In this post, you will find the NCERT solutions for class 10 maths ex 13.2. These solutions are based on the latest syllabus of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 13.2
1. The following table shows the ages of the
patients admitted in a hospital during a year:
|
Age
(in years) |
5 – 15 |
15 – 25 |
25 – 35 |
35 – 45 |
45 – 55 |
55 – 65 |
|
Number
of patients |
6 |
11 |
21 |
23 |
14 |
5 |
Solution:
To calculate the mode:
In the given data, maximum frequency is 23 and it corresponds to the class
interval 35 – 45.
Therefore,
the modal class
= 35 – 45
And l = 35, f1 = 23, f0 = 21, f2 = 14 and h = 10
=
=
= 35 + 20/11
= 35 + 1.8
= 36.8
To
calculate the mean:
Width of the
class (h) =
10
Therefore,
= 43/80
= 0.5375
Using formula,
Mean
= 30 + 10
(0.5375)
= 30 + 5.375
= 35.375
Hence, mode of
the given data is 36.8 years and mean of the given data is 35.37 years.
Therefore, it is
clear from the above result that the maximum number of patients admitted in the
hospital are of the age 36.8 years and on an average the age of a patient
admitted in the hospital is 35.37 years.
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
|
Lifetimes (in hours) |
0 – 20 |
20 – 40 |
40 – 60 |
60 – 80 |
80 – 100 |
100 – 120 |
|
Frequency |
10 |
35 |
52 |
61 |
38 |
29 |
Determine the modal lifetimes of the components.
Solution: In the given data, maximum frequency
is 61 and it corresponds to the class interval 60 – 80.
Therefore,
the modal class
= 60 – 80
And l = 60, f1 = 61, f0
= 52, f2 = 38 and h = 20
Therefore, Mode =
=
= 60 + 9/32 × 20
= 60 + 5.625
= 65.625
Hence, the modal
lifetimes of the components is 65.625 hours.
3. The following data gives the distribution
of total monthly household expenditure of 200 families of a village. Find the
modal monthly expenditure of the families. Also, find the mean monthly
expenditure:
|
Expenditure (in Rs.) |
Number of families |
|
1000 – 1500 |
24 |
|
1500 – 2000 |
40 |
|
2000 – 2500 |
33 |
|
2500 – 3000 |
28 |
|
3000 – 3500 |
30 |
|
3500 – 4000 |
22 |
|
4000 – 4500 |
16 |
|
4500 – 5000 |
7 |
Solution:
To calculate the mode:
In the given data, maximum frequency is 40 and it corresponds to the class
interval 1500 – 2000.
Therefore,
the modal class
= 1500 – 2000
And l = 1500, f1 = 40, f0
= 24, f2 = 33
and h = 500
Since, Mode =
=
= 1500 +
8000/23
= 1500 + 347.83
= 1847.83
To calculate the mean:
Width of the
class (h) =
500
Since,
= –35/200
= –0.175
Using formula,
Mean
= 2750 + 500(–
0.175)
= 2750 – 87.50
= 2662.50
Hence, the modal
monthly expenditure of the families is Rs. 1847.83 and the mean monthly
expenditure is Rs. 2662.50.
4. The following distribution gives the
state-wise teacher-student ratio in higher secondary schools of India. Find the
mode and mean of this data. Interpret the two measures.
|
Number of students per teacher |
Number of states/U.T. |
|
15 – 20 |
3 |
|
20 – 25 |
8 |
|
25 – 30 |
9 |
|
30 – 35 |
10 |
|
35 – 40 |
3 |
|
40 – 45 |
0 |
|
45 – 50 |
0 |
|
50 – 55 |
2 |
Solution: To calculate the mode: In the given data, maximum frequency is 10 and it corresponds to the class interval 30 – 35.
Therefore,
the modal class
= 30 – 35
And l = 30, f1 = 10, f0
= 9, f2 = 3 and h = 5
Since, Mode =
=
=
= 30 + 5/8
= 30 + 0.625
= 30.6 (approx.)
To calculate the mean:
Width of the
class (h) =
5
Therefore,
= –23/35
= –0.65
Using formula,
Mean
= 32.5 + 5(–0.65)
= 32.5 – 3.25
= 29.2 (approx.)
Hence, the mode
and mean of the given data is 30.6 and 29.2.
From the above results,
it is clear that the states/U.T. have a student teacher ratio of 30.6 on an
average, this ratio is 29.2.
5. The given distribution shows the number of
runs scored by some top batsmen of the world in one-day international cricket
matches:
|
Runs scored |
Number of batsmen |
|
3000 – 4000 |
4 |
|
4000 – 5000 |
18 |
|
5000 – 6000 |
9 |
|
6000 – 7000 |
7 |
|
7000 – 8000 |
6 |
|
8000 – 9000 |
3 |
|
9000 – 10000 |
1 |
|
10000 – 11000 |
1 |
Find the mode of the data.
Solution:
In the given
data, maximum frequency is 18 and it corresponds to the class interval 4000 –
5000.
Therefore,
the modal class
= 4000 – 5000
And l = 4000, f1 = 18, f0
= 4, f2 = 9 and h = 1000
Therefore,
Mode =
=
=
= 4000 + 14000/23
= 4000 + 608.6956
= 4608.7 (approx.)
Hence,
the mode of the given data is 4608.7 runs.
6. A student noted the number of cars passing
through a spot on a road for 100 periods each of 3 minutes and summarised it in
the table given below:
|
Number of cars |
Frequency |
|
0 – 10 |
7 |
|
10 – 20 |
14 |
|
20 – 30 |
13 |
|
30 – 40 |
12 |
|
40 – 50 |
20 |
|
50 – 60 |
11 |
|
60 – 70 |
15 |
|
70 – 80 |
8 |
Find the mode of the data.
Solution:
In the given
data, maximum frequency is 20 and it corresponds to the class interval 40 – 50.
Therefore,
the modal class
= 40 – 50
And l = 40, f1 = 20, f0
= 12, f2 = 11 and h = 10
Since, Mode =
=
= 40 + 80/17
= 40 + 4.70588
= 44.7 (approx.)
Hence, the mode of the given data is 44.7 cars.