The number on a die are 1, 2, 3,4, 5, or 6.
(i) ∵ Numbers less than 5 are 1, 2, 3 and 4.
∴ Possible outcomes = 4
⇒  Probability of getting a number less than \(5={4\over6}\ or\ {2\over3}\)
(ii) ∵ The composite numbers are 4 and 6.
∴ Number of outcomes = 2
⇒ Probability getting a composite number \(={2\over6}\ or\ {1\over3}\)
(iii) ∵ The prime numbers are 2, 3 and 5.
∴ Number of outcomes = 3
⇒ Probability of getting a prime number \(={3\over6}={1\over2}\)
(iv) ∵ Numbers more than 3 are 4, 5, and 6.
∴ Number of outcomes = 3
⇒ Probability of getting a number greater than \(3={3\over6}\ or\ {1\over2}\)

Total number of outcomes = 10
(i) ∴ One card is having 7 on it
∴ Number of favourable outcome =
⇒ Probability of getting a card having 7 on it = \(1\over10\)
(ii) ∵ Only one card IS having two digit number (i.e., 10)
∴ Number of favourable outcome =
⇒ Probability of getting a card having 10
(two diizgiit number) on it = 1/10
(iii) ∴ The numbers less than 5 are I, 2, 3,
and 4
∴ Number of favourable outcomes = 4
⇒ Probability of getting a number less then \(5={4\over10}\ or\ {2\over5}\)
(iv)  The numbers more than 5 are 6, 7, 8, 9 and 10.
∴ Number of favourable outcomes = 5
⇒ Probability of getting a number more than \(5={5\over10}\ or\ {1\over5}\)

(i) This histogram represents heights (in cms) of students of Class VIII.
(ii) The group (150-160) contains maximum number of students (which has 8 students).
(iii) 9 students (3 + 2 + 4) have height of 160 cm or more.
(iv) Number of students in the group of 130 - 140 cm = 3.
Number of students in the group of 160 - 170 cm = 3
∴ The required ratio = 3 : 3 = 1 : 1

(i) The class interval = [Upper limit of a class] - [Lower limit of the same class] = 225 - 200 = 25
(ii) The class 300 - 325 is having the lowest frequency (which is 17).
(iii) The class 375 - 400 is having the highest frequency (which is 120).
(iv) The class intervals 275-300 and 350 - 375 have the same frequency (which is 45).
(v) The lower limit of the class interval 300 - 325 is 300.
(vi) The upper limit of the class interval 275 - 300 is 300.

1: 1

Here, the total marks obtained by the student are 540. For 540 marks, the complete central angle is 360°.
(i) Since, for 540 marks, central angle = 360°
So, for 1 mark, central angle = \(\frac{360^o}{540}\)
Then, for 105 marks, central angle = \(\frac{360^o}{540}\times 105=(\frac{36\times 105}{54})^o=70^o\)
Hence, the student scored 105 marks in Hindi.
(ii) Marks obtained by the student in Mathematics = Fraction part X Total marks = \(\frac{90^o}{360^o}\times\) 540 = 135
Marks obtained in Hindi = 105
Difference between marks in Mathematics and Hindi = (135 -105) = 30
Hence, the student obtained 30 marks more in
Mathematics than in Hindi.
(iii) Marks obtained by the student in Mathematics = 135
Marks obtained by the student in Social Science = Fraction part for Social Science X Total marks
= \(\frac{65^o}{360^o}\times 540=\frac{65\times 54}{36}=\frac{65\times 6}{4}\) = 97.5
\(\therefore\) Sum of the marks obtained in Social Science and
Mathematics = 97.5 + 135 = 232.5
Now, marks obtained by the student in Science = Fraction parr of Science X Total marks
= \(\frac{80^o}{360^o}\times 540 = \frac{80\times 6}{4}\) = 120
and marks obtained by the student in Hindi = Fraction part for Hindi X Total marks
= \(\frac{70^o}{360^o}\times 540=\frac{70\times 6}{4}\) = 105
\(\therefore\) Sum of the marks obtained in Hindi and Science = 105 + 120 = 225
Hence, sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.

No, since each event have equal chance of occurring.

No. since outcomes of the experiment are equally likely. So each have an equal chance of occurring.

5 [ \(\therefore\) 15 - 10 = 5]

We know that total angle at the centre of a circle is 360°.
Fraction part for sleep = \(\frac{Hours\ for\ sleep}{Total\ hours}=\frac{8}{24}=\frac{1}{3}\)
\(\therefore\) Central angle of sector for sleep = \(\frac{1}{3}\times 360^o =120^o\)
Similarly, for other central angles, we get the following table:
First, we find the central angle corresponding to given activities

Now, draw a circle and divide it into sectors with corresponding central angle, we get the required pie chart given below:

(i) From the given pie chart, we have the following table:

Since, percentage of entertainment is highest, so entertainment programme are viewed the most.
(ii) Here, number of viewers watching news = 15%
Number of viewers watching informative = 10%
\(\therefore\) Sum of numbers of viewers watching news and informative
= (15 + 10)% = 25%
= Number of viewers watching sports
Hence, news and informative programmes have number of viewers equal to those watching sports channel.

We know that, if given data can be grouped into class interval, then we represent the data by histogram. Here, different areas and five companies cannot be grouped into class interval, whereas height and time can be grouped into class interval, so (b) and (d) can be represented by a histogram.

Here, lowest observation =31
and highest observation = 65
So, we have to take class intervals 30-35, 35-40, 40-45, ... in first column and corresponding frequency i.e. number of students whose weight lie in this class interval, in the second column. Then, frequency distribution table for the above data is given below:

Class Interval   [Weights (in kg)]	Tally marks	Frequency  (Number of students)
30-35	III	3
35-40	IIII	4
40-45	IIII	4
45-50		5
50-55	I	1
55-60	I	1
60-65	I	1
65-70	I	1
Total		20

No, because bar heights give the quantity for each category. Since the height remains unchanged. So, the information conveyed regarding the quantity remain unchanged.

From the given bar graph, we find that,
i) The information given by the bar graph is the number of students in Class VIII in various academic years in the school.
(ii)In the year 2004-05, increase in the number of students is maximum.
(iii) The number of students is maximum in the year 2007 - 08.
(iv) False, because the number of students during 2005-06 is more than twice that of 2003-04.