Important Questions-Solutions Mathematics · Important Questions Topic: Frequency Distribution 1. Observe the given frequency distribution table and answer the following questions

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Grade VIIIMathematics

Exam Preparation Booklet

Chapter WiseImportant Questions-Solutions

Data Handling Important Questions

Topic: Frequency Distribution

1. Observe the given frequency distribution table and answer the following questions.

Class interval (Studying hours per week)

Frequency (Number of students)

04 2

48 6

812 8

1216 15

(a) What is the size of the class intervals? (b) How many students study for 8 hours and more in a week? (c) What is the lower limit of the class interval 1216?

(3 marks)

2. The given table lists the daily maximum temperature (°C) of a certain city for the month of April.

32, 33, 32, 31, 32, 29.5, 30.5, 27.5, 28, 29.7, 29.9, 31.2, 31.7, 32, 32.4, 32.5, 33, 34.7, 34.9, 35, 34.8, 35, 36, 37, 36.5, 34, 35.5, 36.5 32, 29

How can the given information be represented in the form of frequency distribution table using appropriate class size?

(3 marks)

Data Handling Topic: Histograms

3. The following table shows the numbers of hours spent on studying by 30 students of a class:

Number of hours Number of students

12 10

23 15

34 12

45 10

56 5

67 2

Prepare a histogram for the given data.

(3 marks)

4. A nurse measured the blood pressure of each patient admitted in a hospital. The given histogram shows the observations in a particular day.

(3 marks)

Data Handling

(a) What percentage of the patients admitted on that day had blood pressure in the range of 110 mm Hg120 mm Hg? (b) The normal blood pressure of a person ranges from 110 mm Hg to 130 mm Hg. What percentage of patients admitted on that day did not have normal blood pressure?

Topic: Circle Graphs

5. A fruit basket contains 60 fruits. The break up of these fruits is as follows:

Fruit Count

Banana 24

Mango 10

Guava 20

Orange 6

Draw a circle graph for the given data.

(3 marks)

Data Handling 6. The given pie chart shows the crime rate in a certain city at different time

intervals in a day.

Answer the following questions based on the above pie chart. (a) During which time interval is crime rate the least? (b) If a man goes out of his house at 4 pm, then what is the probability that he will be a victim of a crime?

(2 marks)

7. Use the following information to answer the next question.

A survey was conducted among a group of boys to know the preferences of the colour of shirt they would like to wear for encouraging their football team. The given piechart shows the result of the survey.

Study the given piechart and answer the following questions. (a) What is the percentage of boys who like to wear black coloured shirts? (b) Which colour of shirt is the least preferred among the group of boys to wear for encouraging their football team? (c) If there were 180 boys in the group, then how many of them prefer to wear white coloured shirts?

(3 marks)

Data Handling Topic: Probability I

8. What is the sample space when two dices are rolled together? (2 marks)

9. In which of the given experiments are the associated outcomes not equally likely? (a) A card is drawn at random from a bag containing 7 identical cards that are numbered as 1, 2, 3, 4, 5, 6, and 7. (b) Choosing a letter at random from the word SCHOOL

(2 marks)

10. Check whether the given events are certain to happen or impossible or can happen but are not certain. When a card is drawn out of cards numbered 10 to 20, an even prime number is obtained.

(2 marks)

11. State whether the following statements are correct or incorrect. Justify your answers. (a) When a die is rolled, it is equally likely that a multiple of 3 or a number greater than 4 will be obtained. (b) When a ball is drawn at random from a bag containing 4 yellow and 3 black balls, it is equally likely that a yellow or a black ball will be drawn. (c) When a coin is tossed, it is not equally likely that a head or a tail will occur. (d) When a card is drawn at random from a wellshuffled deck of 52 cards, it is equally likely to draw a face card or a queen.

(4 marks)

12. List the outcomes of the following experiments. One ball is drawn from a bag containing 3 white and 2 red identical balls.

(2 marks)

Topic: Probability II

13. When a die is thrown, what is the probability of getting: (a) An odd number (b) A number which is a multiple of 3

(2 marks)

14. A fruit seller keeps 15 dozens oranges everyday for selling them. On a particular day, he was left with 42 oranges that could not be sold. By next morning, the oranges had become rotten. However, he mixed them with the fresh oranges. If an orange is chosen at random, then what is the probability that it is fresh?

(2 marks)

15. There are some pairs of black socks and 24 pairs of white socks in a drawer. (2 marks)

Data Handling When a pair of sock is drawn at random, the probability of drawing a pair of black socks is . What is the total number of pairs of socks in the drawer?3

1

Data Handling Solutions

Topic: Frequency Distribution

1. (a) Size of the class intervals = 4 (b) The students who study for 8 hours and more are the ones who study for 8 − 12 hours or 12 − 16 hours. This means that 8 + 15 = 23 students study for 8 hours and more in a week. (c) Lower limit of class interval 12 − 16 is 12.

2. It can be observed that the given data is quite large. Here, the maximum temperature varies from 27.5°C to 36.5°C. Thus, one can condense it into groups such as 27 − 30, 30 − 33, 33 −

36, and 36 − 39. Then, tally marks table is drawn by marking for the daily maximum

temperature in the respective class interval. Here, five tally marks are represented by , where the fifth tally mark is marked across the remaining four tally marks to bunch them into five. Then, the required frequency table can be constructed as:

Class interval ( ° C)

Tally marks Frequency

2730

6

3033

11

3336

9

3639

4

Total 30

Data Handling Topic: Histograms

3. The required histogram is drawn by taking the frequency, i.e., the number of students, along the vertical axis and the class intervals, i.e., hours, along the horizontal axis. The height of the vertical bars is equal to the corresponding frequency of each class interval. Thus, the given data can be represented by a histogram as:

4. In the given histogram, the horizontal line represents blood pressure (in mm Hg) in the form of class interval 100110, 110120 …, and the corresponding number of patients is shown by the height of the bars. Total number of patients admitted on that day = 15 + 35 + 25 + 15 + 5 + 5 = 100 (a) Frequency corresponding to interval 110120 is 35. ∴Total number of patients having blood pressure in the range 110 mm Hg120 mm Hg is 35. Percentage of patients in the range 110 mm Hg120 mm Hg

Thus, 35% of the patients have blood pressure in the range of 110 mm Hg120 mm Hg. (b) Normal blood pressure of a person ranges from 110 mm Hg to 130 mm Hg. Number of patients whose blood pressure is in the range 100110 = 15

Data Handling Number of patients whose blood pressure is in the range 130140 = 15 Number of patients whose blood pressure is in the range 140150 = 5 Number of patients whose blood pressure is in the range 150160 = 5 ∴Total number of patients that do not have normal blood pressure = 15 + 15 + 5 + 5 = 40 Percentage of patients that do not have normal blood pressure

Thus, 40% of the patients admitted on that day did not have normal blood pressure.

Topic: Circle Graphs

5. Firstly, we find the central angle of each sector. Total number of fruits = 60

Fruit Count Infraction Central angle

Banana 24

Mango 10

Guava 20

Orange 6

Draw a circle of any radius, and then draw the angle of each sector using a protractor to obtain the following pie chart.

Data Handling

6. (a) Crime rate is the least during the time interval 7 am − 3 pm. (b) If a man goes out of his house at 4 pm, then the probability that he will be a victim of a crime

7. (a) The central angle of the sector corresponding to blackcoloured shirt is 108°. Thus, percent of boys who like to wear blackcoloured shirts

(b) In the given piechart, it is clearly seen that the size of the sector corresponding to red colour is the least. Thus, redcoloured shirt is the least preferred among the group of boys to wear for encouraging their football team. (c) The central angle of the sector corresponding to white colour is 144°. Total number of boys in the group = 180 Therefore, number of boys who prefer to wear whitecoloured shirts

Data Handling Topic: Probability I

8. When two dices are rolled together, the following outcomes are obtained: S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

9. The outcomes of an experiment are said to be equally likely if the possibility of occurring of each outcome is the same. (a) The given experiment is drawing of a card at random from a bag containing 7 identical cards that are numbered as 1, 2, 3, 4, 5, 6, and 7. Here, the outcomes are 1, 2, 3, 4, 5, 6, and 7. In this case, the possibility of occurring of each outcome is the same. Thus, the outcomes associated with experiment are equally likely. (b) The given experiment is choosing a letter at random from the word SCHOOL. Here, the outcomes are S, C, H, O, O, and L. In this case, the possibility of choosing the letter O is more than that of other letters. Thus, the outcomes associated with experiment are not equally likely.

10. (a) When a die is tossed, an odd number may or may not turn up. Hence, this event can happen but it is not certain. (b) When a card is drawn out of cards numbered 10 to 20, an even prime number cannot be obtained because 2 is the only prime number. Hence, this is an impossible event.

11. Two outcomes of an experiment are said to be equally likely if the possibility of occurrence of each outcome is the same. (a) Correct A die has six faces, numbered as 1, 2, 3, 4, 5 and 6. Out of these, 3 and 6 are multiples of 3, while 5 and 6 are greater than 4. Therefore, the number of multiples of 3 and the number of digits greater than 4 are the same. Thus, when a die is rolled, it is equally likely that a multiple of 3 or a number greater than 4 will be obtained. (b) Incorrect The bag contains more yellow balls than black balls. Thus, when a ball is drawn atrandom from the bag containing 4 yellow and 3 black balls, it is more likely that a yellow ball will be drawn.

Data Handling (c) Incorrect A coin has two faces: head and tail. When a coin is tossed, it is equally likely to land head or a tail. (d) Incorrect A standard deck of 52 cards contains twelve face cards and four queens. Thus, the number of face cards is more than the number of queens. Hence, when a card is drawn at random from a wellshuffled deck of 52 cards, it is more likely that a face card will be drawn.

12. (a) In drawing one ball from a bag containing 3 white and 2 red balls, the sample space is {W 1 , W 2 , W 3 , R 1 , R 2 } (b) In selecting two girls out of a group of five girls, the sample space is: {(G 1 G 2 ), (G 2 G 3 ), (G 3 G 4 ), (G 4 G 5 ), (G 2 G 4 ), (G 2 G 5 ), (G 3 G 5 ), (G 1 G 3 ), (G 1 G 4 ), (G 1 G 5 )}

Topic: Probability II

13 When a die is thrown, then the possible outcomes are 1, 2, 3, 4, 5 and 6 Out of these numbers, odd numbers are 1, 3 and 5, and multiples of 3 are 3 and 6. (a) Probability of getting an odd number

(b) Probability of getting a number which is a multiple of 3

14. It is known that 1 dozen is equal to 12. Total oranges kept everyday to be sold = 15 dozens = 15 × 12 = 180 ∴ Number of fresh oranges = 180 Number of rotten oranges = 42 ∴ Total number of oranges = 180 + 42 = 212 ∴ Probability of selecting a fresh orange

15. Let the number of pairs of black socks in the drawer be x . Number of pairs of white socks is given as 24. ∴ Total number of pairs of socks in the drawer = x + 24 Probability of drawing a pair of black socks

Data Handling

Thus, total number of pairs of socks in the drawer = 12 + 24 = 36

Comparing Quantities Important Questions

Topic: Percentage

1. The number 12000 is increased by 37%. Find the new number obtained. (2 marks)

2. The price of a car was Rs 3, 70, 000. After a year, it decreased by 14%. What is the present price of the car?

(2 marks)

3. In the previous month, a ticket of a DTC bus for a particular route cost Rs 15 while presently it costs Rs 20. By what percentage has the cost of ticket increased over a month?

(2 marks)

Topic: Sales Tax and Value Added Tax

4. On a car which costs Rs 2,10,000, a discount of 5% is offered. Reema bought the car and paid a sales tax of 10%. What is the total amount paid by Reema for the car?

(3 marks)

5. Reena purchased a crockery set and a tea set for Rs 16,875 including 12.5% VAT. What was the initial bill amount without tax?

(2 marks)

6. Bob purchased a computer for Rs 8,640. If he paid 8% tax, then what was the marked price of the computer?

(2 marks)

7. In a college election, candidate A got 12% more votes than candidate B. If A and B are the only two candidates in the election, then what percent of less votes did B get as compared to A?

(3 marks)

Topic: Concept of Profit Percent and Loss Percent

8. A shopkeeper sells a laptop at Rs 69000 making a profit of 15%. What is the cost price of the laptop?

(2 marks)

9. A shopkeeper sold a handbag at a loss of 4% for Rs 2,400. What was the cost price of the handbag?

(3 marks)

10. A fruit seller bought 16 dozen bananas for Rs 240. He sold 6 dozen bananas at a loss of 10% and the remaining at a profit of 5%. What was his total profit or loss percentage?

(2 marks)

Comparing Quantities 11. Sonu bought a bike and after some years, he sold it to his friend Dimpu

at a profit of 10%. Later on, Dimpu sold the bike to his friend Jammy at a loss of 20%. If jammy paid Rs 44,352 for the bike, then at what price did Sonu bought the bike?

(4 marks)

Topic: Concept of Discount

12. Sheena bought artificial jewellery set for Rs 3800. If the shop owner gave her a discount of 5%, then find the marked price of the set.

(2 marks)

13. During a sale, a shop offered a discount of 25% on each suit price and a discount of 35% on each saree. Vaishali purchased three suits marked at Rs 4500 each and a saree marked at Rs 12500. Find the amount that Vaishali will have to pay in order to buy the three suits and the saree.

(3 marks)

Topic: Simple Interest and Compound Interest I

14. Arvind lends Rs 30,000 to Astha for 2 years at the rate of 5% per annum on simple interest. How much more interest would he get if he lends the same amount at the same rate of interest for 2 years, compounded annually?

( 3 marks)

15. Mohit took a loan of Rs 90,000 from his friend at the rate of 5% per annum, compounded annually for 3 years. What amount would Mohit pay at the end of the 2 nd year?

(3 marks)

16. Shivam takes a loan of Rs 80,000, out of which he receives Rs 40,000 at the rate of 5% per annum compounded annually and Rs 40,000 at the rate of 6% per annum on simple interest. How much money will he return after two years?

(3 marks)

17. A car was bought at Rs 300000. Its value depreciated at the rate of 7% per annum. Find its value after 2 years.

(2 marks)

18. Rajat invested Rs 12,000 for two years and received Rs 14,520 at the end of two years. If the interest is compounded annually, then at what rate of interest had he invested the money?

(3 marks)

19. If Rs 8000 is invested at 15% p.a. compounded annually, then after how many years will it amount to Rs 12167?

(3 marks)

Comparing Quantities Topic: Simple Interest and Compound Interest II

20. Daman took a loan of Rs 27000 for years from a bank. The bank2 2

1 charged him an interest at the rate of 8% per annum compounded half yearly. Find the amount which Daman will have to repay.

(3 marks)

Comparing Quantities Solutions

Topic: Percentage

1. Original number = 12000 Percentage increase = 37% 37% of 12000

= 37 × 120 = 4440 New number = Old number + Increase = 12000 + 4440 = 16440 Thus, the new number is 16440.

2. Original price of car = Rs 3, 70, 000 Percentage decrease = 14%

∴New price = Old price − Decrease = Rs 3, 70,000 − Rs 51,800 = Rs 3,18,200 Thus, after a year, the price of the car got reduced to Rs 3,18,200.

3. Cost of the ticket in the previous month = Rs 15 Present cost of the ticket = Rs 20 ∴ Increase in the cost of the ticket = Rs (20 − 15) = Rs 5

Topic: Sales Tax and Value Added Tax

4. M.P. of the car = Rs 2,10,000 Discount offered = 5%

Price after discount = Rs 2,10,000 − Rs 10,500 = Rs 1,99,500

Comparing Quantities Rate of sales tax = 10%

∴ Amount to be paid = Rs 1,99,500 + Rs 19,950 = Rs 2,19,450 Thus, Reema paid Rs 2,19,450 for the car.

5. Let the initial bill amount without tax be x . The VAT is charged at the rate of 12.5%. ∴ x + 12.5% of x = Rs 16,875

Thus, the initial amount of bill without VAT is Rs 15,000.

6. It is given that the selling price of the computer is Rs 8,640. Let the marked price of the computer be Rs x .

S.P. = M.P. + Sales tax

⇒ x = 8,000 Thus, the marked price of the computer was Rs 8,000.

7. Let candidate B had got x votes. ∴ Number of votes got by candidate A = x + 12% of x

Comparing Quantities

∴ Percentage of votes got by B as compared to A

Thus, candidate B got 10.71% less votes than A.

Topic: Concept of Profit Percent and Loss Percent

8. S.P. of laptop = Rs 69000 Profit % = 15% It means that if C.P. is Rs 100, then profit is Rs 15. ∴ S.P. = Rs 100 + Rs 15 = Rs 115 When S.P. is Rs 115, C.P. = Rs 100

Thus, the cost price of the laptop is Rs 60000.

9. Let C.P. be the cost price of the handbag. Selling price of the handbag, S.P. = Rs 2,400 Loss percent = 4%

Comparing Quantities ∴ S.P. = C.P. − Loss

Thus, the cost price of the handbag was Rs 2,500.

10. C.P. of 16 dozen bananas = Rs 240

Loss percent = 10%

∴ S.P. of 6 dozen bananas = Rs 90 − Rs 9 = Rs 81

∴ S.P. of 6 dozen bananas = Rs 150 + Rs 7.50 = Rs 157.50 ∴ Total S.P. = Rs 81 + Rs 157.50 = Rs 238.50 ∴ Overall loss = Rs 240 − Rs 238.50 = Rs 1.50

Thus, the overall loss percent is 0.625%.

11. When Dimpu sold the bike to jammy: Selling price (S.P.) of the bike = Rs 44,352 Let x be the cost price of the bike, i.e. the price at which Dmipu bought the bike. Loss percent = 20%


S.P. = C.P. − Loss

⇒ Rs 44,352= x − xError ⇒ Rs 44,352 = x5

4

Thus, Dimpu bought the bike for Rs 55,440. When Sonu sold the bike to Dimpu: Selling price (S.P.) = Rs 55,440 Let y be the price at which Sonu bought the bike. ∴ Cost price (C.P.) = y Profit percent = 10%

Profit = S.P. − C.P.

⇒ y = Rs 55,440 − y110

⇒ y + y = Rs 55,440110

⇒ y = Rs 55,4401011

⇒ y = Rs 50,400 Thus, Sonu bought the bike for Rs 50,400.

Topic: Concept of Discount

12. Let the market price ( MP ) of the set be Rs x . We know that Discount = Marked price − Selling price ∴ Discount = Rs x − Rs 3800 (1)

Comparing Quantities Thus, from (1), we have

13. Marked price ( MP ) of each suit = Rs 4500 Discount percentage = 25% We know that:

∴Selling price( SP ) of each suit = MP − Discount = Rs 4500 − Rs 1125 = Rs 3375 ∴Selling price of 3 suits = Rs 3375 × 3 = Rs 10125 Marked price ( MP ) of saree = Rs 12500 Discount percentage = 35% Therefore, we have,

∴ SP of saree = MP − Discount = Rs 12500 − Rs 4375 = Rs 8125 ∴ SP of the three suits and the saree = Rs 10125 + Rs 8125 = Rs 18250 Thus, Vaishali will have to pay Rs 18, 250 in order to buy the three suits and the saree.

Topic: Simple Interest and Compound Interest I

14. It is given that Principal = Rs 30,000 Rate of interest = 5% per annum Time = 2 years


and Amount = P + I Now,

For compound interest,

Amount at the end of 1 st year = Rs 30,000 + Rs 1,500 = Rs 31,500

Total interest at the end of two years = Rs 1500 + Rs 1575 = Rs 3075 Hence, the additional amount that Arvind would get is Rs (3075 − 3000) = Rs 75.

15. Principal( P ) = Rs 90,000 Rate of interest( R ) = 5% p.a. Time( T ) = 3 years

and Amount = P + I

Amount at the end of 1 st year = Rs 90,000 + Rs 4500 = Rs 94,500

Amount at the end of 2 nd year = Rs 94500 + Rs 4725 = Rs 99,225 Hence, Mohit would pay Rs 99,225 at the end of the second year.

16. Amount on compound interest:

Comparing Quantities Principal( P ) = Rs 40000 Rate of interest( R ) = 5% p.a.

and Amount = P + I

Amount at the end of 1 st year = Rs 40,000 + Rs 2000 = Rs 42,000 Now, Interest for 2 nd year

= Rs 2100 Amount at the end of 2 nd year = Rs 42,000 + Rs 2,100 = Rs 44,100 Amount on simple interest: Principal( P ) = Rs 40,000 Rate of interest( R ) = 6% p.a. Time( T ) = 2 years Simple Interest

= Rs 4800 Amount at the end of two years = Rs 40,000 + Rs 4800 = Rs 44,800 Thus, total amount that Shivam will return at the end of two years = Rs 44,100 + Rs 44,800 = Rs 88,900.

17. P = Rs 300000 r = 7% n = 2


Thus, the value of the car will be Rs 259470 after 2 years.

18. It is given that: Principal( P ) = Rs 12,000 Amount( A ) = Rs 14,520 Time( n ) = 2 years Let the rate of compound interest be r .

Thus, Rajat had invested the money at 10% rate of interest.

19. Let the time period be n years. Principal ( P ) = Rs 8000 Amount ( A ) = Rs 12167 Rate of interest ( R ) = 15% p.a.

Comparing Quantities It is known that amount received is given by

Thus, the sum would amount to Rs 12167 after 3 years.

Topic: Simple Interest and Compound Interest II

20. Here, principal, P = Rs 27000

Also, there are 5 half years in years.2 21

Now,

Thus, Daman will have to pay Rs 32, 849.6 back to the bank at the end of years.2 21

Algebraic Expressions and Identities Important Questions

Topic: Multiplication of Monomials and Polynomials

1. Simplify the expression p ( p − 1) − p ( p + 1) + 2( p + 1). (2 marks)

2. Add the following: 4 a ( a − b + c ), c (2 a + cb ), and 2 b (5 a − c 2 )

(2 marks)

3. If the base of a triangle is 5 units more than thrice the height of the triangle, then what is the area of the triangle?

(2 marks)

4. Simplify: ( a + b )( a 2 + b 2 ) − a 2 ( a + 3 b ) − b 2 (3 a + b ) (2 marks)

5. Subtract 5 p ( p + q ) + ( p 2 + 1) ( q + 2) from (2 + q )( p − p 2 ) + 3 p ( q + 1). (2 marks)

6. In a cupboard, there were two rectangular boxes A and B. The length, breadth, and height of the box A were l , b , and h respectively. The length of box B was double than that of box A. The breadth of box B was 3 cm more than the breadth of box A and the height of box B was 2 cm more than three times the height of box A. Find the volume of box B in terms of l , b , and h .

(2 marks)

Topic: Algebraic Expressions and Identities I

7. Simplify the following expressions: (a) (1.6 a + 2.5 b ) 2

(2 marks)

8. If (3 x 2 − 5) 2 + (2 x 2 + 2) 2 = px 4 + q + 2 × r × x 2 , then what is the value of p × q × r ?

(2 marks)

9. Prove that: (2 a 4 − 3 b 2 ) 2 + 24 a 4 b 2 = (2 a 4 + 3 b 2 ) 2

(3 marks)

Algebraic Expressions and Identities Topic: Algebraic Expressions and Identities II

10. Simplify the expression ( a − b + 2)( a − b + 4). (2 marks)

11. Simplify the expression (3 x + 2) (3 x − 3) + 9 ( x 2 − 6) and then evaluate for

(3 marks)

12. If the length, breadth, and height of a cuboid are (2 x 2 + y 2 − 4 x ), (2 x 2 − 3 x + y 2 ) and 1 respectively, then what is the volume of the cuboid?

(3 marks)

13.

(2 marks)

14. Factorize the expression 9 m 3 − 27 m 2 − 16 m + 48. (2 marks)

15.

(3 marks)

Algebraic Expressions and Identities Solutions

Topic: Multiplication of Monomials and Polynomials

1. p ( p − 1) − p ( p + 1) + 2( p + 1) = p × p − p × 1 − p × p − p × 1 + 2 × p + 2 × 1 = p 2 − p − p 2 − p + 2 p + 2 = ( p 2 − p 2 ) + (− p − p + 2 p ) + 2 = 2

2.

3. Let the height of the triangle be x . ∴ Base of the triangle = 3 x + 5

Algebraic Expressions and Identities 4.

5.

Now, we subtract (1) from (2).

Algebraic Expressions and Identities 6. According to the given information,

Length of box B = 2 l cm Breadth of box B = ( b + 3) cm Height of box B = (2 + 3 h ) cm We know that volume of a cuboid = Length × Breadth × Height ∴ Volume of box B = 2 l × ( b + 3) × (2 + 3 h ) = (2 l × b + 2 l ×3) ×(2 + 3 h ) = (2 lb + 6 l ) ×(2 + 3 h ) = 2 lb × 2 + 2 lb × 3 h + 6 l × 2 + 6 l × 3 h = 4 lb + 6 lbh + 12 l + 18 lh

Topic: Algebraic Expressions and Identities I

7. (a) (1.6 a + 2.5 b ) 2

8. The given expression is: (3 x 2 − 5) 2 + (2 x 2 + 2) 2 = px 4 + q + 2 × r × x 2

It can be simplified as: (3 x 2 − 5) 2 + (2 x 2 + 2) 2 = px 4 + q + 2 × r × x 2

⇒ {(3 x 2 ) 2 − 2 × 3 x 2 × 5 + (5) 2 } + {(2 x 2 ) 2 + 2 × 2 x 2 × 2 + (2) 2 } = px 4 + q + 2 × r × x 2 {Using the identities ( a − b ) 2 = a 2 − 2 ab + b 2 and ( a + b ) 2 = a 2 + 2 ab + b 2 } ⇒ 9 x 4 − 30 x 2 + 25 + 4 x 4 + 8 x 2 + 4 = px 4 + q + 2 × r × x 2 ⇒ 13 x 4 − 22 x 2 + 29 = px 4 + q + 2 × r × x 2 Comparing the coefficients of different powers of x from both sides: p = 13, q = 29, and r = −11

Algebraic Expressions and Identities ∴ p × q × r = 13 × 29 × (−11) = − 4147 Thus, the value of p × q × r is −4147.

9. The expression (2 a 4 − 3 b 2 ) 2 + 24 a 4 b 2 can be simplified as: (2 a 4 − 3 b 2 ) 2 + 24 a 4 b 2 = {4 a 8 − 2 × 2 a 4 × 3 b 2 + 9 b 4 } + 24 a 4 b 2 {Using the identity ( a − b ) 2 = a 2 − 2 ab + b 2 } = 4 a 8 − 12 a 4 b 2 + 9 b 4 + 24 a 4 b 2 = 4 a 8 + 9 b 4 + 12 a 4 b 2 The expression (2 a 4 + 3 b 2 ) 2 can be simplified as: (2 a 4 + 3 b 2 ) 2 = (2 a 4 ) 2 + 2 × 2 a 4 × 3 b 2 + (3 b 2 ) 2 {Using the identity ( a + b ) 2 = a 2 + 2 ab + b 2 } = 4 a 8 + 12 a 4 b 2 + 9 b 4 ∴ (2 a 4 − 3 b 2 ) 2 + 24 a 4 b 2 = (2 a 4 + 3 b 2 ) 2 Hence, proved

Topic: Algebraic Expressions and Identities II

10. The given expression can be simplified as: ( a − b + 2) × ( a − b + 4) = {( a − b ) + 2} × {( a − b ) + 4} = ( a − b ) 2 + (2 + 4) ( a − b ) + 2 × 4 {Using the identity ( x + a ) ( x + b ) = x 2 + ( a + b ) x + ab } = a 2 − 2 ab + b 2 + 6( a − b ) + 8 = a 2 − 2 ab + b 2 + 6 a − 6 b + 8

11. The given expression can be simplified as: (3 x + 2) (3 x − 3) + 9 ( x 2 − 6) = (3 x ) 2 + {2 + (−3)} × 3 x + (2) (−3) + 9 x 2 − 54 {Using the identity ( x + a ) ( x + b ) = x 2 + ( a + b ) x + ab } = 9 x 2 − 3 x − 6 + 9 x 2 − 54 = 18 x 2 − 3 x − 60

Algebraic Expressions and Identities

12. It is given that: Length of the cuboid, l = (2 x 2 + y 2 − 4 x ) Breadth of the cuboid, b = (2 x 2 − 3 x + y 2 ) Height of the cuboid, h = 1 It is known that the volume of a cuboid is given by l × b × h . ∴ Volume of the cuboid = (2 x 2 + y 2 − 4 x ) × (2 x 2 − 3 x + y 2 ) × 1 = {(2 x 2 + y 2 ) − 4 x } × {(2 x 2 + y 2 ) − 3 x } = (2 x 2 + y 2 ) 2 + (−4 x − 3 x ) (2 x 2 + y 2 ) + (−4 x ) × (−3 x ) {Using the identity ( x + a ) ( x + b ) = x 2 + ( a + b ) x + ab } = 4 x 4 + y 4 + 4 x 2 y 2 − 7 x (2 x 2 + y 2 ) + 12 x 2 = 4 x 4 + y 4 + 4 x 2 y 2 − 14 x 3 − 7 xy 2 + 12 x 2 Thus, the volume of the cuboid is (4 x 4 + y 4 + 4 x 2 y 2 − 14 x 3 − 7 xy 2 + 12 x 2 ) cubic units.

13.

14.

15. Consider the expression

Algebraic Expressions and Identities

It can be observed that it is same as the other expression. Thus, the given expressions are equal.

Visualising Solid Shapes Important Questions

Topic: Visualising Solid Shapes I

1. The given figure shows the map of a colony:

Based on the given map, answer the following questions: (a) Which landmark lies on the extreme NorthWest side? (b) How many temples are shown in the given map?

(2 mark)

Visualising Solid Shapes 2.

What is the actual distance between Kalkaji and M Block market?

(2 marks)


(a) What is the shortest distance that one has to travel to reach the nearest taxi stand from the Metro Station? (b) What is the shortest distance between Ashoka garden and Crafts shop?

(2 marks)


The given figure shows the map of a college campus.

(a) Among the boys’ hostel and the playground, which one is at the North of the boys’ common room? (b) Which landmark is present in between the playground and the college park? (c) In which direction of commerce block is the cafeteria located? (d) Among the landmarks playground and garden, which one is Southernmost?

(4 marks)


Classify the following solids into convex and concave polyhedrons. (a)

(b)

(c)

(d)

(4 marks)

Visualising Solid Shapes

Topic: Visualising Solid Shapes II

6.

Is the given figure a pyramid? Explain.

(1 mark)

7. How can the given polyhedrons be classified as pyramid or prism? Also, name them. (a)

(b)

(2 marks)


8. Classify the following figure as prism or pyramid. (a)

(b)

(c)

(d)

(4 marks)


9. The following figure shows a solid.

Verify Euler’s formula for the given solid.

(2 marks)

10. For a prism, the number of faces is threefourth times the number of its vertices. If the number of vertices and edges are in the ratio 2: 3, then what is the number of edges of the prism?

(3 marks)

Visualising Solid Shapes Solutions

Topic: Visualising Solid Shapes I

1. (a) The landmark that lies on the extreme NorthWest side is the Mall. (b) There are two temples shown in the map.

2. It is observed that scale used in the map is 1 cm = 800 m. The distance between Kalkaji and M Block market on the map is 8.3 cm. Thus, actual distance between Kalkaji and M Block market = 8.3 × 800 m = 6640 m = 6.640 km

3. (a) It can be seen in the following figure that the distance of different taxi stands from the Metro Station are as follows:

Distance of taxi stand 1 from Metro Station = 9 km Distance of taxi stand 2 from Metro Station = 11 km Distance of taxi stand 3 from Metro Station via route 1 = 13 km Distance of taxi stand 3 from Metro Station via route 2 = 13 km Thus, the shortest distance that one has to travel to reach the nearest taxi stand from the Metro

Visualising Solid Shapes Station is 9 km. (b) There are two possible routes between Ashoka Garden and the Crafts shop. They are shown as:

Distance between Ashoka Garden and Crafts shop via route 1= 14 km Distance between Ashoka Garden and Crafts shop via route 2 = 14 km Thus, the shortest distance between Ashoka garden and Crafts shop is 14 km.

4. (a) It is observed that among the boys’ hostel and the playground, playground is at the North of the boys’ common room. (b) Science block is located in between the playground and college park. (c) The cafeteria is located in the NorthWest direction of the commerce block. (d) Among the landmarks playground and garden, garden is Southernmost.

5. A polyhedron is said to be concave if at least one line joining two points on the polyhedron lies outside the polyhedron. A polyhedron is said to be convex if all the lines joining any two points on the polyhedron always lie on the polyhedron.

Visualising Solid Shapes (a) In the given polyhedron, the line joining two points on the polyhedron lies outside the polyhedron which can be shown as:

Thus, the given polyhedron is a concave polyhedron. (b) All the lines formed by joining any two points on the polyhedron always lie on the polyhedron. Thus, the given polyhedron is a convex polyhedron. (c) In the given polyhedron, the line joining two points on the polyhedron lies outside the polyhedron which can be shown as:

Thus, the given polyhedron is a concave polyhedron. (d) All the lines formed by joining any two points on the polyhedron always lie on the polyhedron. Thus, the given polyhedron is a convex polyhedron.

Topic: Visualising Solid Shapes II

Visualising Solid Shapes 6. Pyramid is a polyhedron in which the base is a polygon and the lateral faces are triangles that

meet at a common vertex. Here the base is a circle, which is not a polygon. Thus, the given figure does not represent a pyramid.

7. Prism is a polyhedron whose base and top are congruent polygons and other faces are parallelogram in shape. Pyramid is a polyhedron whose base is a polygon and other faces are triangles having a common vertex. A prism or a pyramid is named after its base. (a) The given figure consists of two congruent pentagonal faces. The other faces of the polyhedron are in this shape of parallelogram. Thus, the given polyhedron is a prism. Since the base of the prism is a pentagon, the given figure can be named as pentagonal prism. (b) The given polyhedron consists of one hexagon (polygon of 6 sides). The remaining faces of the polyhedrons are triangles. Thus, the given polyhedron is a pyramid. Since the base of the pyramid is hexagon, it can be named as hexagonal pyramid.

8. We know that a prism is a polyhedron whose base and top are congruent polygons and whose lateral faces are parallelogram in shape. A pyramid is a polyhedron whose base is a polygon (of any number of sides) and whose lateral faces are triangles with a common vertex. The given figures are: (a) Prism (b) Pyramid (c) Pyramid (d) Prism

9. It is known that according to Euler’s formula for polyhedron, F + V − E = 2, where F , V and E are the numbers of faces, vertices and edges of the polyhedron. It can be seen in the given figure that, Number of faces, F = 7 Number of vertices, V = 10

Visualising Solid Shapes Number of edges, E = 15 Consider LHS of F + V − E = 2. 7 + 10 − 15 = 17 − 15 = 2 = RHS Thus, the Euler’s formula is verified for the given solid.

10. Let the number of vertices be x . ∴ Number of faces

The number of vertices and edges are in the ratio 2: 3 ∴ Number of edges

According to Euler’s formula for polyhedron, F + V − E = 2, where F , V and E are the numbers of faces, vertices and edges of the polyhedron.

⇒ x = 8 ∴ Number of edges

Thus, there are 12 edges in the prism.

Direct and Inverse Proportions Important Questions

Topic: Direct Proportion

1. The radius and circumference of a circle are in direct proportion. Explain.

(2 marks)

2. A typist can type 1600 words in 40 minutes. How many words can he type in half an hour?

(2 marks)

3. The cost of 2 m of a cloth is Rs 84. Using the following table, find the value of b + c − 5 a .

Length of cloth(in m) 2

4.5 5 c + 1

Cost(in Rs) 84 105 3 b 252

(3 marks)

4. A worker earns Rs 1860 in 12 days.What amount of money can he earn in 20 days and in how many days, can he earn Rs 6045? [Assume that the amount of money earned each day is same.]

(3 marks)

5. A printer can print at a constant rate of 8 prints a minute. Find (a) time taken by the printer to print 36 prints (b) number of prints the printer can print in 6 minutes 15 sec

(3 marks)

Topic: Inverse Proportion

6. State whether the two quantities involved in each of the following cases are in direct proportion or inverse proportion. Justify your answer. (a) Number of days required to complete a work and the time spent each day (b) Number of sides of a regular polygon and measure of its exterior angles (c) Number of people engaged in construction of a road and the length of the constructed road, assuming that the work rate for each person is same

(3 marks)

Direct and Inverse Proportions 7. 25 men were assigned to do a work in 40 days. However, due to some

reasons, 5 men did not turn up for the work. How many more days are required by the remaining men to complete the work?

(2 marks)

8. Observe the following tables and classify the variations in x and y in each of them as direct proportion or inverse proportion. (a)

X 1 2 3 6 15

Y 900 450 300 150 60

(b)

X 2 4 7 25 273

Y 5 9 15 51 547

(c)

X 10 40 90 120 230

Y 25 100 225 300 575

(3 marks)

9. A farmer has enough food to feed 64 cows for 12 days. How long will the food last, if onefourth the number of cows are sold, and are therefore not needed to be fed?

(3 marks)

10. Six women or four men can complete a job in 12 days. How long will four men and three women together take to complete the same job?

(3 marks)

Direct and Inverse Proportions Solutions

Topic: Direct Proportion

1. Circumference ( c ) of a circle with radius ( r ) is given by, c = 2π r

Since 2 and π are constants,

This shows that radius and circumference of a circle are in direct proportion.

2. Suppose that the typist can type x words in half an hour i.e., 30 minutes. The given information can be represented using a table as:

Number of words 1600 x

Time taken to type 40 30

If the words increase, then the time taken to type those words will also increase. Thus, this is a case of direct proportion. Hence,

Hence, the typist can type 1200 words in half an hour.

Direct and Inverse Proportions 3. Cost of cloth and its length are in direct proportion.

Now, b + c − 5 a = 63 + 1 − 5 × 5 = 64 − 25 = 39

4. Let the amount of money that can be earned in 20 days be Rs x and the number of days in which the worker can earn Rs 6045 be y days. The given information can be tabulated as

Number of days 12 20 y

Amount of money earned (in Rs) 1860 x 6045

It can be seen that more the number of days the worker works, more will be the wages earned by him. Therefore, these two quantities are directly proportional to each other.

Here,

Thus, the worker can earn Rs 3100 in 20 days.

Thus, in 39 days, the worker can earn Rs 6045.

5. We know that 1 minute = 60 sec ∴ 6 minutes 15 sec = (6 × 60 + 15) sec = 375 sec

Direct and Inverse Proportions Let time taken by the printer to print 36 prints be x sec and the number of prints that can be printed in 6 minutes 15 seconds i.e., 395 seconds, be y .

Number of prints 8 36 y

Time taken (in sec) 60 x 375

Since the printing rate is constant, the number of prints and time taken to print these prints are in direct variation. (a) We have,

Hence, 36 prints will be printed in 270 seconds i.e., 4 minutes 30 seconds. (b) We have,

Hence, 50 prints will be printed in 6 minutes 15 seconds.

Topic: Inverse Proportion

6. (a) To complete a piece of work, if the number of days is increased, then time taken to complete the work decreases. Thus, this is a case of inverse proportion. (b) If the number of sides of a regular polygon is n , then its each exterior angle is of measure

. It can be observed that: Number of sides in a polygon × Measure of each exterior angle

Direct and Inverse Proportions

= 360° = Constant Thus, this is a case of inverse variation. (c) Keeping the rate of work of each person constant, if the number of people is increased, then they will construct a longer road. Thus, this is a case of direct proportion.

7. Let the number of days required to complete the work be x . It can be seen that lesser the number of workers, more will be the time taken by them to complete the work. Therefore, it is a case of inverse proportion.

Number of workers 25 20

Days 40 x

∴ 25 × 40 = 20 × x ⇒ x = 50 Thus, (50 − 40) = 10 more days are required to complete the work.

8. (a) It can be observed that: 1 × 900 = 2 × 450 = 3 × 300 = 6 × 150 = 15 × 60 = 900 (constant) This means that x varies with y according to the relation: xy = constant Hence, this is a case of inverse proportion. (b) It can be observed that:

Or, 2 × 5 ≠ 4 × 9 ≠ 7 × 15 ≠ 25 × 51 ≠ 273 × 517

This means x neither varies with y according to the relation = constant nor accordingError to the relation xy = constant Hence, this is neither a case of direct proportion nor of inverse proportion. (c) It can be observed that:

which is a constant This means that x varies with y according to the relation:


= constantError

9. Let the food last for t days, if onefourth the number of cows is sold. Original number of cows = 64 Number of cows sold

Number of cows remaining = 64 − 16 = 48 The information can be represented in a table as:

Number of cows 64 48

Number of days food will last 12 t

Lesser the number of cows, more the number of days the food will last. Thus, the two quantities are in inverse proportion.

Thus, if onefourth of the cows are sold, then the food will be sufficient to feed the remaining cows for 16 days.

10. It is given that 6 women are capable of doing the same work as 4 men. ∴ 6 women = 4 men

Let 4 men and 3 women together take d days to complete the job.


∴ 4 men + 3 women = 4 men + 2 men = 6 men This information can be tabulated as:

Number of men 4 6

Number of days 12 d

More men will mean lesser days to complete the same job. Therefore the two quantities are inversely proportional.

Thus, 4 men and 3 women together will take 8 days to complete the same job.

Factorisation Important Questions

Topic: Factorisation of Algebraic Expressions I

1. What are the common factors of the terms 36 a 2 b 5 , 54 a 3 b 4 , and 81 a 4 b 6 ? (2 marks)

2. Factorise the following expressions: (a) 14 pq + 21 pq 3

(b) 2 abxy + 14 a 3 bx 2 y + 7 x 2 y 3 a 2 b 3 − 2 a 2 b 2 x 2 y 2 (c) 16 x + 32 y + 8 xy

(3 marks)

3. Factorise the expression 9 m 3 − 27 m 2 − 16 m + 48. (2 marks)

4. How can the expression m x + y + m x n y − m y n x − n x + y be factorized? (2 marks)

Topic: Factorisation of Algebraic Expressions II

5. If 997 × a = 1003 × 997 − 1000 2 + 2 × 1000 × 3 − 3 2 , then how can the value of a be calculated without actually multiplying the numbers?

(2 marks)

6. If 16 x 4 − 81 y 4 = 28(4 x 2 + 9 y 2 ) and 2 x + 3 y = 14, then what is the value of the expression 4 x 2 − 12 xy + 9 y 2 ?

(3 marks)

7. Factorise the following expressions: (a) x 2 − 10 x + 21 (b) a 2 + 15 a + 56

(2 marks)

8. The volume of a cuboid is 2 x 2 y + 2 xy − 84 y . What is its total surface area?

(3 marks)

Topic: Division of Polynomials I

9. What is the expression obtained on dividing −80 x 3 y 4 z 5 by −5 xy 2 z 3 ? (1 mark)

10. Divide the following expressions: (a) 49 x 2 y 3 z 4 ÷7 xy 2 (b) (25 abp 2 − 35 a 2 bq 2 ) ÷ (− 5 ab ) (c) (46 a 2 b − 32 ab 2 + 2 ab ) ÷ (2 ab )

(3 marks)

Factorisation 11. Manisha spent Rs (15 x 2 + 3 xy ) on buying 3 x toffees. What is the cost of

each toffee? (2 marks)

Topic: Division of Polynomials II

12. What is the quotient obtained when 51 a 3 (72 b 2 − 242) is divided by 34 a 2 (6 b + 11)?

(2 marks)

13. If a cm, b cm, and c cm respectively are the length, breadth, and height of a cuboidal box such that its total surface area is 288 cm 2 and volume

is 288 cm 3 , then what is the value of the expression ?a1 + b

1 + c1

(3 marks)

14. How can the error(s) in the equation (6 x 3 − 5) 2 = 36 x 6 − 25 be corrected? (2 marks)

15. Find and correct the error in the following mathematical expressions:

(b) ( a − 9) 2 = a 2 + 81 + 18 a (c) (2 x 3 ) 2 + 9 = 2 x 6 + 9 (d) 6 q + 19 q = 25 q 2

(4 marks)

Factorisation Solutions

Topic: Factorisation of Algebraic Expressions I

1. The given terms can be factorised as: 36 a 2 b 5 = 2 × 2 × 3 × 3 × a × a × b × b × b × b × b 54 a 3 b 4 = 2 × 3 × 3 × 3 × a × a × a × b × b × b × b 81 a 4 b 6 = 3 × 3 × 3 × 3 × a × a × a × a × b × b × b × b × b × b Thus, the common factors of the given terms are 3 × 3, a × a , b × b × b × b i.e., 9, a 2 , and b 4 .

2. (a) 14 pq + 21 pq 3

(b) 2 abxy + 14 a 3 bx 2 y + 7 a 2 b 3 x 2 y 3 − 2 a 2 b 2 x 2 y 2

(c) 16 x + 32 y + 8 xy

Factorisation 3.

4. The given expression is m x + y + m x n y − m y n x − n x + y

Thus, the factorized form of the given expression is ( m x − n x ) ( m y − n y ).

Topic: Factorisation of Algebraic Expressions II

5. It is given that: 997 × a = 1003 × 997 − 1000 2 + 2 × 1000 × 3 − 3 2

⇒ 997 × a = 1003 × 997 − (1000 2 − 2 × 1000 × 3 + 3 2 ) ⇒ 997 × a = 997 × 1003 − (1000 − 3) 2 {Using the identity a 2 − 2 ab + b 2 = ( a − b ) 2 } ⇒ 997 × a = 997 × 1003 − 997 2 ⇒ 997 × a = 997 × 1003 − 997 × 997 ⇒ 997 × a = 997 × (1003 − 997) ⇒ 997 × a = 997 × 6 Dividing both sides by 997: ⇒ a = 6 Thus, the value of a is 6.

Factorisation 6. It is given that: 16 x 4 − 81 y 4 = 28(4 x 2 + 9 y 2 )

Thus, the value of the expression 4 x 2 − 12 xy + 9 y 2 is 4.

7. (a) Comparing x 2 − 10 x + 21 with x 2 + ( a + b ) x + ab , we obtain

(b) Comparing a 2 + 15 a + 56 with a 2 + ( x + y ) a + xy , we obtain

8. It is known that the volume of a cuboid is given by l × b × h . ∴ l × b × h = 2 x 2 y + 2 xy − 84 y Therefore, to find the dimensions of the cuboid the expression 2 x 2 y + 2 xy − 84 y needs to be factorized. This can be done as:

Factorisation

Therefore, the dimensions of the cuboid are 2 y × ( x − 6) × ( x + 7). It is known that the surface area of a cuboid is given by, 2( lb + bh + hl ). ∴ Surface area of the cuboid = 2 {2 y ( x − 6) + ( x − 6) ( x + 7) + 2 y ( x + 7)}

Thus, the surface area of the cuboid is sq units.

Topic: Division of Polynomials I

9.

Thus, the expression obtained on dividing −80 x 3 y 4 z 5 by −5 xy 2 z 3 is 16 x 2 y 2 z 2 .

10. (a) 49 x 2 y 3 z 4 ÷ 7 xy 2

(b) (25 abp 2 − 35 a 2 bq 2 ) ÷(− 5 ab )

Factorisation

(c) (46 a 2 b − 32 ab 2 + 2 ab ) ÷ (2 ab )

11. Amount spent on toffees = Rs (15 x 2 + 3 xy ) Number of toffees purchased = 3 x Cost of each toffee

Thus, the cost of each toffee is Rs (5 x + y ).

Topic: Division of Polynomials II

12. 51 a 3 (72 b 2 − 242) can be divided by 34 a 2 (6 b + 11) as:

Factorisation

Thus, the quotient obtained is 6 a (6 b − 11)

13. The total surface area ( S ) and volume ( V ) of a cuboid are given by: S = 2( ab + bc + ca ) V = abc According to the given information: 2( ab + bc + ca ) = 288 and abc = 288

Thus, the value of the expression i s .a1 + b

1 + c1 Error

14. The expression (6 x 3 − 5) 2 can be simplified as: (6 x 3 − 5) 2 = (6 x 3 ) 2 − 2(6 x 3 ) (5) + (5) 2 {Using the identity ( a − b ) 2 = a 2 − 2 ab + b 2 } = 36 x 6 − 60 x 3 + 25 Thus, the given equation can be corrected as (6 x 3 − 5) 2 = 36 x 6 − 60 x 3 + 25

Factorisation 15. (a) When the division and dividend are same, the quotient is always 1.

(b) ( a − 9) 2 = a 2 − 18 a + 81 {Using the identity ( a − b ) 2 = a 2 − 2 ab + b 2 } (c) (2 x 3 ) 2 + 9 = 4 x 6 + 9 (d) 6 q + 19 q = 25 q

Introduction to Graphs Important Questions

Topic: Line Graphs

1. The following table represents the radii of circle and their corresponding areas:

Radii (cm) 3.5 7 10.5 14

Circumference (cm)

22 44 66 88

Draw a line graph for the given data.

(2 marks)

2. Sixty students entered their school library at 10:00 am to search for their required study material. Sixteen students were able to find the required study material at 10:05 am, ten students at 10:10 am, sixteen students at 10:15 am, eight students at 10:20 am, six students at 10:25 am, and the remaining students at 10:30 am. How can the given information be represented in form of line graph?

(3 marks)

Introduction to Graphs 3. The following line graph shows the number of people in a particular store at various

times of a day.

Answer the following questions based on the above line graph. (a) What was the busiest time of that day at the store? (b) At what time did business start to slow down? (c) How many people were in the store when it opened? (d) About how many people were there in the store at around 2:30 pm? (e) What was the greatest number of people in the store? (f) What was the least number of people in the store?

(3 marks)

Introduction to Graphs 4. The given line graph shows the number of dolphins found in a particular channel

during a period of six years.

(a) What is the percentage increase or decrease in the number of dolphins in the channel from 1996 to 2001? (b) In which year (s), the number of dolphins is less than the average number of dolphins in the span of six years?

(3 marks)

Introduction to Graphs Topic: Position of a Point on a Graph

5. Observe the following graph and answer the questions:

(a) What are the coordinates of point D? (b) Which points are located by the points (1, 5), (10, 0), (4, 9)?

(2 marks)

6. Plot the points A (3, 6) and B (6, 3) on a coordinate plane. (2 marks)

7. What is the area of the figure obtained by plotting and joining the points (0, 2), (0, 6), and (4, 2) on a graph?

(3 marks)

8. Check whether the points (0, 0), (4, 3), (6, 4.5) and (8, 6) are collinear or not. (2 marks)

Topic: Application of Line Graphs

9. The volume of a cylinder is given by the formula V = π r 2 h , where r is the radius of its base and h is the height of the cylinder. Find the dependent and independent variables among v, r ,and h .

(2 marks)

Introduction to Graphs 10. The given graph shows the cost of CNG (Compressed Natural Gas) in a particular

city corresponding to its quantity.

(a) Mr. Verma’s car is refilled with 7.5 kg of CNG. How much will it cost him to get it refilled with the CNG? (b) How much amount of CNG can be refilled with an amount of Rs 300?

(3 marks)

11. The following table provides the information regarding the simple interest obtained on the given sums, each deposited for a year.

Deposit(in Rs) 4000 8000 12000 16000

Simple interest(in Rs) for a year

100 200 300 400

Draw a graph for the above data by choosing suitable scales for the axes.

(2 marks)


A financial company offers 20% simple interest (S.I.) on deposits under its special scheme.

Draw a graph to illustrate the relation between the sum deposited and simple interest earned. Then from the graph, find

(3 marks)

Introduction to Graphs (a) the annual interest obtainable on deposit of Rs 4500 (b) the investment one has to make to obtain an annual simple interest of Rs 700

13. Roshni is walking around the boundary of a rectangular park of dimensions 10 m × 40 m at uniform speed. She spends 20 minutes to complete one round around the park. (a) How can the given information be represented on a timedistance graph? (b) If she walked for 30 minutes on a particular day, then how can the number of rounds completed by her be found from the obtained graph?

(4 marks)

14. (a) Draw a graph representing the relationship between simple interest earned and time period for the sum of Rs 12,000 deposited by Meena. (b) If Meena had deposited the mentioned sum for 4 years, then how can the simple2

1 interest earned by her be found from the obtained graph? (c) If Meena earns Rs 4,500 as simple interest, then how can the duration of time for which she had deposited the mentioned sum be found from the obtained graph?

(5 marks)

15. A car is riding at a constant speed of 40 km/ hour. Draw a timedistance graph for this situation. Now answer the following questions: (a) How much time does the car take to ride 100 km?

(b) Find the distance covered by the car in 3 hours.21

(c) Find the distance covered by the car in 5 hours.

(5 marks)

Introduction to Graphs Solutions

Topic: Line Graphs

1. The given data is:

Radii (cm) 3.5 7 10.5 14

Circumference (cm) 22 44 66 88

We start by representing radii of circle on the horizontal axis ( x axis) and the corresponding circumference on the vertical axis ( y axis). The given points can be plotted on a graph and the points can then be joined with a line to obtain the required graph as:

2. Number of students that were able to find the study material at 10:30 am = 60 − (16 + 10 + 16 + 8 + 6) = 60 − 56 = 4 The given data can be arranged in a tabular form as:

Time 10:05 am 10:10 am 10:15 am 10:20 am 10:25 am 10:30 am

Number of

students

16 10 16 8 6 4

Introduction to Graphs To represent the given information in the form of line graph, time is marked along the horizontal axis and the number of students along the vertical axis. The appropriate scale along the vertical axis is chosen. It can be taken as 1 unit = 2 students. Then, the number of students is plotted on a graph and the points are then joined by line segments to obtain the required graph as:

3. On studying the line graph, we obtain (a) Busiest time of day at the store was 1 pm. (b) Business started to slow down at 3 pm. (c) There were 2 people in the store when it opened. (d) There were around 10 people in the store at around 2:30 pm. (e) The greatest number of people in the store was 22. (f) The least number of people in the store was 2.

4. In the given graph, years are represented along the horizontal axis and the number of dolphins is represented along the vertical axis. (a) It is observed from the given graph that: Number of dolphins in year 2001 = 6500

Introduction to Graphs Number of dolphins in year 1996 = 2000 ∴Increase in the number of dolphins = 6500 − 2000 = 4500 ∴ Percentage increase in the number of dolphins

Thus, percentage increase in the number of dolphins in the channel from 1996 to 2001 is 225%. (b) It is observed from the given graph that: Number of dolphins in year 1996 = 2000 Number of dolphins in year 1997 = 5000 Number of dolphins in year 1998 = 3000 Number of dolphins in year 1999 = 4500 Number of dolphins in year 2000 = 3000 Number of dolphins in year 2001 = 6500 ∴Average number of dolphins

Therefore, average number of dolphins in the span of six years is 4000. It can be observed that during the years 1996, 1998, and 2000, the number of dolphins is below the average number of dolphins.

Topic: Position of a Point on a Graph

5. (a) D is 5 units from y axis and 2 units from x axis. Hence, the coordinates of D are (5, 2). (b) Point C is located by (1, 5). Point E is located by (10, 0). Point A is located by (4, 9).

6. To plot A (3, 6) on a coordinate plane, we move 3 units to the right of origin and then we move 6 units upwards. To plot B (6, 3) on a coordinate plane, we move 6 units to the right of origin and then we move 3 units upwards.

Introduction to Graphs

7. The given points are (0, 2), (0, 6), and (4, 2). Point (0, 2) can be plotted by moving 2 units vertically upwards from the origin (0, 0). Point (0, 6) can be plotted by moving 6 units vertically upwards from the origin (0, 0). Point (4, 2) can be located by moving 4 units horizontally to the right from origin (0, 0) and then vertically upwards by 2 units. The plotted points are then joined by line segments as:

It is observed that the obtained figure ABC is a right triangle, rightangled at A. ∴AB and AC represent the base and height of ΔABC.

Introduction to Graphs It can be observed that: AC = 4 units, AB = 4 units ∴ Area of ΔABC

= 8 sq units Thus, area of the figure obtained by plotting and joining the given points is 8 sq units.

8. The points (0, 0), (4, 3), (6, 4.5) and (8, 6) can be plotted on a graph as:

It can be seen that the given points lie on the same line. Thus, the given points are collinear.

Topic: Application of Line Graphs

9. If we increase the radius of the base of cylinder i.e., r , then its volume will increase. Similarly, if we increase the height of cylinder i.e., h , then its volume will increase. Thus, r and h are independent variables and V is dependent variable.

10. (a) It can be observed that quantity of CNG is represented on the horizontal axis and the corresponding cost is represented on the vertical axis. To find the cost corresponding to 7.5 kg, a vertical line can be drawn through the 7.5 mark such that it intersects the graph on point X. This can be done as:


From point X, a horizontal line is drawn such that it meets the vertical axis. This can be done as:

As observed in the graph, the horizontal line cuts the vertical axis at 150. Thus, it will cost Mr. Verma Rs 150 to get his car refilled with 7.5 kg of CNG. (b) A horizontal line through the 300 mark on the vertical axis is drawn to meet the line graph at point Y and then a vertical line is drawn from Y to meet the horizontal axis. This can be done as:


It can be observed from the graph that vertical line cuts the x axis at 15. Thus, 15 kg of CNG can be refilled with an amount of Rs 300.

11. We take the deposited sums along x axis and the simple interest obtained along y axis. On x axis, we take scale as 1 unit = Rs 4000 On y axis, we take scale as 1 unit = Rs 100

Introduction to Graphs 12. The simple interest earned on different sums is calculated as follows.

Sum deposited (Rs)

Simple interest for a year (Rs)

1000

2000

3000

4000

5000

Taking simple interest earned along vertical axis, deposits along horizontal axis, and scale as 1 unit = Rs 200 along vertical axis and 1 unit = Rs 1000 along horizontal axis. Plotting these points and joining them by line segments, we obtain the graph as


(a) It can be seen from the graph that on a deposit of Rs 4500, the simple interest earned is Rs 900. (b) Also, a simple interest of Rs 700 is obtained on deposit of Rs 3500.

13. (a) Distance covered to complete one round around the park is equal to the perimeter of the rectangular park. Perimeter of the park = 2 (Length + Breadth) = 2 (10 m + 40 m) = 100 m Time taken to complete one round around the park is given as 20 min. ∴ Distance covered in 20 min = 100 m Therefore, distances covered for different durations of walking can tabulated as:

Time (minutes) 20 40 60 80 100

Distance covered (m) 100 200 300 400 500

By representing time along the horizontal axis and the distance covered along the vertical axis, the points (20, 100), (40, 200), (60, 300), (80, 400), (100, 500) can be plotted on a graph and then joined with line segments to obtain the required graph. This can be done as:


(b) To obtain the number of rounds completed in 30 minutes, firstly distance covered in 30 minutes is to be found. For this, a vertical line has to be drawn from the mark 30 on the x axis to meet the line graph at the point X and then from point X, a horizontal line is drawn to meet the y axis. This can be done as:

It is observed from the graph that the horizontal line cuts the y axis at 150. ∴ In 30 minutes, Roshni covers 150 m. 150 m = 100 m + 50 m, which is the length of one and a half round Thus, Roshni completed 1 rounds in 30 minutes2

1

Introduction to Graphs 14. It is given that principal, P is Rs 12,000, rate of interest, R is 15%.

It is known that: S.I .

For deposit of 1 year, S.I.

Similarly, the S.I. earned on the sum deposited for different time periods can be tabulated as:

Time period (year) 1 2 3 4 5

Simple interest earned (Rs) 1,800 3,600 5,400 7,200 9,000

By representing time period along the x axis and the corresponding simple interest along the y axis, the points (1, 1,800), (2, 3,600), (3, 5,400), (4, 7,200), (5, 9,000) can be plotted and joined by line segments to obtain the required line graph. This can be done as:

(b) To obtain the interest earned on deposit for 4 years, a vertical line through the mark 4.5 is2

1 drawn to meet the line graph at point X and then from point X , a horizontal line is drawn to cut the y axis as:


It is observed that the horizontal line meets the y axis at the mark 8,100. Thus, she can earn an interest of Rs 8,100 for the deposit of 4 years.2

1 (c) To find the duration of time for earning an interest of Rs 4,500, a horizontal line is drawn through the mark 4,500 on the y axis to meet the line graph at point Y and then from this point, a vertical line is drawn to meet the x axis. This can be done as:

It is observed that the vertical line cuts the x axis at the mark 2.5. Thus, the duration of time for which she had deposited the mentioned sum for earning an interest

of Rs 4,500 is 2 years.21

Introduction to Graphs 15.

Hours of ride Distance covered (in km)

1 40

2 2 × 40 = 80

3 3 × 40 = 120

4 4 × 40 = 160

Thus, we obtain the table of values as:

Time (in hours)

1 2 3 4

Distance (in km)

40 80 120 160

We take the time taken by the car along x axis and the distance covered along y axis. Scale: Horizontal axis: 2 units = 1 hour Vertical axis: 2 units = 40 km

Introduction to Graphs (a) By looking at this graph, it can be found that the car takes hours to cover a distance of 1002 2

1 km.

(b) By looking at this graph, it can be found that the car covered a distance of 140 km in 3 hours.2

1

(c) By looking at this graph, it can be found that the car covered a distance of 200 km in 5 hours.


Playing with Numbers Important Questions

Topic: General Form and Properties of Numbers

1. Consider a twodigit number ab . By which number is the number ( ab + ba ) definitely divisible?

(2 marks)

2. If x and y are respectively the largest and smallest threedigit numbers formed by all the nonzero digits a , b , and c where a > b > c . How can the difference between the numbers x and y be written using the digits a , b , and c ?

(2 marks)

3. What is the difference between a threedigit number xyz and a threedigit number having the digit ( x − 3) at hundreds place, the digit ( y − 2) at tens place, and the digit ( z − 1) at ones place?

(3 marks)


By using 3 different nonzero digits, J , K , and L , four numbers are formed such that A = J K L B = L J K C = K L J D = L K J

(a) Write four factors other than 1 for the expression A − D . (b) Is the expression ( A + B + C ) a prime number? Give reasons.

(2 marks)

5. The sum of a twodigit number and the number obtained by reversing its digits is of the form x ( a + b ), where a is an even prime number and the original number is divisible by 13. What is the value of x ? How many such numbers are possible?

(3 marks)

6. The sum or difference of two odd numbers or two even numbers is always even. Show that if both the digits at tens and hundreds places of a threedigit number are odd, then the difference between the threedigit number and the sum of digits of the number is always a multiple of 6.

(3 marks)

Playing with Numbers 7. Find the value of P and Q in the given addition.

(2 marks)

8. The product of two threedigit numbers is given as

What are the values of A , Y , and Z , where A , Y , and Z represent single digits?

(3 marks)

9. The sum of two threedigit numbers is given as:

Here, each of the letters X , Y , and Z represents one digit. What is the value of the expression X × Y × Z ?

(3 marks)

Topic: Divisibility of a Number

10. For what values of x and y is the number 982 x 01 y divisible by 10? (2 marks)

11. If the sixdigit number 469 x 51 is exactly divisible by 3, then what is the maximum value of x ?

(2 marks)

12. For what minimum value of 2 a + b is the number 340 a 924 b divisible by 3? (2 marks)

13. If fivedigit number 2 x 15 y is a multiple of 5 as well as 9, then what are the possible values of the expression x + y + xy ?

(3 marks)

14. For what value of x is the sevendigit number 293654 x exactly divisible by 6? (3 marks)

Playing with Numbers 15. If the number 98 1 is divisible by both 5 and 9, then what can be the

respective values of and ?

(3 marks)

Playing with Numbers Solutions

Topic: General Form and Properties of Numbers

1. ab = a × 10 + b = 10 a + b ba = b × 10 + a = 10 b + a ( ab + ba ) = 10 a + b + 10 b + a = 11 a + 11 b = 11( a + b ) Clearly, ( ab + ba ) is definitely divisible by 11. Hence, the sum of a number and number formed by reversing its digits is always divisible by 11.

2. x and y are respectively the largest and smallest threedigit numbers formed by all the nonzero digits a , b , and c , where a > b > c . ∴ x = abc and y = cba These numbers can be written as:

∴

Thus, the difference between x and y can be written using a , b , and c as 99 ( a − c ).

3. The threedigit number xyz can be written as: xyz = 100 x + 10 y + z The threedigit number having the digit ( x − 3) at hundreds place, the digit ( y − 2) at tens place, and the digit ( z − 1) at ones place is of the form ( x − 3) ( y − 2) ( z − 1). It can be written as:

Thus, required difference between the threedigit numbers

4. A = J K L = 100 J + 10 K + L

Playing with Numbers B = L J K = 100 L + 10 J + K C = K L J = 100 K + 10 L + J D = L K J = 100 L + 10 K + J (a) A − D = (100 J + 10 K + L ) − (100 L + 10 K + J ) = 99 J − 99 L = 99 ( J − L ) Expression A − D is a multiple of 99. Its factors are 3, 9, 11, and 33. Thus, the expression A − D is also a multiple of 3, 9, 11, and 33. (b) A + B + C = (100 J + 10 K + L ) + (100 L + 10 J + K ) + (100 K + 10 L + J ) = 111 ( J + K + L ) It can be seen that 111 = 37 × 3. Therefore, the expression A + B + C is not a prime number.

5. A twodigit number ab can be written in general form as 10 a + b The number obtained by reversing the digits is ba , which can be written in general form as 10 b + a ∴ Sum of the numbers = 10 a + b + 10 b + a = 11 a + 11 b = 11 ( a +b ) Comparing it with the given expression, it is obtained that x = 11. Therefore, the value of x is 11. It is given that a is an even prime number. The only prime number which is even is 2. ∴ a = 2 ⇒ 11 ( a +b ) = 11 (2 + b ) Now, the number ab is divisible by 13. It is known that the only number divisible by 13 with 2 at its tens place is 26. ∴ b = 6 Thus, the only possible number which satisfies the given conditions is 26.

6. Let the threedigit number by xyz , where x and y are odd numbers. This threedigit number xyz can be written as: xyz = 100 x + 10 y + z Sum of the digits of the threedigit number xyz = x + y + z Difference between the threedigit number and the sum of its digits

Playing with Numbers For any digit x , 10 x is an even number. Since x and y are odd numbers, ( x + y ) is an even number. Since 10 x and ( x + 4) are even numbers, 10 x + ( x + y ) is an even number. Let 10 x + ( x + y ) = 2 k, where k is a natural number. ∴ 9[10 x + ( x + y )] = 9 × 2 k = 18 k Thus, the difference is divisible by 18. It is known that if a number is divisible by any number, then it is also divisible by each of the factors of the number. Since the difference is divisible by 18 while 6 is a factor of 18, the difference is divisible by 6. Hence, the difference between the given threedigit number and the sum of its digits is always a multiple of 6.

7. In ones column, it is seen that P + 8 = 1 or 11. We know that there is no digit such that the sum of the digit and 8 is 1. ∴ P + 8 = 11 ⇒ P = 11 − 8 = 3 Therefore, the given addition can be written as

In hundreds column, it is seen that 6 + Q = 7 ∴ Q = 7 − 6 = 1 Therefore, the given addition becomes

Thus, the values of P and Q in the given addition are 3 and 1 respectively.

8. The given multiplication is:

Playing with Numbers

It can be observed that the units digit of the product is 5. It is possible only when A = 5 as the units digit of 25 (= 5 × 5) is 5. Substituting the value of A in the given multiplication:

Comparing 375 with YZ 5: Y = 3, Z = 7 Thus, the values of A , Y , and Z are 5, 3, and 7 respectively.

9. The given sum is:

In the ones place of the given sum, it can be observed that X + 6 = 3. It is possible only when X = 7 because the units digit of 13 (= 7 + 6) is 3. Here, 1 will be carried forward to the tens place of the addition. The given sum can be rewritten as:

In the tens place of the given addition, 1 + Y + 7 = 2 i.e., 8 + Y = 2 This is possible only when Y = 4 because the units digit of (12 = 8 + 4) is 2. Here, 1 will be carried forward to the hundreds place of the addition.

Playing with Numbers The given sum can be rewritten as:

In the hundreds place of the given addition, 1 + Z + 4 = 8 i.e., Z + 5 = 8 This is possible only when Z = 3 Therefore, X = 7, Y = 4, and Z = 3 Thus, value of the expression X × Y × Z = 7 × 4 × 3 = 84

Topic: Divisibility of a Number I

10. A number is divisible by 5, if the digit at ones place is either 0 or 5. It is given that the number leaves 3 as the remainder when divided by 5. Therefore, the ones place digit of the number is either 0 + 3 = 3 or 5 + 3 = 8. It is also given that ones place digit of the number is more than 5. Therefore, ones place digit of the number is 8. A given number is even, if ones place digit of the number is either 0 or 2 or 4 or 6 or 8. Thus, the number is even.

11. A number is divisible by 3, if the sum of the digits of the number is divisible by 3. It is given that the given number 469 x 51 is divisible by 3.

4 + 6 + 9 + x + 5 +1 = x + 25 is divisible by 3 The numbers next to 25, which are divisible by 3, are 27, 30, 33, 36,..

x = 27 − 25 or 30 − 25 or 33 − 25 or 36 − 25,.. ⇒ x = 2 or 5 or 8 or 11 Since x is a digit, x = 2 or 5 or 8 Thus, the maximum value of x for which the number 469 x 51 is divisible by 3 is 8.

12. A number is divisible by 3, if and only if its sum of digits is divisible by 3. Sum of the digits of the given number = 3 + 4 + 0 + a + 9 + 2 + 4 + b = 22 + a + b The nearest multiple of 3 is 24. ∴ 22 + a + b = 24 ∴ a + b = 2 Now, 3 cases arise: Case (i) When a = 0 and b = 2, 2 a + b = 2(0) + 2 = 2 Case (ii)

Playing with Numbers When a = 1 and b = 1, 2 a + b = 2(1) + 1 = 3 Case (iii) When a = 2 and b = 0, 2 a + b = 2(2) + 0 = 4 Out of these possible cases, the minimum value (i.e., 2) of 2 a + b occurs at a = 0 and b = 2 Hence, the minimum value of 2 a + b , for which the given number 340 a 924 b is divisible by 3, is 2.

13. It is known that a given number is a multiple of 5 if the units digit of the number is 0 or 5. Since the number 2 x 15 y is a multiple of 5, units digit y is either 0 or 5. If y = 0, then the given number is 2 x 150. If y = 5, then the given number is 2 x 155. It is known that a number is a multiple of 9, if the sum of the digits of the number is a multiple of 9. Thus, 2 x 150 will be divisible by 9, if 2 + x + 1 + 5 + 0 = 8 + x is divisible by 9. The next numbers to 8 that are divisible by 9 are 9, 18, .... ∴ 8 + x = 9 ⇒ x = 9 − 8 = 1 8 + x = 18 ⇒ x = 18 − 8 = 10, which is not possible as x is a digit ∴ x = 1, when y = 0 Thus, in this case, x + y + xy = 1 + 0 + 1 × 0 = 1 The number 2 x 155 is a multiple of 9, if 2 + x + 1 + 5 + 5 = (13 + x ) is a multiple of 9. The next numbers to 13 that are multiples of 9 are 18, 27. ∴ 13 + x = 18 ⇒ x = 18 − 13 = 5 13 + x = 27 ⇒ x = 27 − 13 = 14, which is not possible as x is a digit ∴ x = 5, when y = 5 Thus, in this case, x + y + xy = 5 + 5 + 5 × 5 = 10 + 25 = 35 Thus, the value of the expression x + y + xy is either 35 or 1.

14. The given sevendigit number is 293654 x . A number is exactly divisible by 6, if the number is exactly divisible by both the numbers 2 and 3. A number is divisible by 2, if the units digit of the number is either 0 or 2 or 4 or 6 or 8. ∴ x = 0 or 2 or 4 or 6 or 8 A number is divisible by 3, if the sum of its digits is divisible by 3. Sum of the digits of the number 293654 x = 2 + 9 + 3 + 6 + 5 + 4 + x = 29 + x When x = 0, sum of the digits of the number 293654 x = 29 + 0 = 29, which is not divisible by

Playing with Numbers 3 When x = 2, sum of the digits of the number 293654 x = 29 + 2 = 31, which is not divisible by 3 When x = 4, sum of the digits of the number 293654 x = 29 + 4 = 33, which is divisible by 3 When x = 6, sum of the digits of the number 293654 x = 29 + 6 = 35, which is not divisible by 3 When x = 8, sum of the digits of the number 293654 x = 29 + 8 = 37, which is not divisible by 3 Therefore, 2936544 is divisible by 3 when x = 4. Thus, the required value of x is 4.

15. It is known that if a number is divisible by 5, then the digit at ones place is either 0 or 5.

Since the number 98 1 is divisible by 5, the value of is either 0 or 5.

Case (i): = 0

The number becomes 98 10. If a number is divisible by 9, then the sum of the digits of the number is divisible by 9.

∴ Sum of the digits of the number 98 10 is divisible by 9.

Sum of the digits of the number 98 10 excluding is 9 + 8 + 1 + 0 = 18. 18 is divisible by 9. The number next to 18 that is divisible by 9 is 27.

It is clear that, value of is 18 − 18 or 27 − 18, i.e., 0 or 9.

Case (ii): = 5

The number becomes 98 15.

Sum of the digits of the number 98 15 is divisible by 9.

Sum of digits of the number 98 15 excluding is 9 + 8 + 1 + 5 = 23 The number next to 23 that is divisible by 9 is 27.

It is clear that when is 5, the value of is 27 − 23 = 4 .

Thus, the possible values of and are 0 and 0, or 9 and 0, or 4 and 5 respectively.

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Important Questions-Solutions Mathematics · Important Questions Topic: Frequency Distribution 1. Observe the given frequency distribution table and answer the following questions