NCERT Solutions Class 8 Maths Chapter 5

Q.1 For which of these would you use a histogram to show the data?

(a) The number of letters for different areas in a postman’s bag.

(b) The height of competitors in an athletics meet.

(c) The number of cassettes produced by 5 companies.

(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a station. Give reasons for each.

Ans-

1. The number of letters for different areas in a postman’s bag: Here, we cannot use the histogram as we do not know about the number of letters of different areas. Therefore, we can’t divide the given data into class intervals.
2. The height of competitors in an athletics meet:

Here, a histogram can be drawn as the given data can be divided into class intervals.

3. The number of cassettes produced by 5 companies: As we are not aware about the number of cassettes produced by the given companies, so we cannot divide the data into class intervals. Therefore, a histogram can’t be drawn.
4. The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a station: The histogram can be drawn here as the given data can be divided into class intervals.

Q.2 The shoppers who come to a departmental store are marked as: man (M), woman(W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning:

W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W

Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.

Ans-

The frequency distribution table is as follows:

$\begin{array}{|ccc|}\hline \mathrm{Shopper}& \mathrm{Tally}\mathrm{marks}& \mathrm{Number}\\ \mathrm{W}& \bcancel{\mathrm{llll}}\bcancel{\mathrm{llll}}\bcancel{\mathrm{llll}}\bcancel{\mathrm{llll}}\bcancel{\mathrm{llll}}\mathrm{lll}& 28\\ \mathrm{M}& \bcancel{\mathrm{llll}}\bcancel{\mathrm{llll}}\bcancel{\mathrm{llll}}& 15\\ \mathrm{B}& \bcancel{\mathrm{llll}}& 5\\ \mathrm{G}& \bcancel{\mathrm{llll}}\bcancel{\mathrm{llll}}\mathrm{ll}& 12\\ \hline\end{array}$

The bar graph is as follows:

Q.3 The weekly wages (in Rs) of 30 workers in a factory are:

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

Using tally marks make a frequency table with intervals as 800–810, 810–820 and so on.

Ans-

The frequency distribution table is as follows:

Q.4 Draw a histogram for the frequency table made for the data in Question 3, 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?

Ans-

Histogram is as follows:

(i) The maximum number of workers is in the group 830 − 840.

(ii) From the graph, we can observe that the workers who earn more than Rs 850 falls in the group of 850 − 860 or 860 − 870 or 870 − 880 or 880 − 890.

Hence, the total number of workers earning more than 850 = 1 + 3 + 1 + 1 + 4 = 10

(iii) It is clear from the graph that the workers who earn less than Rs 850 falls in the group of 800 − 810 or 810 − 820 or 820 − 830 or 830 − 840 or 840 − 850.

Hence, the total number of workers earning less than 850 = 3 + 2 + 1 + 9 + 5 = 20

Q.5 The number of hours for which students of a particular class watched television during holidays is shown through the given graph.

Answer the following.

1. For how many hours did the maximum number of students watch TV?
2. How many students watched TV for less than 4 hours?
3. How many students spent more than 5 hours in watching TV?

Ans-

(i)The maximum number of students watched TV for 4 − 5 hours.

(ii) The students who watched TV for less than 4 hours falls in the group of 1 − 2 hours or 2 − 3 hours or 3 − 4 hours.

Hence, total number of students who watched TV for less than 4 hours

= 4 + 8 + 22

= 34

(iii) The students who watched TV for more than 5 hours falls in the group of 5 − 6 hours or 6 − 7 hours.

Hence, total number of students who watched TV for more than 5 hours

= 8 + 6

= 14

Q.6 A survey was made to find the type of music that a certain group of young people liked in a city. Adjoining pie chart shows the findings of this survey.

From this pie chart answer the following:

(i) If 20 people liked classical music, how many young people were surveyed?

(ii) Which type of music is liked by the maximum number of people?

(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make?

Ans-

(i) From the given pie chart the number of people who like classical music =10%

Let the number of young people who were surveyed be ‘x’.

According to the question, we get

10% of x = 20

$\begin{array}{l}\frac{10}{100}\times \mathrm{x}=20\\ \Rightarrow \mathrm{x}=\frac{20\times 100}{10}\\ \Rightarrow \mathrm{x}=200\end{array}$

Therefore, 200 young people were surveyed.

(ii) From the given pie chart, it can be observed that light music is liked by maximum number of people.

(iii)

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{CD}’\mathrm{s}\mathrm{for}\mathrm{light}\mathrm{music}=40\mathrm{\%}\\ =\frac{40}{100}\times 1000=400\\ \mathrm{Number}\mathrm{of}\mathrm{CD}’\mathrm{s}\mathrm{for}\mathrm{folk}\mathrm{music}=30\%\\ =\frac{30}{100}\times 1000=300\\ \mathrm{Number}\mathrm{of}\mathrm{CD}’\mathrm{s}\mathrm{for}\mathrm{classical}\mathrm{music}=10\mathrm{\%}\\ =\frac{10}{100}\times 1000=100\\ \mathrm{Number}\mathrm{of}\mathrm{CD}’\mathrm{s}\mathrm{for}\mathrm{semi}-\mathrm{classical}\mathrm{music}=20\mathrm{\%}\\ =\frac{20}{100}\times 1000=200\end{array}$

Q.7 A group of 360 people were asked to vote or their favourite season from the three seasons rainy, winter and summer.
(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii)Draw a pie chart to show this information.

Ans-

(i) Winter season got the maximum votes.
(ii) Here, total number of votes = 90 + 120 + 150 = 360.
The central angles of each sector are shown in the following table:

$\begin{array}{|cccc|}\hline \mathrm{Season}& \mathrm{No}.\mathrm{of}\mathrm{votes}& \mathrm{In}\mathrm{Fraction}& \mathrm{Central}\mathrm{angle}\\ \mathrm{Summer}& 90& \frac{90}{360}& \frac{90}{360}\times {360}^{\circ }={90}^{\circ }\\ \mathrm{Rainy}& 120& \frac{120}{360}& \frac{120}{360}\times {360}^{\circ }={120}^{\circ }\\ \mathrm{Winter}& 150& \frac{150}{360}& \frac{150}{360}\times {360}^{\circ }={150}^{\circ }\\ \hline\end{array}$

(iii) The pie chart is as follows:

Q.8 Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Ans-

The central angles for a pie chart are calculated in the following table:

$\begin{array}{|cccc|}\hline \mathrm{Language}& \mathrm{No}.\text{of\hspace{0.33em}}\mathrm{students}& \mathrm{In}\mathrm{Fraction}& \mathrm{Central}\mathrm{angle}\\ \mathrm{Blue}& 18& \frac{18}{36}& \frac{18}{36}\times {360}^{\circ }={180}^{\circ }\\ \mathrm{Green}& 9& \frac{9}{36}& \frac{9}{36}\times {360}^{\circ }={90}^{\circ }\\ \mathrm{Red}& 6& \frac{6}{36}& \frac{6}{36}\times {360}^{\circ }={60}^{\circ }\\ \mathrm{Yellow}& 3& \frac{3}{36}& \frac{3}{36}\times {360}^{\circ }={30}^{\circ }\\ \hline\end{array}$

The pie chart is as follows:

Q.9 The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.

(i) In which subject did the student score 105 marks?(Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)

(ii) How many more marks were obtained by the student in Mathematics than in Hindi?

(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.

Ans-

(Hint: Just study the central angles)

Total marks obtained by the student are 540.

Hence, 540 marks represent 360º.

Now, we have to find the central angle for 105 marks.

$\begin{array}{l}=\frac{105}{540}\times 360°\\ =70°\end{array}$

Therefore, the subject in which the student scored 105 marks is Hindi.

(ii) Marks obtained by student in Mathematics = 90 %

$\begin{array}{l}=\frac{90°}{360°}\times 540\\ =135\end{array}$

Now, the marks obtained by student in Hindi = 70 %

= 105

Therefore, the difference between the marks of Mathematics and Hindi = 135 – 105 = 30.

Hence, 30 more marks were obtained by the student in Mathematics than in Hindi.

(iii) The sum of central angles of Social Science and Mathematics

= 90° + 65°

= 155°

However, the sum of central angles of Science and Hindi

= 80° + 70°

= 150°

We can observe that, the sum of the central angles for Social Science and Mathematics is more than that of Science and Hindi.

Therefore, the student scored more in Social Science and Mathematics than in Science and Hindi.

List the outcomes you can see in these experiments.

(a) Spinning a wheel

(b) Tossing two coins together

(a)The possible outcomes on spinning a wheel are A, B, C, and D.

(b) The possible outcomes when two coins are tossed together are:

HT, TH, HH, TT where, H represents Head and T represents Tail.

Q. When a die is thrown; list the outcomes of an event of getting
(i) (a) a prime number

(b) not a prime number

(ii) (a) a number greater than 5

(b) a number not greater than 5.

Ans-

When a die is thrown, the possible outcomes are: 1, 2, 3, 4, 5 and 6.

Therefore, the outcomes of an event of getting:

(i) (a) a prime number are: 2, 3 and 5

(b) not a prime number are : 1 , 4 and 6

(ii) (a) a number greater than 5 is : 6

(b) a number not greater than 5 are: 1, 2, 3, 4 and 5

Find the

1. Probability of the pointer stopping on D ?

2. Probability of getting an ace from a well shuffled deck of 52 playing cards?

3. Probability of getting a red apple. (See figure below)

Ans-

(i) The possible outcomes on spinning a wheel are :

A, B, C, and D

Therefore, the pointer can stop at one of the regions.

Total number of regions is 5.

The pointer will stop at region D only once.

Hence, probability that the pointer will stop at region D =

$\frac{1}{5}.$

(ii) Total number of cards in a well shuffled deck = 52

Total number of Ace cards = 4.

Therefore, the probability of getting an ace card =

$\frac{4}{52}=\frac{1}{13}.$

(iii)Total number of apples = 7

Total number of red apples = 4

Total number of green apples = 3

Therefore, the probability of getting a red apple =

$\frac{4}{7}$

Q. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of

(i) getting a number 6?

(ii) getting a number less than 6?

(iii) getting a number greater than 6?

(iv) getting a 1-digit number?

Ans-

Total number of slips = 10

(i) The probability of getting a number 6 =

$\frac{1}{10}$

(ii) The numbers less than 6 are: 1, 2, 3, 4 and 5

Therefore, the probability of getting a number less than 6 =

$\frac{5}{10}=\frac{1}{2}$

(iii) The numbers greater than 6 are: 7, 8, 9 and 10

Therefore, the probability of getting a number greater than 6 =

$\frac{4}{10}=\frac{2}{5}$

(iv) The one digit numbers are: 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Therefore, the probability of getting a 1-digit number =

$\frac{9}{10}$

Q. 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? What is the probability of getting a non blue sector?

Ans-

Total number of green sectors =3

Total number of blue sectors = 1

Total number of red sector = 1

Total number of sectors = 5

Therefore, the probability of getting a green sector =

$\frac{3}{5}$

Now, the non blue sectors are the blue and red sectors. Therefore, the probability of getting a non blue sector =

$\frac{4}{5}$

Find the probabilities of the events given in Question 2.

When a die is thrown, the possible outcomes are: 1, 2, 3, 4, 5 and 6

(i) (a) The prime numbers are: 2, 3 and 5

Hence, the probability of getting a prime number =

$\frac{3}{6}=\frac{1}{2}$

(b)The numbers which are not prime are: 1, 4 and 6

Hence, the probability of not getting a prime number =

$\frac{3}{6}=\frac{1}{2}$

(ii) (a) A number greater than 5 is 6.

Therefore, the probability of getting a number greater than 6 =

$\frac{1}{6}$

(b) A number not greater than 5 are: 1, 2, 3, 4 and 5

Therefore, the probability of not getting a number greater than 5 =

$\frac{5}{6}$

Q. The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.

Ans-

The central angle for each subject is as follows:

$\begin{array}{|cccc|}\hline \mathrm{Language}& \mathrm{No}.\text{of\hspace{0.33em}}\mathrm{students}& \mathrm{In}\mathrm{Fraction}& \mathrm{Central}\mathrm{angle}\\ \mathrm{Hindi}& 40& \frac{40}{72}& \frac{40}{72}\times {360}^{\circ }={200}^{\circ }\\ \mathrm{English}& 12& \frac{12}{72}& \frac{12}{72}\times {360}^{\circ }={60}^{\circ }\\ \mathrm{Marathi}& 9& \frac{9}{72}& \frac{9}{72}\times {360}^{\circ }={45}^{\circ }\\ \mathrm{Tamil}& 7& \frac{7}{72}& \frac{7}{72}\times {360}^{\circ }={35}^{\circ }\\ \mathrm{Bengali}& 4& \frac{4}{72}& \frac{4}{72}\times {360}^{\circ }={20}^{\circ }\\ \hline\end{array}$

The pie chart is as follows: