Students can view and download the PDF of RD Sharma Solutions for Class 7 Maths Exercise 23.2 of Chapter 23 Data Handling – II, available here. The BYJU’S subject experts in Maths formulate the solutions to questions present in the textbook based on the understanding abilities of students. Exercise 23.2 explains the exterior and interior angles of a triangle. RD Sharma Solutions for Class 7 PDF can be downloaded by the students and referred to while solving the textbook problems. This exercise also includes the arithmetic mean of grouped data.
Download the PDF of RD Sharma Solutions for Class 7 Maths Chapter 23 – Data Handling – II (Central Values) Exercise 23.2
Access answers to Maths RD Sharma Solutions For Class 7 Chapter 23 – Data Handling – II (Central values) Exercise 23.2
1. A die was thrown 20 times and the following scores were recorded:
5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1
Prepare the frequency table of the scores on the upper face of the die and find the mean score.
Solution:
The frequency table for the given data is as follows:
| x: | 1 | 2 | 3 | 4 | 5 | 6 |
| f: | 2 | 5 | 1 | 4 | 6 | 2 |
To compute arithmetic mean we have to prepare the following table:
| Scores (xi) | Frequency (fi) | xi fi |
| 1 | 2 | 2 |
| 2 | 5 | 10 |
| 3 | 1 | 3 |
| 4 | 4 | 16 |
| 5 | 6 | 30 |
| 6 | 2 | 12 |
| Total | Σ fi = 20 | Σ fi xi |
Mean score = Σ fi xi/ Σ fi
= 73/20
= 3.65
2. The daily wages (in Rs) of 15 workers in a factory are given below:
200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180
Prepare the frequency table and find the mean wage.
Solution:
| Wages (xi) | 130 | 150 | 180 | 200 |
| Number of workers (fi) | 2 | 4 | 6 | 3 |
To compute arithmetic mean we have to prepare the following table:
| xi | fi | xi fi |
| 130 | 2 | 260 |
| 150 | 4 | 600 |
| 180 | 6 | 1080 |
| 200 | 3 | 600 |
| Total | Σ fi = N = 15 | Σ fi xi = 2540 |
Mean score = Σ fi xi/ Σ fi
= 2540/15
= 169.33
3. The following table shows the weights (in kg) of 15 workers in a factory:
| Weight (in Kg) | 60 | 63 | 66 | 72 | 75 |
| Number of workers | 4 | 5 | 3 | 1 | 2 |
Calculate the mean weight.
Solution:
Calculation of mean:
| xi | fi | xi fi |
| 60 | 4 | 240 |
| 63 | 5 | 315 |
| 66 | 3 | 198 |
| 72 | 1 | 72 |
| 75 | 2 | 150 |
| Total | Σ fi = N = 15 | Σ fi xi = 975 |
Mean score = Σ fi xi/ Σ fi
= 975/15
= 65 kg
4. The ages (in years) of 50 students of a class in a school are given below:
| Age (in years) | 14 | 15 | 16 | 17 | 18 |
| Number of students | 15 | 14 | 10 | 8 | 3 |
Find the mean age.
Solution:
Calculation of mean:
| xi | fi | xi fi |
| 14 | 15 | 210 |
| 15 | 14 | 210 |
| 16 | 10 | 160 |
| 17 | 8 | 136 |
| 18 | 3 | 54 |
| Total | Σ fi = N = 50 | Σ fi xi = 770 |
Mean score = Σ fi xi/ Σ fi
= 770/50
= 15.4 years
5. Calculate the mean for the following distribution:
| x: | 5 | 6 | 7 | 8 | 9 |
| f: | 4 | 8 | 14 | 11 | 3 |
Solution:
| xi | fi | xi fi |
| 5 | 4 | 20 |
| 6 | 8 | 48 |
| 7 | 14 | 98 |
| 8 | 11 | 88 |
| 9 | 3 | 27 |
| Total | Σ fi = N = 40 | Σ fi xi = 281 |
Mean score = Σ fi xi/ Σ fi
= 281/40
= 7.025
6. Find the mean of the following data:
| x: | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
| f: | 13 | 15 | 16 | 18 | 16 | 15 | 13 |
Solution:
| xi | fi | xi fi |
| 19 | 13 | 247 |
| 21 | 15 | 315 |
| 23 | 16 | 368 |
| 25 | 18 | 450 |
| 27 | 16 | 432 |
| 29 | 15 | 435 |
| 31 | 13 | 403 |
| Total | Σ fi = N = 106 | Σ fi xi = 2650 |
Mean score = Σ fi xi/ Σ fi
= 2650/106
= 25
7. The mean of the following data is 20.6. Find the value of p.
| x: | 10 | 15 | p | 25 | 35 |
| f: | 3 | 10 | 25 | 7 | 5 |
Solution:
| xi | fi | xi fi |
| 10 | 3 | 30 |
| 15 | 10 | 150 |
| P | 25 | 25p |
| 25 | 7 | 175 |
| 35 | 5 | 175 |
| Total | Σ fi = N = 50 | Σ fi xi = 530 + 25p |
Mean score = Σ fi xi/ Σ fi
20.6 = 530 + 25p/50
530 + 25 p = 20.6 x 50
25 p = 1030 – 530
p = 500/25
p = 20
8. If the mean of the following data is 15, find p.
| x: | 5 | 10 | 15 | 20 | 25 |
| f: | 6 | p | 6 | 10 | 5 |
Solution:
| xi | fi | xi fi |
| 5 | 6 | 30 |
| 10 | P | 10p |
| 15 | 6 | 90 |
| 20 | 10 | 200 |
| 25 | 5 | 125 |
| Total | Σ fi = 27 + p | Σ fi xi = 445 + 10p |
Mean score = Σ fi xi/ Σ fi
15 = 445 + 10p/27 + p
445 + 10 p = 405 + 15p
5 p = 445 – 405
p = 40/5
p = 8
9. Find the value of p for the following distribution whose mean is 16.6
| x: | 8 | 12 | 15 | p | 20 | 25 | 30 |
| f: | 12 | 16 | 20 | 24 | 16 | 8 | 4 |
Solution:
| xi | fi | xi fi |
| 8 | 12 | 96 |
| 12 | 16 | 192 |
| 15 | 20 | 300 |
| P | 24 | 24p |
| 20 | 16 | 320 |
| 25 | 8 | 200 |
| 30 | 4 | 120 |
| Total | Σ fi = N = 100 | Σ fi xi = 1228 + 24p |
Mean score = Σ fi xi/ Σ fi
16.6 = 1228 + 24p/100
1228 + 24 p = 16.6 x 100
24 p = 1660 – 1228
p = 432/24
p = 18
10. Find the missing value of p for the following distribution whose mean is 12.58
| x: | 5 | 8 | 10 | 12 | p | 20 | 25 |
| f: | 2 | 5 | 8 | 22 | 7 | 4 | 2 |
Solution:
| xi | fi | xi fi |
| 5 | 2 | 10 |
| 8 | 5 | 40 |
| 10 | 8 | 80 |
| 12 | 22 | 264 |
| P | 7 | 7p |
| 20 | 4 | 80 |
| 25 | 2 | 50 |
| Total | Σ fi = N = 50 | Σ fi xi = 524 + 7p |
Mean score = Σ fi xi/ Σ fi
12.58 = 524 + 7p/50
524 + 7 p = 12.58 x 50
7 p = 629 – 524
p = 105/7
p = 15
11. Find the missing frequency (p) for the following distribution whose mean is 7.68
| x: | 3 | 5 | 7 | 9 | 11 | 13 |
| f: | 6 | 8 | 15 | p | 8 | 4 |
Solution:
| xi | fi | xi fi |
| 3 | 6 | 18 |
| 5 | 8 | 40 |
| 7 | 15 | 105 |
| 9 | P | 9p |
| 11 | 8 | 88 |
| 13 | 4 | 52 |
| Total | Σ fi = N = 41 + p | Σ fi xi = 303 + 9p |
Mean score = Σ fi xi/ Σ fi
7.68 = 303 + 9p/41 + p
303 + 9 p = 314.88 + 7.68p
1.32 p = 314.88 – 303
p = 11.88/1.32
p = 9
12. Find the value of p, if the mean of the following distribution is 20
| x: | 15 | 17 | 19 | 20 + p | 23 |
| f: | 2 | 3 | 4 | 5p | 6 |
Solution:
| xi | fi | xi fi |
| 15 | 2 | 30 |
| 17 | 3 | 51 |
| 19 | 4 | 76 |
| 20 + p | 5P | (20 + p) 5p |
| 23 | 6 | 138 |
| Total | Σ fi = 15 + 5p | Σ fi xi = 295 + (20 +p) 5p |
Mean score = Σ fi xi/ Σ fi
20 = [(295 + (20 + p) 5p)]/ 15 + 5p
295 + 100 p + 5p2 = 300 + 100p
5p2 = 300 – 295
5p2= 5
p2 = 1
p = 1
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