Average: An average or an arithmetic mean of given data is the sum of the given observations divided by number of observations.

Properties of Average:

1.     Average of a given data is less than the greatest observation and greater than the smallest observation of the given data.

Ex: Average of 3, 7, 9, and 13 = ((3 + 7 + 9 + 13)/4) = (32/4) = 8

Here Clearly, 8 is less than 13 and greater than 3.

2.      If the observations of given data are equal, then the average will also be the same as observations.

Ex: Average of 6, 6, 6, and 6 = ((6 + 6 + 6 + 6)/4) = (24/4) = 6

3.      If 0 (zero) is one of the observations of a given data, then that 0 (zero) will also be included while calculating average.

Ex: Average of 3, 6, and 0 = ((3 + 6 + 0)/3) = (9/3) = 3

NOTE:

·       If all the numbers get increased by a, then their average must be increased by a

·       If all the numbers get decreased by a, then their average must be decreased by a

·       If all the numbers are multiplied by a, then their average must be multiplied by a

·       If all the numbers are divided by a, then their average must be divided by a

Important Formulae Related to Average of Numbers:

1. Average of first n natural numbers - 
2. Average of first n even numbers = (n + 1)
3. Average of first n odd numbers = n 
4. Average of consecutive numbers =  
5. Average of 1 to n odd numbers =  
6. Average of 1 to n even numbers =  
7. Average of squares of first n natural numbers = 
8.  Average of the cubes of first n natural numbers =  
9. Average of n multiples of any number = 

 Examples:

1.      Find out the average of 4, 7,10,13, ..., 28, 31.

    Solution. Here, the difference between any two numbers written in continuous sequence is 3.

Hence, this is a series of consecutive numbers.

As, we know, average of consecutive numbers = 

Here, first number = 4 and last number = 31

Therefore, required average = 

2.      Find the average of all the odd numbers and average of all the even numbers from 1 to 45.

Solution: According to the formula,

Average of 1 to n odd numbers = 

Here, the last odd number = 45

Therefore, Average of 1 to 45 odd numbers = 

Again, according to the formula,

Average of 1 to n even numbers = 

Here, the last odd number = 44

Therefore, Average of 1 to 44 odd numbers = 

 

3.      The average salary of the entire staff in an office is $200 per day. The average salary of officers is $550 and that of non-officers is $120. If the number of officers is 16, then find the numbers of non-officers in the office.

Solution: Let number of non-officers = x

Then, 120x + 550 x 16 = 200 (16 + x)

=> 12x + 55 x 16 = 20 (16 + x)

=> 3z + 55 x 4 = 5 (16 + x)

=> 3x + 220 = 80 + 5x

=> 5x - 3x = 220 – 80

=> 2x = 140

=> x = 70

4.      The average runs scored by a cricketer in 42 innings, is 30. The difference between his maximum and minimum scores in an innings is 100. If these two innings are not taken into consideration, then the average score of remaining 40 innings is 28. Calculate the maximum runs scored by him in an innings?

Solution: Let the minimum score = x

Maximum score = x + 100

x + (x + 100) = (30 x 42) – (40 x 28)

2x + 100 = 1260 – 1120

2x + 100 = 140

2x = 140 – 100

2x = 40

x = 20

Hence, the maximum score = x + 100 = 20 + 100 = 120

5.      The average weight of the students in four sections A, B, C and D is 60 kg. The average weight of the students of A, B, C and D individually are 45 kg, 50 kg, 72 kg and 80 kg, respectively. If the average weight of the students of section A and B together is 48 kg and that of B and C together is 60 kg, what is the ratio of the number of students in sections A and D?

Solution:

Let number of students in the sections A, B, C and D be a, b, c and d, respectively.

Then, total weight of students of section A = 45a

Total weight of students of section B = 50b

Total weight of students of section C = 72c

Total weight of students of section D = 80d

According to the question, Average weight of students of sections A and B = 48 kg

 

And average weight of students of sections B and C=60kg

 50b + 72c = 60b + 60c

 10b = 12c

Now, average weight of students of A, B, C and D = 60 kg

45a + 50b + 72c + 80d = 60(a + b + c + d)

 => 15a + 10b - 12c - 20d = 0

 => 15a = 20d

 

=> a: d = 4 :3 

 

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Percentage:

 

The term per cent means 'for every hundred'.

 

It can be defined as follows

 

 "A per cent is a fraction whose denominator is 100 and the numerator of the fraction is called the rate per cent." Per cent is denoted by the sign '%’.

 

Conversion of Per Cent into Fraction:

 

Expressing per cent (x%) into fraction.

Conversion of fraction into Percentage:

Expressing One Quantity as a Per Cent with Respect to Other:

 

To express a quantity as a per cent with respect to other quantity following formula is used

Note: To apply this formula, both the quantities must be in same metric unit

Example: 70 kg is what per cent of 280 kg?

 

Solution: According to formula,

Formula to calculate Per Cent:

Some quick Results:

Examples:

1.     The price of a computer is $ 20000. What will be the price of computer after reduction of 25%?

Solution: Here, x = $ 20000 and y = 25%

 

According to the formula,

2.     The salary of a worker is first increased by 5% and then it is decreased by 5%. What is the change in his salary?

Solution: Let the initial salary of the worker be $ 100.

 

Firstly, the salary of worker is increased by 5%.

 

Now, the salary is reduced by 5% after the increase.

Therefore, required change is a decrease i.e., 100 – 99.75 = 0.25

3.     Population of a city in 2014 was 1000000. If in 2015 there is an increment of 15%, in 2016 there is a decrement of 35% and in 2017 there is an increment of 45%, then find the population of city at the end of year 2017.

Solution:

 

Given that, P = 1000000, R₁ = 15%, R₂ = 35% (decrease) and R₃ = 45%

4.     A student was asked to measure the length and breadth of a rectangle. By mistake, he measured the length 20% less and the breadth 10% more. If its original area is 200 sq cm, then find the area after this measurement?

 

Solution:

5.       Due to an increase of 30% in the price of eggs, 6 eggs less are available for $7.80. The present rate of eggs per dozen is?

 

Solution: Let the original price per egg be x.

 

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Profit and Loss: Profit and loss are the terms related to monetary transactions in trade and business. Whenever a purchased article is sold, then either profit is earned or loss is incurred.

Cost Price (CP) This is the price at which an article is purchased or manufactured.

Selling Price (SP) This is the price at which an article is sold.

Overhead Charges

Such charges are the extra expenditures on purchased goods apart from actual cost price. Such charges include freight charges, rent, salary of employees, repairing cost on purchased articles etc.

Note:

 If overhead charges are not specified in the question, then they are not considered

 Profit (SP>CP) When an article is sold at a price more than its cost price, then profit is earned,

 Loss (CP>SP) When an article is sold at a price lower than its cost price, then loss is incurred.

Basic Formulae Related to Profit and Loss:

Things to Keep in Mind:

·       Profit and loss are always calculated on cost price unless otherwise stated in the question.

·        If an article is sold at a certain gain (say 45%), then SP = 145% of CP

 

·        If an article is sold at a certain loss (say 25%), then SP =75% of CP 

Examples:

          1.    Find the SP, when CP is $100 and gain is 20%.

Sol. Given, CP = $100 and gain = 20%

2.     A person sold a table at $2000 and got a loss of 20%. At what price should he sell it to gain 20%?

Sol. Given SP = $2000 and loss = 20%

Now, CP = $ 2500 and gain = 20%

3.     A woman bought eggs at $ 30 per dozen. The selling price per hundred so as to gain 12% will be (in $)

Sol. Given 12 eggs cost = $ 30

 

Therefore, Cost Price of 100 eggs = 2.5 x 100 = $ 250

Now, let the SP of 100 eggs be $ u.

The selling price per hundred so as to gain 12% will be $280.

4.     A person sold his watch for $ 75 and got a percentage profit equal to the cost price. The cost price of the watch is 

Sol: Let CP of the watch be $ v.

Therefore, CP of the watch = $ 50

5.     Two boxes of onions with equal quantity, one costing $ 10 per kg and the other costing $ 15 per kg, are mixed together in to a bag and whole bag is sold at $ 15 per kg. What is the profit or loss?

Sol: Let each box contains x kg onion,then total cost price of these two boxes together(bag) = 10x + 15x = 25x

Selling price of whole bag = 15(x + x) = 15(2x) = 30x

Therefore, profit percentage = 20%

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Discount is defined as the amount of rebate given on a fixed price (called as marked price) of an article. It is given by merchants/ shopkeepers to increase their sales by attracting customers.

Discount = Marked Price - Selling Price

Marked Price (List Price):

The price on the label of an article/product is called the marked price or list price.

This is the price at which product is intended to be sold.

However, there can be some discount given on this price and actual selling price of the product may be less than the marked price.

 It is generally denoted by MP.

Selling Price:

Selling price = Marked price - Discount

where, r% is the rate of discount allowed

Note: Discount is always calculated with respect to marked price of an article

Successive Discount:

When a series of discounts (one after the other) are allowed on marked price of an article, then these discounts are called successive discounts.

Let r1%, r2%, r3%......be the series of discounts on an article with marked price of $ P, then the selling price of the article after all the discounts is given as

Basic Formulae Related to Discount:

Examples:

1. What will be a single equivalent discount for successive discounts of 10% and 5% on marked price of an article?

Sol: Given r₁ = 10% and r₂ = 5%

Therefore, single equivalent discount = 14.5%

  2.  A man bought an article listed at $ 1500 with a discount of 20% offered on the list price. What additional discount must be offered to the man to bring the net price to $ 1104?

Sol: Listed price of an article = $ 1500

Now, second discount = 1200 – 1104 = $ 96

  3.  A man purchased a shirt and pant with a discount of 25% on its marked price. He sold them at a price 40% more than the price at which he bought them. How much per cent the new selling price to its marked price?

Sol:  Let the original price of pant and shirt to be $ p

  4.  A dozen pair of socks quoted at $ 80 are available at a discount of 10%. How many pair of socks can be bought for $ 24?

Sol:  Since MP of one dozen of pairs of socks = $ 80

  5.  A seller marks his goods 30% above their cost price but allows 15% discount for cash payment. His percentage of profit when sold in cash, is

Sol: Let CP of the goods = $ p

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When a person borrows some amount of money from another person or organisation (bank), then the person borrowing money (borrower) pays some extra money during repayment, that extra money during repayment is called interest

Principal (P) Principal is the money borrowed or deposited for a certain time.

Amount (A) The sum of principal and interest is called amount

Amount = Principal + Simple Interest

Rate of Interest (R) It is the rate at which the interest is charged on principal. It is always specified in percentage terms.

Time (T) The period, for which the money is borrowed or deposited, is called time.

Simple Interest (SI):

If the interest is calculated on the original principal for any length of time, then it is called simple interest

Basic Formulae Related to Simple Interest:

Where, SI = Simple Interest, P = Principal, R = Rate of interest, T = Time, A = Amount.

Things to remember:

• To calculate interest, the day on which amount is deposited, is not counted but the day on which amount is withdrawn is counted.

     Instalments:

When a borrower paid the total money in some equal parts (i.e., not in a single amount), then we say that he/she is paying in instalments

The important point is that borrower has to also pay the interest for using the borrowed sum or purchased article.

In general, the value of each instalment is kept constant even when the interest charged on each instalment vary for each instalment for n equal instalments we only calculate up to (n -1) term.

For simple interest

Where, A = Total amount paid;

x = Value of each instalment.

Where, P is the principal n is the number of instalments R is the rate of interest

Examples:

1. A private finance company A claims to be lending money at simple interest. But the company includes the interest every 6 months for calculating principal. If company A is charging an interest of 10%, the effective rate of interest after 1 yr becomes

Sol: Let the sum be $100

Now, principal becomes 100 + 5 = 105

Hence, amount at the end of 1 year = 105 + 5.25 = $ 110.25

Therefore, Effective SI = 110.25 – 100 = $ 10.25%

2. Jack lent out $ 8750 at 7% annual interest. Find the simple interest in 3 yr.

Sol: Given, t = 3 yr, r = 7%, P = $ 8750

 SI = $ 1837.50

3. Simple interest for the sum of $ 1230 for 2 yr is $ 10 more than the simple interest for $ 1130 for the same duration. Find the rate of interest

Sol: According to the question,

  4.  A principal amount to $ 944 in 3 yr and to $ 1040 in 5 yr, each sum being invested at the same simple interest. The principal was

Sol: Let the principal be $ p

Rate of interest = R%

Case I:

P = $ p, T = 3 yr

R = R%, SI = $(944 – p)

Case II:

P = $ p, T = 5 yr

R = R%, SI = $(1040 – p)

From Eq (1) and (2), we get

Therefore, p = $ 800

5. A person invests $ 12000 as fixed deposit at a bank at the rate of 10% per annum simple interest. But due to some pressing needs, he has to withdraw the entire money after 3 yr, for which the bank allowed him a lower rate of interest. If he gets $ 3320 less than, what he would have got at the end of 5 yr, the rate of interest allowed by bank is

Sol: Let the rate of interest allowed by bank be r%

According to the question,

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As we know that when we borrow some money from bank or any person, then we have to pay some extra money at the time of repaying. This extra money is known as interest.

If interest accrued on principal, it is known as simple interest.

Sometimes it happens that we repay the borrow money some late.

After the completion of specific period, interest accrued on principal as well as interest due of the principal. Then, it is known as compound interest.

Compound Interest = Amount – Principal

Basic Formulae Related to Compound Interest

Let principal = P, rate = R% pa and time = n yr

1.  If interest is compounded annually,

Compound interest = Amount – Principal

  5. If rates of interest are R₁%, R₂% and R₃% for 1^st, 2^nd and 3^rd years respectively, then

Instalments:

When a borrower pays the sum in parts, then we say that he/she is paying in instalments.

x = value of each instalment

n = Number of instalments

Examples:

1. What will be the present worth of $ 169 due in 2 yr at 4% pa compound interest?

Sol: Given, R = 4%, n = 2 years and A = $ 169 and P =?

  2. A sum of $ 400 amounts to $ 441 in 2 yr. What will be its amount, if the rate of interest is increased by 5%?

Sol: According to the given condition,

New rate = 5 + 5 = 10%

  3. A sum, at the compound rate of interest, becomes 5/2 A times in 6 yr. The same sum becomes what times in 18 yr?

  1.  A borrowed sum was paid in the two annual instalments of $ 121 each. If the rate of compound interest is 10% pa, what sum was borrowed?

Sol: According to the question,

  5.  The population of a particular area A of a city is 5000. It increases by 10% in 1st yr. It decreases by 20% in the 2nd yr because of some reason. In the 3rd yr, the population increases by 30%. What will be the population of area A at the end of 3 yr?

Sol: Given that, P = 5000, R₁ = 10%, R₂ = -20%(decrease) and R₃ = 30%

Therefore, Population at the end of 3^rd year

Population at the end of 3^rd year = $ 5720

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True Discount If a person borrows certain money from another person for a certain period and the borrower wants to clear off the debt right now, then for paying back the debt, the borrower gets certain discount which is called True Discount (TD)

Present Worth The money to be paid back is called the Present Worth (PW).

 Amount Sum due is called Amount (A). Amount (A) = PW + TD

 True Discount It is the difference between the Amount (A) and the Present Worth (PW).

Discount (TD) = A-PW

Things to remember:

1. True discount is the interest on Present Worth (PW).

2. Interest is reckoned on PW and TD is reckoned on amount.

 According to the definition, we have TD = A – PW

Formulae:

1. If rate of interest is R% per annum, time is T yr and present worth is PW, then True

  2. If rate of interest is R% and money due for amount is A after T yr, then

  3. If money due A, rate of interest R% and time T are given, then

  4. If the true discount on a certain sum of money due certain year hence and the simple interest on the same sum for the same time and at the same rate is given, then

  5. If the true discount on a certain sum of money due T yr hence and the simple interest on the same sum for the same time and at the same rate of interest R% per annum are given, then

Examples:

1. What will be the true discount for the present worth of $ 6000 for a period of 9 months at 12% per annum?

  2. The true discount on a certain sum of money due 4 yr hence is X 75 and the simple interest on the same sum for the same time and at the same rate of interest is X 225. Find the rate per cent.

Sol: Given that, SI = $ 225, TD = $ 75, T = 4 yr and R =?

From PW we can say that $10000 cash is better offer

  4. The true discount on a certain sum of money due 10 yr hence is $ 68 and the simple interest on the same sum for the same time and at the same rate of interest is $ 102. Find the sum due.

Sol: Given that, TD = $ 68 and SI = $ 102

  5. What will be the present worth of $ 4840 due 2 yr hence, when the interest is compounded at 10% per annum? Also, find true discount.

Sol: Given that, A = $4840, T = 2 years, R = 10%, PW =? and TD =?

                                                    TD = A – PW = 4840 – 4000 = $ 840

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Ratio

When two or more similar quantities are compared, then to represent this comparison, ratios are used.

or

Ratio of two quantities is the number of times one quantity contains another quantity of same kind.

The ratio between x and y can be represented as x: y, where x is called antecedent and y is called consequent.

Type of Ratio:

The different types of ratio are explained as under

1. Duplicate Ratio If two numbers are in ratio, then the ratio of their squares is called duplicate ratio. If x and y are two numbers, then the duplicate ratio of x and y would be xr: jr.

 For example: Duplicate ratio of 3 :4 = 3 :4 =9:16

2. Sub-duplicate Ratio If two numbers are in ratio, then the ratio of their square roots is called sub-duplicate ratio. If x and y are two numbers, then the sub-duplicate ratio of x and y would be Vx: Jy.
3. Triplicate Ratio If two numbers are in ratio, then the ratio of their cubes is called triplicate ratio. If x and y are two numbers, then the triplicate ratio of x and y would be x³: y³.

For example: Triplicate ratio of 2 :3 = 2³ :3³ = 8 :27

  4. Sub-triplicate Ratio If two numbers are in ratio, then the ratio of their cube roots is called sub-triplicate ratio. If x and y are two numbers, then the sub-triplicate ratio of x and y would be 

  5. Inverse Ratio If two numbers are in ratio, then their antecedent and consequent are interchanged and the ratio obtained is called inverse ratio, If x and y are two numbers and their ratio is x: y, then its inverse ratio will be y: x.

For example, Inverse ratio of 4: 5 is 5: 4

  6. Compound Ratio If two or more ratios are given, then the antecedent of one is multiplied with antecedent of other and respective consequents are also multiplied.

If a: b, c: d and e: f are three ratios, then their compound ratio will be ace: bdf.

Note:

1. If the antecedent is greater than the consequent, then the ratio is known as the ratio of greater inequality, such as 7: 5.
2. If the antecedent is less than the consequent, then the ratio is called the ratio of less inequality, such as 5: 7.

Comparison of Ratios:

Rules used to compare different ratios are as follows

1. If the given ratios are a: b and c: d, then (i) a: b> c: d, if ad > be (ii) a: b< c: d, if ad < be (iii) a: b = c: d, if ad = bc.
2. if two ratios are given for comparison, convert each ratio in such a way that both ratios have same denominator, then compare their numerators, the fraction with greater numerator will be greater.
3. If two ratios are given for comparison, convert each ratio in such a way that both ratios have same numerator, then compare their denominators, the fraction with lesser denominator will be greater.

Proportion:

An equality of two ratios is called the proportion.

If a/b = c/d  or a: b = c: d, then we can say that a, b, c and d are in proportion and can be written as a: b::c: d, where symbol '::' represents proportion and it is read as 'a is to b' as 'c is to d'.

Here, a and d are called 'Extremes' and b and c are called as 'Means'.

Basic Rules of Proportion:

1. if a:b::b:c, then c is called third proportional to a and b, which are in continued proportion, c will be calculated as a:b::b:c => a: b = b: c => a x c = b x b => b² = ac => c = (b²/a)
2. if a: b :: c: d, then d is called the 4th proportional to a, b and c, d will be calculated as

  3. Mean proportional between a and b is . If mean proportional is x, then a: x :: x: b

Examples:

1. If P: Q = 8:15 and Q: R = 3:2, then find P: Q: R.

Sol: Given that, P: Q = 8: 15, Q: R = 3: 2

P: Q: R = (8 x 3): (15 x 3): (15 x 2) = 24: 45: 30

Therefore, P: Q: R = 8: 15: 10

Here, consequent of first ratio should be equal to the antecedent of second ratio.

2. Salary of Mr. X is 80% of the salary of Mr. Y and the salary of Mr. Z is 120% of the salary of Mr. X. What is the ratio between the salaries of X, Y and Z, respectively?

Sol: Let Y’s salary = 100

Therefore, X’s salary = 80

Therefore, required ratio = 80: 100: 96 = 20: 25: 24

3. $ 710 were divided among A, B and C in such a way that A had $ 40 more than B and C had $ 30 more than A. How much was C’s share?

Sol: Let B gets x.

Then, A gets (x + 40) and C gets (x + 70).

According to the question,

x + 40 + x + x + 70 = 710

3x = 710 – 110

3x = 600

x = 200

C’s share = 200 + 70 = $ 270

  4. What will be the mean proportional between 4 and 25?

Sol: Let mean proportional be x.

Then, 4: x :: x: 25 => 4: x :: x: 25 => 4 x 25 = x²

  5. The ratio between the number of passengers travelling by 1st and 2nd class between the two railway stations is 1: 50, whereas the ratio of 1st and 2nd class fares between the same stations is 3: 1. If on a particular day, $ 1325 were collected from the passengers travelling between these stations, then what was the amount collected from the 2nd class passengers?

Sol: Let the number of passengers in 1^st class be x and number of passengers in 2^nd class be 50x.

Then, total amount of 1^st class = 3x and total amount of 2^nd class = 50x.

Ratio of the amounts collected from 1^st class and the 2^nd class passengers = 3: 50

Where, x = total amount a = 3, b = 50

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Mixture

The new product obtained by mixing two or more ingredients in a certain ratio is called a mixture. or Combination of two or more quantities is known as mixture.

Mean Price

The cost price of a unit quantity of the mixture is called the mean price. It will always be higher than cost price of cheaper quantity and lower than cost price of dearer quantity.

Rule of Mixture or Alligation:

 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.

is the ratio, in which two quantities should be mixed, while A₁, A₂ and A_w are the cheaper price, dearer price and mean price, respectively.

Remember: A₁ < A_w < A₂

The rule is also applicable for solving questions based on average i.e., speed, percentage, price, ratio etc., and not for absolute values. In other words, we can use this method whenever per cent, per hour, per kg etc., are being compared.

Examples:

1. 600g of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?

Solution: 40% sugar is in 600g of sugar solution.

Let x g sugar be added.

  2.  A container is filled with liquid, 6 parts of which are water and 10-part milk. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half milk?

Solution: Let the container initially contains 16 L of liquid.

Let a L of liquid be compressing water.

  3.  How many kilograms of tea worth $ 25 per kg must be blended with 30 kg of tea worth $ 30 per kg, so that by selling the blended variety at $ 30 per kg, there should be a gain of 10%?

Solution: Let the quantity of tea worth $ 25 be x kg.

According to the question,

  4.  In a mixture of 60 L the ratio of acid and water is 2: 1. If the ratio of acid and water is to be 1: 2, then the amount of water (in litres) to be added to the mixture is

x = 60 L

1. In a container, milk and water are present in the ratio 7: 5. If 15 L water is added to this mixture, the ratio of milk and water becomes 7: 8. Find the quantity of water in the new mixture.

Sol. Let the quantity of milk and water in initial mixture be 7x and 5x L.

Therefore, quantity of water in initial mixture = 5 x 5 = 25 L

And quantity of water in new mixture = 25 + 15 = 40 L

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A linear equation is an equation for a straight line. So, the equation which has degree 1, i.e., which has linear power of the variables, is called a linear equation.

It is written as ax + by + c = 0, where a, b and c are real numbers and a and b both are not zero

For example, y = 2x +1 is a linear equation. The different values of x and y are

All these values of (x, y) as (1,3), (2,5), (0,1) etc., are the solutions of the given linear equation.

If we are given two equations in x and y, then we are to find those values of x and y which satisfy both the given equations.

Linear Equation in One Variable

A linear equation in which number of unknown variables is one, is known as linear equation in one variable. For example, 3x + 5=10, y + 3=5 etc

Linear Equation in Two Variables

A linear equation in which number of unknown variables are two, is known as linear equation in two variables. For example, 2x + 5y = 10, x + 4y = 8 etc

Linear Equation in Three Variables

A linear equation in which number of unknown variables are three, is known as linear equation in three variables. For example, 4x + 6y + 7z = 20, x + y + 2z = 5 etc.

Note:

1. Linear equation in one variable represents a point in number line.

2. Linear equation in two variable represents a line in XY-plane (cartesian plane).

3. Linear equation in three variables represents a plane in XVZ-coordinate system.

Methods of Solving Linear Equations

There are following methods which are useful to solve the linear equations

Substitution Method

In this method, first the value of one variable must be represented in the form of another variable and put this value in another equation and solve it. Thus, a value of one variable is obtained and this value is used to find the value of another variable.

Elimination Method

In this method, the coefficients of one of the variables of each equation become same by multiplying a proper multiple. Solve these equations and by which we get the value of another variable and thus with the help of this value, we can find the value of another variable.

Cross Multiplication Method

Let a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are two equations

Therefore, By cross multiplication method,

Consistency of the System of Linear Equations

A set of linear equations is said to be consistent, if there exists at least one solution for this equation.

A set of linear equations is said to be inconsistent, if there are no solution for this equation. Let us consider a system of two linear equations as shown, a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0

Consistent System:

Inconsistent System:

The above system will be inconsistent, if  and do not have any solution. It represents a pair of parallel lines.

Examples:

1. Solve the following equations with cross multiplication method.

2x – 3y + 1 = 0, 3x + 4y – 5 = 0

Solution: Given 2x – 3y + 1 = 0, 3x + 4y – 5 = 0

By cross multiplication method,

  2. For what value of K, the system of equations 2x + 4y +16 = 0 and 3x + Ky + 24 = 0 has an infinite number of solutions.

Therefore, K = 6

  3. The cost of 21 pencils and 9 clippers is $ 819. What is the total cost of 7 pencils and 3 clippers together?

Solution:  Let the cost of 1 pencil and 1 clipper be p and c respectively.

Now, according to the question,

21p + 9c = $ 819

3(7p + 3c) = $ 819

7p + 3c = $ 273

Cost of 7 pencils and 3 clippers = $ 273

  4. David has some hens and some dogs. If the total number of animal heads is 100 and the total number of animal feet is 248, what is the total number of dogs David has?

Solution: Let hens = H, dogs = D

 According to the question, H + D = 100 ...(i)

2H + 4D = 248 ...(ii)

On multiplying Eq. (i) by 2 and subtracting from Eq. (ii), we get

2H + 2D = 200

  5. In an examination, a student scores 4 marks for every correct answer and losses 1 mark for every wrong answer. A student attempted all the 200 questions and scored 200 marks. Find the number of questions, he answered correctly.

Solution: Let the number of correct answers be x and number of wrong answers be y.

Then, 4x – y = 200 -> (1) and x + y = 200 -> (2)

On adding Equations (1) and (2), we get

Therefore, number of correct answers = 80

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Square

If a number is multiplied with itself, then the result of this multiplication is called the square of that number.

For example

(i) Square of 7 = 7 x 7 = 49

(ii) Square of 11 = 11 x 11 = 121

(iii) Square of 100 = 100 x 100 = 10000

Methods to Find Square:

Different methods to calculate the square of a number are as follows

Multiplication Method

In this method, the square of any 2-digit number can be calculated by the following given steps.

• Step I: Square the unit's digit {If the square has two digits, then write ten's digit as carry.}
• Step II: 2 x Ten's digit x Unit's digit + Carry
• Step III: (Ten's digit)² + Carry from step II
• Step IV: Now arrange the numbers first write step III number, then unit’s place number of step II and at unit's place step I’s unit’s place number.

For example:  Find the square of 74:

Step I: (4)² = 16 {Carry = 1}

Step II: 2 x 7 x 4 + 1 = 57 {Carry = 5}

Step III: (7)² + 5 = 49 + 5 = 54

Step IV: (74)² = 5476

Algebraic Method:

To calculate square by this method, two formulae are used.

(i) (a + b)² = a² + b² + 2ab (ii) (a-b)² = a² + b² -2ab

For example: The square of 34 is (34)² = (30 + 4)² = (30)² + (4)² + 2 x 30 x 4 = 900+ 16+240; (34)² = 1156

Square of Decimal Numbers:

To find the square of any decimal number, write the square of the number ignoring the decimal and then place the decimal twice the place of the original number starting from unit's place.

For example: The square of 3.5 is as follows (35)² = 1225

• Here, the decimal is after one-digit in 3.5.
• Hence, the decimal will be placed twice the place of original number in the result. (3.5)² = 12.25

Square Root:

The square root of a number is that number, the square of which is equal to the given number.

There are two types of square roots of a number, positive and negative.

It is denoted by the sign '√'.

For example: 49 has two square roots 7 and - 7, because (7)² = 49 and (- 7)² = 49. Hence, we can write 

Methods to Find Square Root:

Different methods to calculate the square root of a number are as follows

Prime Factorisation Method

This method has the following steps

Step I: Express the given number as the product of prime factors.

Step II: Arrange the factors in pairs of same prime numbers.

Step III: Take the product of these prime factors taking one out of every pair of the same primes. This product gives us the square root of the given number.

Division Method

If it is not easy to evaluate square root using prime factorisation method, then we use division method.

The steps of this method can be easily understood with the help of following examples.

Example: Find the square root of 18769.

Solution.

Step I: In the given number, mark off the digits in pairs starting from the unit digit. Each pair and the remaining one-digit (if any) are called a period.

Step II: Now, 1=1; On subtracting, we get 0 (zero) as remainder.

Step III: Bring down the next period, i.e., 87. Now, the trial divisor is 1 x 2 = 2 and trial dividend is 87. So, we take 23 as divisor and put 3 as quotient. The remainder is 18 now.

Step IV: Bring down the next period, which is 69. Now, trial divisor is 13 x 2 = 26 and trial dividend is 1869. So, we take 267 as dividend and 7 as quotient. The remainder is 0.

Step V The process (processes like III and IV) goes on till all the periods (pairs) come to an end and we get remainder as 0 (zero) now.

Hence, the required square root = 137

Properties of Squares and Square Roots

• The difference of squares of two consecutive numbers will always be equal to the sum of the number i.e., (a² -b²) = (a + b) (a-b). Here, a>b and (a, b) being consecutive (a — b) =1.

Example: If a = 12 and b =11, then (12² -11²) = (12 + 11) (12-11) = 23

• If the square of any number ends with 1, then its square root will end with 1 or 9.
• If the square of any number ends with 4, then its square root will end with 2 or 8.
• If the square of any number ends with 5, then its square root will end with 5.
• If the square of any number ends with 6, then its square root will end with 7 or 6.
• If the square of any number ends with 9, then its square root will end with 3 or 7.
•  The square of any number always ends with 0,1,4, 5,6 or 9 but will never end with 2,3, 7 or 8,
• Square root of negative number is imaginary

Important Relations:

Square Root of Decimal Numbers:

If in a given decimal number, the number of digits after decimal are not even, then we put a 0 (zero) at the extreme right, So, that these are even number of digits after the decimal point. Now, periods are marked as marked in previous explanation starting from right hand side before the decimal point and from the left hand after the decimal digit.

For example: 156.694

There are odd number of digits after decimal.

So, we put a zero after the digit, so that there are even digits after the decimal 156.6940

Now, periods are marked as

After the periods are marked, then previous method is used to find the square root

Square Root of a Fraction:

To find square root of a fraction, we have to find the square roots of numerators and denominators, separately.

Note:

Sometimes, numerator and denominator are not a complete square. In these types of cases, it is better to convert the given fraction into decimal fraction to find the square root.

Examples:

  3. A shop-keeper has 1000 boxes. He wants to keep them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of boxes that he needs more for this purpose?

Solution: Let the number of rows and columns be m.

Then, total boxes should be m x m.

Now, 1000 is not a square of any number.

Let m = 30 Then, m x m = 30 x 30 = 900 Which is less than total boxes.

Now, let m = 32 Then, m x m = 32 x 32 = 1024 Which is greater than 1000.

So, the minimum number of boxes that he needs for this purpose = 1024 - 1000 = 24 boxes

  5. What is the least number to be added to 8200 to make it a perfect square?

Solution: Given number = 8200

Now, lets find the nearest square values of given number

-> (90)² = 8100 and (91)² = 8281

-> (90)² < 8200 > (91)²

Therefore, least number to be added to 8200 to make it a perfect square = 8281 – 8200 = 81

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Area:

• Total space enclosed by the boundary of a plane figure is called the area of that particular figure.
• In another words, the area of a figure is a measure associated with the part of plane enclosed in the figure.
• Area is measured in square unit like, square metre, square centimetre etc.

Perimeter:

• Perimeter is the length of border around any enclosed plane.
• Therefore, sum of the sides of a plane figure is the perimeter of that particular figure.
• Unit of perimeter is same as the unit of sides like metre, centimetre etc.

Triangle:

• A figure enclosed by three sides is known as a triangle.
• A triangle has three angles with total sum of 180°.
• Adjoining figure represents a triangle with sides, AB, BC, CA and ZA, ZB, ZC are the three angles of the triangle. Various types of triangles are discussed below.

Equilateral Triangle:

It has all three sides equal and each angle equal to 60^o

Isosceles Triangle:

It has any two sides and two angles equal and altitude drawn on non-equal side bisects it.

where, a = Each of two equal sides b = Third side

Scalene Triangle:

It has three unequal sides.

Right Angled Triangle:

 It is a triangle with one angle equal to 90°

where, p = Perpendicular, b =Base and h = Hypotenuse

h² = p² + b²

Isosceles Right-Angled Triangle:

It is a triangle with one angle equal to 90° and two sides containing the right angle are equal.

Properties of Triangle:

 1. Sum of any two sides of a triangle is greater than the third side

 2. Side opposite to the greatest angle will be the greatest and side opposite to the smallest angle will be the smallest

3. Among all the triangles that can be formed with a given perimeter, the equilateral triangle will have the maximum area

4. The lines joining the mid-points of sides of a triangle to the opposite vertex are called medians. In the given figure, AF, BE and CD are medians

5. The point where the three medians of a triangle meet are called centroid. In the given figure, O is the centroid. The centroid divides each of the median in the ratio of 2: 1

6. The median of a triangle divides it into two triangles of equal areas

7. The incentre and circumcentre lies at a point that divides the height in the ratio 2:1. i.e., the circumradius is always twice the median

10. The area of the triangle formed by joining the mid-points of the sides of a given triangle is  of the area of the given triangle

Examples:

1. The perimeter of an equilateral triangle is 51 cm. Find its area.

 Sol. Given, perimeter of an equilateral triangle is 51 cm.

Let each side of triangle be a cm, then sum of sides = 51 cm

  2. The area of a right-angled triangle is 10 sq cm. If its perpendicular is equal to 20 cm, find its base.

Solution: Given, area = 10 sq cm

Perpendicular = 20 cm

  3. Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively. Find the cost of sowing seeds in the field at the rate of $ 5 per sqm.

Solution: Let’s check if it is right angled triangle

Hence it is a right-angled triangle

So, the triangular filed is right angled at B.

From this the cost of sowing seed is $ 5 per sq m.

Therefore, cost of sowing seed for 150 m² = 150 x 5 = $ 750

  4. A ?DEF is formed by joining the mid-points of the sides of ?ABC. Similarly, a ?PQR is formed by joining the mid-points of the sides of the ?DEF. If the sides of the ?PQR are of lengths 1, 2 and 3 units, what is the perimeter of the ?ABC?

Solution: Given lengths are 1, 2 and 3 units

• Perimeter of ?PQR = 1 + 2 + 3 = 6 units

• So, 2PQ = FE
• Similarly, DF = 2PR and DE = 2QR
• Therefore, perimeter of ?DEF = 2 x 6 = 12 units
• Similarly, Perimeter of ?ABC = 2 x perimeter of ?DEF

= 2 x 12 = 24 units

  5. Two isosceles triangles have equal vertical angles and their corresponding sides are in the ratio of 3: 7. What is the ratio of their areas?

Solution: Here, given that triangles are equiangular and hence they are similar.

Ratio of their areas = ratio of squares of corresponding sides

                                    = (3)²: (7)²

                                    = 9: 49

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Quadrilateral:

• A figure enclosed by four sides is called a quadrilateral.
• A quadrilateral has four angles and sum of these angles is equal to 360°.
• Various types of quadrilaterals are discussed below

Square:

It is a parallelogram with all 4 sides equal and each angle is equal to 90°.

1. Area = (side)² = a²
2. Perimeter = 4 x side = 4a
3. Diagonal(d) = 

Where, a = side and d = diagonal

Properties of Square:

 1. Diagonal of a square are equal and bisect each other at right angles (90°)

2. All square are rhombus but converse is not true

3. Diagonal is the diameter of the circumscribing circle that circumscribes the square and circumradius 

4. If area of two squares is in the ratio of A₁: A₂ then ratio of their perimeter is given 

Rectangle:

• It is a parallelogram with opposite sides equal and each angle is equal to 90°.

1. Area = Length x Breadth = L x B
2. Perimeter = 2 (L + B)
3. 

Properties of Rectangle:

 1. The diagonals of a rectangle are of equal lengths and they bisect each other

 2. All rectangles are parallelograms but reverse is not true.

Parallelogram:

 A quadrilateral, in which opposite sides are parallel is called a parallelogram.

1. Area = Base x Height = b x h
2. Perimeter = 2 (a + b)

Note: Opposite angles are equal in a parallelogram but they are not right angle.

Properties of Parallelogram:

1. Diagonals of a parallelogram bisect each other.
2. A parallelogram inscribed in a circle is a rectangle.
3. A parallelogram circumscribed about a circle is a rhombus.
4. Each diagonal of a parallelogram divides it into two triangles of equal area
5. A parallelogram and a rectangle have equal areas if they are on the same base and between the same parallel lines
6. The opposite angles of parallelogram are equal
7. The sum of the squares of the four sides is equal to the sum of squares of diagonal

Trapezium

 It is a quadrilateral with any one pair of opposite sides parallel

1.  

   Where a and b are parallel sides and h is the height or perpendicular distance between a and b.

2. Perimeter =AB + BC + CD + AD 
3.  Area of trapezium, when the lengths of parallel and non-parallel sides are given

Rhombus:

• It is a parallelogram with all 4 sides equal.
• The opposite angles in a rhombus are equal but they are not right angle.

Where, a = side, d₁ and d₂ are diagonals.

Properties of Rhombus:

1. A rhombus has unequal diagonal and they bisect each other at right angles (90°)
2. A rhombus may or may not be a square but all squares are rhombus
3. All rhombus is parallelogram but reverse is not true

Examples:

1. A square field has its area equal to 289 sq m. Find its side and perimeter.

Solution: Given area of square = 289 sq m.

Perimeter = 4 x s = 4 x 17 = 68 cm

  2. The length and breadth of a rectangle are 6 cm and 4 cm, respectively. What will be its diagonal?

Solution: Given that, Length(L) = 6 cm and Breadth(B) = 4 cm and diagonal =?

  3. The base of a parallelogram is thrice of its height. If the area of the parallelogram is 2187 sq cm, find its height.

Solution: Area of parallelogram = base x height

Let height = p and base = 3p

According to the question, 3p x p = 2187

  4. The area of a trapezium is 384 cm. If its parallel sides are in the ratio 3: 5 and the perpendicular distance between them is 12 cm, the smaller of the parallel sides is

Solution: Given parallel sides are in the ratio 3: 5

Let the sides of trapezium be 5x and 3x respectively.

Length of smaller of the parallel sides = 8 x 3 = 24 cm

  5. If the diagonals of a rhombus are 4.8 cm and 1.4 cm, then what is the perimeter of the rhombus?

perimeter of rhombus = 10 cm

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It is a plane figure enclosed by a line on which every point is equidistant from a fixed point (centre) inside the curve

Sector: Sector is a part of area of circle between two radii and  is the angle enclosed between two radii

Semi-circle:

A circle when separated into two parts along its diameter, then each half part is known as semicircle.

Circular Ring:

(i)  Area = R² – r²

(ii)  Difference in circumference of both rings = 

Examples:

1. If the area of a semi-circle be 77 sq m, find its perimeter?

Solution: According to the question, Area of semi-circle = 77 m

\

1. The area of a circle is increased by 22 sq cm when its radius is increased by 1 cm. Find the original radius of the circle.

 

Solution: Let original radius be r.

Therefore, original radius of the circle = 3 cm

 

  3. The ratio of the areas of the circumcircle and the incircle of a square is

 

  4. AB and CD are two diameters of a circle of radius r and they are mutually perpendicular. What is the ratio of the area of the circle to the area of the triangle ACD?

Solution:

 

Let’s, draw a figure from the given data

 

  5. The largest triangle is inscribed in a semi-circle of radius 4 cm. Find the area inside the semi-circle which is not occupied by the triangle

 

Solution: Let’s draw a figure from given data

 

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• Volume and surface area are related to solids or hollow bodies.
• These bodies occupy space and have usually three dimensions length, breadth and height.

Volume:

 

• Space occupied by an object is called the 'volume' of that particular object.

 

• It is always measured in cube unit like cubic meter, cubic centimetre etc

 

Surface Area:

• Surface area of a solid body is the area of all of its surfaces together.
• Surface area is measured in square unit like square centimetre, square metre etc.

 

Cube:

 

• A solid body having 6 equal faces with equal length, breadth and height is called a cube.
• In fact, each face of a cube is a square

• Volume of the cube = a³
• Lateral surface area of the cube = 4a²
• Total surface area of the cube = 6a²
• Diagonal of the cube 

where, a is Side (edge) of the cube

Cuboid:

• A rectangular solid body having 6 rectangular faces is called a cuboid.

• Volume of the cuboid = bh
• Lateral surface area of a cuboid = 2 (l + b) h
• Total surface area of the cuboid = 2 (lb + bh + lh)
• Diagonal of cuboid 

where, = Length, b - Breadth and h – Height

Some other cube or cuboidal shaped object are as follows.

Room

 

• A rectangular room has 4 walls (surfaces) and opposite walls have equal areas.

• Total area of walls = 2 (l + b) x h
• Total volume of the room = lbh

 

where, = Length, b = Breadth and h = Height

Box

• A box has its shape like cube or cuboid.
• The amount that a box can hold or contain, is called the capacity of the box.
• In fact, capacity is the internal volume

• Surface area of an open box = 2 (length + breadth) x Height + Length x Breadth

        = 2 x (l+ b) x h + l x b

• Capacity of box = (l- 2t) (b – 2t) (h – 2t) where, t = Thickness of the box
• Volume of the material of the box = External volume - Internal volume (capacity) = lbh-{l -2t)(b - 2t) {h - 2t) where, = Length, b = Breadth and h = Height

 

Note For calculation of any of the parameter, length, breadth and height should be in same unit.

Examples:

1. Find the volume and surface area of a cuboid 18 m long, 14 m broad and 7 m high.

Solution:

• Volume of the cuboid = Length x Breadth x Height
• Here, length = 18 m, breadth = 14 m and height = 7 m = 18 X 14 X 7= 1764m³
• Surface area = 2(lb + bh + lh) = 2(18 x 14 + 14 x 7 + 18 x 7) = 2(252+ 98+ 126) = (2 x 476) = 952 sq m

 

  2. A wooden box measures 10 cm x 6 cm x 5 cm. Thickness of wood is 2 cm. Find the volume of the wood required to make the box.

Solution:

External volume = 10 x 6 x 5 = 300 cm³

Internal volume (l - 2t) (b - 2t) (h - 2t) = (10 - 4) X (6 - 4) X (5 - 4) = 6 x 2 x 1= 12 cm³

Volume of the wood = External volume - Internal volume = 300 - 12 = 288 cm³

 

  3. The surface area of a cube is 486 sq cm. Find its volume.

Sol. Let the edge of cube = a

Volume = a³ = 9³ = 9 x 9 x 9 = 729 cm³

 

  4. Three cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new cube. Find half of the surface area of the new cube.

Solution: Volume (new cube) = (1³ + 6³ + 8³) = 729 cm³

ð a³ = 729

ð a = 9 cm

 

  5. A metal box measures 20 cm x 12 cm x 5 cm. Thickness of the metal is 1 cm. Find the volume of the metal required to make the box.

Solution: External volume = 20 x 12 x 5 = 1200 cm³

Internal volume = (20 – 2) x (12 – 2) x (5 – 2) = 18 x 10 x 3 = 540 cm³

 

Therefore, volume of the metal = External volume – internal volume = 1200 – 540 = 660 cm³

 

 

 

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Theory:

 i.            Types of Numbers:

                                    I.            Natural numbers: Natural numbers are counting numbers. For example N = {1,2, 3...}.  All-natural numbers are positive. Zero is not a natural number.

                                  II.            Whole Numbers: All-natural numbers and zero form the set of whole numbers. For example W = {0,1,2,3,...}

                                III.            Integers: Whole numbers and negative numbers form the set of integers. For example / = {...,-4,-3,-2,-1,0,1,2,3,4,...}. Integers are of two types.

·         Positive Integers: Natural numbers are called as positive integers. For example I + = {1,2,3,4,...}

·         Negative Integers: Negative of natural numbers are called as negative integers. For example I~ ={-1,-2,-3,-4,...}. 

                               IV.            Even Numbers: A counting number which is divisible by 2, is called an even number. For example: 2, 4, 6, 8, 10, 12, ... etc.

                                 V.            Odd Numbers: A counting number which is not divisible by 2, is known as an odd number. For example: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ... etc.

                               VI.            Prime Numbers: A counting number is called a prime number when it is exactly divisible by, 1 and itself. For example: 2, 3, 5, 7, 11, 13, ... etc.

                             VII.            Composite Numbers: Composite numbers are non-prime natural numbers. They must have atleast one factor apart from 1 and itself. For example: 4, 6, 8, 9, etc.

                           VIII.            Coprimes: Two natural numbers are said to be coprimes, if their HCF is 1. For example (7, 9), (15, 16). Coprime numbers may or may not be prime

                                IX.            Rational numbers: A number that can be expressed as p/q is called a rational number, where p and q are inteqers and q is not equal to zero

                                  X.            Irrational numbers:The numbers that cannot be expressed in the form of p/q are called irrational numbers, where p and q are integers and q is not equal to 0.

                                XI.            Real numbers: Real numbers include rational and irrational numbers both. For example: 

ii.            Divisibility Tests:

                                    I.            Divisibility by 2: When the last digit of a number is either 0 or even, then the number is divisible by 2. For example 12, 86, 472, 520, 1000 etc., are divisible by 2.

                                  II.            Divisibility by 3: When the sum of the digits of a number is divisible by 3, then the number is divisible by 3. For example (i) 1233: 1 + 2 + 3 + 3 = 9, which is divisible by 3, so 1233 must be divisible by 3. (ii) 156:1 + 5 + 6 = 12, which is divisible by 3, so 156 must be divisible by 3.

                                III.            Divisibility by 4: When the number made by last two-digits of a number is divisible by 4, then that particular number is divisible by 4. Apart from this, the number having two or more zeroes at the end, is also divisible by 4. For example (i) 6428 is divisible by 4 as the number made by its last two digits i.e., 28 is divisible by 4. (ii) The numbers 4300, 153000, 9530000 etc., are divisible by 4 as they have two or more zeroes at the end.

                               IV.            Divisibility by 5: Numbers having 0 or 5 at the end are divisible by 5. For example 45, 4350, 135, 14850 etc., are divisible by 5 as they have 0 or 5 at the end.

                                 V.            Divisibility by 6: When a number is divisible by both 3 and 2, then that particular number is divisible by 6 also. For example 18, 36, 720, 1440 etc., are divisible by 6 as they are divisible by both 3 and 2.

                               VI.            Divisibility by 7: A number is divisible by 7 when the difference between twice the digit at ones place and the number formed by other digits is either zero or a multiple of 7. For example 658 is divisible by 7 because 65 - 2 X 8 = 65 - 16 = 49. As 49 is divisible by 7, the number 658 is also divisible by 7

                             VII.            Divisibility by 8: When the number made by last three digits of a number is divisible by 8, then the number is also divisible by 8. Apart from this, if the last three or more digits of a number are zeroes, then the number is divisible by 8. For example (i) 2256 As 256 (the last three digits of 2256) is divisible by 8, therefore 2256 is also divisible by 8. (ii) 4362000 As 4362000 has three zeroes at the end. Therefore it will definitely divisible by 8.

                           VIII.            Divisibility by 9:When the sum of all the digits of a number is divisible by 9, then the number is also divisible by 9. For example (i) 936819 9+3 + 6 + 8 + 1 + 9= 36 which is divisible by 9. Therefore, 936819 is also divisible by 9. (ii) 4356 4 + 3 + 5 + 6 = 18 which is divisible by 9. Therefore, 4356 is also divisible by 9.

                                IX.            Divisibility by 10: When a number ends with zero, then it is divisible by 10. For example 20, 40, 150, 123450, 478970 etc., are divisible by 10 as these all end with zero.

                                  X.            Divisibility by 11:When the sums of digits at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11. For example (i) 2865423: Let us see Sum of digits at odd places (A) = 2 + 6+4 + 3 = 15 Sum of digits at even places (B) = 8 + 5 + 2 = 15 =>A = B Hence, 2865423 is divisible by 11. (ii) 217382 Let us see Sum of digits at odd places (A) = 2 + 7 + 8 = 17 Sum of digits at even places (B) = 1 + 3 + 2 = 6 A- B = 17-6 = 11 Clearly, 217382 is divisible by 11.

                                XI.            Divisibility by 12: A number which is divisible by both 4 and 3 is also divisible by 12. For example 2244 is divisible by both 3 and 4. Therefore, it is divisible by 12 also.

                              XII.            Divisibility by 14: A number which is divisible by both 7 and 2 is also divisible by 14. For example 1232 is divisible by both 7 and 2. Therefore, it is divisible by 14 also.

                            XIII.            Divisibility by 15: A number which is divisible by both 5 and 3 is divisible by 15 also. For example 1275 is divisible by both 5 and 3. Therefore, it is divisible by 15 also.

                           XIV.            Divisibility by 16: A number is divisible by 16 when the number made by its last 4-digits is divisible by 16. For example 126304 is divisible by 16 as the number made by its last 4-digits i.e., 6304 is divisible by 16.

                             XV.            Divisibility by 18: A number is divisible by 18 when it is even and divisible by 9. For example 936198 is divisible by 18 as it is even and divisible by 9.

                           XVI.            Divisibility by 25: A number is divisible by 25 when its last 2-digits are either zero or divisible by 25. For example 500, 1275, 13550 are divisible by 25 as last 2-digits of these numbers are either zero or divisible by 25.

                         XVII.            Divisibility by 125: A number is divisible by 125 when the number made by its last 3-digits is divisible by 125. For example 630125 is divisible by 125 as the number made by its last 3-digits are divisible by 125.

 

Basic Formulae:

1. (a+ b) (a -b) = (a²-b²)
2. (a + b)² = a²+ b²+ 2ab
3. (a - b)² = a²+ b²- 2ab
4. (a + b+ c)² = a²+ b²+ c²+ 2ab + 2bc +2ca
5. (a³ + b³) = (a + b) (a²- ab+ b²)
6. (a³ - b³) = (a - b) (a² + ab+ b²)
7. a³ + b³ + c³ - 3abc = (a + b+ c) (a²+ b²+ c²- ab – bc - ca)
8. when a + b +c = 0, then a³ + b³ + c³ = 3abc
9. (a + b)²-(a - b)² = 4ab
10. (a + b)² + (a - b)² = 2(a²+ b²)
11. Division algorithm: Dividend = (Divisor * Quotient) + Remainder
12. (x^m – a^m) is divisible by (x - a) for all values of m.
13. (x^m - a^m) is divisible by (x + a) for even values of m.
14. (x^m + a^m) is divisible by (x + a) for odd values of m.
15. Number of prime factors of a^p + b^q + c^r+ d^s is p + q + r + s, where a, b, c and d are prime number.
16. Arithmetic Progression:

An A.P with first term a and common difference d is given by a, (a+ d), (a+ 2d), (a + 3d).

The nth term of the A.P is given by T_n = a (n – 1) d.

Sum of n terms in the A.P is given by S_n= (n/2) (2a+ (n-1) d) which is (n/2) (first term + last term).

1. Sum of first n natural numbers = (n(n+1))/2
2. Sum of first n odd numbers = n²
3. Sum of first n even numbers = n(n+1)
4. Sum of square of first n natural numbers = (n(n+1) (2n+1))/6
5. Sum of cube of first n natural numbers = (n(n+1))²/4

17. Geometrical Progression:

A G.P with first term a and common ratio r is a, ar, ar², ar³,…..

The nth term of the G.P is given by T_n = ar^n-1.

Sum of n terms in the G.P is given by S_n= (a(1-rⁿ)/(1-r)).

Example Problems

1. A computer program was tested 300 times before its release. The testing was done in three stages of 100 tests each. The software failed 15 times in Stage I, 12 times in Stage II, 8 times in Stage III, 6 times in both Stage I and Stage II, 7 times in both Stage II and Stage III, 4 times in both Stage I and Stage III, and 4 times in all the three stages. How many times the software failed in a single stage only?

Solution:

 Assume that the software fails a, b, and c times in a single stage, in two stages, and in all stages respectively.

 Therefore b + 3c = 6+ 7 + 4 = 17 but c = 4, hence b = 5

Similarly, we have a + 2b + 3c = 15 + 12 + 8 = 35

a = 35 – 12 – 10 = 35 – 22 = 13

 

2. What is the sum of all positive integers lying between 200 and 400 that are multiples of 7? 

Solution: Least number divisible by 7 and above 200 is 203.

Greatest number divisible of 7 and below 400 is 399.

Total numbers divisible by 7 between 200 to 400 are 29

Now, sum of n terms of AP = (n/2) (first term + last term) where, first term = 203, last term = 399 and n = 29

sum of n terms of AP = (29/2) (203+399) = 8729

3. A number when divided by the sum of 555 and 445 gives two times their differences as quotient and 30 as the remainder. The number is:

Solution:

Required number = ((555+ 445) *2*110) + 30 = 220030

Hence option (d) is the answer

Dividend = (divisor * quotient) + Remainder.

4. What is the unit digit in (7⁹⁵- 3⁵⁸)?

Solution: Unit digit in 7⁹⁵ = unit digit in [(7⁴)²³ x 7³]

                                            = unit digit in [(unit digit in (2401))²³ x (343)]

                                            = unit digit in (1²³ x 343)

                                            = unit digit in (343)

                                            = 3

              Unit digit in 3⁵⁸ = unit digit in [(4⁴)¹⁴ x 3²]

                                            = unit digit in [(unit digit in (81))¹⁴ x (9)]

                                            = unit digit in (1¹⁴ x 9)

                                            = unit digit in (1 x 9)

                                            = 9

unit digit in (7⁹⁵- 3⁵⁸) = unit digit in (343-9) = unit digit in (334) = 4.

5. What will be the remainder when 17²⁰⁰ is divided by 18?

(a) 5

(b) 3

(c) 2

(d) 1

Solution: When m is even, (x^m - a^m) is completely divisible by (x + a).

                 (17²⁰⁰-1²⁰⁰) is completely divisible by (17+1) i.e. 18

                 (17²⁰⁰ – 1) is divisible by 18.

                 On dividing 17²⁰⁰ by 18  be get 1 as remainder

                Hence, option (d) is the correct answer.

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Theory:

1. Factors and Multiples: If a number x divides another number y exactly (without leaving any remainder), then x is a factor of y and y is a multiple of x.

Factors Set of numbers which exactly divides the given number.

Multiples Set of numbers which are exactly divisible by the given number.

Common Multiple: A common multiple of two or more numbers is a number which is completely divisible (without leaving remainder) by each of them.

For example,we can obtain common multiples of 4, 6 and 12 as follows

 Multiples of 4 = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, ...}

 Multiples of 6 = {6, 12, 18, 24, 30, 36, 42, 48, ...}

 Multiples of 12 = {12, 24, 36, 48, 60, ...}

∴ Common multiples of 4, 6 and 12 = {12, 24, 36, 48, ...}

2. Least Common Multiple (L.C.M): The LCM of two or more given numbers is the least number to be exactly divisible by each of them.

 For example, we can obtain LCM of 4 and 12 as follows:

Multiples of 4 = 4,8,12,16,20, 24,28,32,36, .........

Multiples of 12 =12, 24,36,48,60,72..........

 Common multiples of 4 and 12 =12,24,36, .........

∴ LCM of 4 and 12 =12

There are two methods of finding the L.C.M of a given set of numbers:

1. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M is the product of the highest powers of all the factors.

Ex. 1:

Find the LCM of 8,12 and 15.

Factors of 8 = 2x2x2 = 2³, Factors of 12 = 2 x 2 x 3 = 2² x 3¹, Factors of 15 = 3 x 5 = 3¹ x 5¹. Here, the prime factors that occur in the given numbers are 2, 3 and 5 and their highest powers are 3, 1 and 1 .∴ Required LCM = 2³ X 3¹ X 5¹ = 8 X 3 X 5 = 120.

2. Division Method: Arrange the given numbers in a row in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M of the given numbers.

Ex. 1: What will be the LCM of 15, 24, 32 and 45?

LCM of 15, 24,32 and 45 is calculated as

 

∴ Required LCM = 2x2x2x3x5x4x3= 1440 Note Start division with the least prime number

3. Highest Common Factor (HCF):

HCF of two or more numbers is the greatest number which divides each of them exactly. For example, 6 is the HCF of 12 and 18 as there is no number greater than 6 that divides both 12 and 18. Similarly, 3 is the highest common factor of 6 and 9.

There are two methods to calculate the HCF of two or more numbers which are explained below:

1. Prime Factorisation Method: Express the each one of he given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.

Ex. 1: Find the HCF of 24, 30 and 42.

 Resolving 24,30 and 42 into their prime factors,

∴ Factors of 24 = 2x2x2x3 = (2³ x 3¹)

 Factors of 30 = 2 X 3 X 5 = (2¹ X 3¹ X 5¹)

Factors of 42 = 2 x 3 x 7 = (2¹ x 3¹ x 7¹).

The product of common prime factors with the least powers = 2 X 3 =6. So, HCF of 24, 30 and 42 = 6.

2. Division Method:Suppose we have to find the H.C.F of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

Finding the H.C.F of more than two numbers :Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers.

4. Method to Calculate LCM and HCF of Fractions:

 The LCM and HCF can be obtained by the following formulae:

              LCM of fractions = (LCM of numerators)/ (HCF of denominators)

              HCF of fractions = (HCF of numerators)/ (LCM of denominators)

Note:

 1. All the fractions must be in their lowest terms. If they are not in their lowest terms, then conversion in the lowest form is required before finding the HCF or LCM

 2. The required HCF of two or more fractions is the highest fraction which exactly divides each of the fractions

 3. The required LCM of two or more fractions is the least fraction/integer which is exactly divisible by each of them

 4. The HCF of numbers of fractions is always a fraction but this is not true in case of LCM

5. Product of two numbers =Product of their H.C.F and L.C.M

Examples:

1. In a fire range, 4 shooters are firing at their respective targets. The first, the second, the third and the fourth shooter hit the target once in every 5 s, 6 s, 7 s and 8 s, respectively. If all of them hit their targets at 9:00 am, when will they hit their targets together again?

Solution: Time after which they will hit the target again together = LCM (5,6,7 and 8)

= 5X3X7X2X4 =840s.

 They will hit together again at 9: 14 am

1. What is the HCF of 8 (x⁵ - x³ + x) and 28(x⁶ +l)?

Solution: Let p(x) = 8(x⁵ – x³ + x) = 4 * 2 * x (x⁴ – x² + 1) and

q(x) = 28(x⁶ + 1) = 7 X 4 [(x²)³ + (l)³] = 4 * 7 * (x² + 1) (x⁴ – x² + 1)

∴HCF of p(x) and q(x) = 4 (x⁴ – x² + 1)

2. The H.C.F of (9/10), (12/25), (18/35) and (21/40) is:

Solution: Required H.C.F = (H.C.F OF 9, 12, 18, 21)/ (L.C.M OF 10, 25, 35, 40)

                                             = 3/2800

3. Two numbers, both greater than 29, have H.C.F. 29 and L.C.M. 4147. The sum of the numbers is:

Solution: Product of numbers = 29 x 4147

Let the numbers be 29a and 29b. Then, 29a X 29b = (24 x 4147) -> ab = 143.

Now, co-primes with product 143 are (1, 143) and (11,13).

So, the numbers are (29 x 1, 29 x 143) and (29 x 11, 29 x13).

Since both numbers are greater than 29, the suitable pair is (29 x 11, 29 x 13) i.e., (319, 377).

The required sum = (319 + 377) = 696.

4. How many numbers are there between 4000 and 6000 which are exactly divisible by 32, 40, 48 and 60?

Solution: LCM of 32, 40, 48 and 60 = 480.

The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760.

Hence, required number of numbers are 4.

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Statistics is a branch of mathematics that deals with numbers and analysis of the data. Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data.

In short, Statistics deals with collecting, classifying, arranging, and presenting collected numerical data in simple comprehensible ways.

With the help of statistics, we are able to find various measures of central tendencies and the deviation of data values from the center.

Basic Formulae:

• Mean is defined as the sum of all the elements of a set divided by the number of elements.

• Median is the middle value of a dataset.
• If a set consists of an odd number of values, then the middle value will be the median of the set.
•  If the set consists of an even number of sets, then the median will be the average of the two middle values.

Median(M) = If n is odd, then 

If n is even, then 

• The mode in a dataset is the value that is most frequent in the dataset.

Mode = The value which occurs most frequently

• 
• The standard deviation is defined as the square rooting of the variance of the data.

Standard Deviation(S) = 

Where, x = observations given

 = Mean

 = Total number of observations

 

1. In a group of 10, 4 students were selected at random and their total marks in the final assessments are recorded, which are 81, 83, 98, 76. Find the standard deviation of their marks.

Solution: Here N = 4

 2. Weight of girls = {40, 45, 50, 45, 55, 45, 60, 45}. Find the mode?

Solution: Here we have 45 as repeating value

Since only on value is repeating it is a unimodal list.

S0, mode = 45

 3. Find the median of the data: 24, 36, 45, 18, 20, 26, 38

Solution: Arrange them in ascending order 18, 20, 24, 26, 36, 38, 45

Median = middle most observation or  term when n is odd

So, median = 26

 4. All the students in a mathematics class took a 100-point test. Eight students scored 100, each student scored at least 55, and the mean score was 75.

What is the smallest possible number of students in the class?

Solution:

• Let the number of students be 
• Then the sum of their scores is at least 
• we need to achieve the mean 75, which is equivalent to achieving the sum 75n.

• we must have 
• The smallest integer n for which this is true is n = 18.
• To finish our solution, we now need to find one way how 18 students could have scored on the test.
• We have 18 x 75 = 1350 points to divide among them.
• The eight 100s make 800, hence we must divide the remaining 550 points among the other 10 students.
• This can be done by giving 55 points to each of them.
• Hence the smallest possible number of students is 18.

 5. A manager has given a test to his team in which 20% are women and 80% are men. The average score on the test was 70. Women all received the same score, and the average score of the men was 60. What score did each of the woman receive on the test?

• Solution:

  • Let Total students = 100, Men = 80, Women = 20.
  • Sum of all the scores = 100 x 70 = 7000.
  • Sum of all the scores of men = 60 * 90 = 5400.
  • Difference in scores = 7000 - 5400= 1600.
  • 1600 is the scores obtained by women.
  • It's stated in the question that all Girls got the same score
  • so, score each of the woman received on the test 1600/20 = 80.

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Permutation

Each of the different arrangements which can be made by taking some or all of a given number of things or objects at a time, is called a permutation.

Permutation implies arrangement, where order of the things is important.

For example:

 The permutations of three items a, b and c taken two at a time are ab, ba, ac, ca, cb and be.

Since, the order in which the items are taken, is important, ab and ba are counted as two different permutations.

Cases of Permutation

There are several cases of permutation

  1. Formation of numbers with given digits

In these types of question, it is asked to form numbers with some different digit. These digits can be used with repetition or without repetitions.

  2. Formation of words with given letters

These questions are very much similar to previous case questions but here in place of numbers, word is formed from a set of English alphabets given in the form of a word.

Important point:

Number of permutations of n objects out o which p are alike and are of one type, q are alike and are of second type and r are alike and are of third type  

  3. Arrangement of persons in a row or at a round table

 These types of question are based on arrangement of person (boy or girls etc) in a straight line facing some direction or around some circular object like table etc.

Note: Number of permutations of n objects taken all at a time is n! when repetition is not allowed.

  4. Arrangement of books on a shelf, etc

 In such questions arrangement of books is done into a shelf in a row or one over the other.

Note:

Questions based on sending invitation to different persons are similar to questions based on arrangement of books.

Number of permutations of n different objects taken i at a time, when repetition is allowed = nⁱ

Examples:

1. How many numbers of four digits can be formed with the digits 1, 2, 3, 4 and 5? (Repetition of digits is not allowed.)

Solution: There are five numbers and number of places to be filled up = 4

2. In how many ways, can the letters of the word 'DIRECTOR' be arranged, so that the three vowels are never together?

Solution; Total number of letters = 8 and total number of vowels = 3

Here, R occurs two times,

Therefore, total number of arrangements when there is no reaction = = 20160, but when three vowels are together, regarding them as one letter, we have only 5 + 1 = 6

These 6 letters can be arranged in  ways, since R occurs twice.                                                                             

Also, three vowels can be arranged among themselves in 3! ways.

Hence, number of arrangements when the three vowels are together = 

Number of arrangements, so that the three vowels are never together = 20160 - 2160= 18000

3. In how many different ways 8 girls can be seated in a row?

Solution: Number of ways in which 8 girls can be seated in a row = 8! = 8x7x6x5x4x3x2x1= 40320

  4. Find the number of permutations that can be made from the letters of the word 'OMEGA' in such a way that Vowels occupying odd places.

Solution:

 Three vowels (O, E, A) can be arranged in the odd places in 3! ways

i.e., 1st position, 3rd position, 5th position

And two consonants (M, G) can be arranged in the even places in 2! ways

i.e., 2nd place and 4th place

.^..  Total number of ways = 3! X 2! = 12

  5. A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?

Solution: The first marble can be put into the pockets in 4 ways,

 In the same way second and third.

Thus, the number of ways in which the child can put the marbles = 4X4X4= 64 ways

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Probability means the chances of happening/occurring of an event. So, in this chapter we discuss about the predictability of an event to happen/occur. We usually predict about many events based on certain parameters.

For example:

• Getting a head or tail, when a coin is tossed
• Getting a number from 1 to 6, when a die is rolled

The better we know about the parameters related to an event better will be the accuracy of the result predicted.

Mathematically, we can say that probability of happening an event is equal to the ratio of number of favourable outcomes to number of possible outcomes.

It is represented as shown below Probability happening of an event P

Terms Related to Probability

 Various terms related to probability are as follows

Experiment:

An action where the result is uncertain even though the all-possible outcomes related to it is known in advance. This is also known as random experiment, e. g., Throwing a die, tossing a coin etc.

Sample Space

A sample space of an experiment is the set of all possible outcomes of that experiment. It is denoted by S.

 For example: If we throw a die, then sample space S = {1, 2, 3, 4, 5, 6} If we toss a coin, then sample space S = {Head, Tail}

Possible outcomes

All possibilities related to an event are known as possible outcomes.

Tossing a Coin When a coin is tossed, these are two possible outcomes.

 So, we say that the probability of getting H is 1/2 or the probability of getting T is 1/2,

Throwing a Die When a single die is thrown, there are six possible outcomes 1, 2, 3, 4, 5 and 6.

The probability of getting any one of these numbers is 

Event

Event is the single result of an experiment, e. g., Getting a head is an event related to tossing of a coin.

Types of Events

Various types of events are as follows

Certain and Impossible Events

A certain event is certain to occur, i.e., S (sample space) is a certain event.

Probability of certain event is 1, i.e., P(S) = 1.

An impossible event has no chance of occurring.

Probability of impossible event is 0, i. e., P (0) = 0.

Equally Likely Events

• Events related to an experiment are said to be equally likely events, if probability of occurrence of each event is same.

 For example:

When a dice is rolled the possible outcome of getting an odd number = possible outcome of getting an even number = 3.

So, getting an even number or odd number are equally likely events.

Complement of an Event

• The complement of an event A is the set of all outcomes in the sample space that are not included in the outcomes of event A.
• The complement of event A is represented by A (read as A bar).

The probability of complement of an event can be found by subtracting the given probability from 

 

Mutually Exclusive and Exhaustive events:

 

• Mutually exclusive: refers to two (or more) events that cannot both occur when the random experiment is formed. A ∩ B = Ø

 

Two events, A and B, are said to be mutually exclusive if the occurrence of A prohibits the occurrence of B (and vice versa)

 

• Exhaustive Events: Two events, A and B, are said to be exhaustive if at least one of them will definitely occur.

 

Dependent Events:

 

• Two events are called dependent, if the outcome or occurrence of the first affects the outcome or occurrence of the second, so that the probability is changed.

 

Independent Events

 

• Two events A and B are called independent, if occurring or non-occurring of A does not affect the occurring or non-occurring of B.
• If A and B are independent events, then P (A and B) = P (A ∩ B) = P(A) -P (B)

 

For example: getting head after tossing a coin and getting a 5 on a rolling single 6-sided die are independent events.

 

Rules/Theorems Related to Probability

 

The various theorems related to probability are discussed below

 

Addition Rule of Probability:

 

When two events A and B are mutually exclusive, the probability that A or B will occur, is the sum of the probability of each event.

 

 P (A or B) = P(A) + P(B) and P (A U B) = P (A) + P (B)

 

But when two events A and B are non-mutually exclusive, the probability that A or B will occur, is

 

 P (A or B) = P (A) + P(B) – P (A and B)

 

 P (A U B) = P(A) + P(B)- P (A ∩ B).

 

Multiplication Theorem of Probability

 

When two events A and B are mutually exclusive, the probability that A and B will occur simultaneously is given as P (A ∩ B) = P(A) * P (B/A) and P (A∩ B) = P(A) * P(B) (A and B are independent event).

 

Law of Total Probability:

 

The rule states that if the probability of an event is unknown, it can be calculated using the known probabilities of several distinct events.

 

Mathematically, the total probability rule can be written in the following equation:

 

Where: n = number of events

 

B_n = the distinct event.

 

For Example: There are three events: A, B, and C. Events B and C are distinct from each other while event A intersects with both events.

 

We do not know the probability of event A.

 

 However, we know the probability of event A under condition B and the probability of event A under condition C.

 

 The total probability rule states that by using the two conditional probabilities, we can find the probability of event A.

 

Conditional Probability

 

The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already been occurred.

 

The notation for conditional probability is P (B/A), It is pronounced as the probability of happening of an event B given that A has already been happened.

Points to be noted while we find probability of cards

 

i)                 So, there are 13 cards of each suit Clubs, Diamonds, Hearts and Spades.

 

ii)                There are 4 Aces, 4 Jacks, 4 Queens and 4 Kings.

iii)              There are 26 red and 26 black cards.

iv)               There are 12 face cards

 

Examples:

1.    A coin is tossed and a single 6-sided die is rolled. Find the probability of getting the head side of the coin and getting a 3 on the die

Solution: Probability of getting a head when a coin is tossed 

Probability of getting a 3 when a die is rolled = 

 

2.     A Mathematics teacher conducted two tests in her class. 25% of the students passed both tests and 42% of the students passed the first test. What per cent of the students passed the second test given that they have already passed the first test?

 

Solution:

 

This problem describes a conditional probability, since it asks us to find the probability that the second test was passed given that the first test was passed.

According to the formula,

3.    A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a king or a club?

 Sol. There are 4 kings in a standard deck and 13 club cards.

Also 1 king is of club, so probability of getting a king = 

Probability of getting a club = 

Probability of getting a king club = 

4.    A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles. If a single marble is chosen at random from the jar, what is the probability that it is yellow or green?

Solution: Total marble = 1 + 3 + 2 + 4 = 10, i.e., n(s) = 10

Now, probability of getting a yellow marble = 

Probability of getting a green marble = 

 

Since, the events are mutually exclusive

5.    A person can hit a target 4 out of 7 shots. If he fixes 10 shots, what is the probability that he hit the target twice?

 

Solution: Here, n = 10 and r = 2

 

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A Venn diagram is a diagram that helps us visualize the logical relationship between sets and their elements.

It shows logical relations between two or more sets

Venn diagrams are also called logic or set diagrams

They are widely used in set theory, logic, math, teaching, business data science and statistics.

 A Venn diagram typically uses circles other closed figures can also be used to denote the relationship between sets.

In general, Venn diagrams shows how the given items are similar and different

In Venn diagram 2 or 3 circles are most used one, there are many Venn diagrams with larger number of circles (5, 6, 7, 8, 10….).

Union: When two or more sets intersect, all different elements present in sets are collectively called as union.

It is represented by U

Union includes all the elements which are either present in Set A or set B or in both A and B

i.e., A ∪ B = {x: x ∈ A or x ∈ B}.

The union of set corresponds to logical OR

For example: If we have A = {1, 2, 3, 4, 5} and B = {3, 5, 7}

A U B = {1, 2, 3, 4, 5, 7}

Intersection: When two or more sets intersect, overlap in the middle of the Venn diagram is called intersection.

This intersection contains the common elements in all the sets that overlap.

It is denoted by ∩

All those elements that are present in both A and B sets denotes the intersection of A and B. So, we can write as A ∩ B = {x: x ∈ A and x ∈ B}.

The intersection of set corresponds to the logical AND

For example: If we have A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}

A ∩ B = {4, 5}

Cardinal Number of Set:

The number of different elements in a finite set is called its cardinal number of a set

It is denoted as n(A)

A = {1, 2, 3, 4, 5, 7}

n(A) = 6

Formula:

• n (A ∪ B) = n(A) + n(B) – n (A ∩ B)
• n (A ∪ B ∪ C) = n(A) + n(B) + n(C) – n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C)

Examples:

1. Out of 20 boys, 5 boys have ice-cream only and 10 boys have chocolate only, 3 boys have both chocolate and ice-cream, how many boys has only one of ice-cream or chocolate.

Solution: Given total boys= 20

Number of boys having ice-cream = 5

number of boys having chocolate = 10

number of boys who has only one of ice-cream or chocolate = 5 + 10 = 15

2. In a class of 60 students, 21 play Tennis, 13 play Cricket and 14 play Basketball. 6 plays both Tennis and Cricket, 5 play Cricket and Basketball and 7 play Tennis and Basketball. If 22 students do not play any of these given sports, how many students play exactly two of these sports?

Solution: Given Total = 60;

 T = 21, C=13, and B=14; 

T ∩ C=6, C ∩ B = 5, and T ∩ B = 7.

Neither=19.

[Total] = Tennis + Cricket + Basket Ball – (TC+CB+TB) + (All three) + (Neither)

55 = 21 + 13 + 14 - (6+5+7) + (All three) + 22

(All three) = 3;

Students play only Tennis and Cricket are 6-3=3;

Students play only Cricket and Basketball are 5-3=2;

 Students play only Tennis and Basketball are 7-3 = 4;

 Hence, 3 + 2 + 4 = 9 students play exactly two of these sports.

 3. Last month 30 students of a certain college travelled to Egypt, 30 students travelled to India, and 36 students travelled to Italy. Last month no students of the college travelled to both Egypt and India, 10 students travelled to both Egypt and Italy, and 17 students travelled to both India and Italy. How many students of the college travelled to at least one of these three countries last month?

Solution: Given

Students travelled to Egypt n(A) = 30

Students travelled to India n(B) = 30

Students travelled to Italy n(C) = 36

Egypt and India travellers n (A∩ B) = 0

Egypt and Italy travellers n (A ∩ C) = 10

India and Italy travellers n (B ∩ C) = 17

From all the information we can determine that 0 people travelled to all 3 countries because 0 people travelled to both Egypt and India.

To know how many students travelled to at least one country,

Total travellers = Egypt + India + Italy - sum of (travelled exactly two countries) - 2 times (travelled all three countries)

Total travellers = 30 + 30 + 36 - (10 + 17 + 0) - 2(0)

Total travellers = 96 - 17 - 0 = 69

Thus, 69 people travelled to at least one country.

 4. Each person who attended a conference was either a client of the company, or an employee of the company or both. If 56 percent of these who attended the conference were clients and 49 percent were employees. What percent were clients, who were not employees?

Solution: Total = Stockholders + Employees - Both;

100 = 56 + 49 – Both

 Both = 5;

Percent of clients, who were not employees is: Clients - Both = 56 - 5 = 51.

 5. There are 45 students in PQR College. Of these, 20 have taken an accounting course, 20 have taken a course in finance and 12 have taken a marketing course. 7 of the students have taken exactly two of the courses and 1 student has taken all three of the courses. How many of the 40 students have taken none of the courses?

Solution:  Given Total= 45; Let P = 20, Q = 20, and R = 12; sum of EXACTLY 2 - group overlaps = 7;

P ∩ Q ∩ R =1;

Total = P + Q + R – (sum of exactly 2 – group overlaps) – 2 * P ∩ Q ∩ R + None

45 = 20 + 20 + 12 – 7 – (2 * 1) + none

None = 2

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