *9 = 936x = 936/8 = 117Problem - 7A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th innings?Solution17th innings avg. = x, Runs = 17x16th innings avg. = x -3, Runs = 16 (x -3)16 (x-3) + 87 = 17x16x 48 +87 = 17xX = 39

Problem - 7There are 24 students in a class. One of them, who was 18 yrs old left the class and his place was filled up by the newcomer. If the average of the class thereby was lowered by one month, what is the age of the newcomer?SolutionAverage reduced by 1 month,24 * 1 = 2 yearsSo, the newcomers age is 18 -2 = 16 years

Problem - 8The average of marks in mathematics for 5 students was found to be 50. Later, it was discovered that in the case of one student the mark 48 was misread as 84. What is the correct average?SolutionDifference = 84 48 = 3636 /5 = 7.2 (Increased)The corrected average = 50 7.2 = 42.8Problem - 9The average salary of all the workers in a factory is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. What is the total number of workers in the factory?SolutionMembers Avg. 12000X 60006x = 7*12 X = 7812/6 = 14Total no. of workers = 7 + 14 = 21

Problem - 10Average salary of all the 50 employees including 5 officers of the company is Rs. 850. If the average salary of the officers is 2500, find the average salary of the remaining staff of the company.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)Problem - 11Find the average of 8 consecutive odd numbers 21,23,25,27,29,31,33,35Solution1st number + last Number /2= 21 + 35 /2 = 28Problem - 12A train covers 50% of the journey at 30 km/hr, 25% of the journey at 25 km/hr, and the remaining at 20 km/hr. Find the average speed of the train during entire journey.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 Problem - 13The average of 10 numbers is 7. What will be the new average if each number is multiplied by 8? 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

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?SolutionBoys = x/10 = 70 = 700Girls = x/15 = 60 = 90010 + 15 = 700 + 90025 = 16001600/25 = 64%Problem - 15Of the three numbers the first is twice the second and the second is thrice the third. If the average of the three numbers is 10, what are the numbers?

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

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

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

PercentageProblem -1 When 75% of the Number is added to 75%, the result is the same number. What is the number?SolutionPercentage Number 75x+75100 x100x + 7500 = 75x25x = 7500x = 300

Problem - 2A tank is full of milk. Half of the milk is sold and the tank is filled with water. Again half of the mixture is sold and the tank is filled with water. This operation is repeated thrice. Find the percentage of milk in the tank after the third operation?SolutionMilk Water 0 50 50(1st) 25 75 (2nd) 12.587.5 (3rd)After 3 operation Milk 12.5%

Problem 3A large water-melon weighs 20kg with 96% of its weight being water. It is allowed to stand in the sun and some of the water evaporates so that now, only 95% of its weight is water. What will be its reduced weight?Solution 20 *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.Problem 4The population of a city is 155625. For every1000 men, there are 1075 women. If 40% of men and 24% of women be literate, then what is the percentage of literate people in the city?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%Problem 5300 grams of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?

Solution Grams Sugar 40%X 50%50x = 40*300x = 40*300/50 = 240300 240 =60 KgProblem - 6A man lost 12% of his money and after spending 70% of the remainder, he has Rs. 210 left. How much did the man have at first?SolutionLet the amount be 100Then, 100.00 12.50 = 87.50

70% of 87.50 = 87.50 *70/100 =61.25

The remaining amount will be Rs. 26.25InitialFinal100 26.25X 21026.25x = 21000; x = 21000/26.25 = 800 Problem - 7During one year the population of a town increases by 10% and during next year it diminished by 10%. If at the end of the second year, the population was 89,100, what was the Population at the beginning of first year?SolutionLet the population be 1001st Year = 100 + 10 = 1102nd Year = 110 * 10/100 = 110 -11 = 99Percentage Population 89100 x99x = 89100*100;x = 8910000/99 = 90000

Problem - 8When a number is first increased by 20% and then again 20% by what percent should the increase number be reduced to get back the original number?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%

Problem - 9The number of students studying Arts, Commerce and Science in an institute were in the ratio 6 : 5 : 3 respectively. If the number of students in Arts, Commerce and science were increased by 10%, 30% and 15% respectively, what was the new ratio between number of students in the three streams?SolutionA : C : S6 : 5 : 36x : 5x : 3x6x*110/100 : 5x*130/100 : 3x*115/1006x*110 : 5x*130 : 3x*115660 : 650 : 345132 : 130 : 69Problem - 10In measuring the sides of rectangle errors of 5% and 3% in excess are made. What is the error percent in the calculated area?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% ExcessProblem - 11In a certain examination there were 2500 candidates. Of them 20% of them were girls and rest were boys. If 5% of boys and 40% of girls failed, what was the Percentage of candidates passed?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%Problem - 12A person saves every year 20% of his income. If his income increases every year by 10% then his saving increases by?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%Problem - 13On a test containing 150 questions carrying 1 mark each, meena answered 80% of the first answers correctly. What percent of the other 75 questions does she need to answer correctly to score 60% on the entire exam?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

Problem - 14A boy after giving away 80% of his pocket money to one companion and 6% of the remainder to another has 47 paise left with him. How much pocket money did the boy have in the beginning?

SolutionLet the amount be 100To the first companion = 100*80/100 = 80 Remaining = 100 80 = 20To the 2nd person = 20*6/100 = 1.20The remaining = Rs.18.80 or 1880 paiseInitial Final 1880X 471880x = 47*100x = 4700/1880 = 2.5Problem - 15The length of a rectangle is increased by 10% and breath decreased by 10%. Then the area of the new rectangle?SolutionI D I*D /10010 -10 10*10/1000 1 = -1 Decrease by 1%Profit and LossGain =(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 LossProblem - 1A trade man allows two successive discount of 20% and 10%. If he gets Rs.108 for an article. What was its marked price?SolutionI1 + I2 I1*I2/10020 + 10 20*10 /100= 28%Discount = 28%, 72 Percent Cost is 108Then 100percent cost = 72 108 100x100*108/72 = 150

Problem - 2A trade man bought 500 metres of electric wire at 75 paise per metre. He sold 60% of it at profit of 8%. At what gain percent should he sell the remainder so tas to gain 12% on the wholeSolution 500* 60/100 = 3008X12300200300 : 200 = 6 : 481812 4Remainder at 18% ProfitProblem - 3A man purchased a box full of pencils at the rate of 7 for Rs. 9 and sold all of them at the rate of 8 for Rs. 11. in this bargains he gains Rs. 10. How many pencils did the box contains.Solution LCM = 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 112Problem - 4A cloth merchant decides to sell his material at the cost price, but measures 80cm for a metre. His gain % is?Solution100 80 = 20 cm differenceActual = 8020/80*100 = 25% GainProblem - 5Sales of a book decrease by 2.5% when its price is hiked by 5%. What is the effect on sales?SolutionLet the sales be 100 2.5 = 97.5Profit = 100+5 = 105Sales Profit97.5105 X100x = 97.5*105x = 97.5*105/100 = 102.375100 102.375 = 2.375 = 2.4 profit (app)

Problem - 6A dealer buys a table listed at Rs.1500 and gets successive discount of 20% and 10%. He spends Rs. 20 on transportation and sells it at a profit of 10%. Find the selling price of the table.SolutionDiscount = 20+10 20*10/100 = 28%Actual price = 100 28 = 72 1001500 72x72*1500/100 = 1080Transport = 1080 +20 = 1100 1001100 110 x1100*110/100 = 1210Problem - 7A fridge is listed at Rs. 4000. due to the off season, a shopkeeper announces a discount of 5%. What is the S.P?Solution= 4000*95/100 = 3800Problem - 8If the cost price of 9 pens is equal to the S.P of 11 pens. What is the gain or loss?Solution= 11 9 = 2= 2/11*100 = 18 2/11% lossProblem - 9A machine is sold for Rs.5060 at a gain of 10% what would have been the gain or loss percent if it had been sold Rs.4370?SolutionS.P = Rs.5060 = Gain = 10%C.P = 100/110*5060 = 4600IF S.P = Rs.4370 and C.P = Rs.4600 Loss = 230Loss % = 230/4600 * 100 = 5% lossProblem - 10A person purchased two washing machines each for Rs.9000. he sold one at a loss of 10% and other at a gain of 10%. What is his gain or loss?SolutionEach Rs.9000. one is 10% profit and other is 10% loss. So No profit and No lossProblem - 11Four percent more is gained by selling an article for Rs.180, then by selling if for Rs.175. then its C.P is?SolutionLet the cost price = Rs. X4% of x = 180 175 = 4x/100 = 54x = 500; x = 500/4 = 125Problem - 12An article is sold at a profit of 20%. If it had been sold at a profit of 25%. It would have fetched Rs.35% more. The Cost Price of the article is?SolutionLet C.P = Rs. X125% of x 120% of x = 355% of x =Rs.35 = x = 35*100/5 = 700C.P = Rs. 700Problem - 13A reduction of 20% in the price of orange enables a man to buy 5 oranges more for Rs. 10. The price of an orange before reduction was,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 PaiseProblem - 14A man sells two horses for Rs.1475. The cost price of the first is equal to the S.P of the second. If the first is sold at 20% loss and the second at 25% gain. What is his total gain or loss? ( in rupees)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 gainProblem - 15Rekha sold a watch at a profit of 15%. Had he bought it at 10% less and sold it for Rs. 28 less, he would have gained 20%. Find the C.P of the Watch.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 ProbabilityProbabilityProbability: P() = n() / n(s) (Addition theorem on probability: n(AUB) = n(A) + n(B) - n(AB)Mutually Exclusive: P(AUB) = P(A) + P(B)Independent Events: P(AB) = P(A) * P(B)

Problem - 1Four cards are drawn at random from a pack of 52 playing cards. Find the probability of getting all face cards?

Solutionn(E) = 52C4 n(S) = 12C4 = 12C4/52C4Problem - 2Four persons are to be chosen at random from a group of 3 men, 2 women and 4 children. Find the probability of selecting 1 man, 1 woman or 2 children?

SolutionTotal 3 M + 2 W + 4 C = 9 C 4 = 126n (E) = 3C1 * 2C1 * 4C2 = 3636/126 = 2/7Problem - 3A word consists of 9 letters, 5 consonants and 4 vowels. Three letters are chosen at random. What is the probability that more than one vowels will be selected?

Solutionn(E) = 9C3 = 84More than one Vowels. So,2V +1C or 3 V4C2 *5C1 + 4C3 = 34= 34/84 = 17/42Problem - 4A bag contains 10 mangoes out of which 4 are rotten. Two mangoes are taken out together. If one of them was found to be good, then what is the probability that the other one is also good?

Solution10 mangoes 4 are rotten = 6 good mangoesGetting good mangoes = 6C1/10C1 = 6/10Getting second mango to be good = 5/9 1st and 2nd mangoes6/10 *5/9 = 1/3Problem - 5Out of 13 applicants for a job there are 5 women and 8 men. It is desired to select 2 persons for the job. What is the probability that at least one of the selected person will be a woman?

Solutionn(E) = 13C2 = 78n(S) = 1m and 1 w or 2 w = 8C1*5C1 + 5C2 = 50= 50/78 = 25/39Problem - 6Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black or both are queen?

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 = 55/221Problem -7A man and his wife appear in an interview for two vacancies in the same post. The probability of husbands selection is 1/7 and the probability of wifes selection is 1/5. Find the probability that only one of them is selected?

SolutionHusbands Selection = 1/7; Not getting selected = 1 1/7 = 6/7Wifes selection = 1/5; Not getting selected = 1 1/5 = 4/5Only one of them is selected =(Husbands Selection + Wife Not getting selected) or (Wifes selection + Husbands Not getting selected)= (1/7*4/5) + 1/5*6/7) = 2/7Problem - 8Four persons are chosen at random from a group of 3 men, 2 women and 4 children. What is the chance that exactly 2 of them are children?

Solution3 + 2 + 4 = 9C4 = 1264 members 2(M and W) + 2(boy)5C2 + 4C2 = 60 = 60 / 126 = 10/21Problem - 9Prakash can hit a target 3 times in 6 shots, Priya can hit the target 2 times in 6 shots and Akhilesh can hit the target 4 times in 4 shots. What is the probability that at least 2 shots hit the target?

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 = Problem - 10There are two boxes A and B. A contains 3 white balls and 5 black balls and Box B contains 4 white balls and 6 black balls. One box is taken at random and what is the probability that the ball picked up may be a white one?

Solution(Box A is selected and a ball is picked up ) or (Box B is selected and a ball is picked up)*3/8 + *4/10 = 31/80Problem - 11A bag contains 6 white balls and 4 black balls. Four balls are successively drawn without replacement. What is the probability that they are alternately of different colour?

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 = 360/5040 360/5040 +360/5040 = 1/7 Problem - 12A problem in statistics is given to four students A, B, C and D. Their chances of solving it are 1/3, , 1/5 and 1/6 respectively. What is the probability that the problem will be solved?

SolutionA is not solving problem = 2/3,B is not solving problem = C not solving problem = 4/5D not solving problem = 5/62/3*3/4*4/5*5/6 = 1/3All together the probability of solving the problem = 1 -1 /3 = 2/3Problem - 13There are 8 questions in an examination each having only 2 answers choices Yes or No. All the questions carry equal marks. If a student marks his answer randomly, what is the probability of scoring exacting 50%?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 = 35/128A 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.Problem - 14Let the number of men is 100 and women be 100Employed men and women = (100-10)+(100-45)= 145Probability = 145 / 200 = 29 / 40SolutionProblem - 15 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? 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

Permutation and CombinationPermutation and CombinationPermutation means ArrangementCombination means SelectionPermutation and CombinationPermutations: Each of the arrangements which can be made by taking some (or) all of a number of items is called permutations.npr = n(n-1)(n-2)(n-r+1)=n!/(n-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. nCr = n!/(r!)(n-r)!

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!

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!)

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!

Types5. How many ways of arrangement possible by using word ARRANGEMENTLetters 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.Problem - 1A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways this can be done, when at least 2 ladies are included?Solution2 ladies * 3 Gents 4C2 * 6 C3 = 120b. 3 ladies * 2 Gents4C3 * 6C2 = 60c. 4 ladies * 1 Gent4C4 *6C1 = 1*6 = 6Total ways = 120 +60 +6 = 186

Problem - 2It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?SolutionTotal places = 9Odd places = 5Even places = 4 4 even places occupied by 4 women = 4P4 = 4! = 245 odd places occupied by 5 men = 5P4 = 5! = 120Total ways = 120*24 = 2880 waysProblem - 3A set of 7 parallel lines is intersected by another set of 5 parallel lines. How many parallelograms are formed by this process?SolutionTwo parallel lines from the first set and any two from the second set will from a parallelogram.7C2 *5C2 = 21 * 10 = 210

Problem - 4There are n teams participating in a football championship. Every two teams played one match with each other. There were 171 matches on the whole. What is the value of n?SolutionTotal number of matches played = nC2nC2 = 171n(n-1)/2= 171n2 n 342 = 0(n+18) (n-19) = 0 n = 19Problem - 5In an examination, a candidate has to pass in each of the 6 subjects. In how many ways can he fail?Solution6C1 + 6C2 + 6C3 + 6C4+6C5+6C61 + 6 + 15 + 20 + 15 + 6 = 63 waysProblem - 6In how many ways can a pack of 52 cards be distributed to 4 players, 17 cards to each of 3 and one card to the fourth player?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!)3A 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 Problem - 7n = 10r = 3n P r = n!/(n-r)! = 10! / (10-3)! = 10! / 7! = 8*9*10 = 720 Number of ways is 720.Solution 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 ?Problem - 8It is selection so use combination formulaProgrammers and managers = 6C3 * 4C2 = 20 * 6 = 120Total number of ways = 120 ways.SolutionProblem - 9A man has 7 friends. In how many ways can he invite one or more of them to a party?SolutionIn 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 Problem - 9Find the number of different 8 letter words formed from the letters of the word EQUATION if each word is to start with a vowelSolutionFor the words beginning with a vowel, the first letter can be any one of the 5 vowels, the remaining 7 places can be filled by7P7= 5040The number of words = 5 * 5040 = 25200Problem - 10In how many different ways can the letters of the word TRAINER be arranged so that the vowels always come together?SolutionA,I,E can be arranged in 3! Ways

(5! * 3!) / 2! = 360 waysProblem - 11In how many different ways can the letters of the Word DETAIL be arranged so that the vowels may occupy only the odd positions?Solution___ ___ ___ ___ ___ ___ 3P3 = 3! = 1*2*3 = 63P3 = 3! = 1*2*3 = 6 = 6*6 = 36Problem - 12There are 5 red, 4 white and 3 blue marbles in a bag. They are taken out one by one and arranged in a row. Assuming that all the 12 marbles are drawn, find the number of different arrangements?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!= 27720Problem - 135 men and 5 women sit around a circular table, the en and women alternatively. In how many different ways can the seating arrangements be made?

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 waysProblem - 14In a chess board there are 9 vertical and 9 horizontal lines. Find the number of rectangles formed in the chess board.SolutionSolution:

9C2 * 9C2 = 1296Problem - 15 In how many ways can a cricket team of 11 players be selected out of 16 players, If one particular player is to be excluded?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 waysArea and VolumeArea and VolumeCube:Let each edge of the cube be of length a. then,Volume = a3cubic unitsSurface area= 6a2 sq.units.Diagonal = 3 a units.Cylinder:Let each of base = r and height ( or length) = h. Volume = r2hSurface area = 2 r h sq. unitsTotal Surface Area = 2 r ( h+ r) units. Area and VolumeCone:Let radius of base = r and height=h, thenSlant height, l = h2 +r2 unitsVolume = 1/3 r2h cubic unitsCurved surface area = r l sq.unitsTotal surface area = r (l +r)Area and VolumeSphere:Let the radius of the sphere be r. then,Volume = 4/3 r3Surface area = 4 r2sq.unitsArea and VolumeCircle: A= r 2 Circumference = 2 r Square: A= a 2 Perimeter = 4aRectangle: A= l x bPerimeter= 2( l + b)Area and VolumeTriangle:A = 1/2*base*heightEquilateral = 3/4*(side)2 Area of the Scalene Triangle S = (a+b+c)/ 2 A = s*(s-a) * (s-b)* (s-c)Area and VolumeProblem - 1A rectangular sheet of size 88 cm * 35 cm is bent to form a cylindrical shape with height 35 cm. What is the area of the base of the cylindrical shape?

SolutionThe circumference of the circular region = 88 cm2r = 88 r = 88*7/22*2 = 14 cmArea of the base = r2 = 22/7*14*14 v= 616 cm2Problem - 2The radius of the base of a conical tent is 7 metres. If the slant height of the tent is 15 metres, what is the area of the canvas required to make the tent?SolutionR = 7 mL = 15 mArea of Canvas required = Curved Surface Area of conerl = 22/7*7*15 = 330 sq.mProblem - 3Three spherical balls of radius 1 cm, 2 cm and 3 cm are melted to form a single spherical ball. In the process, the material loss was 25%. What would be the radius of the new ball?

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 * = 36 V = 4/3r3 = 36r3 = 36*3/4 = 27r = 3 cmProblem - 4A rectangular room of size 5m(l)*4m(w)*3m(h) is to be painted. If the unit of painting is Rs. 10 per sq.m, what is the total cost of painting?SolutionArea of 4 walls = 2h(l+b)The area to be painted includes the 4 walls and the 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.740Problem - 5The radius of a sphere is r units. Each of the radius of the base and the height of a right circular cylinder is also r units. What is the ratio of the volume of the sphere to that of the cylinder?SolutionVol. of sphere = 4/3r3 and Vol. of Cylinder = r2h = r3

Required Ratio = 4/3 r3: r3 = 4/3 : 1 = 4 : 3Problem - 6A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each, are dropped in it and they sink down in the water completely. What will be the increase in the level of water in the jar?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 = 16/3 H = 16/3*25 = 16/75 cmWhat 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 / m2Problem - 7Area of triangular field = * 3.2 * 2.8 m2 = 4.48 m2 Cost = Rs.100 * 4.48 = Rs.448..SolutionProblem - 8Find the length of the longest pole that can be placed in a room 14 m long, 12 m broad, and 8 m high.SolutionLength of the longest pole = Length of the diagonal of the room = (142 + 122 + 82) = 404= 20.09 m Area of a rhombus is 850 cm2. If one of its diagonal is 34 cm. Find the length of the other diagonal.Problem - 9850 = * d1 * d2 = * 34 * d2 = 17 d2 d2 = 850 / 17 = 50 cm Second diagonal = 50cm SolutionA 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 ?Problem - 10 No. of Boxes = 25*42*60 / 7*6*5 = 300300 boxes of cereal box can be placed.SolutionProblem - 11If the radius of a circle is diminished by 10%, what is the change in area in percentage?Solution= x + y + xy/100= -10 - 10 + 10*10/100= -19% Diminished area = 19%.

Problem - 12A 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?Solutionlength of wire = 2 r = (22/7*14*14)cm = 264cmPerimeter of Rectangle = 2(6x+5x) cm = 22xcm 22x =264 x = 12 cm Smaller side = (5*12) cm = 60 cmProblem - 13A beam 9m long, 40cm wide and 20cm deep is made up of iron which weights 50 kg per cubic metre. Find the weight of the Beam.SolutionVol. of the Beam = lbh = 9*40/100*10/100 = 72 m3Weight of the iron beam is given as lm3 = 50 kg

72/100 m3 = 72/100*50 = 36 kgProblem - 14If the length of a rectangle is reduced by 20% and breadth is increased by 20%. What is the percentage change in the area?

Solutionx + y + (xy/100)%= - 20 + 20 400/100= -4The area would decrease by 4%Problem - 15Find the number of bricks measuring 25 cm in length, 5 cm is breadth and 10 cm in height for a wall 40 m long, 75 cm broad and 5 metres in height?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

Calendar CalendarOdd days:0 = Sunday1 = Monday2 = Tuesday3 = Wednesday4 = Thursday5 = Friday6 = Saturday 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. CalendarOrdinary year = (A + B + C + D )-2 -----------------------take remainder 7

Leap year = (A + B + C + D) 3 ------------------------- take remainder 7Problem - 111th January 1997 was a Sunday. What day of the week on 7th January 2000?Solution11th Jan 1997 = Sunday11th Jan 1998 = Monday11th Jan 1999 = Tuesday11th Jan 2000 = Wednesday7th Jan 2000 is on SaturdayProblem - 2What day of the week was on 5th June 1999?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 = SaturdayProblem - 3On what dates of August 1988 did Friday fall?SolutionA = 1988 / 7 = 0B = 1988/4 = 497/7 = 0C = 3D = x0+0+3+x+3/7 = x/7 = 5(Friday)Friday falls on = 5,12,19,26Problem - 4India got independence on 15 August 1947. What was the day of the week?SolutionA = 1947/7 = 1B = 1947/4 = 486/7 = 3C = 15/7 = 1D = 21+3+1+2 2 /7 = 5/7 = FridayProblem - 57th January 1992 was Tuesday. Find the day of the week on the same date after 5 years. i.e on 7th January 1997.Solution7th January 1992 = Tuesday7th January 1993 = Thursday (Leap)7th January 1994 = Friday7th January 1995 = Saturday7th January 1996 = Monday ( Leap)7th January 1997 = TuesdayProblem - 6The first Republic day of India was celebrated on 26th January 1950. What was the day of the week on that date?SolutionA = 1950/7 = 4B = 1950/4 = 487/7 = 4C = 0D = 26/7 = 54+4+0+5 2/7 = 11/7 = 4 = ThursdayProblem - 7Find the Number of times 29th day of the month occurs in 400 consecutive year?Solution1 year = 1 (Ordinary Year)1 year = 12 (Leap Year)400 years = 97 leap year97 * 12 = 1164303*11 = 3333 = 1164+3333 = 4497 timesProblem - 8If 2nd March 1994 was on Wednesday, 25 Jan 1994 was on,SolutionA = 1994/7 = 6B = 1994/4 = 498/7 = 1C = 0D = 25/7 = 4= 6 + 1 + 0 + 4 2 / 7 = 3 = TuesdayProblem - 9Calendar for 2000 will serve also?Solution= 2000 + 2001 + 2002 + 2003 + 2004 = 2 + 1 + 1 + 1 + 2 = 7 (Complete Week) 2005Problem - 10If Pinkys 1st birthday fell in Jan 1988 on one of the Mondays, the day on which are was born is,SolutionJan = 1988 = MondayJan = 1987 = SundayProblem - 11Akshaya celebrated her 60th birthday on Feb 24, 2000. What was the day?SolutionA = 2000 /7 = 7B = 2000/4 = 500/7 = 3C = 3D = 24/7 = 0= 7+3+3+0-3/7 = 10/7 = 3 = WednesdayProblem - 12On what dates of April 2008 did Sunday Fall?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.Problem - 13Today is Friday. After 62 days it will be,Solution62 / 7 = 6 days after Friday then it will be TuesdayProblem - 14What will be the day of the week on 1st Jan 2010?SolutionA = 1B = 5C = 0D = 1= 1+5+0+1 2/ 7 = 5/7 = 5 = Friday What is the day of the week on 30/09/2007?Problem - 15Solution: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 7CalendarClockClocksClock:

Angle between hour hand and minute hand = (11m/2) 30h

Angle between minute hand and hour hand =30h (11m/2)

Problem - 1What is the angle between the minute hand and hour hand when the time is 2.15?Solution= 11 m/2 30(h)= 11 15/2 30(2)= 11(7.5) 60= 82.5 60 = 22 1/2Problem - 2At what time between 5 and 6 oclock the hands of a clock coincide?

SolutionCoinciding Angle = 0Min. hand to hour hand = 25 min apart60/55*25 = 12/11 * 25 = 300/11= 27 3/11min past 5 Problem - 3At what time between 12 and 1 oclock both the hands will be at right angles?SolutionRight angle = 90 degrees= 30(h) 11 m/290 = 30(12) 11 m/2180 = 360 11m11m = 360 180M = 180/1116 4/11 past 12Problem - 4Find at what time between 7 and 8 oclock will the hands of a clock be in the same straight line but not together?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 Problem - 5At what time between 5 and 6 are the hands of the clock 7 minutes apart?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 5Problem - 6A clock strikes 4 and takes 9 seconds. In order to strike 12 at the same rate what will be the time taken?SolutionStrikeSec 3 (interval)9 11x3x = 11*9X = 11*9/3 = 33 Sec