In this post, you will find the NCERT solutions for class 10 maths ex 13.1. These solutions are based on the latest curriculum of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 13.1
1. A survey was conducted by a group of
students as a part of their environment awareness programme, in which they
collected the following data regarding the number of plants in 20 houses in a
locality. Find the mean number of plants per house.
|
Number
of plants |
0 – 2 |
2 – 4 |
4 – 6 |
6 – 8 |
8 – 10 |
10 – 12 |
12 – 14 |
|
Number
of houses |
1 |
2 |
1 |
5 |
6 |
2 |
3 |
Solution:
Since the number
of plants and the number of houses are small in their values, so we use direct
method.
Mean = 8.1
Hence, the mean
number of plants per house is 8.1.
2. Consider the following distribution of
daily wages of 50 workers of a factory.
|
Daily
wages (in Rs) |
100 – 120 |
120 – 140 |
140 – 160 |
160 – 180 |
180 – 200 |
|
Number
of workers |
12 |
14 |
8 |
6 |
10 |
Solution:
Width of the
class (h) =
20
Therefore,
Using formula,
Mean
= 150 + 20(–0.24)
= 150 – 4.8
= 145.2
Hence, the mean
daily wages of the workers of the factory is Rs. 145.20.
3. The following distribution shows the daily
pocket allowance of children of a locality. The mean pocket allowance is Rs.18.
Find the missing frequency (f).
|
Daily
pocket allowance (in Rs.) |
11 – 13 |
13 – 15 |
15 – 17 |
17 – 19 |
19 – 21 |
21 – 23 |
23 – 25 |
|
Number
of children |
7 |
6 |
9 |
13 |
f |
5 |
4 |
Solution:
Therefore,
⇒ 
⇒ 
⇒ 2f
= 40
⇒ f
= 40/2
⇒ f
= 20
Hence, the
missing frequency is 20.
4. Thirty women were examined in a hospital by
a doctor and the number of heart beats per minute were recorded and summarised
as follows:
|
Number
of heart beats per minute |
65 – 68 |
68 – 71 |
71 – 74 |
74 – 77 |
77 – 80 |
80 – 83 |
83 – 86 |
|
Number
of women |
2 |
4 |
3 |
8 |
7 |
4 |
2 |
Find the mean heart beats per minute for these women, choosing a suitable method.
Solution:
From given data,
let assumed mean (a) = 75.5
Width of the
class (h) =
3
Therefore,
Using formula,
Mean
= 75.5 + 0.39
= 75.89
= 75.9 (approx.)
Hence, the mean
heart beats per minute for these women is 75.9.
5. In a retail market, fruit vendors were
selling mangoes kept in packing boxes. These boxes contained varying number of
mangoes. The following was the distribution of mangoes according to the number
of boxes.
|
Number
of mangoes |
50 – 52 |
53 – 55 |
56 – 58 |
59 – 61 |
62 – 64 |
|
Number
of boxes |
15 |
110 |
135 |
115 |
25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution: Since the number of mangoes and the number
of boxes are large numerically, so we use step-deviation method.
Width of the
class (h) =
3
Therefore,
= 25/40
= 0.0625 (approx.)
Using formula,
Mean
= 57 + 3(0.0625)
= 57 + 0.1875
= 57.1875
= 57.19 (approx.)
Hence, the mean
number of mangoes kept in a packing box is 57.19.
6. The table below shows the daily expenditure
on food of 25 households in a locality.
|
Daily
expenditure (in
Rs.) |
100 – 150 |
150 – 200 |
200 – 250 |
250 – 300 |
300 – 350 |
|
Number
of households |
4 |
5 |
12 |
2 |
2 |
Find the mean daily expenditure on food by a suitable method.
Solution:
Width of the
class (h) =
50
Therefore,
= –7/25
= –0.28
Using formula,
Mean
= 225 + 50(–
0.28)
= 225 – 14
= 211
Hence, the mean
daily expenditure on food is Rs. 211.
7. To find out the concentration of SO2 in
the air (in parts per million, i.e., ppm), the data was collected for 30
localities in a certain city and is presented below:
|
Concentration of SO2 (in
ppm) |
Frequency |
|
0.00 –
0.04 |
4 |
|
0.04 –
0.08 |
9 |
|
0.08 –
0.12 |
9 |
|
0.12 – 0.16 |
2 |
|
0.16 – 0.20 |
4 |
|
0.20 – 0.24 |
2 |
Solution:
From given data,
let assumed mean (a) = 0.10
Width of the
class (h) =
0.04
Therefore, 
(approx.)
Using formula,
Mean
= 0.10 + 0.04(–
0.033)
= 0.10 – 0.00132
= 0.09868
= 0.099 (approx.)
Hence,
the mean concentration of SO2 in the air
is 0.099 ppm.
8. A class teacher has the following absentee
record of 40 students of a class for the whole term. Find the mean number of
days a student was absent.
|
Number
of days |
0 – 6 |
6 – 10 |
10 – 14 |
14 – 20 |
20 – 28 |
28 – 38 |
38 – 40 |
|
Number
of students |
11 |
10 |
7 |
4 |
4 |
3 |
1 |
Solution:
Therefore,
mean
= 17 + (–181)/40
= 17 – 4.52
= 12.48
Hence, the mean
number of days a student was absent is 12.48 days.
9. The following table gives the literacy rate
(in percentage) of 35 cities. Find the mean literacy rate.
|
Literacy
rate (in %) |
45 – 55 |
55 – 65 |
65 – 75 |
75 – 85 |
85 – 95 |
|
Number
of cities |
3 |
10 |
11 |
8 |
3 |
Solution:
From given data,
let assumed mean (a) = 70
Width of the
class (h) =
10
Therefore,
= –2/35
= –0.057
Using formula,
Mean
= 70 + 10(–
0.057)
= 70 – 0.57
= 69.43
Hence, the mean literacy rate is 69.43%.