Uncategorized

Exercise 23.1

Question 1.
Define the following terms :
(i) Observations
(ii) Raw data
(iii) Frequency of an observation
(iv) Frequency distribution
(v) Discrete frequency distribution
(vi) Grouped frequency distribution
(vii) Class-interval
(viii) Class-size
(ix) Class limits
(x) True class limits
Solution:
(i) Observations : Each entry in the given data is called an observation. ‘
(ii) Raw data: A collection of observations by an observer, is called raw data.
(iii) Frequency of an observation : The number of times an observation occurs in the given data is called its frequency.
(iv) Frequency distribution : The presentations of given data in order of magnitude ascending or descending, is called the frequency distribution.
(v) Discrete frequency distribution: When the given data is represented by tally marks after arranging it in an order. This kind of distribution is called discrete frequency distribution.
(vi) Grouped frequency distribution: If the number of data is large, and the difference between the greatest and the smallest observation is large, then we represent then in groups or classes. Such representation of data is called grouped frequency distribution.
(vii) Class intervals: The difference between the upper limit and lower limit of a class is called class interval.
(viii) Class-size : Class intervals are also called the class size. Each size of the same intervals
(ix) Class limits : Every class has two limits : upper limit and lower limit.
(x) True class limits or Exclusive limits : When the upper limit of one is the lower limit of the next interval then these are call true class limits.

Question 2.
The final marks in mathematics of 30 students are as follows :
53, 61, 48, 60, 78, 68, 55, 100, 67, 90, 75, 88,77,37,84,58,60,48,62,56,44,58,52,64, 98, 59, 70, 39, 50, 60
(i) Arrange these marks in the ascending order, 30 to 39 one group, 40 to 49 second group etc.
Now answer the following:
(ii) What is the highest score ?
(iii) What is the lowest score ?
(iv) What is the range ?
(v) If 40 is the pass mark how many have failed ?
(vi) How many have scored 75 or more ?
(vii) Which observations between 50 and 60 have not actually appeared ?
(viii) How many have scored less than 50 ?
Solution:
(i) Arranging the given data in ascending order.
30 to 39 : 37,39
40 to 49 : 44, 48, 48
50 to 59 : 50, 52, 53, 55, 56, 58, 58, 59
60 to 69 : 60, 60, 60, 61, 62, 64, 67, 68
70 to 79 : 70, 75, 77, 78
80 to 89 : 84, 88
90 to 99 : 90, 98
100 to 109 : 100
(ii) Highest score is 100
(iii) Lowest score is 37
(iv) Range is 100 – 37 = 63
(v) If 40 is pass marks then number of failed candidates will be = 2
(vi) Number of students who scored 75 or more = 8
(vii) Between 50 and 60, the observations 51, 54, 57 do not appear.
(viii) Number of students who scored less than 50 = 5

Question 3.
The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6,3.0, 2.5, 2.9, 2-8,3.1, 2.5, 2.8, 2.7, 2.9, 2.4.
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight:
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 kg ?
(vii) How many babies weigh more than 2.8 kg ?
(viii) How many babies weigh 2.8 kg ?
Solution:
(i) Arranging the given weights in descending order.
3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.4, 2.3, 2.2, 2.1
(ii) Highest weight = 3.1 kg
(iii) Lowest weight = 2.1 kg
(iv) Range : 3.1 – 2.1 = 1 kg.
(v) Number of babies born on that day = 15
(vi) Number of babies having weight below 2.5 kg = 4
(vii) Number of babies having weight more than 2.8 kg = 4
(viii) Number of babies weigh 2.8 kg = 2

Question 4.
Following data gives the number of children in 41 families:
1,2,6,5,1,5,1,3,2,6,2,3,4,2,0,0,4,4, 3, 2, 2, 0, 0,1, 2, 2, 4,3, 2,1, 0, 5,1, 2,4, 3, 4, 1, 6, 2, 2.
Represent it in the form of a frequency distribution.
Solution:
Frequency distribution table

Question 5.
Prepare a frequency table of the following scores obtained by 50 students in a test:
42, 51, 21, 42, 37. 37, 42, 49, 38, 52, 7, 33, 17, 44, 39, 7, 14, 27, 39, 42, 42, 62, 37, 39, 67, 51, 53, 53, 59, 41, 29, 38, 27, 31, 54,19, 53, 51, 22, 61, 42, 39, 59, 47, 33, 34, 16, 37, 57, 43
Solution:
Frequency distribution table

Question 6.
A die was thrown 25 times and following scores were obtained:
1,5,2,4,3,6,1,4,2,5,1,6,2,6,3,5,4,1, 3, 2, 3, 6,1, 5, 2
Prepare a frequency table of the scores.
Solution:
Frequency distribution table

Question 7.
In a study of number of accidents per day, the observations for 30 days were obtained as follows:
6,3,5,6,4,3, 2,5,4,2,4,2,1,2, 2,0,5,4,6,1,6,0, 5, 3, 6,1, 5, 5, 2, 6
Prepare a frequency distribution table.
Solution:
Frequency distribution table

Question 8.
Prepare a frequency table of the following ages (in years) of 30 students of class VIII in your school:
13,14,13,12,14,13,14,15,13,14,13,14, 16,12,14,13,14,15,16,13,14,13,12,17,13, 12,13,13,13,14
Solution:
Frequency distribution table

Question 9.
Following figures relate to the weekly wages (in Rs) of 15 workers in a factory :
300,250,200,250,200,150,350,200,250, 200,150, 300,150, 200, 250 Prepare a frequency table.
(i) What is the range in wages (in Rs) ?
(ii) How many workers are getting Rs 350 ?
(iii) How many workers are getting the minimum wages ?
Solution:
Frequency distribution table

(i) Range = 350- 150 = 200
(ii) Number of workers getting Rs 350 = 1
(iii) Number of workers getting minimum wages = 3

Question 10.
Construct a frequency distribution table for the following marks obtained by 25 students in a history test in class Vin of a school:
9,17,12, 20,9,18, 25,17,19,9,12,9,12, 18, 17,19, 20, 25, 9,12,17,19, 19, 20, 9
(i) What is the range of marks ?
(ii) What is the highest mark ?
(iii) Which mark is occurring more frequently ?
Solution:
Frequency distribution table

(i) Range = 25 – 9 = 16
(ii) Highest marks = 25
(iii) Marks occurring more frequently = 9

Exercise 23.2

Question 1.
The marks obtained by 40 students of class VIII in an examination are given below :
16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13,
21, 13, 15, 19, 24, 16, 3, 23, 5, 12, 18, 8, 12, 6,
8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23.
Divide the data into five groups, namely 0-5,5-10,10-15,15-20 and 20-25 and prepare a grouped frequency table.
Solution:
Frequency distribution table

Question 2.
The marks scored by 20 students in a test are given below :
54, 42, 68, 56, 62, 71, 78, 51, 72, 53, 44, 58, 47, 64, 41, 57, 89, 53, 84, 57.
Complete the following frequency table :

What is the class interval in which the greatest frequency occurs ?
Solution:

The class in which the greatest frequency is 50-60

Question 3.
The following is the distribution of weights (in kg) of 52 persons :


(i) What is the lower limit of class 50-60 ?
(ii) Find the class marks of the classes 40-50, 50-60.
(iii) What is the class size ?
Solution:
(i) Lower limit of class 50-60 = 50
(ii) Class marks of 40-50 = 40+502 = 902
= 45 and of 50-60 = 50+602 = 1102 =55
(iii) Class size is 10

Question 4.
Construct a frequency table for the following weights (in gm) of 35 mangoes using the equal class intervals, one of them is 40-45 (45 not included):
30,40,45,32,43,50,55,62,70,70,61,62, 53,52, 50,42,35,37,53,55,65,70, 73, 74,45, 46, 58, 59, 60, 62, 74, 34, 35, 70 ,68.
(i) What is the class mark of the class interval 40-45 ?
(ii) What is the range of the above weights ?
(iii) How many classes are there ?
Solution:
Smallest observation = 30
Greatest observation = 74
Range = 74 – 30 = 44
Now forming the distribution table

(i) Class mark of 40-45
= 40+452 = 852 = 42.5
(ii) Range = 74 – 30 = 44
(iii) Number of classes are 9

Question 5.
Construct a frequency table with class-intervals 0-5 (5 not included) of the following marks obtained by a group of 30 students in an examination :
0, 5, 7,10, 12,15, 20, 22, 25, 27, 8, 11, 17,3, 6, 9,17,19, 21, 29, 31,35,37,40,42, 45, 49, 4, 50, 16.
Solution:
Frequency distribution table.

Question 6.
The marks scored by 40 students of class VIII in mathematics are given below:
81,55, 68, 79,85,43,29,68,54,73,47, 35, 72,64,95,44,50, 77,64,35,79, 52, 45,54,70,83, 62′, 64,72,92,84,76,63, 43, 54, 38, 73, 68, 52, 54.
Prepare a frequency distribution with class size of 10 marks.
Solution:
Largest marks = 95
Lowest marks = 29
Range = 95 – 29 = 66
Frequency distribution table

Question 7.
The heights (in cm) of 30 students of class VIII are given below :
155.158.154.158.160.148.149.150.153, 159,161,148,157,153,157,162,159,151, 154,156,152,156,160,152,147,155,163,155,157,153.
Prepare a frequency distribution table with 160-164 as one of the class intervals.
Solution:
Largest height =163
Lowest height =147
Range = 163- 147 = 16

Question 8.
The monthly wages of 30 workers in a factory are given below :
830,835,890,810,835,836,869,845,898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.
Solution:
Highest wage = 898
Lowest wage = 804
Range = 898 – 804 = 94
Frequency distribution table

Question 9.
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) at 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included):
220, 268, 258, 242, 210, 268, 272,242, 311, 290, 300, 320,319,304,302,318,306,292, 254, 278, 210,240, 280,316,306, 215, 256, 236.
Solution:
Highest wages = 320
Lowest wages = 210
Range = 320-210= 110
Frequency distribution table

Question 10.
The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows :
-12.5, -10.8, -18.6, -8.4, -10.8, -4.2, -4.8, -6.7, -13.2, -11.8, -2.3,1.2, 2.6, 0, -2.4, 0, 3.2, 2.7,3.4,0, -2.4, -2.4, 0,3.2, 2.7,3.4, 0,2.4, -5.8, -8.9, -14.6, -12.3, -11.5, -7.8, – 2.9
Represent them as frequency distribution table taking -19.9 to -15 as the first class interval.
Solution:
Lowest temperature = -19.9
Highest temperature = 3.4
Frequency distribution table

Exercise 25.1

Question 1.
The number of hours, spent by a school boy on different activities in a working day, is given below:

Present the information in the form of a pie-chart .
Solution:
Total = 24

Now we draw a circle and divide it in the sectors having above central angles as shown in the figure.

Question 2.
Employees of a company have been categorized according to their religions as given below:

Draw a pie-chart to represent the above information.
Solution:
Total =1080

Now draw a circle and divided it into sectors having the above central angles as shown in the figure.

Question 3.
In one day the sales (in rupees) of different items of a baker’s shop are given below :

Draw pie-chart representing the above sales.
Solution:
Total = 480


Now draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 4.
The following data shows the expenditure of a person on different items during a month. Represent the data by a pie-chart.

Solution:
Total = 10800

Now we draw a circle and divide it into sector having the above central angles as shown in figure.

Question 5.
The percentages of various categories of workers in a state are given in the following table:

Present the information in the form of a pie-chart.
Solution:
Total = 100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 6.
The following table shows the expenditure incurred by a publisher in publishing a book:

Present the above data in the form of a pie-chart.
Solution:
Total = 100% = 100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure

Question 7.
Percentage of the different products of a village in a particular district are given below. Draw a pie-chart representing this information.

Solution:
Total = 100

Now, we draw a circle and divided it into sectors having the above central angles as shown in the figure.

Question 8.
Draw a pie-diagram for the following data of expenditure pattern in a family.

Solution:
Total =100


Now we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 9.
Draw a pie-diagram of the areas of continents cf the world given in the following table :

Solution:
Total = 133.3

Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 10.
The following data gives the amount spent of the construction of a house. Draw a pie diagram.

Solution:
Total = 300 (in thousands)

Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 11.
The following table shows how a student spends his pocket money during the course of a month. Represent it by a pie-diagram.

Solution:
Total expenditure = 100


Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 12.
Represent the following data by a pie-diagram :

Solution:
1. For family A
Total = 10000

Now, we draw a circle and divide it in sectors having the above central angles as shown in the figure.
(2) For family B
Total = 11680

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 13.
Following data gives the break up of the cost of production of a book:

Draw a pie – diagram depicting the above information.
Solution:
Total = 100

Now, we draw a circle and divide it into sectors of above central angles as shown in the figure.

Question 14.
Represent the following data with the help of pie-diagram.

Solution:
Total = 6000 tons

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 15.
Draw a pie-diagram representing the relative frequencies (expressed as percentage) of the eight classes as given below :
12.6,18.2,17.5,20.3,2.8,4.2,9.8,14.7
Solution:
Total =100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure given.

Question 16.
Following is the break up of the expenditure of a family on different items of consumption :

Draw a a pie – diagram to represent the above data.
Solution:
Total = Rs. 3000

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 17.
Draw a pie-diagram for the following data of the investment pattern in five year plan :

Solution:
Total = 100


Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Exercise 25.2

Question 1.
The pie-chart given in figure represents the expenditure on different items in constructing a flat in Delhi. If the expenditure incurred on cement is Rs 1,12,500, find the following:

(i) Total cost of the flat,
(ii) Expenditure incurred on labour.
Solution:
Expenditure on cement = Rs 1,12,500
and its central angle = 75°

Question 2.
The pie-chart given in the figure shows the annual agricultural production of an Indian state. If the total production of all the commodities is 81000 tonnes, find the production (in tonnes) of:

(i) Wheat
(ii) Sugar
(iii) Rice
(iv) Maize
(v) Gram
Solution:
Total production = 81000 tonnes

Question 3.
The following pie-chart shows the number of students admitted in different faculties of a college. If 1000 students are admitted in Science answer the following:

(i) What is the total number of students ?
(ii) What is the ratio of students in science and arts ?
Solution:
Students admitted in science = 1000
Central angle = 100°
(i) Total number of students

∴ Ratio in science and arts = 1000 : 1200 = 5:6

Question 4.
In the figure, the pie-chart shows the marks obtained by a student in an examination. If the student secures 440 marks in all, calculate his marks in each of the given subjects.

Solution:
Total marks secured by a student = 440

Question 5.
In the figure, the pie-charts shows the marks obtained by a student in various subjects. If the student scored 135 marks in mathematics, find the total marks in all the subjects. Also, find his score in individual subjects.

Solution:
Marks obtained in mathematics =135
Central angle = 90°

Question 6.
The following pie-chart shows the monthly expenditure of Shikha on various items. If she spends Rs 16,000 per month, answer the following questions:

(i) How much does she spend on rent ?
(ii) How much does she spend on education ?
(iii) What is the ratio of expenses on food and rent ?
Solution:
Total expenditure per month = Rs 16,000

Question 7.
The pie-chart (as shown in the figure) represents the amount spent on different sports by a sports club in a year. If the total money spent by the club on sports is Rs 1,08,000, find the amount spent on each sport.

Solution:
total amount spent on sports = Rs 1,08,000

Exercise 27.1

Question 1.
Plot the points (5,0 ), (5,1), (5, 8). Do they lie on a line ? What is your observation ?
Solution:
Draw XOX’ and YOY’ the co-ordinates axis on the graph.
Take 1 cm = 1 unit
Point A (5, 0), B (5, 1) and C (5, 8) have been plotted on the graph. By joining A, B and C, we see that these points lie on the same line which is 5 units from y-axis.

Question 2.
Plot the points (2, 8), (7, 8) and (12, 8). Join these points in pairs. Do they lie on a line ? What do you observe ?
Solution:
Draw XOX’ and YOY’, the co-ordinate axis on the graph.
Take 0.5 cm = 1 unit.
Now points A (2, 8), B (7, 8), and C (12, 8) have been plotted on the graph. By joining them, we see that these points lie on the same line which is at a distance of 8 unit from x-axis.

Question 3.
Locate the points :
(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2,3), (2,4)
(iii) (1,3), (2,3), (3,3), (4,3)
(iv) (1,4), (2,4), (3,4), (4,4).
Solution:
The points given in (i) and (ii) are locates in first graph.

(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2, 3), (2, 4)
Points of (iii) and (iv) are located in the adjoining graph.
(iii) (1, 3), (2, 3), (3, 3), (4, 3)
(iv) (1,4), (2, 4), (3, 4), (4,4)
2C1270%2C681%2C1270%2C610&vis=1&rsz=%7C%7Cs%7C&abl=NS&fu=128&bc=31&jar=2022-02-13-13&ifi=4&uci=a!4&btvi=4&fsb=1&xpc=Crc36rR6Fm&p=https%3A//www.learninsta.com&dtd=47483

54&bih=610&scr_x=0&scr_y=0&eid=42531398%2C44750

Question 3.
Locate the points :
(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2,3), (2,4)
(iii) (1,3), (2,3), (3,3), (4,3)
(iv) (1,4), (2,4), (3,4), (4,4).
Solution:
The points given in (i) and (ii) are locates in first graph.

(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2, 3), (2, 4)
Points of (iii) and (iv) are located in the adjoining graph.
(iii) (1, 3), (2, 3), (3, 3), (4, 3)
(iv) (1,4), (2, 4), (3, 4), (4,4)

Question 4.
Find the coordinates of points A, B, C, D in the figure

Solution:
Draw perpendicular from A, B, C and D on x-axis and also on y-axis.
A is 1 unit from y-axis and 1 unit from x-axis.
∴ Co-ordinates of A are (1,1)
B is 1 unit fr onr y-axis and 4 units from x-axis
∴ Co-ordinates of B are (1,4)
C is 4 units from y-axis and 6 units from x-axis
∴ Co-ordinates of C are (4, 6)
D is 5 units fromy-axis and 3 units from x-axis
∴ Co-ordinates of D are (5,3)

Question 5.
Find the coordinates of points P, Q, R and S in Fig.

Solution:
Through P, Q, R and S, draw perpendiculars on x-axis and also on y-axis.
(i) P is 10 units form >’-axis and 70 units from y-axis
∴ Co-ordinates of P are (10, 70)
(ii) Q is 12 unit from x-axis and 80 units from y-axis
∴ Co-ordinates of Q are (12, 80)
(iii) R is 16 units from x-axis and 100 units from y-axis
∴ Co-ordinates of R are (16, 100)
(iv) S is 20 units from x-axis and 120 units from y-axis
∴ Co-ordinates of S are (20, 120)

Question 6.
Write the coordinates of each of the vertices of each polygon in the figure.

Solution:
(i) In figure OXYZ
Co-ordinates of O are (0, 0) ∵It is the origin
Co-ordinates of X are (0, 2) ∵ It lies on y – axis
Co-ordinates of Y are (2, 2)
Co-ordinates of Z are (2, 0) ∵It lies on x-axis
(ii) In figure ABCD, draw perpendicular from A, B, C, D on x-axis and y-axis.
A is 4 unit from y-axis and 5 units from x- axis
∴ Co-ordinates of A are (4, 5)
B is 7 units from y-axis and 5 units from x – axis
∴ Co-ordinates of B are (7, 5)
C is 6 units from y-axis and 3 units from x- axis
∴ Co-ordinates of C are (6, 3)
D is 3 units from y-axis as well x-axis
∴ Co-ordinates of D are (3, 3)
(iii) In figure PQR, perpendiculars for P, Q, R are drawn on x-axis and also on y-axis.
∴ Co-ordinates of P are (7, 4), of Q are (9, 5) and of R are (9, 3).

Exercise 27.2

Question 1.
The following table shows the number of patients discharged from a hospital with HIV diagnosis in different years :

Represent this information by a graph.
Solution:
Represent years along x-axis and number of patients along y-axis. Now, plot the points (2002, 150), (2003, 170), (2004, 195), (2005, 225) and (2006, 230) on the graph and join them in order. We get the required graph as shown on the graph.

Question 2.
The following table shows the amount of rice grown by a farmer in different years :

Plot a graph to illustrate this information.
Solution:
We represent years along x-axis and rice (in quintals) along y-axis. Now we plot the points (2000,200), (2001, 180), (2002,240), (2003,260), (2004,250), (2005,200) and (2006,270) on the graph and join them in order as shown on the graph.

Question 3.
The following table gives the information regarding the number of persons employed to a piece of work and time taken to complete the work:

Plot a graph of this information.
Solution:
We represent number of person along x-axis and time taken in day along y-axis. Now plot the points (2,12), (4,6), (6,4) and (8, 3) on the graph and join them in order as shown on the graph.

Question 4.
The following table gives the information regarding length of a side of a square and its area :

Draw a graph to illustrate this information.
Solution:
We represent length of a side (in cm) along x-axis and Area of square (in cm2) along the y-axis. Now plot the points (1,1), (2, 4), (3, 9), (4, 16) and (5, 25) on the graph and join them in order. We get the required graph as shown on the graph.

Question 5.
The following table shows the sales of a commodity during the years 2000 to 2006.

Draw a graph of this information.
Solution:
We represent years on x-axis and sales (in lakh rupees) along y-axis. Now plot the points
(2000, 1.5), (2001, 1.8), (2002, 2.4), (2003, 3.2), (2004, 5.4), (2005, 7.8) and (2006, 8.6) on the graph and join them in order we get the required graph as shown.

Question 6.
Draw the temperature-time graph in each of the following cases :

Solution:
(i) We represent time along x-axis and temperature (in °F) alongy-axis. Now plot the points (7:00, 100), (9:00, 101), (11:00, 104), (13:00, 102), (15:00, 100), (17:00, 99), (19:00, 100) and (21:00, 98) on the graph and join them in order to get the required graph as shown on the graph.


(ii) We represent time along x-axis and temperature (in °F) along j-axis. Now plot the points (8:00, 100), (i0:00, 101), (12:00, 104), (14:00, 103), (16:00, 99), (18:00,98), (20:00,100) on the graph and join them in order to get the required graph as shown on the graph.

Question 7.
Draw the velocity-time graph from the following data :

Solution:
We represent time (in hours) along x- axis and speed (in km/hr) along y-axis. Now plot the points (7:00,30), (8:00,45), (9:00,60), (10:00, 50), (11:00, 70), (12:00, 50), (13:00, 40) and (14:00,45) on the graph and join them in order to get the required graph as shown.

Question 8.
The runs scored by a cricket team in first 15 overs are given below :

Draw the graph representing the above data in two different ways as a graph and as a bar chart.
Solution:
We represent overs along.t-axis and runs along v-axis. Now plot the points (1,2), (II, 1), (III, 4), (IV, 2), (V, 6), (VI, 8), (VII, 10), (VIII, 21), (IX, 5), (X, 8), (XI, 3), (XII, 2), (XIII, 6), (XIV, 8), (XV, 12) on the graph and join them to get the graph, as shown bar graph of the given data is given below:

Question 9.
The runs scored by two teams A and B in first 10 overs are given below :

Draw a graph depicting the data, making the graphs on the same axes in each case in two different ways as a graph and as a bar chart.
Solution:
We represent overs along x-axis and runs scored by team A and team B with different types of lines along y-axis.
Plot the points for team A : (I, 2), (II, 1), (III,8), (IV, 9), (V, 4), (VI, 5), (VII, 6), (VIII, 10), (IX, 6) and (X, 2)
and for team B, the points will be : (I, 5), (II, 6), (III, 2), (IV, 10), (V, 5), (VI, 6), (VII, 3), (VIII, 4), (IX, 8), (X, 10).
Then join them in order to get the required graph for team A and team B as shown.
Note : For team A ………
for team B …………

Exercise 26.1

Question 1.
The probability that it will rain to morrow is 0.85. What is the probability that it will not rain tomorrow ?
Solution:
Total number of possible events = 1
∴ P (A¯ ) = 0.85
∴ P (A¯ ) = 1-0.85 = 0.15

Question 2.
A die is thrown. Find the probability of getting (i) a prime number (ii) 2 or 4 (iii) a multiple of 2 or 3.
Solution:
Total number of possible events = 6 (1 to 6)
(i) Let A be the favourable occurrence which are prime number i.e., 2,3,5
∴ P(A) = 36 = 12
(ii) Let B be the favourable occurrence which are 2 or 4
∴ P(B) = 26 = 13
(iii) Let C be the favourable occurrence which are multiple of 2 or 3 i.e., 2, 3, 4, 6.
∴ P(C) = 46 = 23

Question 3.
In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet, of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution:
By throwing of a pair of dice, total number of possible events = 6 × 6 = 36
(i) Let A be the occurrence of favourable events whose sum is 8 i.e. (2,6), (3,5), (4,4), (5,3) , (6,2) which are 5
∴ P(A) = 536
(ii) Let B be the occurrence of favourable events which are doublets i.e. (1, 1), (2, 2), (3, 3), (4,4), (5, 5) and (6, 6).
∴ P(B) = 636 = 16
(iii) Let C be the occurrence of favourable events which are doublet of prime numbers which are (2, 2), (3,3), (5, 5)
∴ P(C) = 336 = 112
(iv) Let D be the occurrence of favourable events which are doublets of odd numbers which are (1, 1), (3, 3) and (5, 5)
∴ P(D) = 336 = 112
(v) Let E be the occurrence of favourable events whose sum is greater than 8 i.e, (3,6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6) which are 6 in numbers
∴ P(E) = 636 = 16
(vi) Let F be the occurrence of favourable events in which is an even number is on first i.e (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1),(4,2), (4,3) (4,4), (4,5), (4,6), (6,1), (6,2), (6, 3), (6,4 ), (6, 5), (6,6) which are 18 in numbers.
∴ P(F) = 1836 = 12
(vii) Let G be the occurrence of favourable events in which an even number on the one and a multiple of 3 on the other which are (2,3), (2,6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6,2), (6,4) = which are 11th number
∴ P(G) = 1136
(viii) Let H be the occurrence of favourable events in which neither 9 or 11 as the sum of the numbers on the faces which are (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) , (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (5, 1), (5,2), (5,3), (5,5), (6,1), (6,2), (6,4), (6,6) which are 30
∴ P(H) = 3036 = 56
(ix) Let I be the occurrence of favourable events, such that a sum less than 6, which are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4,1) which are 10
∴ P(I) = 1036 = 518
(x) Let J be the occurrence of favourable events such that a sum is less than 7, which are
(1.1) , (1,2), (1,3), (1,4), (1,5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5.1) which are 15
∴ P(J) = 1536 = 56
(xi) Let K be the occurrence of favourable events such that the sum is more than 7, which are (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) which 15
∴ P(K) = 1536 = 512
(xii) Let L be the occurrence of favourable events such that at least P (L) one is black card
∴ P(L) = 2652 = 12
(xiii) Let M is the occurrence of favourable events such that a number other than 5 on any dice which can be (1,1), (1,2), (1,3), (1,4), (1,6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3,1), (3,2), (3, 3), (3,4), (3,6), (4,1), (4, 2), (4, 3), (4,4), (4,6), (6, 1), (6,2), (6, 3), (6,4), (6, 6) which are 25
∴ P(M) = 2536

Question 4.
Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
(iv) no tails
Solution:
Total number of events tossed by 3 coins each having one head and one tail = 2x2x2 = 8
(i) Let A be the occurrence of favourable events which is exactly two heads, which can be 3 in number which are HTH, HHT, THH.
∴ P(A) = 38
(ii) Let B be the occurrence of favourable events which is at least two heads, which will be 4 which are HHT, HTH, THH, and HHH.
∴ P(B) = 48 = 12
(iii) Let C be the occurrence 6f favourable events which is at least one head and one tail which are 6 which can be HHT, HTH, THH, TTH, THT, HTT
∴ P(C) = 68 = 34
(iv) Let D be the occurrence of favourable events in which there is no tail which is only 1 (HHH)
∴ P(D) = 18

Question 5.
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) neither a red card nor a queen
(ix) other than an ace
(x) a ten
(xi) a spade
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xv) the ace of spades
(xvi) a queen
(xvii) a heart
(xviii) a red card
Solution:
A pack of cards have 52 cards, 26 black and 26 red and four kinds each of 13 cards from 2 to 10, one ace, one jack, one queen and one king.
∴ Total number of possible events = 52
(i) Let A be the occurrence of favourable events which is a black king which are 2.
∴ P(A) = 252 = 126
(ii) Let B be the occurrence of favourable events such that it is either a black card or a king.
Total = number of black cards = 26 + 2 red kings = 28
∴ P(B) = 2852 = 713
(iii) Let C be the occurrence of favourable events such that it is black and a king which can be 2.
∴ P(C) = 252 = 126
(iv) Let D be the occurrence of favourable events such that it is a jack, queen or a king which will be4 + 4 + 4 = 12
∴ P(D) = 1252 = 313
(v) Let E be the occurrence of favourable events such that it is neither a heart nor a king.
∴ Number of favourable event will be 13 x 3 -3 = 39 – 3 = 36
∴ P(E) = 3652 = 913
(vi) Let F be the occurrence of favourable events such that it is a spade or an ace.
∴ Number of events = 13 + 3 = 16
∴ P(F) = 1652 = 413
(vii) Let G be the occurrence of favourable events such that it neither an ace nor a king.
∴Number of events = 52 – 4 – 4 = 44
∴ P(G) = 4452 = 1113
(viii) Let H be the occurrence of favourable events such that it is neither a red card nor a queen.
∴Number of events = 26 – 2 = 24,
∴ P(H) = 2452 = 613
(ix) Let 1 be the occurrence of favourable events such that it is other than an ace.
∴ Number of events = 52 – 4 = 48
∴ P(I) = 4852 = 1213
(x) Let J be the occurrence of favourable event such that it is ten
∴ Number of events = 4
∴ P(J) = 452 = 113
(xi) Let K be the occurrence of favourable event such that it is a spade.
∴ Number of events =13
∴ P(K) = 1352 = 14
(xii) Let L be the occurrence of favourable event such that it is a black card.
∴ Number of events = 26
∴ P(L) = 2652 = 12
(xiii) Let M be the occurrence of favourable event such that it is the seven of clubs.
∴ Number of events = 1
∴ P(M) = 152
(xiv) Let N be the occurrence of favourable event such that it is a jack.
∴ Number of events = 1
∴ P(N) = 452 = 113
(xv) Let O be the occurrence of favourable event such that it is an ace of spades.
∴ Number of events = 1
∴ P(O) = 152
(xvi) Let Q be the occurrence of favourable event such that it is a queen.
∴ VNumber of events = 4
∴ P(P) = 452 = 113
(xvii) Let R be the occurrence of favourable event such that it is a heart card.
∴ Number of events =13
∴ P(P) = 1352 = 14
(xviii) Let S be the occurrence of favourable event such that it is a red card
∴ Number of events = 26
∴ P(P) = 2652 = 12

Question 6.
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Solution:
Number of possible events = 10 + 8 = 18
Let A be the occurrence of favourable event such that it is a white ball.
∴ Number of events = 8
∴ P(A) = 818 = 49

Question 7.
A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) white ? (ii) red ? (iii) black ? (iv) not red ?
Solution:
Number of possible events = 3 + 5 + 4 = 12 balls
(i) Let A be the favourable event such that it is a white ball.
∴ P(A) = 412 = 13
(ii) Let B be the favourable event such that it is a red ball.
∴ P(B) = 312 = 14
(iii) Let C be the favourable event such that it is a black ball.
∴ P(C) = 512
(iv) Let D be the favourable event such that it is not red.
∴ Number of favourable events = 5 + 4 = 9
∴ P(D) = 912 = 34

Question 8.
What is the probability that a number selected from the numbers 1,2,3,…………, 15 is a multiple of 4 ?
Solution:
Number of possible events =15
Let A be the favourable event such that it is a multiple of 4 which are 4, 8, 12
∴ P(A) = 315 = 15

Question 9.
A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black ?
Solution:
Number of possible events = 6 + 8 + 4 = 18 balls
Let A be the favourable event such that it is not a black
∴ Number of favourable events = 6 + 4=10
∴ P(A) = 1018 = 59

Question 10.
A bag contains 5 white and 7 red balls. One ball is drawn at random, what is the probability that ball drawn is white ?
Solution:
Number of possible events = 5 + 7 = 12
Let A be the favourable event such that it is a while which are 5.
∴ P(A) = 512

Question 11.
A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is (i) white (if) red (iii) not black (iv) red or white.
Solution:
Number of possible events = 4 + 5 + 6 = 15
(i) Let A be the favourable events such that it is a white.
∴ P(A) = 615 = 25
(ii) Let B be the favourable event such that it is a red
∴ P(B) = 415
(iii) Let C be the favourable event such that it is not black.
∴ Number of favourable events = 4 + 6=10
∴ P(C) = 1015 = 23

Question 12.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) red (ii) black.
Solution:
Number of possible events = 3 + 5 = 8
(i) Let A be the favourable events such that it is red.
∴ P(A) = 38
(ii) Let B be the favourable event such that it is black.
∴ P(B) = 58

Question 13.
A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
(i) red
(ii) white
(iii) not green.
Solution:
Total number of possible events = 5 + 8+4=17
(i) Let A be the favourable event such that it is red.
∴ P(A) = 57
(ii) Let B be the favourable event such that it is white
Then P (B) = 87
(iii) Let C be the favourable event such that it is not green.
∴ Number of favourable events = 5 + 8 = 13
∴ P(C) = 1317

Question 14.
If you put 21 consonants and 5 vowels in a bag. What would carry greater probability ? Getting a consonant or a vowel ? Find each probability.
Solution:
Total number of possible events = 21 + 5 = 26
(i) Probability of getting a consonant is greater as to number is greater than the other.
(ii) Let A be the favourable event such that it is a consonant.
∴ P(A) = 2126
(iii) Let B be the favourable event such that it is a vowel.
∴ P(B) = 526

Question 15.
If we have 15 boys and 5 girls in a class which carries a higher probability ? Getting a copy belonging to a boy or a girl ? Can you give it a value ?
Solution:
Number of possible outcome (events) = 15 + 5 = 20
∵ The number of boys is greater than the girls
∴ The possibility of getting a copy belonging to a boy is greater.
Let A be the favourable outcome (event) then
P(A) = 1520 = 34

Question 16.
If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is (i) white ? (if) black ?
Solution:
Total number of possible outcomes =6+3=9
(i) Let A be the favourable outcome which is white pair.
∴ P(A) = 69 = 23
(ii) Let B be the favourable outcome which is a black pair.
∴ P(B) = 39 = 13

Question 17.
If you have a spinning wheel with 3 green sectors, 1-blue sector and 1-red sector, what is the probability of getting a green sector ? Is it the maximum ?
Solution:
Total number of possible outcomes = 3 + 1 + 1 =5
Let A be the favourable outcome which is green sector
∴ P(A) = 35
∴ Number of green sectors is greater.
∴ It’s probability is greater.

Question 18.
When two dice are rolled :
(i) List the outcomes for the event that the total is odd.
(ii) Find probability of getting an odd total.
(iii) List the outcomes for the event that total is less than 5.
(iv) Find the probability for getting a total less than 5.
Solution:
∵ Every dice has 6 number from 1 to 6.
∴ Total outcomes = 6 x 6 = 36.
(i) List of outcomes for event then the total is odd will be (1,2), (1,4), (1,6), (2,1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6,1), (6, 3), (6, 5)
(ii) Probability of getting an odd total
Let A be the favourable outcomes which are
P(A) = 1836 = 12
(iii) List of outcomes for the event that total is less than 5 are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2,1), which are 6.
(iv) Probability of getting a total less than 5 Let B be the favourable outcome,
Then P (B) = 636 = 16

Exercise 24.1

Question 1.
Given below is the frequency distribution of the heights of 50 students of a class :

Draw a histogram representing the above data.
Solution:
We represent class intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis and y-axis we construct the rectangles as shown in the figure. This is the required histogram.

Question 2.
Draw a histogram of the following data :

Solution:
We represent class-intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis andy-axis, we construct rectangles as shown in the figure. This is the required histogram.

Question 3.
Number of workshops organized by a school in different areas during the last five years is as follows :

Draw a histogram representing the above data.
Solution:
We represent years along x-axis and number of workshops along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 4.
In a hypothetical sample of 20 people the amounts of money with them were found to be as follows :
114, 108,100, 98, 101,109,117,119, 126, 131, 136, 143, 156, 169, 182, 195, 207, 219, 235, 118.
Draw the histogram of the frequency distribution (taking one of the class intervals as 50-100).
Solution:
Highest sample = 235
Lowest sample = 98
Range = 235-98 = 137
Now frequency distribution table will be as under:

We represent class intervals along x-axis and frequency along j’-axis. Taking suitable intervals, we construct a rectangles as shown in the figure. This is the required histogram.

Question 5.
Construct a histogram for the following data:

Solution:
We represent monthly school fee (in Rs) along x-axis and number of schools along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 6.
Draw a histogram for the daily earnings of 30 drug stores in the following table :

Solution:
We represent daily earnings (in Rs) along x-axis and number of stores along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 7.
Draw a histogram to represent the following data:


Solution:
We represent monthly salary (in Rs) along x-axis and number of teachers along y-axis. Taking suitable intervals we construct rectangles as shown in the figure. This is the required histogram

Question 8.
The following histogram shows the number of literate females in the age group of 10 to 40 years in a town :
(i) Write the age group in which the number of literate female is highest.
(ii) What is the class width ?

(iii) What is the lowest frequency ?
(iv) What are the class marks of the classes ?
(v) In which age group literate females are least ?
Solution:
(i) The age group in which the number of literate females is 15-20.
(ii) The class width is 5.
(iii) Lowest frequency is 320.
(iv) The class marks of the classes are
10+152 = 252 =12.5, similarly other class marks will be 17.5,22.5,27.5,32.5,37.5
(v) The least literate females is in the class 10-15

Question 9.
The following histogram shows the monthly wages (in Rs) of workers in a factory:

(i) In which wage-group largest number of workers are being kept ? What is their number ?
(ii) What wages are the least number of workers getting ? What is the number of such workers ?
(iii) What is the total number of workers ?
(iv) What is the factory size ?
Solution:
(i) The largest number of workers are in wage group 950-1000 and is 8.
(ii) The least number of workers are in the wage group 900-950 and is 2.
(iii) Total number of workers is 40 (3 + 7 + 5 + 4 + 2 + 8 + 6 + 5)
(iv) The factory size is 50.

Question 10.
Below is the histogram depicting marks obtained by 43 students of a class :
(i) Write the number of students getting highest marks.
(ii) What is the class size ?

Solution:
(i) The number of students getting highest marks is 3.
(ii) The class size is 10.

Question 11.
The following histogram shows the frequency distribution of the ages of 22 teachers in a school:

(i) What is the number of eldest and youngest teachers in the school ?
(ii) Which age group teachers are more in the school and which least ?
(iii) What is the size of the classes ?
(iv) What are the class marks of the classes?
Solution:
(i) The number of eldest teacher is 1 and the number of youngest teacher is 2.
(ii) The teachers in age group 35-40 is most.
(iii) Size of classes is 5.
(iv) Class marks of class 20-25 is 20+252= 452 = 22.5
and similarly others will be 27.5, 32.5, 37.5, 42.5, 47.5, 52.5.

Question 12.
The weekly wages of 30 workers in a factory are given:
830,835,890,810,835,836,869,845,898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Mark a frequency table with intervals as 800-810,810-820 and so on, using tally marks.
Also, draw a histogram and answer the following questions:
(i) Which group has the maximum number of workers ?
(ii) How many workers earn Rs 850 and more ?
(iii) How many workers earn less than Rs 850?
Solution:
The frequency table will be as given below:

We represent wages (in Rs) along x-axis and number of workers along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
(i) Maximum workers are in the wage group 830-840.
(ii) Number of workers getting Rs 850 and more are 1 + 3 + 1 + 1 + 4 = 10.
(iii) Number of workers getting less than Rs 850 are 3 + 2 + 1 + 9 + 5 = 20

Exercise 19.1

Question 1.
What is the least number of planes that can enclose a solid ? What is the name of the solid ?
Solution:
The least number of planes that can enclose a solid is called a Tetrahedron.

Question 2.
Can a polyhedron have for its faces :
(i) three triangles ?
(ii) four triangles ?
(iii) a square and four triangles ?
Solution:
(i) No, polyhedron has three faces.
(ii) Yes, tetrahedron has four triangles as its faces.
(iii) Yes, a square pyramid has a square as its base and four triangles as its faces.

Question 3.
Is it possible to have a polyhedron with any given number of faces ?
Solution:
Yes, it is possible if the number of faces is 4 or more.

Question 4.
Is a square prism same as a cube ?
Solution:
Yes, a square prism is a cube.

Question 5.
Can a polyhedron have 10 faces, 20 edges and 15 vertices ?
Solution:
No, it is not possible as By Euler’s formula
F + V = E + 2
⇒ 10 + 15 = 20 + 2
⇒ 25 = 22
Which is not possible

Question 6.
Verify Euler’s formula for each of the following polyhedrons :

Solution:
(i) In this polyhedron,
Number of faces (F) = 7
Number of edges (E) = 15
Number of vertices (V) = 10
According to Euler’s formula,
F + V = E + 2
⇒ 7 + 10 = 15 + 2
⇒ 17 = 17
Which is true.
(ii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.
(iii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) =18
Number of vertices (V) = 11
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 11 = 18 + 2
⇒ 20 = 20
Which is true.
(iv) In this polyhedron,
Number of faces (F) = 5
Number of edges (E) = 8
Number of vertices (V) = 5
According to Euler’s formula,
F + V = E + 2
⇒ 5 + 5 = 8 + 2
⇒ 10 = 10
Which is true.
(v) In the given polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.

Question 7.
Using Euler’s formula, find the unknown:

Solution:
We know that Euler’s formula is
F + V = E + 2
(i) F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8
Faces = 8
(ii) F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V = 11 – 5 = 6
Vertices = 6
(iii) F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2 = 30
Edges = 30

Exercise 19.2

Question 1.
Which among the following are nets for a cube ?

Solution:
Nets for a cube are (ii), (iv) and (vi)

Question 2.
Name the polyhedron that can be made by folding each net:

Solution:
(i) This net is for a square

(ii) This net is for triangular prism.

(iii) This net is for triangular prism.

(iv) This net is for hexagonal prism.

(v) This net is for hexagon pyramid.

(vi) This net is for cuboid.

Question 3.
Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice ?


Solution:
Figure (i) shows the net of cube or dice.

Question 4.
Draw nets for each of the following polyhedrons:


Solution:
(i) Net for cube is given below :

(ii) Net of a triangular prism is as under :

(iii) Net of hexagonal prism is as under :

(iv) The net for pentagonal pyramid is as under:

Question 5.
Match the following figures:


Solution:
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)

Exercise 16.1

Question 1.
Define the following terms:
(i) Quadrilateral
(ii) Convex Quadrilateral.
Solution:
(i) Quadrilateral: A closed figure made of four line segments is called a quadrilateral such that:

(a) no three points of them are collinear
(b) the line segments do not intersect except at their ends points.
(ii) Convex quadrilateral: A quadrilateral is called a convex quadrilateral of the line containing any side of the quadrilateral has the remaining vertices on the same side of it. In the figure, quadrilateral ABCD is a convex quadrilateral.

Question 2.
In a quadrilateral, define each of the following:
(i) Sides
(ii) Vertices
(iii) Angles
(iv) Diagonals
(v) Adjacent angles
(vi) Adjacent sides
(vii) Opposite sides
(viii) Opposite angles
(ix) Interior
(x) Exterior
Solution:
(i) Sides: In a quadrilateral ABCD, form line segments AB, BC, CD and DA are called sides of the quadrilateral.
(ii) Vertices : The ends points are called the vertices of the quadrilateral. Here in the figure, A, B, C and D are its vertices.
(iii) Angles: A quadrilateral has four angles which are at their vertices. In the figure, ∠A, ∠B, ∠C and ∠D are its angles.
(iv) Diagonals: The line segment joining the opposite vertices is called diagonal. A quadrilateral has two diagonals.
(v) Adjacent Angles : The angles having a common arm (side) are called adjacent angles.
(vi) Adjacent sides : If two sides of a quadrilateral have a common end-point, these are called adjacent sides.
(vii) Opposite sides: If two sides do not have a common end-point of a quadrilateral, they are called opposite sides.
(viii) Opposite angles : The angles which are not adjacent are called opposite angles.
(ix) Interior: The region which is surrounded by the sides of the quadrilateral is called its interior.
(x) Exterior : The part of the plane made up by all points as the not enclosed by the quadrilateral, is called its exterior.

Question 3.
Complete each of the following, so as to make a true statement:
(i) A quadrilateral has ………… sides.
(ii) A quadrilateral has ………… angles.
(iii) A quadrilateral has ……….. vertices, no three of which are …………
(iv) A quadrilateral has …………. diagonals.
(v) The number of pairs of adjacent angles of a quadrilateral is ………….
(vi) The number of pairs of opposite angles of a quadrilateral is ……………
(vii) The sum of the angles of a quadrilateral is …………
(viii) A diagonal of a quadrilateral is a line segment that joins two ………. vertices of the quadrilateral.
(ix) The sum of the angles of a quadrilateral is …………. right angles.
(x) The measure of each angle of a convex quadrilateral is …………. 180°.
(xi) In a quadrilateral the point of intersection of the diagonals lies in ………….. of the quadrilateral.
(xii) A point is in the interior of a convex quadrilateral, if it is in the ……….. of its two opposite angles.
(xiii) A quadrilateral is convex if for each side, the remaining …………. lie on the same side of the line containing the side.
Solution:
(i) A quadrilateral has four sides.
(a) A quadrilateral has four angles.
(iii) A quadrilateral has four vertices, no three of which are collinear .
(iv) A quadrilateral has two diagonals.
(v) The number of pairs of adjacent angles of a quadrilateral is four .
(vi) The number of pairs of opposite angles ot a quadrilateral is two.
(vii) The sum of the angles of a quadrilateral is 360°.
(viii) A diagonal of a quadrilateral is a line segment that join two opposite vertices of the quadrilateral.
(ix) The sum of the angles of a quadrilateral is 4 right angles.
(x) The measure of each angle of a convex quadrilateral is less than 180°.
(xi) In a quadrilateral the point of intersection of the diagonals lies in interior of the quadrilateral.
(xii) A point is in the interior of a convex quadrilateral, if it is in the interior of its two opposite angles.
(xiii) A quadrilateral is convex if for each side, the remaining vertices lie on the same side of the line containing the side.

Question 4.
In the figure, ABCD is a quadrilateral.
(i) Name a pair of adjacent sides.
(ii) Name a pair of opposite sides.
(iii) How many pairs of adjacent sides are there?
(iv) How many pairs of Opposite sides are there ?

(v) Name a pair of adjacent angles.
(vi) Name a pair of opposite angles.
(vii) How many pairs of adjacent angles are there ?
(viii) How many pairs of opposite angles are there ?
Solution:
In the figure, ABCD is a quadrilateral
(i) Pairs of adjacent sides are AB, BC, BC, CD, CD, DA, DA, AB.
(ii) Pairs of opposite sides are AB and CD; BC and AD.
(iii) There are four pairs of adjacent sides.
(iv) There are two pairs of opposite sides.
(v) Pairs of adjacent angles are ∠A, ∠B; ∠B, ∠C; ∠C, ∠D; ∠D, ∠A.
(vi) Pairs of opposite angles are ∠A and ∠C; ∠B and ∠D.
(vii) There are four pairs of adjacent angles.
(viii) There are two pairs of opposite angles.

Question 5.
The angles of a quadrilateral are 110°, 72°, 55° and x°. Find the value of x.
Solution:
Sum of four angles of quadrilateral is 360°
110° + 12° + 55° + x° = 360°
⇒ 237° + x° = 360°
⇒ x° = 360° – 237° = 123°
x = 123°

Question 6.
The three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Solution:
The sum of four angles of a quadrilateral = 360°
Three angles are 110°, 50° and 40°
Let fourth angle = x
Then 110° + 50° + 40° + x° = 360°
⇒ 200° + x° = 360°
⇒ x = 360° – 200° = 160°
x = 160°

Question 7.
A quadrilateral has three acute angles each measures 80°. What is the measure of fourth angle ?
Solution:
Sum of four angles of a quadrilateral = 360°
Sum of three angles having each angle equal to 80° = 80° x 3 = 240°
Let fourth angle = x
Then 240° + x = 360°
⇒ x° = 360° – 240°
⇒ x° = 120°
Fourth angle = 120°

Question 8.
A quadrilateral has all its four angles of the same measure. What is the measure of each ?
Solution:
Let each equal angle of a quadrilateral = x
4x° = 360°
⇒ x° = 3604 = 90°
Each angle will be = 90°

Question 9.
Two angles of a quadrilateral are of measure 65° and the other two angles are equal. What is the measure of each of these two angles ?
Solution:
Measures of two angles each = 65°
Sum of these two angles = 2 x 65°= 130°
But sum of four angles of a quadrilateral = 360°
Sum of the remaining two angles = 360° – 130° = 230°
But these are equal to each other
Measure of each angle = 2302 = 115°

Question 10.
Three angles of a quadrilateral are equal. Fourth angle is of measure 150°. What is the measure of equal angles ?
Solution:
Sum of four angles of a quadrilateral = 360°
One angle = 150°
Sum of remaining three angles = 360° – 150° = 210°
But these three angles are equal
Measure of each angle = 2103 = 70°

Question 11.
The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles.
Solution:
Sum of four angles of a quadrilateral = 360°
and ratio in angles = 3 : 5 : 7 : 9
Let first angles = 2x
Then second angle = 5x
third angle = 7x
and fourth angle = 9x
3x + 5x + 7x + 9x = 360°
⇒ 24x = 369°
⇒ x = 36024 = 15°
First angle = 3x = 3 x 15° = 45°
second angle = 5x = 5 x 15° = 75°
third angle = 7x = 7 x 15° = 105°
and fourth angle = 9x = 9 x 15° = 135°

Question 12.
If the sum of the two angles of a quadrilateral is 180°, what is the sum of the remaining two angles ?
Solution:
Sum of four angles of a quadrilateral = 360°
and sum of two angle out of these = 180°
Sum of other two angles will be = 360° – 180° = 180°

Question 13.
In the figure, find the measure of ∠MPN.

Solution:
In the figure, OMPN is a quadrilateral in which
∠O = 45°, ∠M = ∠N = 90° (PM ⊥ OA and PN ⊥ OB)
Let ∠MPN = x°
∠O + ∠M + ∠N + ∠MPN = 360° (Sum of angles of a quadrilateral)
⇒ 45° + 90° + 90° + x° = 360°
⇒ 225° + x° = 360°
⇒ x° = 360° – 225°
⇒x = 135°
∠MPN = 135°

Question 14.
The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles ?
Solution:
The sides of a quadrilateral ABCD are produced in order, forming exterior angles ∠1, ∠2, ∠3 and ∠4.

Now ∠DAB + ∠1 = 180° (Linear pair) ……(i)
Similarly,
∠ABC + ∠2 = 180°
∠BCD + ∠3 = 180°
and ∠CDA + ∠4 = 180°
Adding, we get
∠DAB + ∠1 + ∠ABC + ∠2 + ∠BCD + ∠3 + ∠CDA + ∠4 = 180° + 180° + 180° + 180° = 720°
⇒ ∠DAB + ∠ABC + ∠CDA + ∠ADC + ∠1 + ∠2 + ∠3 + ∠4 = 720°
But ∠DAB + ∠ABC + ∠CDA + ∠ADB = 360° (Sum of angles of a quadrilateral)
360° + ∠1 + ∠2 + ∠3 + ∠4 = 720°
⇒ ∠l + ∠2 + ∠3 + ∠4 = 720° – 360° = 360°
Sum of exterior angles = 360°

Question 15.
In the figure, the bisectors of ∠A and ∠B meet at a point P. If ∠C = 100° and ∠D = 50°, find the measure of ∠APB.
Solution:
In quadrilateral ABCD,

∠D = 50°, ∠C = 100°
PA and PB are the bisectors of ∠A and ∠B.
In quadrilateral ABCD,
∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ ∠A + ∠B + 100° + 50° = 360°
⇒ ∠A + ∠B + 150° = 360°’
⇒ ∠A + ∠B = 360° – 150° = 210°
and 12 ∠A + 12 ∠B = 2102 = 105°
(PA and PB are bisector of ∠A and ∠B respectively)
∠PAB + ∠PBA = 105°
⇒ ∠PAB + ∠PBA + ∠APB = 180° (Sum of angles of a triangle)
⇒ 105° + ∠APB = 180°
⇒ ∠APB = 180° – 105° = 75°
∠APB = 75°

Question 16.
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral.
Solution:
Sum of angles A, B, C and D of a quadrilateral = 360°
i.e. ∠A + ∠B + ∠C + ∠D = 360°
But ∠A = ∠B = ∠C = ∠D = 1 : 2 : 4 : 5

Let ∠A = x,
Then ∠B = 2x
∠C = 4x
∠D = 5x
x + 2x + 4x + 5x = 360°
⇒ 12x = 360°
⇒ x = 36012 = 30°
∠A = x = 30°
∠B = 2x = 2 x 30° = 60°
∠C = 4x = 4 x 30° = 120°
∠D = 5A = 5 x 30° = 150°

Question 17.
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 12 (∠A + ∠B).
Solution:
In quadrilateral ABCD,

Question 18.
Find the number of sides of a regular polygon when each of its angles has a measures of
(i) 160°
(ii) 135°
(iii) 175°
(iv) 162°
(v) 150°.
Solution:
In a n-sided regular polygon, each angle

Question 19.
Find the number of degrees in each exterior angle of a regular pentagon.
Solution:
In a pentagon or a polygon, sum of exterior angles formed by producing the sides in order, is four right angles or 360°
Each exterior angle = 3605 = 72°

Question 20.
The measure of angles of a hexagon are x°, (x – 5)° (x – 5)°, (2x – 5)°, (2x – 5)°, (2x + 20)°. Find the value of x.
Solution:
We know that the sum of interior angels of a hexagon = 720° (180° x 4)
⇒ x + x – 5 + x – 5 + 2x – 5 + 2x – 5 + 2x + 20 = 720°
⇒ 9x – 20 + 20 = 720
⇒ 9x = 720
⇒ x = 7209 = 80°
x = 80°

Question 21.
In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of its exterior angles formed by producing the sides in the same order.
Solution:
In a convex hexagon ABCDEF, its sides AB, BG, CD, DE, EF and FA are produced in order forming exterior angles a, b, c, d, e, f

∠a + ∠b + ∠c + ∠d + ∠e + ∠f = 4 right angles (By definition)
By joining AC, AD, and AE, 4 triangles ABC, ACD, ADE and AEF are formed
In ∆ABC,
∠1 + ∠2 + ∠3 = 180° = 2 right angle (Sum of angles of a triangle) …… (i)
Similarly,
In ∆ACD,
∠4 +∠5 + ∠6 = 180° = 2 right angles
In ∆ADE,
∠1 + ∠8 + ∠9 = 2 right angles …(iii)
In ∆AEF,
∠10 + ∠11 + ∠12 = 2 right angles …(iv)
Joining (i), (ii), (iii) and (iv)
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 + ∠11 + ∠12 = 8 right angles
⇒ ∠2 + ∠3 + ∠5 + ∠6 + ∠8 + ∠9 + ∠11 + ∠12 + ∠1 + ∠4 + ∠7 + ∠10 = 8 right angles
⇒ ∠B + ∠C + ∠D + ∠E +∠F + ∠A = 8 right angles
⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 2 (∠a + ∠b + ∠c + ∠d + ∠e + ∠f)
Sum of all interior angles = 2(the sum of exterior angles)
Hence proved.

Question 22.
The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon.
Solution:
Let number of sides of a regular polygon = n
Each interior angle = 2n–4n right angles
Sum of all interior angles = 2n–4n x n
right angles = (2n – 4) right angles
But sum of exterior angles = 4 right angles
According to the condition,
(2n – 4) = 3 x 4 (in right angles)
⇒ 2n – 4 = 12
⇒ 2n = 12 + 4 = 16
⇒ n = 8
Number of sides of the polygon = 8

Question 23.
Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.
Solution:
Ratio in exterior angle and interior angles of a regular polygon = 1 : 5
But sum of interior and exterior angles = 180° (Linear pair)

By cross multiplication:
6n – 12 = 5n
⇒ 6n – 5n = 12
⇒ n = 12
Number of sides of polygon is 12

Question 24.
PQRSTU is a regular hexagon. Determine each angle of ∆PQT.
Solution:
In regular hexagon, PQRSTU, diagonals PT and QT are joined.

In ∆PUT, PU = UT
∠UPT = ∠UTP
But ∠UPT + ∠UTP = 180° – ∠U = 180° – 120° = 60°
∠UPT = ∠UTP = 30°
∠TPQ = 120° – 30° = 90° (QT is diagonal which bisect ∠Q and ∠T)
∠PQT = 1202 = 60°
Now in ∆PQT,
∠TPQ + ∠PQT + ∠PTQ = 180° (Sum of angles of a triangle)
⇒ 90° + 60° + ∠PTQ = 180°
⇒ 150° + ∠PTQ = 180°
⇒ ∠PTQ = 180° – 150° = 30°
Hence in ∆PQT,
∠P = 90°, ∠Q = 60° and ∠T = 30°

Exercise 15.1

Question 1.
Draw rough diagrams to illustrate the following:
(i) Open curve
(ii) Closed curve
Solution:

Question 2.
Classify the following curves as open or closed.

Solution:
Open curves : (i), (iv) and (v) are open curves.
(ii) , (iii), and (vi) are closed curves.

Question 3.
Draw a polygon and shade its interior. Also draw its diagonals, if any.
Solution:
In the given polygon, the shaded portion is its interior region AC and BD are the diagonals of polygon ABCD.

Question 4.
Illustrate, if possible, each one of the following with a rough diagram:
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
Solution:
(i) Close curve but not a polygon.

(ii) An open curve made up entirely of line segments.

(iii) A polygon with two sides. It is not possible. At least three sides are necessary

Question 5.
Following are some figures : Classify each of these figures on the basis of the following:


(i) Simple curve
(ii) Simple closed curve
(iii) Polygon
(iv) Convex polygon
(v) Concave polygon
(vi) Not a curve
Solution:
(i) It is a simple closed curve and a concave polygon.
(ii) It is a simple closed curve and convex polygon.
(iii) It is neither a curve nor polygon.
(iv) it is neither a curve not a polygon.
(v) It is a simple closed curve but not a polygon.
(vi) It is a simple closed curve but not a polygon.
(vii) It is a simple closed curve but not a polygon.
(viii) It is a simple closed curve but not a polygon.

Question 6.
How many diagonals does each of the following have ?
(i) A convex quadrilateral
(ii) A regular hexagon
(iii) A triangle.
Solution:
(i) A convex quadrilateral
Here n = 4

Question 7.
What is a regular polygon ? State the name of a regular polygon of:
(i) 3 sides
(ii) 4 sides
(iii) 6 sides.
Solution:
A regular polygon is a polygon which has all its sides equal and so all angles are equal,
(i) 3 sides : It is an equilateral triangle.
(ii) 4 sides : It is a square.
(iii) 6 sides : It is a hexagon.

Exercise 12.1

Question 1.
Write each of the following as percent: Solution—
(i) 725
(ii) 16625
(iii) 58
(iv) 0.8
(v) 0.005
(vi) 3 : 25
(vii) 11 : 80
(viii) 111 : 125
(ix) 13 : 75
(x) 15 : 16
(xi) 0.18
(xii) 7125
Solution:

Question 2.
Convert the following percentages to fractions and ratios :
(i) 25%
(ii) 2.5%
(iii) 0.25%
(iv) 0.3%
(v) 125%
Solution:

Question 3.
Express the following as decimal fractions :
(i) 27%
(ii) 6.3%
(iii) 32%
(iv) 0.25%
(v) 7.5%
(vi) 18 %
Solution:

Exercise 12.2

Question 1.
Find :
(i) 22% of 120
(ii) 25% of Rs. 1000
(iii) 25% of 10 kg
(iv) 16.5% of 5000 metres
(v) 135% of 80 cm
(vi) 2.5% of 10000 ml
Solution:

Question 2.
Find the number ‘a’, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3
(iii) 12 % of a is 50
(iv) 100% of a is 100
Solution:

Question 3.
x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.
Solution:
x = 5% of y, y = 24% of z.
x = 480

Question 4.
A coolie deposits Rs. 150 per month in his post office Saving Bank account. If this is 15% of his monthly income, find his monthly income.
Solution:
Let his monthly income = Rs. x
15% of x = Rs. 150

Question 5.
Asha got 86.875% marks in the annual examination. If she got 695 marks, find the number of marks of the Examination.
Solution:
Let total marks of the examination = x
86.875% of x = 695
=> 86.875 x 1100 x x = 695

Question 6.
Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened ?
Solution:
Let the school opened for = x days = 90% of x = 216

Question 7.
A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees.
Solution:
Number of total trees = 2000

Rest trees = 2000 – (240 + 360) = 2000 – 600 = 1400
Number of orange trees = 1400

Question 8.
Balanced diet should contain 12% of protein, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food in take.
Solution:
Balance diet contains
Protein = 12%
Fats = 25%
Carbohydrates = 63%
Number of total calories = 2600
Number of calories of proteins = 12% of 2600 = 12100 x 2600 = 312
Number of calories of fats = 25% of 2600 = 25100 x 2600 = 650
Number of calories of carbohydrates = 63% of 2600 = 63100 x 2600 = 1638

Question 9.
A cricketer scored a total of 62 runs in 96 balls. He hits 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in :
(i) Sixes
(ii) Fours
(iii) Twos
(iv) Singles
Solution:
Total score of a cricketer = 62 runs
(z) Number of sixes = 3
Run from 3 sixes = 3 x 6 = 18
Percentage = 1862 x 100 = 29.03%
(ii) Number of fours = 8
Total run from 8 fours = 4 x 8 = 32
Percentage = 3262 x 100 = 51.61%
(iii) Number of twos = 2
Total score from 2 twos = 2 x 2 = 4
Percentage = 462 x 100 = 40062 = 6.45%
(iv) Number of single run = 8
Percentage = 862 x 100 = 80062 = 12.9%

Question 10.
A cricketer hits 120 runs in 150 balls during a test match. 20% of the runs came in 6’s, 30% in 4’s, 25% in 2’s and the rest in 1’s. How many runs did he score in :
(i) 6’s
(ii) 4’s
(iii) 2’s
(iv) singles
What % of his shots were scoring ones ?
Solution:
Total runs scored by a cricketer =120
(i) Number of runs from sixes (6’s) = 20% of 120

Question 11.
Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest ?
Solution:
Total earning from investment = Rs. 187
Percent earning = 22%
Let his investment = x
Then 22% of x = Rs. 187

Question 12.
Rohit deposits 12% his income in a bank. He deposited Rs. 1440 in the bank during 1997. What was his total income for the year 1997 ?
Solution:
Deposit in the bank = Rs. 1440
Percentage = 12% of his total income
Let his total income = Rs. x

Question 13.
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur ?
Solution:
(i) In gunpowder,
Nitre = 75%
Sulphur = 10%
Let total amount of gunpowder = x kg
Nitre = 9 kg

Question 14.
An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy ?
Solution:
In an alloy,
Number of parts of tin = 15
and number of parts of copper = 105
Total parts = 15 + 105 = 120
Percentage of copper in the alloy = 105120 x 100 = 87.5%

Question 15.
An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.
Solution:
In an alloy,
Copper = 32%
Nickel = 40%
Rest is zinc = 100 – (32 + 40) = 100 – 72 = 28%
Mass of zinc in 1 kg = 28% of 1 kg = 28100 x 100 gm = 280 gm.

Question 16.
A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride ?
Solution:
Distance travelled before first stop = 122 km
Let total journey = x km
10% of x = 122

Question 17.
A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female ?
Solution:
Total numbers of teachers = 30
Number of male teachers = 12
Number of female teacher = 30 – 12 = 18
Percentage of female teachers = 18×10030 = 60%

Question 18.
Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income ?
Solution:
Let Anil’s income = Rs. 100
Then Aman’s income = Rs, 100 – 20 = Rs. 80
Now, difference of both’s incomes = 100 – 80 = Rs. 20
Anil income is Rs. 20 more than that of Aman’s
Percentage = 20×10080 = 25%

Question 19.
The value of a machine depreciates every year by 5%. If the present value of the machine be Rs. 100000, what will be its value after 2 years ?
Solution:
Present value of machine = Rs. 100000
Rate of depreciation per year = 5%
Period = 2 years
Value of machine after 2 years

Question 20.
The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years ?
Solution:
Present population of the town = 60000
Increase annually = 10%
Period = 2 years
Population after 2 years will be

Question 21.
The population of a town increases 10% annually. If the present population is 22000, find its population a year ago.
Solution:
Let the population of the town a year ago was = x
Increase in population = 10%

Question 22.
Ankit was given an increment of 10% on his salary. His new salary is Rs. 3575. What was his salary before increment ?
Solution:
Let the salary of Ankit before increment = x
Increment given = 10% of the salary
Salary after increment will be

Question 23.
In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase ?
Solution:
Let price of petrol before budged = Rs. 100
Increase = 10%
Price after budget = Rs. 100 + 10 = Rs. 110
Let the consumption of petrol before budget = 100 l
Price pf 100 l = Rs. 110
Now of new price is Rs. 110, consumption = 100 l
are of new price will be 100, then

Question 24.
Mohan’s income is Rs. 15500 per month. He saves 11% of his income. If his income increases by 10% then he reduces his saving by 1%, how much does he save now ?
Solution:
Mohan’s income = Rs. 15500

We see that the savings is same
There is no change in savings.

Question 25.
Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s ?
Solution:
Let Shalu’s income = Rs. 100
Then Shikha’s income will be = Rs. 100 + 60 = Rs. 160
Now difference in their incomes = Rs. 160 – 100 = Rs. 60
Shalu’s income is less than Shikha’s income by Rs. 60
Percentage less = 60×100160 = 752 % = 37.5%

Question 26.
Rs. 3500 is to be shared among three people so that the first person gets 50% of the second who in turn gets 50% of the third. How much will each of them get ?
Solution:
Let the third person gets = Rs. x
Then second person will get

Question 27.
After a 20% hike, the cost of Chinese Vase is Rs. 2000. What was the original price of the object ?
Solution:
Let the original price of the vase = Rs. x
Hike in price = 20%

Original price of the vase = Rs. 1666.66

Exercise 11.1

Question 1.
Rakesh can do a piece of work in 20 days. How much work can he do in 4 days ?
Solution:
Rakesh can do it in 20 days = 1
his 1 day’s work = 120
and his 4 days work = 120 x 4 = 15 th work

Question 2.
Rohan can paint 13 of a painting in 6 days. How many days will he take to complete the painting ?
Solution:
Rohan can paint 13 of painting in = 6 days
he will complete the painting in = 6×31 = 18 days

Question 3.
Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together ?
Solution:
Anil’s 1 day’s work = 15

Question 4.
Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone ?
Solution:

Question 5.
Sita can finish typing a 100 page document in 9 hours, Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together?
Solution:
Sita can do a work in 1 hour = 19

Question 6.
A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work ?
Solution:

Question 7.
A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together ?
Solution:

Question 8.
A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work ?
Solution:

Question 9.
A, B and C can reap a field in 1534 days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it ?
Solution:

Question 10.
A and B can polish the floors of a building in 10 days A alone can do 14 th of it in 12 days. In how many days can B alone polish the floor ?
Solution:

Question 11.
A and B can finish a work in 20 days. A alone can do 15 th of the work in 12 days. In how many days can B alone do it ?
Solution:

Question 12.
A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work ?
Solution:

Question 13.
A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work ?
Solution:

Question 14.
Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll ?
Solution:

Question 15.
A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.
Solution:

Question 16.
6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job ?
Solution:
6 men can complete the work in = 7 days
1 man will complete the same work in = 7 x 6 days (Less men, more days)
21 men will finish the work in = 7×621 days (More men, less days) = 2 days

Question 17.
8 men can do a piece of work in 9 days. In how many days will 6 men do it ?
Solution:
8 men can do a work in = 9 days
1 men will do the work in = 9 x 8 days (Less men, more days)
6 men will do the work in = 9×86 days (More men, less days)
= 726 = 12 days

Question 18.
Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?
Solution:
Reema can weave 35 baskets in = 25 days

Question 19.
Neha types 75 pages in 14 hours. How many pages will she type in 20 hours ?
Solution:
Neha types pages in 14 hours = 75 pages

Question 20.
If 12 boys earn Rs. 840 in 7 days, what will 15 boys earn in 6 days ?
Solution:
12 boys in 7 days earn an amount of = Rs. 840

Question 21.
If 25 men earn Rs. 1000 in 10 days, how much will 15 men earn in 15 days ?
Solution:
25 men can earn in 10 days = Rs. 1000

Question 22.
Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days ?
Solution:
Ashu can copy a book in 18 days working in a day = 8 hours
He will copy the book in 1 day working = 8 x 18 hours a day (Less days, more hours a day)
He will copy the book in 12 days working in a day = 8×1812 hours
(More days, less hours a day)
= 14412 = 12 hours a day

Question 23.
If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour.
Solution:
135 garlands in 3 hours are prepared by = 9 girls
1 garland in 3 hours will be prepared by

Question 24.
A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together ?
Solution:
First tap’s 1 hour work to fill the cistern = 18

Question 25.
Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank ?
Solution:

Question 26.
A pipe can fill a cistern in 10 hours. Due to a leak in the bottom, it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?
Solution:

Question 27.
A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely ?
Solution:

Question 28.
A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern of both the tap and the pipe are opened together ?
Solution: