Numerical Reasoning

Problem on Numbers

Arithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d Sum of n terms of A.P S = n/2 *[2a+(n-1)d)]

Geometrical Progression:

Tn = arn – 1.

Sn = a(rn – 1) / (r-1)

Problems on Numbers

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)

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 = 9

Problem - 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 remainder

Problem - 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 = 24

Problem - 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 + ¼)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 =40

Problem - 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 = 50

Problem - 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 days

Problem - 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 days

Problem - 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

80 120 20 x80x = 120*20X = 120*20/80 = 30Total No.of students = 120+30 = 150

Ratio and Proportion

Ratio and ProportionRatio : Relationship between two variables.

= a : bProportion : Relationship between two ratios.

= a : b : : c : dProportion Calculation = a*d : b*c

Problem - 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 = 15360

Problem - 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 = 100

Problem - 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 = 8:13

Problem - 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.

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

Problem - 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 = 45

Problem - 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 eachA’s 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. Wife’s hare is equal to 6/7th of son’s share and daughter share is equal of 4/7th of Son’s. If the son and daughter together receives Rs.1,02,300. How much does his wife get?

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

Problem - 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 : Water

Initial = 81 : 01/3 removed = 54 : 271/3 mixture = 36 : 45

Ratio of Milk and Water = 4 : 5

Problem - 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

Ratio Water 2 162 3 x

2x = 3*162/2 = 243243 – 161 = 81 ml water is to be added

Problem - 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 = 1200

Problem - 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 = 3000

Problem - 13In a class room, ¾ 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?

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*1/3 x = 24/2 = 12 Girls

Problem - 14

Rs. 770 was divided among A, B and C such that

A receives 2/ 9th of what B and C together

receive. Find A’s 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 Mixture

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)

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 = 28/10

3rd Vessel:= Milk = 5*4/15 = 20/15= Water = 5*11/15 = 55/15

Milk : 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

Problem - 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 = 300/1125 30

300/11

30/11 25/1130 : 256 : 536 : 30kg

Problem - 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 6 7.5

07.5 615 125 : 4

1350*5/9 = 7501350 *4/9 = 600

Problem - 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.

SolutionGirls Boys30 80

6020 302 : 365*2/5 = 2665*3/5 = 39

Problem - 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 = 10

7 1210

2 : 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 km

A merchant has 100 kg of salt, part of which

he sells at 7% profit and the rest at 17% profit.

He gains 10% on the whole. Find the quantity

sold at 17% profit?

Problem - 6

7 17 10 (17-10) (10-7) 7 : 3The quantity of 2nd kind = 3/10 of 100kg = 30kg

Solution

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

Problem - 7

27 3230

2 3Quantity of cheaper tea = 2Quantity of superior tea 3

Quantity of cheaper tea =2*x/5 = 60 , x=150Quantity of superior tea = 3 * 150/5 = 90 kg

Solution

A 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 - 8

R : S2 : 175% : 25%3 : 1

1 gallon of R should be added.

Solution

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

Problem - 9

B+C/2 A85 95

905 5

Ratio is 1:1 so A = 50 , B + C = 50

The quantity would be 50 : 25 : 25

Solution

In what proportion water must be added to

spirit to gain 20% by selling it at the cost price?

Problem - 10

Profit%=20%Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3

0 10 25/3

5/3 25/3

The ratio is 1: 5

Solution

In 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 - 11

SolutionSolution:70 85 7510 5

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

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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 = 5

Problem - 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?

SolutionMilk Water

10 lit(1st) 50 : 50 45 : 45 45 : 55

2nd 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 Sliver

Percen 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

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?

Solution50 30

355 : 15

5 : 15 = 1 : 3Replaced = 3/4

Partnership

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.

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.

Cont…

Type - 2

A : B = 5000 : 3000 = 5*12 : 3*9 = 60 : 27 = 20 : 9

A’s share = 3500*20/29 = 2413.7

B’s Share = 3500*9/29 = 1086.3

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

Type - 3

A’s share = 5000* 47/119 = 1974.8

B’s share = 5000*72/119 = 3025.2

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

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 B’s 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 : 2

B’s share = 6/17*8500 = 3000

Problem - 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 B’s share?

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 – 560 X = 140B’s 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 Prakash’s share?

SolutionAkilesh : Jaga : Prakash 8 : 10 : 6

4 : 5 : 34*36 : 5836 : 3*30144 : 180 : 908 : 10 : 5

Prakash’s 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 Vijay’s income is one half of Priya’s 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 : 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

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(A’s capital) = 6(B’s capital) = 10 (C’s capital) then out of a profit of rs.4650. Find C’s share?

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

Problem - 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 A’s 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 = 14,700

Problem - 9A and B are partners in a business, A contributes ¼ of he capital for 15 months and B received 2/3 of the profit. For how long B’s 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 months

Problem - 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 ¼ 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 : 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

Problem - 12Rs. 1290 is divided between A, B and C. So, that A’s share is 1 ½ times B’s and B’s share is 1 ¾ times C. What is C’s share?

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

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 ½ 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 ½ of investment = 9/2/100*90000 = Rs.4050Profit = 30,000 – 4050 = 25,95035/100*25,950 =9082.50

Problem - 14A and B invest in a business in the ratio 3 : 2. If

5% of the total profit goes to charity and A’s

share is Rs. 855, what is the total profit %?

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

Problem - 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 = 47200

Initial investment of A = 47200 * 3 = Rs. 141600

Problems on Age

Problem - 1

The age of the Father is 4 times the age of his Son. If 5 years ago, Father’s age was 7 times the age of his Son, what is the Father’s present age?

SolutionF = 4S F - 5 = 7(S - 5)4S – 5 = 7S – 353S = 30S = 10Father’s age = 4* 10 = 40 years

Problem - 2

The 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. Gupta’s Son.

SolutionG = 4SG + 10 = 2 ( S + 10)4S + 10 = 2S + 202S = 10 S = 5Son’s Age = 5 years

Problem - 3

10 years ago Anu’s 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: Anu’s Age = 20

Problem – 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 - 5

The 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 – 6

The 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 - 7

The ages of Ram and Shyam differ by 16 years. Six years ago, Shyam’s age was thrice as that of Ram’s. 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 Shyam’s Age = 14 + 16 = 30.

Problem - 8

A man’s 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 years

Problem - 9

3 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, Sushil’s age was triple the age of Snehal. Six years later, Sushil’s 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 years

Problem - 12

Susan got married 6 years ago. Today her age is 1¼ 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 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

Problem - 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 bornFather’s age when brother was born

= 28 + 4 = 32 yearsMother’s age when brother was born

= 26 – 3 = 23 years

Problem - 14

If 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 Gagan’s present age?

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

Problem - 15

Ramu’s grandfather says, “ Ram, I am now 30 years older than your father. 15 years ago, I was 2½ times as old as your father”. How old is the grandfather now?

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

Average

AverageAverage = Sum of Quantities

Number of QuantitiesSum of quantities

= Average*Number of Quantities.

Number of quantities = Sum of Quantities

Average

Problem - 1

The average age of a class of 22 students is 21 years. The average increases by 1 when the teacher’s age is also included. What is the age of the teacher?

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

Problem - 2

The 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 - 3

The 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 - 4

The 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?

Solution

Average 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

Problem - 5

9 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 = 117

Problem - 7

A 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 newcomer’s 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 = 42.8

Problem - 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.

7 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 - 11

Find the average of 8 consecutive odd numbers 21,23,25,27,29,31,33,35

Solution

1st number + last Number /2= 21 + 35 /2 = 28

Problem - 12

A 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 - 13

The 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

Percentage

Percentage

• 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

Percentage

Problem -1

When 75% of the Number is added to 75%, the result is the same number. What is the number?

SolutionPercentage Number 75 x+75

100 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 Water100 0 50 50(1st) 25 75 (2nd) 12.5 87.5 (3rd)After 3 operation Milk 12.5%

Problem 3

A 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 5

300 grams of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?

Solution Grams Sugar300 40%X 50%50x = 40*300x = 40*300/50 = 240300 – 240 =60 Kg

Problem - 6

A man lost 12½% 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. 26.25Initial Final100 26.25X 210

26.25x = 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 Population99 89100100 x99x = 89100*100;x = 8910000/99 = 90000

Problem - 8

When 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 : 69

Problem - 10

In measuring the sides of rectangle errors of 5% and 3% in excess are made. What is the error percent in the calculated area?

Solution

Area = 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

Problem - 11

In 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 - 12

A 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 - 14

A 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 Final100 1880X 471880x = 47*100x = 4700/1880 = 2.5

Problem - 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 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.

Profit and Loss

Problem - 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

100 x100*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 whole

Solution 500* 60/100 = 3008 X

12300 200300 : 200 = 6 : 48 18

126 4Remainder at 18% Profit

Problem - 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 112

Problem - 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% Gain

Problem - 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 Profit97.5 105100 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 = 72 100 1500 72 x72*1500/100 = 1080Transport = 1080 +20 = 1100 100 1100 110 x1100*110/100 = 1210

Problem - 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 = 3800

Problem - 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% loss

Problem - 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% loss

Problem - 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 loss

Problem - 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 = 125

Problem - 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. 700

Problem - 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 Paise

Problem - 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 gain

Problem - 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

Probability

Probability• Probability: P(є) = n(є) / 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 = 12C4/52C4

Problem - 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 = 2/7

Problem - 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 = 17/42

Problem - 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 = 1/3

Problem - 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 = 25/39

Problem - 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 = 55/221

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

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

Problem - 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 = 10/21

Problem - 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 = ½

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)½*3/8 + ½*4/10 = 31/80

Problem - 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, ¼, 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 = 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 the problem = 1 -1 /3 = 2/3

Problem - 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 = 35/128

A group consists of equal number of men and

women. Of them 10% of men and 45% of

women are unemployed. If a person is randomly

selected from the group find the probability for

the selected person to be an employee.

Problem - 14

Let the number of men is 100 and women be 100

Employed men and women = (100-10)+(100-45)

= 145

Probability = 145 / 200 = 29 / 40

Solution

Problem - 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?

Solution

It 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 Combination

Permutation and Combination

Permutation means Arrangement

Combination means Selection

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.

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 places

Vowels = 4P3 = 4!Consonant = 4P4 = 4!Total Number of arrangement = 4!*4!

Types5. How many ways of arrangement possible by

using word ARRANGEMENTLetter’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.

Problem - 1

A 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?

Solutiona. 2 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 - 2

It 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 ways

Problem - 3

A set of 7 parallel lines is intersected by another set of 5 parallel lines. How many parallelograms are formed by this process?

Solution

Two parallel lines from the first set and any two from the second set will from a parallelogram.7C2 *5C2 = 21 * 10 = 210

Problem - 4

There 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 = nC2

nC2 = 171n(n-1)/2= 171n2 – n – 342 = 0(n+18) (n-19) = 0 n = 19

Problem - 5

In an examination, a candidate has to pass in each of the 6 subjects. In how many ways can he fail?

Solution

6C1 + 6C2 + 6C3 + 6C4+6C5+6C6

1 + 6 + 15 + 20 + 15 + 6 = 63 ways

Problem - 6

In 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 C17

2nd player = 35C17

3rd player = 18C17

4th player = 1= 52C17*35C17*18C17

= 52!/17!35! * 35!/17!*18! * 18!/17!*1!= 52!/(17!)3

A foot race will be held on Saturday. How many

different arrangements of medal winners are

possible if medals will be for first, second and

third place, if there are 10 runners in the race …

Problem - 7

n = 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 - 8

It is selection so use combination formula

Programmers and managers = 6C3 * 4C2

= 20 * 6 = 120

Total number of ways = 120 ways.

Solution

Problem - 9

A 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 - 9

Find the number of different 8 letter words

formed from the letters of the word EQUATION

if each word is to start with a vowel

SolutionFor the words beginning with a vowel, the first letter can be any one of the 5 vowels, the remaining 7 places can be filled by

7P7 = 5040The number of words = 5 * 5040 = 25200

Problem - 10

In 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 ways

Problem - 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 = 36

Problem - 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!= 27720

Problem - 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 ways

Problem - 14

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

SolutionSolution:

9C2 * 9C2 = 1296

Problem - 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 ways

Area and Volume

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.

Cylinder:

• 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.

Area and Volume

Cone:• 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)

Area and Volume

Sphere:

• Let the radius of the sphere be r. then,

• Volume = 4/3 πr3

• Surface area = 4 π r2sq.units

Area and Volume

Circle: A= π r 2

Circumference = 2 π r

Square: A= a 2

Perimeter = 4aRectangle: A= l x bPerimeter= 2( l + b)

Area and Volume

Triangle:

A = 1/2*base*height

Equilateral = √3/4*(side)2

Area of the Scalene Triangle

S = (a+b+c)/ 2

A = √ s*(s-a) * (s-b)* (s-c)

Area and Volume

Problem - 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 cm2

Problem - 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.m

Problem - 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 r3

Vol. 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 = 4/3r3 = 36r3 = 36*3/4 = 27r = 3 cm

Problem - 4

A 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.740

Problem - 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 : 3

Problem - 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 cm3

Vol. 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 cm

What is the cost of planting the field in the form

of the triangle whose base is 2.8 m and height

3.2 m at the rate of Rs.100 / m2

Problem - 7

Area of triangular field = ½ * 3.2 * 2.8 m2 = 4.48 m2 Cost = Rs.100 * 4.48 = Rs.448..

Solution

Problem - 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 - 9

850 = ½ * d1 * d2 = ½ * 34 * d2 = 17 d2 d2 = 850 / 17 = 50 cm Second diagonal = 50cm

Solution

A grocer is storing small cereal boxes in large cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each small cereal box is 7 inches by 6 inches by 5 inches then what is maximum number of small cereal boxes that can be placed in each large carton ?

Problem - 10

No. of Boxes = 25*42*60 / 7*6*5 = 300

300 boxes of cereal box can be placed.

Solution

Problem - 11

If 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 - 12

A circular wire of radius 42 cm is bent in the

form of a rectangle whose sides are in the ratio

of 6:5. Find the smaller side of the rectangle?

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

Problem - 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.

Solution

Vol. of the Beam = lbh = 9*40/100*10/100 = 72 m3

Weight of the iron beam is given as lm3 = 50 kg

72/100 m3 = 72/100*50 = 36 kg

Problem - 14

If 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= -4

The 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 m3

Vol. of 1 bricks = 25/100*5/100*10/100 = 1/80 m3

Number 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 7

Problem - 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 Saturday

Problem - 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 = Saturday

Problem - 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 = 5,12,19,26

Problem - 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 = Friday

Problem - 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 = Tuesday

Problem - 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 = Thursday

Problem - 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 times

Problem - 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 = Tuesday

Problem - 9Calendar for 2000 will serve also?

Solution= 2000 + 2001 + 2002 + 2003 + 2004 = 2 + 1 + 1 + 1 + 2 = 7 (Complete Week) 2005

Problem - 10If Pinky’s 1st birthday fell in Jan 1988 on one of the Monday’s, the day on which are was born is,

SolutionJan = 1988 = MondayJan = 1987 = Sunday

Problem - 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 = Wednesday

Problem - 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

Tuesday

Problem - 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 - 15

Solution: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

Calendar

Clock

ClocksClock:

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

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

Problem - 1

What is the angle between the minute hand and hour hand when the time is 2.15?

Solutionq = 11 m/2 – 30(h)

= 11 15/2 – 30(2)= 11(7.5) – 60= 82.5 – 60 = 22 1/2

Problem - 2

At what time between 5 and 6 o’clock 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 - 3

At what time between 12 and 1 o’clock both the hands will be at right angles?

SolutionRight angle = 90 degreesq = 30(h) – 11 m/290 = 30(12) – 11 m/2180 = 360 – 11m11m = 360 – 180M = 180/1116 4/11 past 12

Problem - 4

Find at what time between 7 and 8 o’clock will the hands of a clock be in the same straight line but not together?

Solution

Minute 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 - 5

At 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 5

Problem - 6

A clock strikes 4 and takes 9 seconds. In order to strike 12 at the same rate what will be the time taken?

SolutionStrike Sec 3 (interval) 9

11 x3x = 11*9X = 11*9/3 = 33 Sec