Ncert Solutions For Class 9 Maths Ex 14.4

Ncert Solutions For Class 9 Maths Chapter 14 Ex 14.4

Q1)In a state assembly election, the seats won by different political parties through a polling outcome is as shown in the figure below.

 

Political party A B C D E F
Seats won 80 70 65 43 18 9

 

(i)         Represent the results of the polling by the help of a bar graph.

(ii)        Predict the winner with maximum number of seats.

 

Solution:

(i)

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(ii) Party ‘A’ has won the maximum seats.

 

Q2)The length of a plant with 40 leaves are measured to one millimeter correctly, and the data is obtained which is shown in the following table;

 

SL.NO LENGTH (IN MM) NUBER OF LEAVES
1. 118 – 126 3
2. 127 – 135 5
3. 136 – 144 9
4. 145 – 153 12
5. 154 – 162 5
6. 163 – 171 4
7. 172 – 180 2

(i) Figure out the given data using histogram.

(ii) Find out the suitable graphical representation for the same data if any.

(iii)Can we conclude that the maximum number of leaves as 153 mm long? If yes, justify?

 

Solution:

(i) Firstly, make the data continuous as the data given is in a discontinuous class interval. The difference is 1. So we need to subtract the lower limit by 0.5 and add 0.5 to the upper limit.

SL.NO Length (in mm) Number of leaves
1. 118 – 126 3
2. 127 – 135 5
3. 136 – 144 9
4. 145 – 153 12
5. 154 – 162 5
6. 163 – 171 4
7. 172 – 180 2

 

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(ii)  Yes, Frequency polygon can also be represented using the data.

(iii)  No, we cannot conclude the maximum number of leaves as 153mm because the length of the maximum number of leaves is in the range of 144.5 – 153.5.

 

Q3)The given table shows the life times of 400 neon lamps:

Life Time (in hours) Number of lamps
300 – 400 14
400 – 500 56
500 – 600 60
600 – 700 86
700 – 800 74
800 – 900 62
900 – 1000 48

 

 

(i)         Represent the following diagram above with the help of a histogram.

(ii)        Predict the number of lamps that have a life time of more than 700 hours.

 

Solution:

(i)

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(ii) The lamps that have more than 700 hours of life time 700 hours is = 74 + 62 + 48 = 184

3) The following table gives the distribution of students in two sections according to the marks obtained by them.

 

Class A Class B
Marks Frequency Marks Frequency
0 – 10 3 0 – 10 5
10 – 20 9 10 – 20 19
20 – 30 17 20 – 30 15
30 – 40 12 30 – 40 10
40 – 50 9 40 – 50 1

 Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons, compare the performance of the two sections.

Solution:

The class mark can be identified by (Lower limit + Upper limit)/2.
For section A,

Marks Class Mark Frequency
0-10 5 3
10-20 15 9
20-30 25 17
30-40 35 12
40-50 45 9

For section B,

Marks Class Mark Frequency
0-10 5 5
10-20 15 19
20-30 25 15
30-40 35 10
40-50 45 1

 

Now, we draw a frequency polygon for the given data.
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4) A random survey of the number of kids of various age groups playing in a park was found as follows:

 

Age(in years) Number of Children
1 – 2 5
2 – 3 3
3 – 5 6
5 – 7 12
7 – 10 9
10 – 15 10
15 – 17 4

Draw a histogram to represent the data above.

Solution:

The class interval in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram .The class interval with minimum class size 1 is selected and the length of the rectangle is proportionate to it.

 

Age (in years) Number of kids (frequency) Width of class Length of rectangle
1-2 5 1 (5/1)×1 = 5
2-3 3 1 (3/1)×1 = 3
3-5 6 2 (6/2)×1 = 3
5-7 12 2 (12/2)×1 = 6
7-10 9 3 (9/3)×1 = 3
10-15 10 5 (10/5)×1 = 2
15-17 4 2 (4/2)×1 = 2

Taking the age of kids on x-axis and proportion of children per 1 year interval on y-axis, the histogram can be drawn

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