Data Handling – II (Central Values Exercise 23.4) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download

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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: 
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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. 
 
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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: 
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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 
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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.