RD Sharma Solutions for Class 7 Maths Chapter 23 Data Handling - II (Central Values) Exercise 23.4

Students can view and download the RD Sharma Solutions for Class 7 Maths Exercise 23.4 of Chapter 23 Data Handling – II from the links provided below. The syllabus given by the CBSE for all the classes is completely based on the RD Sharma textbook. Hence, preparation of exams using the RD Sharma Solutions for Class 7 will definitely help students to score well for the final exam. This exercise explains the mode of a group of observations.

RD Sharma Solutions for Class 7 Maths Chapter 23 – Data Handling – II (Central Values) Exercise 23.4

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Access answers to Maths RD Sharma Solutions for Class 7 Chapter 23 – Data Handling – II (Central values) Exercise 23.4

1. Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14

By using the empirical relation also find the mean.

Solution:

Arranging the data in ascending order such that same numbers are put together, we get:

12, 12, 13, 13, 14, 14, 14, 16, 19

Here, n = 9.

Therefore median = ((n+1)/2)^th term

Median = value of 5^th term

Median = 14

Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.

Now,

Mode = 3 Median – 2 Mean

14 = 3 x 14 – 2 Mean

2 Mean = 42 – 14 = 28

Mean = 28 ÷ 2

= 14.

2. Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34

Solution:

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34, 35, 35, 38, 42

Here, n = 7

Therefore median = ((n+1)/2)^th term

Median = value of 4^th term

Median = 35

Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

3.  Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Solution:

Arranging the data in ascending order such that same values are put together, we get:

0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6

Here, 2, 3 and 4 occur three times each. Therefore, 2, 3 and 4 are the three modes.

4. The runs scored in a cricket match by 11 players are as follows:

6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10

Find the mean, mode and median of this data.

Solution:

Arranging the data in ascending order such that same values are put together, we get:

6, 8, 10, 10, 15, 15, 50, 80, 100, 120

Here, n = 11

Therefore median = ((n+1)/2)^th term

Median = value of 6^th term

Median = 15

Here, 10 occur three times. Therefore, 10 is the mode of the given data.

Now,

Mode = 3 Median – 2 Mean

10 = 3 x 15 – 2 Mean

2 Mean = 45 – 10 = 35

Mean = 35 ÷ 2

= 17.5

5. Find the mode of the following data:

12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Solution:

Arranging the data in ascending order such that same values are put together, we get:

10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 18

Here, clearly, 14 occurs the most number of times.

Therefore, 14 is the mode of the given data.

6. Heights of 25 children (in cm) in a school are as given below:

168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 164, 163, 160, 165, 163, 162

What is the mode of heights?

Also, find the mean and median.

Solution:

Arranging the data in tabular form, we get:

Height of Children (cm )	Tally marks	Frequency
160	|||	3
161	|	1
162	||||	4
163		10
164	|||	3
165	|||	3
168	|	1
Total		25

Therefore median = ((n+1)/2)^th term

Median = value of 13^th term

Median = 163 cm

Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.

Mode = 3 Median – 2 Mean

163 = 3 x 163 – 2 Mean

2 Mean = 326

Mean = 163 cm.

7. The scores in mathematics test (out of 25) of 15 students are as follows:

19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20

Find the mode and median of this data. Are they same?

Solution:

Arranging the data in ascending order such that same values are put together, we get:

5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

Here, n = 15

Therefore median = ((n+1)/2)^th term

Median = value of 8^th term

Median = 20

Here, clearly, 20 occurs most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.

Yes, the median and mode of the given data are the same.

8. Calculate the mean and median for the following data:

Using empirical formula, find its mode.

Solution:

Calculation of mean

Mean = Σ f_i x_i/ Σ f_i

= 332/25

= 13.28

Here, n = 25, which is an odd number. Therefore,

Therefore median = ((n+1)/2)^th term

Median = value of 13^th term

Median = 13

Now, by using empirical formula we have,

Mode = 3Median – 2 Mean

Mode = 3 (13) – 2 (13.28)

Mode = 39 – 26.56

Mode = 12.44.

9. The following table shows the weights of 12 persons.

Find the median and mean weights. Using empirical relation, calculate its mode.

Solution:

Calculation of mean

Mean = Σ f_i x_i/ Σ f_i

= 612/12

= 51 kg

Here n = 12

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 6^th term + value of 7^th term)/2

= (50 + 50)/2

= 50

Now by empirical formula we have,

Now,

Mode = 3 Median – 2 Mean

Mode = 3 x 50 – 2 x 51

Mode = 150 – 102

Mode = 48 kg.

Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.