Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.4)
Data Handling – II (Central Values Exercise 23.4) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
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Page 1
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:
Page 2
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.
Page 3
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:
Page 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:
Marks 10 11 12 13 14 16 19 20
Number of students 3 5 4 5 2 3 2 1
Using empirical formula, find its mode.
Solution:
Calculation of mean
Mean = S f
i
x
i
/ S 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.
Weight (in kg) 48 50 52 54 58
Number of persons 4 3 2 2 1
Find the median and mean weights. Using empirical relation, calculate its mode.
Solution:
x
i
f
i
x
i
f
i
48 4 192
50 3 150
52 2 104
54 2 108
58 1 58
Total S f
i
= 12 S f
i
x
i
= 612
Calculation of mean
Mean = S f
i
x
i
/ S f
i
= 612/12
= 51 kg
Here n = 12
Page 5
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:
Marks 10 11 12 13 14 16 19 20
Number of students 3 5 4 5 2 3 2 1
Using empirical formula, find its mode.
Solution:
Calculation of mean
Mean = S f
i
x
i
/ S 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.
Weight (in kg) 48 50 52 54 58
Number of persons 4 3 2 2 1
Find the median and mean weights. Using empirical relation, calculate its mode.
Solution:
x
i
f
i
x
i
f
i
48 4 192
50 3 150
52 2 104
54 2 108
58 1 58
Total S f
i
= 12 S f
i
x
i
= 612
Calculation of mean
Mean = S f
i
x
i
/ S 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.
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