NCERT Solutions for Class 10 Maths Ex 13.2

# NCERT Solutions for Class 10 Maths Ex 13.2

## 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

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

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

Therefore, Mode =

=

=

= 35 + 20/11

= 35 + 1.8

= 36.8

To calculate the mean:

From given data, assumed mean (a) = 30

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:

From given data, assumed mean (a) = 2750

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:

From given data, assumed mean (a) = 32.5

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.