Age is defined as a period of time that a person has lived or a thing has existed. Age is measured in months, years, decades and so on.

Problem based on ages generally consists of information of ages of two or more persons and a relation between their ages in present/future/past.

Using the information, it is asked to calculate the ages of one 01 more persons in present/future/ past.

Important Rules for Problem Based on Ages:

Rule1:

Rule2:

If ratio of present ages of A and B is x: y and after n yr, the ratio of their ages will be p: q,

Note:

Mostly questions on ages can be solved with the use of linear equations.

 

Examples:

1. Present age of Kevin is 5 times the age of Steve. After 10 yr, Kevin will be 3 times as old as Steve. What are the present ages of Kevin and Steve?

Solution: Let the present age of Steve be x yr. Then, present age of Kevin = 5x yr

After 10 yr, the ratio of ages will be 3: 1.

According to the question,

Therefore, Kevin’s present age = 5 x 10 = 50 yr and Steve’s present age = 10 yr

2. The average age of a family of five members is 24. If the present age of the youngest member is 8 yr, what was the average age of the family at the time of the birth of the youngest member?

Solution: Total age of five members of a family = 24 x 5 = 120

Therefore, total age of four members at the time of birth of youngest = 120 – (8 x 5)

                                                                                                            = 120 – 40 = 80 yr

3. Before 7 yr, the ratio of ages of A and B was 3: After 9 yr, ratio of their ages will be 7: 8. The present age of B will be

Solution: Let the ages of A and B before 7 yr were 3x yr and 4x yr, respectively.

Therefore, present age of A = 3x + 7 and present age of B = 4x + 7

Hence, present age of B = 4 x 4 + 7 = 16 + 7 = 23 yr

 

  4.  The present ages of two persons are 36 and 50 yr, respectively. If after n yr the ratio of their ages will be 3: 4, then the value of n is

  5.  The present age of Peter's father is four times Peter's present age. Five years back, Peter's father was seven times as old as Peter was at that time. What is the present age of Peter's father?

Solution: Let present age of Peter be x.

Then, present age of Peter’s father = 4x

Now, 5 yr ago, Peter's father's age = 7x

Peter's age => 4x - 5 = 7(x - 5)

 => 4x - 5 = 7x - 35

=> 3x = 30

=> x = 10

Peter's present age = x = 10 yr

Peter's father's present age 4x = 4 * 10 = 40 yr

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Unitary method is a fundamental tool to solve arithmetic problems based on variation in quantities.

The method endorses a simple technique to find the amount related to unit quantity.

This method can be applied in questions based on time and work, speed and distance, work and wages etc.

Direct proportion:

Two quantities are said to be in direct proportion to each other, if on increasing (decreasing) a quantity, the other quantity also increases (decreases) to the same extent

i.e., (Quantity 1) ∞ (Quantity 2)

For example: Number of men ∞ Volume of work done (time constant)

i.e., if number of men increases, then the volume of work done also increases.

Similarly, if volume of work increases, then number of men required to finish the work also increases.

 

Indirect Proportion:

Two quantities are said to be in indirect proportion to each other, if on increasing (or decreasing) a quantity, the other quantity decreases (or increases) to the same extent

For example:

The time taken by a vehicle in covering a certain distance is inversely proportional to the speed of the vehicle.

Note: If M₁ persons can do W₁ work in D₁ days and M₂ persons can do W₂ work in D₂ days, then we have a general formula, M₁ W₂ D₁ = M₂ W₁ D₂

1. If 30 men working 18 h per day can reap a field in 32 days, in how many days can 36 men reap the field working 16 h per day?

Solution: Let the required number of days be ‘d’.

More men, less days (Indirect proportion)

Less hours, more days (Indirect proportion)

Therefore, Required number of days = 30

2. Steve completes 5/8 of a job in 20 days. At this rate, how many more days will he take to finish the job?

Solution: Let the required number of days be ‘d’.

Less work, Less days (Direct proportion)

3. If in a hostel, food is available for 45 days for 50 students. For how many days will this food be sufficient for 75 students?

Solution: For 50, food is sufficient for 45 days.

Therefore, for 1 student food is sufficient for 45 x 50 days

  4. If 12 engines consume 30 metric tonnes of coal when each is running 18 h per day, how much coal will be required for 16 engines, each running 24 h per day, it being given that 6 engines of former type consume as much as 8 engines of latter type?

Solution:

Let the required quantity of coal consumed be x.

More engines, More coal consumption (Direct proportion)

More hours, More coal consumption (Direct proportion)

Less rate of consumption, Less coal consumption (Direct proportion)

Engines                        12: 16

Working hours             18: 24

5. A worker makes a toy in every 2 h. If he works for 80 h, then how many toys will he make?

Solution: Let number of toys be x.

More hours, more toys (Direct proportion)

2: 80:: 1: x

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In this, we will study techniques to solve problems based on work and its completion time as well as number of persons required to finish the given work in stipulated time.

Suppose that you are a contractor and you got a contract to construct a flyover in a certain time. For this, you need to calculate the number of men required to finish the work according to their work efficiency.

Important Relations:

 1. Work and Person Directly proportional (more work, more men and conversely more men, more work).

2. Time and Person Inversely proportional (more men, less time and conversely more time, less men).

3. Work and Time Directly proportional (more work, more time and conversely more time, more work).

Basic Rules Related to Work and Time:

Rule1: If a person can do a piece of work in n days, then that person’s 1 day’s (hour’s) work = (1/n)

Rule2: If a person’s 1 day’s(hour’s) work = (1/n), then the person will complete the work in n days(hours).

Rule3: If a person is n times efficient than the second person, then work done by

First person: Second person = n: 1 and time taken to complete a work by

First person: Second person = 1: n

Rule4: If ratio of numbers of men required to complete a work is m: n, then the ratio of time taken by them will be n: m

Examples:

1. A and B together can do a piece of work in 12 days, while B alone can finish it in 30 days. A alone can finish the work in

  2. A can do a piece of work in 10 days and B in 20 days. They begin together but A leaves 2 days before the completion of the work. The whole work will be done in

Solution: Let the required days be x.

A works for (x-2) days, while B works for x days.

According to the question,

  3. 6 boys can complete a piece of work in 16 h. In how many hours will 8 boys complete the same work?

Solution: Given, M₁ = 120, D₁ = 45, M₂ = 120 + 30 = 150 and D₂ = x

Then, using M₁ D₁ = M₂ D₂

  4. If 3 men or 4 women can build a wall in 43 days, in how many days can 7 men and 5 women build this wall?

Solution: 3 men = 4 women

According to the formula, M₁ D₁ W₂ = M₂ D₂ W₁

Therefore, D₂ = 3 x 4 = 12 days

  5. A, B and C can do a piece of work individually in 8, 12 and 15 days, respectively. A and B start working but A quits after working for 2 days. After this, C joins B till the completion of work. In how many days will the work be completed?

After 2 day’s A left the work

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Activity involving physical efforts, done in order to achieve a result is known as work.

Money received by a person for a certain work is called the wages of the person for that particular work,

In other words, we can find the entire wages of any person by the following formula

Entire wages = Total number of days x Wages of 1 day of any person

 

For example:

 If Arjun's monthly wages$ 4200 and he worked for all 30 days, then his daily wages will be calculated as Total wages = Number of days x Daily wages 4200 = 30 x Daily wages

Important Points:

• Wages is directly proportional to the work done. It means, more money will be received for more work and less money will be received for less work.
• Wages is indirectly proportional to the time taken by the individual.
• Wages is directly proportional to 1 day work of each individual.

Examples:

1. Alex can do a piece of work in 6 days, while Kevin can do the same work in 5 days. If the total amount to be given for this work is $ 660, then what will be the share of Kevin, if both work together

Solution: Time taken by Alex = 6 days

time taken by Kevin = 5 days

Total amount earned = $660

  2. Wages of 45 women for 48 days amount to $ 31050. How many men must work for 16 days to receive $ 11500, if the daily wages of a man being double those of a woman?

  3. Men, women and children are employed to do a work in the proportion of 3: 2: 1 and their wages as 5:3:2. When 90 men are employed, total daily wages of all amounts to $ 10350. Find the daily wages of a man.

Solution:

Let the numbers of men, women and children are 3y, 2y and y, respectively.

Given, 3y = 90 => y = 30

Number of women = 60 and number of children = 30

Let the men's, women's and children's wages be $ 5x, $ 3x and $ 2x, respectively.

 According to the question,

Total daily wages = $ 10350

=> 90 x (5x) + 60 x (3x) + 30 x (2x) = 10350

=> x (450 + 180 + 60) = 10350

Therefore, Daily wages of a man = 15 x 5 = $ 75

  4. A alone can finish a work in 2 days, while B alone can finish it in 3 days. If they work together to finish it, then out of total wages of $ 6000, what will be the 20% of A's share?

  5. A man and a boy received $ 1400 as wages for 10 days for the work they did together. The man's efficiency in the work was six times that of the boy. What is the daily wages of the boy?

Solution: The ratio of efficiency of man to boy = 6: 1

Efficiency ∞ Wages

Now, they worked for 10 days.

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Problems on Pipes and Cisterns are based on the basic concept of time and work

Pipes are connected to a tank or cistern and are used to fill or empty the tank or cistern.

In pipe and cistern, the work is done in form of filling or emptying a cistern/tank

Inlet pipe It fills a tank/cistern/reservoir. Outlet pipe It empties a tank/cistern/reservoir

 

Important Points:

• If a pipe can fill/empty a tank in m h, then the pat of tank filled/emptied in 1 hr = 
• If a pipe can fill/empty  part of a tank in 1 h, then it can fill/empty the whole tank in ‘m’ h
• Generally, time taken to fill a tank is taken(+ve) and time taken to empty a tank is taken negative(-ve).
• If a pipe fills a tank in m h and another pipe fills in n h. Then, part filled by both pipes in 

Examples:

1. If a pipe can fill a tank in 2 h and another pipe can fill the same tank in 6 h, then what part of a tank will be filled by both the pipes in 1 h, if they are opened simultaneously?

2. A pipe can fill a tank in 10 h. Due to a leak in the bottom, it fills the tank in 20 h. If the tank is full, how much time will the leak take to empty it?

Solution: Let the leak empties full tank in x h, then part emptied in 1 h by leak = 

Therefore, leak will empty the full tank in 20 h.

3. There are two tanks A and B to fill up a water tank. The tank can be filled in 40 min, if both taps are on. The same tank can be filled in 60 min, if tap A alone is on. How much time will tap B alone take, to fill up the same tank?

Solution:  Part filled by tap A in 1 min = 1/60

Let tap B fills the tank in x min

Then, part filled by tap, B in 1 min = 1/x

Therefore, Tap B can fill the tank in 120 min.

4. Inlet A is four times faster than inlet B to fill a tank. If A alone can fill it in 15 min, how long will it take if both the pipes are opened together?

Solution: Time taken by A to fill the tank, m = 15 min

Therefore, time taken by B to fill the tank, n = 15 x 4 = 60 min

  5. A tap having diameter 'd' can empty a tank in 40 min. How long another tap having diameter '2d' take to empty the same tank?

Solution: Area of tap ∞ work done by pipe

When diameter is doubled, area will be four times. So, it will work four times faster.

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Clock

A clock is an instrument which displays time divided into hours, minutes and seconds.

 A clock mainly consists of four components.

Dial

A clock is a circular dial. The periphery of the dial is numbered 1 through 12 indicating the hours in a 12 h cycle. The circumference of a dial is divided into 60 equal spaces.

Every clock has mainly two hands, one is smaller and other is bigger. The smaller hand is slower the and the bigger hand is faster.

Hour Hand

The smaller or slower hand of a clock is called the hour hand. It makes two revolutions in a day. Minute Hand

The bigger or faster hand of a clock is called the minute hand. It makes one revolution in every hour.

 

Second Hand

Second hand indicates seconds on a circular dial. It makes one revolution per minute.

 Note:

In 1 h minute hand covers 60 min spaces whereas the hour hand covers 5 min spaces Therefore, minute hand gains (60 - 5) = 55 min in 1 h

Important Points Related to Clock

1. In 1 h, both hands coincide once (i.e., 0° apart)

For example: Between 3 and 4'o clock, hands are together as shown in adjacent figure

  2. In 12 h, both hands coincide 11 times (between 11 and 1'o clock they coincide once) and in a day both hands coincide 22 times.

For example, between 11 and 1'o clock, hands are together as shown in adjacent figure.

  3. If two hands are at 90 ° they are 15 min spaces apart. This happens twice in 1 h. In a period of 12 h, the hands are at right angle 22 times (2 common positions) and in a day both hands are at right angle 44 times.

  4. If two hands are in opposite direction. (i.e., 180°apart), then they are 30 min spaces apart. This happens once in 1 h. In a period of 12 h both hands are in opposite direction 11 times and in a day both hands are in opposite direction 22 times.

  6. Angle covered by hour hand in 1 min.

  7. From point 5 and 6, we can say that the minute hand goes ahead by   in comparison to hour band.

Concept of Slow or Fast Clocks

If a watch/clock indicates 9: 15, when the correct time is 9, then it is said to be 15 min too fast.

On the other hand, if the watch/clock indicates 6: 45, when the correct time is 7, then it is said to be 15 min too slow.

Examples:

1.  A clock gains 10 s in every 3 h. If the clock was set right at 4:00 am on Monday morning, then the time it will indicate on Tuesday evening at 7:00 pm.

Solution: Difference of time between 4.00 am on Monday to 7.00 pm Tuesday

                                              24 + 12 + 3 = 39 h

2. At what time between 4 o'clock and 5 o'clock, will the hands of a clock be together?

Solution: At 4 o'clock, the hour hand is at 4 and the minute hand is at 12.

 It means that they are 20 min spaces apart.

To be together, the minute hand must gain 20 min over the hour hand.

As we know, 55 min is gained by minute hand in 60 min

  3. At what time between 3 o'clock and 4 o'clock, will the hands of a clock be in opposite directions?

 Solution: At 3 o'clock, the hour hand is at 3 and the minute hand is at 12.

It means that the two hands are 15 min spaces apart. But to be in opposite directions, the hands must be 30 min spaces apart.

Therefore, the minute hand will have to gain (30 + 15) = 45 min spaces over the hour hand.

Therefore 55 min spaces are gained in 60 min

  4. At what point of time after 3 O' clock, hour hand and the minute hand are at right angles for the first time?

Given that, n = 3

  5. At what time between 9 o'clock and 10 o'clock, will the hands of a clock be in the same straight line but not together?

Solution: At 9 o' clock, the hour hand is at 9 and the minute hand is at 12.

 It means that the two hands are 15 min spaces apart.

To be in the same straight line (but not together), they will be 30 min space apart.

.^.. The minute hand will have to gain (30-15) = 15 min spaces over the hour hand.

As we know, 55 min spaces are gained in 60 min.

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Calendar

 A calendar is chart or series of pages showing the days, weeks and months of a particular year. A calendar consists of 365 or 366 days divided into 12 months.

Ordinary Year

A year having 365 days is called an ordinary year (52 complete weeks + 1 extra day = 365 days)

Leap Year

 A leap year has 366 days (the extra day is 29th of February) (52 complete weeks + 2 extra days =366 days.)

 A leap year is divisible by 4 except for a century. For a century to be a leap year it must be divisible by 400. e.g.,

• Years like 1988, 2008 are leap year (divisible by 4).
•  Centuries like 2000, 2400 are leap year (divisible by 400).
•  Years like 1999, 2003 are not leap year (not divisible by 4).
•  Centuries like 1700, 1800 are not leap year (not divisible by 400).
• In a century, there is 76 ordinary year and 24 leap years.

Odd Days

• Extra days, apart from the complete weeks in a given period are called odd days.
• Number of days in an ordinary year = 365 = (52 x 7) + 1 = 52 weeks + 1 odd day
•  An ordinary year has 1 odd day
• Number of days in a leap year = 366 = (52 x 7) + 2 = 52 weeks + 2 days
• while a leap year has 2 odd days.
•  Number of days in a century (100 yr) = 76 ordinary years + 24 leap years = 76x1+ 24x2) =124 =17x7+5 =17 week+ 5 odd days
• Therefore, 100yrs has 5 odd days.
• Number of odd days in 200 yr = 5x2=10 days = 1 week + 3 days = 3 odd days
• Number of odd days in 300 yr = 5 x 3 = 15 days = 2 weeks + 1 day = 1 odd day
• Number of odd days in 400 yr = (5 x 4 +1) days =21 days = 3 weeks = 0 odd days
• As 400th is a leap year, therefore 1 more day has been taken
•  Similarly, each one of 800 yr, 1200 yr, 1600 yr, 2000 yr, 2400 yr etc., has no odd days.
• Remember the adjacent table for the number of odd days in different months of an year.
• In an ordinary year, February has no odd days, but in a leap year, February has one odd day.
•  The 1 st day of a century must be Tuesday, Thursday or Saturday.
• The last day of a century cannot be Tuesday, Thursday or Saturday.

Day Gain/Loss

Ordinary Year (± 1 day)

• When we proceed forward by 1 yr, then 1 day is gained.

 For example: 9th August 2013 is Friday, then 9th August 2014 has to be Friday +1 = Saturday.

• When we move backward by 1 yr, then 1 day is lost.
•  For example: 24th December 2013 is Tuesday, then 24th December 2012 has to be Tuesday -1 = Monday.

Leap Year (+ 2 days)

• When we proceed forward by 1 leap year, then 2 days are gained.

For example: If it is Wednesday on 25th December 2011, then it would be Friday on 25th December 2012 [Wednesday + 2] because 2012 is a leap year.

• When we move backward by 1 leap year, then 2 days are lost.

For example: If it is Wednesday on 18th December 2012, then it would be Monday on 18th December 2011. [Wednesday -2] because 2012 is a leap year.

Exception

• The day must have crossed 29th February for adding 2 days otherwise 1 day.

For example: If 26th January 2011 is Wednesday, 26th January 2012 would be Wednesday + 1 = Thursday (even if 2012 is leap year, we have added + 1 day because 29 February is not crossed).

If 23rd March 2011 is Wednesday, then 23rd March 2012 would be Wednesday + 2 = Friday (+ 2 days 29th February of leap year is crossed)

To Find a Particular Day on the Basis of Given Day and Date

 Following steps are taken into consideration to solve such questions

 Step I: Firstly, you have to find the number of odd days between the given date and the date for which the day is to be determined.

 Step II: The day (for a particular date) to be determined, will be that day of the week which is equal to the total number of odd days and this number is counted forward from the given day, in case the given day comes before the day to be determined.

 But, if the given day comes after the day to be determined, then the same counting is done backward from the given day.

To Find a Particular Day without Given Date and Day

 Following steps are taken into consideration to solve such questions

 Step I: Firstly, you have to find the number of odd days up to the date for which the day is to be determined.

Step II: Your required day will be according to the following conditions

(a) If the number of odd days = 0, then required day is Sunday.

 (b) If the number of odd days = 1, then required day is Monday.

(c) If the number of odd days = 2, then required day is Tuesday.

 (d) If the number of odd days = 3, then required day is Wednesday.

(e) If the number of odd days = 4, then required day is Thursday.

(f) If the number of odd days = 5, then required day is Friday. (g) If the number of odd days = 6, then required day is Saturday.

Examples:

1. 14 How many days are there in x weeks x days?

Solution: Number of days in x weeks = 7x + x

.^.. Total number of days is x week x days = 7x + x = 8x days

2. Find the day of the week on 26th January 1950.

Solution: Number of odd days up to 26th January, 1950

= Odd days for 1600 yr + Odd days for 300 yr + Odd days for 49 yr + Odd days of 26 days of January 1950 = 0 + 1 + (12 X2 +37) + 5 = 0 + 1 + 61 + 5 = 67 days = 9 weeks + 4 days = 4 odd days

Therefore, It was Thursday on 26th January 1950.

3. January 3, 2007 was Wednesday. What day of the week fell on January 3, 2008?

Solution: The year 2007 is an ordinary year, so it has 1 odd day.

 3rd day of the year 2007 was Wednesday.

.^.. 3rd day of the year 2008 will be one day beyond the Wednesday.

Hence, it will be Thursday.

4. What was the day of the week on 17th August, 2010?

Solution: Period up to 17th August, 2010 = (2009yr + Period from 1.1.2010 to 17.8.2010)

Counting of odd days odd days in 1600 yr = 0 Odd days in 400 yr = 0

9 yr = (2 leap years + 7 ordinary years) = (2x2+7xl) = l week + 4 days = 4 odd days

Number of days between 1.1.2010 to 17.8.2010

January + February + March + April + May + June + July + August

 = (31 +28 + 31+30 + 31 + 30 + 31 + 17) days

 = 229 days = 32 weeks + 5 odd days

Total number of odd days = (0 + 0 + 4 + 5) days = 9 days = 1 week + 2 odd days

 Hence, the required day is Tuesday

5. If 5th March, 1999 was Friday, what day of the week was it on 9th March 2000?

Solution: 5th March, 1999 is Friday.

Then, 5th March 2000 = Friday + 2 = Sunday.

{2000 is leap year and it crosses 29th Feb 2000, so 2 is taken as odd day}

.^..  5th March 2000 = Sunday. Then, 9th March 2000 = Thursday.

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Boats and streams

It is an application of concepts of speed, time and distance. Speed of river flowing either aides a swimmer (boat), while travelling with the direction of river or it opposes when travelling against the direction of river.

Still Water:

 If the speed of water of a river is zero, then water is considered to be still water.

 Stream Water

 If the water of a river is moving at a certain speed, then it is called as stream water.

Speed of Boat:

 Speed of boat means speed of boat (swimmer) in still water. In other words, if the speed of a boat (swimmer) is given, then that particular speed is the speed in still water.

 Downstream Motion:

 If the motion of a boat (swimmer) is along the direction of stream, then such motion is called downstream motion.

Upstream Motion:

 If the motion of a boat (swimmer) is against the direction of stream, then such motion is called upstream motion.

Formulae related to Boats and Strems:

• If the speed of a boat in still water is a miles/h and speed of the stream is b miles/h, then

• Speed downstream = (a + b) miles/h
• Speed upstream = (a – b) miles/h

Examples:

1. If the speed of a boat in still water is 10 miles/h and the rate of stream is 5 miles/h, then find upstream speed of the boat.

Sol. Given,

speed of a boat = a = 10 miles/h Speed of stream = b = 5 miles/h

 Hence, Speed upstream = a – b = 10 - 5 = 5 miles/h.

 

 

2. Kevin can row a certain distance downstream in 24 h and can return the same distance in 36 h. If the stream flows at the rate of 12 miles/h, then find the speed of Kevin in still water.

Solution:

• Let the Kevin’s speed in still water = a
•  Then, Kevin's speed downstream = (a + 12)
• Kevin's speed upstream = (a - 12)

Given that,

• Distance travelled downstream = Distance travelled upstream
• 24 (a + 12) = 36 (a - 12)
• 2 (a + 12) = 3 (a – 12)
•  2a + 24 = 3a – 36
•  3a – 2a = 36 + 24
•  a = 60 miles/h

3. A boat's speed in still water is 12 miles/h, while river is flowing with a speed of 4 miles/h and time taken to cover a certain distance upstream is 6 h more than time taken to river the same distance downstream. Find the distance.

Solution: Let the distance = d

Boat’s rate downstream = 12 + 4 = 16 miles/h

Boat’s rate upstream = 12 – 4 = 8 miles/h

Difference between the time = Time taken by boat to travel upstream - Time taken by boat to travel downstream

  4.  Andrew can row 36 miles/h in still water. It takes him twice as long to row up as to row down the river. Find the rate of stream.

Solution:

• Let rate of stream be a miles/h.

According to the question,

• 36 + a = 2(36 - a)
• 36 + a = 72 - 2a
•  3a = 72 - 36 = 36
•  a = 36/3 = 12 miles/h

  5. A boatman takes twice as long to row a distance against the stream as to row the same distance with the stream. Find the ratio of speeds of the boat in still water and the stream.

Solution:

• Let boatman's speed upstream p
•  His speed downstream = 2p

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When two or more persons make an association and invest money for running a certain business and after certain time receive profit in the ratio of their invested money and time period of investment, then such an association is called partnership and the persons involved in the partnership are called partners.

Partnership is of Two Types

 Simple Partnership

If all partners invest their different capitals (money) for the same time period or same capital for different time period then their profit or loss is in the ratio of their investments or time period of investment then such a partnership is called simple partnership

Compound Partnership

If all partners invest their different capitals (money) for different time period, then their profit not only depends on their investments but also on the time period of their investment, then such a partnership is called compound partnership

Partners are of Two Types

 Active or Working Partner

A partner who not only invests money, but also take part in the business activities for which he draws a defined salary or gets some share from profit before its division is called an active partner.

Sleeping Partner

A partner who only invests money and does not take part in business activities is called sleeping partner.

Ratio of division of gains:

• When the investments made by all the partners are for the same time period, then gain or loss is distributed amongst them in the ratio of their investments,

Suppose A and B invest $ x and $ y respectively for a year in a business, then at the end of the year: (A’s share of profit): (B’s share of profit) = x: y

• When investments are of different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x number of units of time). Now, gain or loss is divided in the ratio of these capitals.

Suppose A invests $ x for p months and B invests $ y for q months, then (A’s share of profit): (B’s share of profit) = xp: yq.

Examples:

1. P and Q entered a partnership for 3 yr. At the start of the business, they invested $ 13000 and $ 25000, respectively. At the end of 3 yr their total profit was $ 76000. What will be share of Q out of this profit?

Solution: P’s share: Q’s share = Ratio of their investments

                                    = 13000: 25000

                                    = 13: 25

  2. A, B and C invested their capitals in the ratio of 5: 6: 8. At the end of the business, they received the profits in the ratio of 5: 3: 1. Find the ratio of time for which they contributed their capitals.

Solution: Here P₁: P₂: P₃ = 5: 3: 1 and x₁: x₂: x₃ = 5: 6: 8

  3. A and B together start a business by investing in the ratio of 4: 3. If 9% of the total profit goes to charity and A's share is $ 1196, find the total profit.

Solution: Let total profit = x

  4. A and B started a business with $ 20000 and $ 35000 respectively. They agreed to share the profit in the ratio of their capital. C joins the partnership with the condition that A, B and C will share profit equally and pays $ 220000 as premium for this, to be shared between A and B. This is to be divided between A and B in the ratio of

Solution: Ratio of total capital of A and B = 20000 x 12: 35000 x 12

                                                                   = 240000: 420000

Now, C gives $220000 to both to make the capital equal.

If A takes $ 200000 and B takes $ 20000 from C, then both have the equal capital

Therefore, required ratio of divided amount = 200000: 20000 = 20: 2 =10: 1

  4. P, Q and R hire a meadow for $ 2920. P puts 10 cows for 20 days; Q puts 30 cows for 8 days and R puts 16 cows for 9 days. Find the rent paid by R.

Solution: Ratio of rents to be paid by P, Q and R

Ratio of monthly equivalent = (10 x 20): (30 x 8): (16 x 9)

                                                = 200: 240: 144

                                                = 25: 30: 18

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

For example (i) Cube of 6 = 6 x 6 x 6=216 (ii) Cube of 8 = 8 x 8 x 8=512

Methods to Find Cube

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

Algebraic Method

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

1. (a + b)³ = a³ + 3ab (a +b) + b³
2.  (a -b)³ = a³ - 3ab (a-b)- b³

For example:

The cube of 16 is (16)³ = (10 + 6)³ = (10)³ + 3 x 10 x 6 (10 + 6) + (6)³ = 1000 + 2880 + 216 = 4096

Shortcut Method:

Step I: The answer consists of 4 parts each of which has to be calculated separately,

Step II: First write down the cube of ten's digit to the extreme left. Write the next two terms to the right of it by creating; GP (Geometric Progression) having common ratio which is equal to  and the fourth number will be cube of unit’s digit.

Step III: Write the double of 2nd and 3rd number below them.

Step IV:  Now, add the number with numbers written below it and write the unit's place digit in a straight line and remaining number is carried forward to the next number.

Example: Find the cube of 35.

Sol. Here, unit's digit is 5 and ten's digit is 3.

Step I: Write the cube of ten's digit at extreme left i.e., (3)³ =27

Step II: Now, the next two terms on the right will be in a GP of common ratio equals to 

 and last term will be cube of unit’s digit i.e., (5)³ = 125, So, they are arranged as 27 45 75 125

Step III: Twice the second and third terms are written under it and are added.

(35)³ = 42875

Cube Root:

The cube root of a given number is the number whose cube is the given number. The cube root is denoted by the sign 

Methods to Find Cube Root

Method to calculate the Cube root of a number is as follow

Prime Factorisation Method

This method has following steps.

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

Step II: Arrange the factors in a group of three of same prime numbers.

Step III: Take the product of these prime factors picking one out of every group (group of three) of the same primes.

This product gives us the cube root of given number

Example: Find the cube root of 9261.

Sol: Prime factors of 9261 = (3 x 3 x 3) x (7 x 7 x 7)

Properties of Cubes and Cube Roots

• If the cube of a number is of 2 or 3-digits, then its cube root will be of 1-digit
•  If the cube of a number is of 4, 5 or 6 digits, then its cube root will be of 2 digits.
•  If the cube of a number has 0,1, 2,3,4, 5, 6, 7,8,9 in its unit's place, then its cube root will have 0, 1, 8, 7, 4, 5, 6, 3, 2 or 9 in their unit's place, respectively.

Example Problems:

1. Find the cube root of -5832

  2. What least number should be subtracted from 6862, so that 19 be the cube root of the result from this subtraction?

Solution: Given,

• number = 6862
• Let’s find out the nearest cube values of 6862

• (18)³ = 5832
• (19)³ = 6859

• (18)³ < 6862 > (19)³
• Therefore, required number = 6862 – 6859 = 3

Solution:

• On squaring both sides, we get

• and now cubing both sides, we get
• (x²)³ = 8*4x
• x⁵ = 32
• x⁵ = 2⁵
• x = 2

  4. Find the cube value of 102

Solution: Given number = 102

It can be written as (100 + 2)³ = 100³ + 2³ + 3 x 100 x 2 (100 + 2)

                                                = 1000000 + 8 + 600(102)

                                                = 1000000 + 8 + 61,200

                                                = 1061208

  5. Find the cube root of 1259712

Solution: Given number = 1259712

Let’s write factors of the give number = 3 x 3 x 3 x 4 x 4 x 4 x 9 x 9 x 9

Therefore, cube root of 1259712 = 108

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Indices:

When a number 'P' is multiplied by itself 'n' times, then the product is called nth power of 'P' and is written as Pⁿ. Here, P is called the base and 'n' is known as the index of the power.

 Therefore, Pⁿ is the exponential expression.

 Pⁿ is read as 'P raised to the power n’ or ‘P to the power n’.

Rules:

Surds

Numbers which can be expressed in the form √p + √q, where p and q are natural numbers and not perfect squares.

Note

 1. All surds are irrational numbers

2. All irrational numbers are not surds

 Order of Surds

• The order of a surd indicates the index of root to be extracted.
• 
• 

• A surd with index of root 2 is called a second order surd or quadratic surd.

• For example: √2, √3, √5, √7, √x are the surds of order 2. Example: √2, √5, √10, √a, √m, √x, √ (x + 1) are second order surd or quadratic surd (since the indices of roots are 2).

• A surd with index of root 3 is called a third order surd or cubic surd.

• For example: ?2, ?3, ?10, ?17, ?x are the surds of order 3 or cubic surds. Example: ?2, ?5, ?7, ?15, ?100, ?a, ?m, ?x, ? (x - 1) are third order surd or cubic surd (since the indices of roots are 3).

Types of Surds

Pure Surds

Those surds which do not have factor other than 1, are known as pure surds

Mixed Surds

Those surds which have factor other than 1. are known as mixed surds

Like and Unlike Surds

• When the radicands of two surds are same, then those are known as like surds.

Example: 2√2, 3√2, 4√2

• When radicands are different, then they are called unlike surds

Example: 2√2, 2√3, 2√5

Properties of Surds

• A quadratic surd cannot be equal to the sum and difference of a rational number and a quadratic surd.

Example:  a + b ≠ √c or √a – b ≠ √c

To Arrange the Surds in Increasing or Decreasing Order

• 
• First of all, take the LCM of a, b and c and use it to make the denominator of the powers the same.
•  Then, easily we can find the required order.

Operations on Surds

Addition and Subtraction of Surds

Only like surds can be added or subtracted. Therefore, to add or subtract two or more surds, first simplify them and add or subtract them respectively like surds

Note:

Multiplication and Division of Surds

To multiply or divide the surds, we make the denominators of the powers equal to each other. Then, multiply or divide as usual.

Comparison of Surds

To compare two or more surds, first of all the denominators of the power of given surds are made equal to each other and then the radicand of the new surds is compared.

Rationalisation of Surds

The method of obtaining a rational number from a surd by multiplying it with another surd is known as rationalisation of surds. Both the surds are known as rationalising factor of each other

Examples:

2. If a^x = b, b^y = c and xyz = 1, then what is the value of c^z?

Solution: Given, a^x = b, b^y = c and xyz = 1

Let us take b = a^x

• b^y = a^xy
• b^yz = a^xyz
• c^z = a

Solution: LCM of 4, 6, 12 = 12

  5. If 3^x – 3^x-1 = 18, then x^x is equal to?

Solution: Given 3^x – 3^x-1 = 18

• 3^x-1(3 – 1) = 18
• 3^x-1 = 9
• 3^x-1 = 3² [Since bases are equal exponents are also equal]
• x – 1 = 2
• x = 3

Therefore, x^x = 3³ = 27

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It is a system of geometry, where the position of points on the plane is described by using an ordered pair of numbers.

 Rectangular Coordinate Axes

The lines XOX' and YOY' are mutually perpendicular to each other and they meet at point O which is called the origin.

 

Line XOX' represents X-axis and line YOY' represents Y-axis and together taken, they are called coordinate axes.

 

Any point in coordinate axis can be represented by specifying the position of x and y-coordinates

Quadrants

The X and Y-axes divide the cartesian plane into four regions referred to quadrants

• The coordinates of point O (origin) are taken as (0,0).
• The coordinates of any point on X-axis are of the form (x, 0).
• The coordinates of any point on Y-axis are of the form (0, y)

Formulae:

Distance Formula

Distance between Two Points If A (x₁, y₁) and B (x₂, y₂) are two points, then 

Distance of a Point from the Origin

The distance of a point A (x, y) from the origin O (0, 0) is given by 

Area of triangle

If A (x₁, y₁) B (x₂, y₂) and C (x₃, y₃) are three vertices of a Triangle ABC, then its area is given by

 Area of triangle  (x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))

Collinearity of Three Points

Three points A (x₁, y₁) B (x₂, y₂) and C (x₃, y₃) are collinear, if

(i)   Area of triangle ABC is 0

(ii)   Slope of AB = Slope of BC = Slope of AC

(iii)   Distance between A and B + Distance between B and C = Distance between A and C

Centroid of a Triangle

 

Centroid is the point of intersection of all the three medians of a triangle. If A (x₁, y₁) B (x₂, y₂) and C (x₃, y₃) are the vertices of triangle ABC, then the coordinates of its centroid are

Circumcentre

The circumcentre of a triangle is the point of inter section of the perpendicular bisectors of its sides and is equidistance from all three vertices.

 

 If A (x₁, y₁) B (x₂, y₂) and C (x₃, y₃) are the vertices of triangles and O (x, y) is the circumcentre of triangle ABC, then OA = OB= OC

Incentre

The centre of the circle, which touches the sides of a triangle, is called its incentre.

 

 Incentre is the point of intersection of internal angle bisectors of triangle.

If A (x₁, y₁) B (x₂, y₂) and C (x₃, y₃) are the vertices of a triangle ABC such that BC = a, CA = b and AB = c, then coordinates of its incentre I are 

Section formulae

• Let A (x₁, y₁) B (x₂, y₂) be two points on the cartesian plane.
• Let point P (x, y) divides the line AB in the ratio of m: n internally.

If P divides AB externally, then 

If P is the mid-point of AB, then  

Basic Points Related to Straight Lines

1. General form of equation of straight line is ax + by + c = 0. Where, a, b and c are real constants and x and y are two unknowns.

2. The equation of a line having slope m and intersects at c on x-axis is y = mx + c.

 

3. Slope (gradient) of a line ax + by + c = 0, by = - ax – c

Comparing with y = mx + c, where m is slope, therefore m = tan θ = 

Slope of the line is always measured in anti-clockwise direction.

 

4. Point slope form A line in terms of coordinates of any two points on it, if (x₁, y₁) and (x₂, y₂) are coordinates of any two points on a line, then its slope is 

5. Two-point form a line the equation of a line passing through the points A (x₁, y₁) and B (x₂, y₂) is 

 

6. Condition of parallel lines

If the slopes of two lines i.e., m₁ and m₂ are equal then lines are parallel.

Equation of line parallel to ax + by + c = 0 is ax + by + q =

 

7. Condition of perpendicular lines

If the multiplication of slopes of two lines i.e., m₁ and m₂ is equal to -1 then lines are perpendicular.

m₁ x m₂ = -1

Equation of line perpendicular to ax + by + c = 0 is bx - ay + q =0

 

 8. Angle between the two lines 

 

9. Intercept form Equation of line L intersects at a and b on x and y-axes, respectively is 

 

10. Condition of concurrency of three lines:

Let the equation of three lines are a₁x + b₁y + c₁ = 0,

a₂x + b₂y + c₂ = 0, and a₃x + b₃y + c₃ = 0.

Then, three lines will be concurrent, if

Distance of a point from the line:

Let ax + by + c = 0 be any equation of line and P (x₁, y₁) be any point in space. Then the perpendicular Distance(d) of a point P from a line is given by 

 

12. The length of the perpendicular from the origin to the line ax + by + c = 0, is 

 

 13. Area of triangle by straight line ax + by + c = 0 where a ≠ 0 and b ≠ 0 with coordinate axes is 

 

 14. Distance between parallel lines ax + by + c = 0 and ax + by + d = 0 is equal to 

 

15. Area of trapezium, between two parallel lines and axes,

Area of trapezium ABCD = Area of OCD

Examples:

Find the area of triangle ABC, whose vertices are A (8, - 4), B (3, 6) and C (- 2, 4).

Solution: Here, A (8, - 4) so, x₁ = 8, y₁ = - 4

                        B (3, 6) so, x₂ = 3, y₂ = 6

 

                           C (-2, 4) so, x₃ = -2, y₃ = 4

Therefore, area of triangle ABC (x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))

(8(6 – 4) + 3(4 – (- 4)) + (-2) (-4-6))

 (16 + 24 + 20)

= 30 sq units

If A (-2,1), B (2, 3) and C (-2, -4) are three points, then find the angle between AB and BC.

 

Solution: Let m₁ and m₂ be the slopes of line AB and BC, respectively.

 

Let θ be the angle between AB and BC

 

 3. In what ratio, the line made by joining the points A (- 4, - 3) and B (5,2) intersects x-axis?

Solution: We know that y-coordinate is zero on x-axis,

 

Given, y₁ = - 3, y₂ = 2

Therefore, 

 

2m – 3n = 0

 

 4. Coordinates of a point is (0, 1) and ordinate of another point is - 3. If distance between both the points is 5, then abscissa of second point is

Solution: Let abscissa be x.

So, (x – 0)² + (-3 -1)² = 5²

x² + 16 = 25

x² = 9

 5. Do the points (4, 3), (- 4, - 6) and (7, 9) form a triangle? If yes, then find the longest side of the triangle

Solution: Let P (4, 3), Q (-4, -6) and R (7, 9) are given points.

Since, the sum of 12.04 and 6.7 is greater than 18.6.

 

So, it will form a triangle, whose longest side is 18.6

 

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Cylinder

• A cylinder is a solid or hollow body that is formed by keeping circles of equal radii one on another.

• A part from this, a cylinder is formed by rolling a rectangular sheet also.

 

• A cylinder has three surfaces

• Curved surface
• Bottom
• Top

Solid Cylinder:

• Volume of cylinder = Area of base x Height = 
• Curved surface area = Perimeter of base x Height = rh
• Total surface area = Curved surface area + Area of both the circles (top and bottom surfaces) = r (h + r) where, r = Radius of base and h = Height

 

Hollow Cylinder:

 

If cylinder is hollow, then

• Volume of hollow cylinder = Outer volume - Inner volume = π(R² – r²) h cubic units.
• If Curved surface area = Curved surface area of outer surface + Curved surface area of inner surface = 2πRh + 2πrh = 2πrh (R + r)
• Total surface area of hollow cylinder = Curved surface area + Area of both top and bottom surface = 2πh (R + r) + 2π(R² – r²)

 

 where, R = External radius of base, r = Internal radius of base and h = Height

 

Examples:

 

1. How many iron rods each of length 14 m and diameter 4 cm can be made out of 0.88 m³ of iron?

Volume of iron rod = 0.88 m³

 

  2. What will be the curved surface area of a right circular cylinder having length 160 cm and radius of the base is 7 cm?

 

  3. A rod of 1 cm diameter and 30 cm length is converted into a wire of 3 m length of uniform thickness. The diameter of the wire is

Solution: Given, r₁ = 1 cm, h₁ =30 cm, h₂ = 300 cm

 

Volume of rod = volume of wire

 

  4. A hollow cylinder made of wood has thickness 1 cm while its external radius is 3 cm. If the height of the cylinder is 8 cm, then find the volume, curved surface area and total surface area of the cylinder.

 

Solution: Radius r = Inner radius = External radius – Thickness = 3 – 1 = 2 cm

 

  5. The ratio of the radii of two cylinders is 2: 3 and the ratio of their heights is 5: 3. The ratio of their volumes will be

 

Solution: Let the radii be 2r and 3r and heights be 5h and 3h.

 

 

 

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Cone:

Cone is a solid or hollow body with a round base and pointed top. It is formed by the rotation of a triangle along any of the side.

 

 

Frustum of Cone:

• If a cone is cut by a plane parallel to the base so as to divide the cone into two parts upper part and lower part, then the lower part is called frustum.

 

Sphere:

• A sphere is a three-dimensional solid figure, which is made up of all points in the space, which lie at a constant distance from a fixed point.
• That constant distance and fixed point are respectively called the radius and centre of the sphere.
• In fact, a sphere is like a solid ball.

 

Hollow Sphere or Spherical Shell:

• It’s both external and internal surfaces are spherical and both the surfaces have a common central point.

Hemisphere:

 

 

• It is the half part of a sphere.

 

Examples:

 

 

  1. The diameter of a right circular cone is 14 m and its slant height is 10 m. Find its curved surface area, total surface area.

 

  2. The frustum of a right circular cone has the diameters of base 10 cm, of top 6 cm and a height of 5 cm. Find its slant height.

 

Solution: Lets draw a figure from given data

 

  3. The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What is the ratio (approximate) of their volumes?

Solution:

 

Given that the diameter of the Moon is approximately one-fourth of the diameter of the Earth.

 

  4. What will be the difference between total surface area and curved surface area of a hemisphere having 2 cm diameter?

 

Solution: Given diameter = 2 cm, so, radius = 1 cm.

 

 

  5. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

 

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Prism:

• A right prism is a solid whose top and bottom face are parallel to each other and are identical polygons that are parallel.
• The faces joining the top and bottom faces are rectangles and are called lateral faces.
• The distance between the base and the top is called height or length of the right prism.

 

• Volume of prism = Area of base x Height of the prism
• Lateral (vertical) surface area = Perimeter of base x Height of the prism
• Total surface area = Lateral surface area + 2 x Area of base 

Pyramid:

• A solid whose base is a polygon and whose faces are triangles, is called a pyramid. The triangular faces meet at a common point called vertex,
• "A pyramid whose base is regular polygon and the foot of the perpendicular from the vertex of the base, coincides with the centre of the base, is called a right pyramid."
• 

• Total surface area = Lateral surface area + Area of the base

Examples:

  1. The perimeter of the triangular base of a right prism is 60 cm and the sides of the base are in the ratio 5: 12: 13. Then, its volume will be

Solution: Let the sides of the base are 5s, 12s and 13s respectively.

Given, perimeter of base = 60 cm

So, 5s + 12s + 13s = 60 cm

The sides of base are 5s = 5 x 2 = 10 cm, 12s = 12 x 2 = 24 cm, 13s = 13 x 2 = 26 cm.

2. A prism and a pyramid have the same base and the same height. Find the ratio of the volumes of the prism and the pyramid.

Solution:

We know that,

Volume of the prism = (Area of the base) x (Height)

Therefore, Ratio of the volumes of the prism and the pyramid = 3: 1

3. The base of a right prism is a square having side of 20 cm. If its height is 8 cm, then find the total surface area and volume of the prism.

Solution:

Given, side = 20 cm and height = 8 cm

• Now, Total surface area = Lateral surface area + 2 x Area of base
• Now, lateral surface area = Perimeter of the base x Height = [20 + 20 + 20 + 20] x 8 = 80 x 8 = 640 cm²
• Area of base = Area of square = 20 x 20 = 400 cm²

On putting, these values in formula,

• we get Total surface area = 640 + 2 x 400 = 640 + 800 = 1440 cm²
• Volume of the prism = Area of base x Height = 20 x 20 x 8 = 3200 cm³

4. A prism has the base a right-angled triangle whose sides adjacent to the right angle are 12 cm and 15 cm long. The height of the prism is 20 cm. The density of the material of the prism is 4 g/cu cm. The weight of the prism is

Solution: Volume of prism = Area of base x height

Therefore, Weight of prism = 1800 x 4

                                                   = 7200 g

                                                   = 72 kg

  5. Find the total surface area of a pyramid having a slant height of 10 cm and a base which is a square of side 2 cm (in cm)?

Solution: Total surface area of pyramid

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• Geometry is a branch of Mathematics which deals with the questions and concepts of shape, size, relative position of figures, their angles etc.

It can be broadly divided into two parts

Plane Geometry

• Plane geometry is about flat shapes like line, circle and triangle.
• These are two-dimensional figure which can easily be drawn on paper.

Solid Geometry

• Solid geometry is about three-dimensional objects like cube, prism and sphere.
• These are three-dimensional figure

Point:

A figure of which length, breadth and height cannot be measured is called a point. It is infinitesimal.

Line:

• A line is defined by its length but has no breadth.
• A line contains infinite points.
• Three or more points are said to be collinear, if there is a line which contains all of them.

Plane:

 It is a flat surface having length and breadth both but no thickness. It is a 2-dimensional figure.

Parallel Lines:

• Two lines in the same plane are said to be parallel, if they don't have any intersection point, however how far they are extended in either direction.
• They remain at same distance for the whole length. The sign of parallel is '| |'.
• Here, 1 and m are called parallel lines.

Transversal Lines:

• A straight line that cuts two or more straight lines at distinct points is called a transversal.
• In the figure given below 1 and m are parallel lines and p is a transversal.

Angle:

• An angle is formed by two rays with a common initial point.
• Let O is the initial point, then O is called the vertex. e.g., ∠AOB = θ

Types of Angles:

1. Right angle: The angle whose value is 90° is called a right angle. θ = 90°)

2. Acute angle: The angle whose value lies between 0° and 90° is called an acute angle. (0°< θ < 90°)

 

 3. Obtuse angle: The angle whose value lies between 90° and 180° is called an obtuse angle. (90°<θ<180°)

 

 4. Straight angle: The angle whose value is 180° is called a straight angle. (θ =180°)

 

 5. Reflex angle: The angle whose value lies between 180° and 360° is called a reflex angle. (180°< θ <360°)

 

 6. Complete angle: The angle whose value is 360° is called a complete angle. (θ =360°)

 

 7. Supplementary angle: If the sum of two angles is 180°, then they are called supplementary angles. Let θ₁ and θ₂ be two angles, then θ₁ + θ₂ =180°

 

 8. Complementary angle: If the sum of two angles is 180°, then they are called supplementary angles. Let θ₁ and θ₂ be two angles, then θ₁ + θ₂ = 90°

Angle Bisector:

• A line which cuts an angle into two equal angles is called an angle bisector
• An angle bisector can be internal or external.

Internal angle bisector:

Here, two angles are formed ∠AOC and ∠BOC. Both angles are equal (θ) because OB is the internal angle bisector.

External angle bisector:

Here, ∠A'OB and ∠BOC are equal and external bisector is OB.

Examples:

1. In the given figure, find x.

Solution: From figure

∠ABC + ∠DBC = 180⁰

(3x + 15)⁰ + (x + 5)⁰ = 180⁰

4x = 160⁰

x = 40⁰

 2. In the given figure, straight lines AB and CD intersect at O. If  = 3∠p, then ∠p is equal to

∠p + 3∠p = 180⁰

4∠p = 180⁰

∠p = 45⁰

 3. In the given figure, AB||CD. If ∠CAB = 80⁰ and ∠EFC = 25°, then ∠CEF is equal to

Solution: Given, ∠CAB = 80⁰ and ∠EFC = 25° and AB||CD

Let ∠CEF = x^o

Here AB||CD and AF is transversal

So, ∠DCF = ∠CAB = 80⁰ [since, corresponding angles]

In triangle CEF, side EC has been produced to D.

x + 25 = 80^o

x = 55^o

 4. In the given figure, AOB is straight line If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then ∠COD is equal to

Solution: Given AOB is a straight line.

So, ∠AOC + ∠EOB + ∠BOD = 180^o

40 + 4x + 3x = 180

7x = 180 – 40

7x = 140

x = 20

Therefore, ∠COD = 4x° = 80^o

 5. If every interior angle of regular octagon is 135°, then find the external angle of it.

Solution: Every external angle of octagon = 180^o – Interior angle

                                                                             = 180^o – 135^o

                                                                             = 45^o

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Fraction:

A digit which can be represented in p/q form, where q * 0, is called a fraction. Here, p is called the numerator and q is called the denominator.

For example, 3/5 is a fraction, where 3 is called numerator and 5 is called denominator.

Simple Fraction:

The fraction which has a denominator other than the power of 10 is called a simple fraction.

For example: 3/7, 5/11, etc.

Simple fraction is also known as vulgar fraction

   i.  Types of Simple Fractions

There are following types of fractions:

1. Proper Fraction When the numerator of a fraction is less than its denominator, then fraction is called proper fraction

Example: ½, 21/43.

2. Improper Fraction When the numerator of a fraction is greater than its denominator, then fraction is called improper fraction

Example: 17/13, 18/14,

3. Compound Fraction A fraction, in which numerator or denominator or both are in fraction, then it is called compound fraction.

Example: 1/ (7/9), (13/11)/17,

4. Inverse Fraction If we inverse the numerator and the denominator of a fraction, then the resultant fraction will be the inverse fraction of the original fraction.

Example: let given function is 3/7, then inverse function is 7/3.

5. Mixed Fraction The fraction, which is the combination of integer and fraction, is called mixed fraction.

6. Continuous Fraction It has no certain definition but only say that a fraction contains additional fractions in its denominators, is called continuous fraction.

   ii.  Operations on Simple Fractions

 Addition of Simple Fractions

1. When Denominators are Same If denominators of fractions are same, then numerators of fractions are added and their addition is divided by denominator.

Example: (1/4) + (2/4) = (1+2) (1/4) = (3/4)

2.  When Denominators are Different If denominators of fractions are not same, then make their denominators equal (by taking their LCM) and then add their numerators.

Example: (1/2) + (1/3) + (1/4) = ((1 x 6) + (1 x 4) + (1 x 3))/12 = (6 + 4 + 3)/12 = 13/12

Subtraction of Simple Fractions

1. When Denominators are Same If denominators of fractions are same, then numerators of fractions are subtracted and their subtraction is divided by the denominator.

Example:(3/4) -(1/4) = (3-1) (1/4) = (2/4) = 1/2

2. When Denominators are Different If denominators of fractions are not same, then make their denominators equal and then subtract their numerators.

Example: (2/3) - (1/2) = ((2 x 2) - (3 x 1))/6 = (4 - 3)/6 = 1/6

Multiplication of Simple Fractions

1. To multiply two or more simple fractions, multiply their numerators and denominators.

Example:(1/2) x (3/4) = (1x3) / (2 x 4) = (3/8)

  2. If fractions are given in mixed form, first convert them into improper fraction and then multiply.

Division of Simple Fractions

To divide two fractions, first fraction is multiplied by the inverse of second fraction

Example: (2/3) + (3/5) = (2/3) x (5/3) = 10/9

   iii.  Comparison of Simple Fractions

 Following are some techniques to compare fractions.

1. Cross Multiplication Method:

Example: If (a/b) and (c/d) are two fractions, then i) if ad>bc, then (a/b) > (c/d)

ii) if ad<bc, then (a/b) < (c/d), iii) if ad=bc, then (a/b) =(c/d)

2. By Changing Fractions in Decimal Form: To compare two or more fractions, first convert fractions into decimal form and then compare.

Example: Between (1/7) and (2/9), which fraction is bigger?

Solution: (1/7) = 0.14 and (2/9) = 0.22. It is clear that 0.22 > 0.14. Therefore (2/9) > (1/7).

3. By Equating Denominators of Given Fractions: For comparison of fractions, take LCM of the denominators of all fractions, so that the denominators of all fractions are same. Now, the fraction having largest numerator is the largest fraction.

Example: Arrange the following fractions in decreasing order (3/5), (7/9), (11/13).

LCM of 5,9 and 13 = 5 x 9 x 13 = 585

(3/5) = (3 x 117)/ (5 x 117) = 351/585;

(7/9) = (7 x 65)/ (9 x 65) = 455/585;

(11/13) = (11 x 45)/ (13 x 45) = 495/585

Now, the fraction having largest numerator will be largest.

Therefore, decreasing order will be (495/585), (455/585), (351/585)

Hence, order will be (495/585), (455/585), (351/585).

Hence, order is (11/13), (7/9), (3/5).

4. By Equating Numerators of Given Fractions: For comparison of fractions, take LCM of the numerator of all fractions, so that numerators of all the fractions are same. Now, the fraction having smallest numerator will be largest.

Example: Which fraction is largest among (3/13), (2/15), (4/17)?

Solution: LCM of 2, 3 and 4 = 2 x 2 x 3 = 12

(3/13) = (3 x 4)/ (13 x 4) = 12/52 = (2 x 6)/ (15 x 6) = (12/90) and (4/17) = (3 x 4)/(3 x 17) = (12/51)

Now, the fraction having smallest denominator will be largest.

Hence, (4/17) is the biggest number

Important Facts Related to Simple Fractions:

1. If in a fraction, numerator is equal to denominator, then the value of fraction is equal to 1
2.  If the numerator of a fraction is always non-zero and denominator is zero, then the value of fraction is infinity (°°).
3. If the numerator of a fraction is zero and denominator is not equal to zero, then the value of fraction is zero.
4.  If the numerator or denominator of any fraction is either multiplied or divided by same number, then the value of fraction remains unchanged.
5.  If the numerator and denominator have no common factor other than 1, then the fraction is said to be in its lowest form
6.  A fraction is a rational number as it can be expressed in the form of p/q and q * 0.

Basic Formulae:

• To represent any fraction in simplified form, divide its numerator and denominator by their HCF.
• If in the given fractions, the difference between numerator and denominator are same, then fraction having larger numerator is the largest and fraction having smaller numerator is the smallest.
•  Formula 3 if in the given fractions, the numerators are increasing by a definite value and the denominator is also increasing by a definite value but the value of denominator is greater than numerator, then the fraction having smaller numerator will be the smallest fraction and the fraction having larger numerator will be the largest fraction.
• If any number is divided by (a/b) instead of multiplying by (a/2), then the obtained value will be x greater than original value and the given number will be (abx)/ (b² – a²).

Examples:

1. 13 Jack was to find 6/7 of a fraction. Instead of multiplying, he divided the fraction by 6/7 and the result obtained was 13/70 greater than original value. Find the fraction given to Jack?

Solution: Given, a = 6, b = 7 and x = (13/70)

Therefore, required fraction = (abx)/ (b² – a²) = (6 x 7 x (13/70))/ (7² – 6²) = (6 x 13)/ (10 x 13) = (3/5)

Hence, fraction given to Jack is (3/5).

2. Arrange the given fractions in increasing order, (4/5), (5/6), (6/7)

Solution: Since, all the fractions have difference in numerator and denominator are same.

Therefore, increasing order (4/5), (5/6), (6/7).

  3. Out of the fractions (5/7), (4/9), (6/11), (2/5) and (3/4), what is the difference between the largest and the smallest fractions?

Solution: (5/7) = 0.71, (4/9) = 0.44, (6/11) = 0.54, (2/5) = 0.40, (3/4) = 0.75

Here, the largest fraction = (3/4) and the smallest fraction = (2/5)

So required difference = (3/4) – (2/5) = (15 – 8)/20 = 7/20

  4. The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator becomes eight times the numerator, then find the fraction.

Solution: Let denominator of fraction = x

Then, numerator = x – 4.

Therefore Fraction = (x-4)/x. Now, according to the question,

8[(x -4) – 2] = (x + 1)

 -> (x -4) – 2 = ((x +1)/8)

-> x – 6 = (x+1)/8

 -> 8(x – 6) = x + 1

-> 8x – 48 = x + 1

-> 7x = 49,

x = 49/7 -> x = 7, Therefore fraction = (7 – 4)/ 7 = (3/7).

  5. 4/7 of a pole is in the mud. When 1/3 of it is pulled out, 250 cm of the pole is still in the mud. Find the full length of the pole.

Solution: Total length of pole = (Length of pole in mud)/ (Remaining part of pole in mud)

                                                     = 250/ ((4/7) – (1/3))

                                                     = 1050

                Therefore, length of pole = 1050.

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Fraction:

A digit which can be represented in p/q form, where q * 0, is called a fraction. Here, p is called the numerator and q is called the denominator.

For example, 3/5 is a fraction, where 3 is called numerator and 5 is called denominator.

Decimal Fraction:

If the fraction has denominator in the powers of 10, then fraction is called decimal fraction.

Example: 10^th part of unit = (1/10) = 0.1, 10^th part of 6 = (6/10) = 0.6

To convert a decimal fraction into a vulgar fraction, place 1 in the denominator under the decimal point. Then, after removing the decimal point, place as many zeroes after it as the number of digits after the decimal point. Finally, reduce the fraction to its lowest terms.

Example: 0.23 = (23/100), 0.0035 = 35/10000 = 7/2000

Note:  Placing zeroes to the right of a decimal fraction, it does not make any change in value Hence, 0.5, 0.50, 0.500 and 0.5000 are equal.

If numerator and denominator of a fraction have same number of decimal places, then each of the decimal points be removed

Thus, (1.84/2.99) = (184/299) = (8/13)

Types of Decimal Fractions

1.     Recurring Decimal Fraction: The decimal fraction, in which one or more decimal digits are repeated again and again, is called recurring decimal fraction. To represent these fractions, a line is drawn on the digits which are repeated.

2.    Pure Recurring Decimal Fraction: When all the digits in a decimal fraction are repeated after the decimal point, then the decimal fraction is called as pure recurring decimal fraction.

To convert pure recurring decimal fractions into simple fractions (vulgar form), write down the repeated digits only once in numerator and place as many nines in the denominator as the number of digits repeated.

Since, there is only 1 repeated digit. Therefore, only single 9 is placed in denominator.

Since, there are only 2 repeated digits. Therefore, two 9's are placed in denominator.

3.     Mixed Recurring Decimal Fraction: A decimal fraction in which some digits are repeated and some are not repeated after decimal is called as mixed recurring decimal fraction.

Example: 0.1733333… = 

To convert mixed recurring decimal fractions into simple fractions, in the numerator, take the difference between the number formed by all the digits after decimal point (repeated digits will be taken only once) and the number formed by non-repeating digits. In the denominator, place as many nines as there are repeating digits and after nine put as many zeroes as the number of non-repeating digits.

Operations on Decimal Fractions:

Addition and Subtraction of Decimal Fractions:

To add or subtract decimal fractions, the given numbers are written under each other such that the decimal points lie in one column and the numbers so arranged can now be added or subtracted as per the conventional method of addition and subtraction.

Example: (i) 353.5 + 2.32 + 43.23 =? (ii) 1000 - 132.23 =?

Solution: 

Multiplication of Two or More Decimal Fractions: Given fractions are multiplied without considering the decimal points and then in the product, decimal point is marked from the right-hand side to as many places of decimal as the sum of the numbers of decimal places in the multiplier and the multiplicand together.

Example: (i) 4.3 x 0.13 =? (ii) 1.12 x 2.3 x 4.325 =?

Sol. (i) 43 x 13 = 559

Sum of the decimal places = (1 + 2) = 3; Therefore, required product = 0.559 (ii) 112 x 23 x 4325 = 11141200 Sum of the decimal places = (2 + 1 + 3) = 6 ∴ Required product = 11.141200

Multiplication of Decimal Fraction by an Integer:

Given integer is multiplied by the fraction without considering the decimal point and then in the product, decimal is marked as many places before as that in the given decimal fraction.

Example: Find the value of the following. (i) 19.72x4 (ii) 0.0745x10 (iii) 3.52x14

Sol. (i) 19.72 x 4

Multiplying without taking decimal point into consideration 1972 x 4 = 7888

So, 19.72 x 4= 78.88

Since, in the given decimal fraction, decimal point is two places before. So, in the product, decimal point will also be put two places before.

Similarly, (ii) 0.0745 x 10 = 0.745 (iii) 3.52 x 14 = 49.28

Dividing a Decimal Fraction by an Integer:

Do simple division i.e., divide the given decimal number without considering the decimal point and place the decimal point as many places of decimal as in the dividend.

Example: Suppose we have to find the quotient (0.0204 ^. 17). Now, 204 ^. 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 ^. 17 = 0.0012

Division of Decimal Fractions:

In such divisions, dividend and divisor both are multiplied first by a suitable multiple of 10 to convert divisor into a whole number and then above-mentioned rule of division is followed.

Example: Thus, (0.00066/0.11) = ((0.00066 x 100)/ (0.11 x 100)) = (0.066/11) = 0.006

Basic Examples

1.      when 0.252525…... is converted into a fraction, then find the result.

Solution: 27 x ((12-1)/9) x ((5.5262-5526)/9000) x (6/9)

                 = 27 x (11/9) x (49736/9000) x (6/9)

                 = (1094192/9000)

 

                 = 121.576888….

                 

3.     In the year 2020, Henry gets $3832.5 as his pocket allowance. Find his pocket allowance per day?

Solution: Henry’s pocket allowance = $3832.50

Total days in 2020 (general year) = 365 days

Allowance per day = (3832.5/365) = $10.5

4.     When 52416 is divided by 312, the quotient is 168. What will be the quotient when 52.416 is divided by 0.0168?

Solution: Given, (52416/312) = 168 -> (52416/168) = 312.

Now, (5.376/16.8) = (53.76/168) = ((53.76/168) x (1/100)) = (32/100) = 0.32

5.     ((36.54)² – (3.46)²)/? = 40

Solution: ((36.54)² – (3.46)²)/x = 40. Then, x = ((36.54)² – (3.46)²)/40 = ((36.54)² – (3.46)²)/ (36.54 + 3.46) = (36.54 – 3.46) = 33.08

 

Since ((a)² – (b)²/ (a + b)) = (a – b).

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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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Combination of things means selection of things. Here, order of things has no importance.

For example: The combination of two letters from the group of three letters A, B and C would be as follows AB, BC, AC.

Here, we make groups. So, AB or BA as a group is same.

Obviously, if order matters, then AB and BA are not same.

It signifies number of groups formed from n different things, when r things are taken into consideration.

Important Points:

• ⁿC_n = ⁿC₀ = 1
• ⁿC_{r -1} + ⁿC_r = ⁿ⁺¹C_r
• ⁿC₁ + ⁿC₂ + …. + ⁿC_n = 2ⁿ – 1
• ⁿC_r = ⁿC_n-r or ⁿC_r = (ⁿP_r)/r!
• ⁿC₀ + ⁿC₁ + ⁿC₂ + …. + ⁿC_n = 2ⁿ
• if there are n points in a plane out of which m are collinear, then (i) Number of straight lines formed = ⁿC₂ - ^mC₂ + 1 (ii) Number of triangles formed = ⁿC₃ - ^mC₃

Cases of Combination

There are several cases of Combination,

1. Formation of committee from a given set of persons

 These questions are based on formation of a committee consisting of some members (male and/ or female) from a group of persons following a certain condition.

2. Selection of questions from question paper etc

In such question, a question paper is given with one or more parts and the different ways in which some specified number of questions can be attempted is asked.

Factorial

Factorial of a number can be defined as the product of all natural numbers up to that number i. e.,

n! = n x (n-l) x (n-2) x (n-3) x (n-4) x..... x 1= n x (n-1)!

4! = 4x3x2x1=4x3!

11! = 11x10x9x8x7x6x5x4x3x2x1

Note: Factorial of negative number and integers is not defined ⁿP_n = n, ⁿP₀ = 1

Fundamental Principles of Counting

 Multiplication Principle

• If an operation can be performed in m different ways, following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m x n ways.
• This can be extended to any finite number of mutually exclusive operations.

Addition Principle

If an operation can be performed in m different ways and another operation, which is independent of the first operation, can be performed in n different ways, then either of the two operations can be performed in (m + n) ways.

This can be extended to any finite number of mutually exclusive operations.

Examples:

1. In how many ways can 5 members form a committee out of 10 be selected so that (i) two particular members must be included. (ii) two particular members must not be included.

Solution: (i) When two particular members are included then, we have to select 5—2 = 3 members out of 10 – 2 = 8

  2. A question paper has two parts, part A and part B, each containing 10 questions. If the student has to choose 8 from part A and 5 from part B, in how many ways can he choose the question?

= 5 x 9 x 3 x 2 x 7 x 6 = 11340

  3. A hall has 12 gates. In how many ways, can a man enter the hall through one gate and come out through a different gate?

Solution: Since, there are 12 ways of entering into the hall, the man come out through a different gate in 11 ways.

Hence, by the fundamental principle of multiplication, total number of ways is 12 x 11 = 132.

  4. In a plane, there are 11 points, out of which 5 are collinear. Find the number of tri angles made by these points.

Solution: Here n = 11, m = 5

Then, required number of triangles = ⁿC₃ - ^mC₃

  5. In how many ways, can 24 persons be seated around a circular table, if there are 13 seats?

Solution: First, we select 13 persons out of 24 persons in ²⁴C₁₃ ways.

 Now, these 13 persons can be seated in 12! Ways around a table.

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Probability means the chances of happening/occurring of an event.

• There are total of 52 cards in a deck of playing cards
•  It consists of 4 suits.
• There are 13 cards of each suit Clubs, Diamonds, Hearts and Spades.
• Spades, clubs are black in colour and hearts, diamonds are red in colour.
• Each suit contains an ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, and 2.
• King, Queen, and Jack are called the face cards.
• In the deck of 52 playing cards, there are 12 face cards and there are 4 Aces, 4 Jacks, 4 Queens and 4 Kings.
• Whole together there are 26 red and 26 black cards.
• The four different suits of cards are shown below in the picture.

• In the above four suits we have 13 cards in each suit
• Picture showing all the 52 cards in the pack of playing cards.

• Here we have

• Club - 13 cards
• Diamond - 13 cards
•  Heart - 13 cards
• Spade - 13 cards
• No. of black cards - 26
• No. of red cards - 26
• No. of Ace cards - 4
• No. of Jack cards - 4
•  No. of Queen cards - 4
• No. of King cards - 4
• No. of face cards – 12

• Probability 

Examples:

1. From a pack of 52 cards two are drawn with replacement. The probability, that the first is a spade and the second is a queen, is

Solution: Given two cards are drawn with replacement.

The probability that card is a spade P(s) 

The probability that card is a queen P(q) 

Required probability = P(s) x P(q) 

 2. Find the probability of drawing a king or an ace from a pack of playing cards.

Solution:

As there are four kings and four aces, the number of favourable cases = 8

The required probability 

 3. What is the probability of drawing a red card from a pack of cards?

Solution: The total number of outcomes = 52

The number of favourable outcomes = 26

Therefore, required probability 

 4. One card is drawn from a well-shuffled pack of 52 cards. What is the probability, that it is not the king of diamonds?

Solution: The king of diamonds can be drawn in only 1 way

 Since in a pack of cards there is only one king of diamonds

P(A) = Probability of drawing the king of diamonds 

Hence the probability of not drawing an ace of hearts 

 5. From a pack of 52 cards, two cards are drawn, what is the probability that both are hearts or both are jacks?

Solution:

Total number of ways = ⁵²C₂

Both are hearts = ¹³C₂

Both are jacks = ⁴C₂

So, required probability = (¹³C₂ + ⁴C₂)/ ⁵²C₂

 

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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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