# Ncert Solutions For Class 10 Maths Ex 14.2

## Ncert Solutions For Class 10 Maths Chapter 14 Ex 14.2

Question 1:

The age of patients admitted in a hospitals are given in the following table for a complete 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

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

Solution:

Modal class = 35-45, length = 35, height = 10, a1=23,a0=21anda3=14$a_{1} = 23, a_{0} = 21\, and\, a_{3} = 14$

Therefore, Mode = l+a1a02a1a0a2×h$l +\frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}}\times h$

= 35+(23212×232114)×10$35+\left ( \frac{23-21}{2\times 23-21-14} \right )\times 10$

= 35+211×10=36.8$35+\frac{2}{11}\times 10=36.8$

Calculating Mean

 CLASS INTERVAL ai zi aizi 5-15 6 10 60 15-25 11 20 220 25-35 21 30 630 35-45 21 30 920 45-55 14 50 700 55-65 5 60 300 Sum ai = 80 Sum aizi= 2830

p¯=aixiai=283080=35.37$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}=\frac{2830}{80}=35.37$

therefore$\ therefore$The mode of data displays that the maximum number of patients is in the age group of 26.8, whereasthe average age of all the patients is 35.37.

Question 2:

The data given below provides the information for the observed lifetime (in hours) of 225 electrical components:

 Lifetime ( in hours) 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100  – 120 Frequency 10 35 52 61 38 29

Find the modal lifetimes of the components.

Solution:

Modal class =  60-80, l = 60,a1=61,a0=52,a2=38$a_{1} = 61, a_{0} = 52, a_{2} = 38$ and h = 20

Mode=l+(a1a02a1a0a2)×h$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

=60+(61522×615238)×20$= 60 + \left ( \frac{61 – 52}{2 \times 61 – 52 – 38 } \right ) \times 20$

=60+932×20=65.62$= 60 + \frac{9}{32} \times 20 = 65.62$

Question 3:

The total monthly household expenses of 200 families of a town are given in the following table. Calculate the modal monthly expenses of the families.  Also, calculate the mean monthly expenses.

 Expenses 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:

Modal class = 1500-2000, l = 1500,a1=40,a0=24,a2=33$a_{1} = 40, a_{0} = 24, a_{2} = 33$ and h = 500

Mode=l+(a1a02a1a0a2)×h$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

=1500+(40242×402433)×500$= 1500 + \left ( \frac{40 – 24}{2 \times 40 – 24 – 33} \right ) \times 500$

=1500+(1623)×500=1847.82$= 1500 + \left ( \frac{16}{23} \right ) \times 500 = 1847.82$

Calculation of mean:

 Class Interval ai zi di =zi – b ui= di/h aiui 1000 – 1500 24 1250 -1500 -3 -72 1500 – 2000 40 1750 -1000 -2 -80 2000 – 2500 33 2250 -500 -1 -33 2500 – 3000 28 2750 0 0 0 3000 – 3500 30 3250 500 1 30 3500 – 4000 22 3750 1000 2 44 4000 – 4500 16 4250 1500 3 48 4500 – 5000 7 4750 2000 4 28 ai = 200 aiui = -35

p¯=aixiai×h$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a{i}}\times h$ =275035200×500$= 2750 – \frac{35}{200} \times 500$

= 2750 – 87.5 = 2662.50

Question 4:

The state – wise teacher- student ratio in a college in India are given in the following table. Calculate the mean and mode of the given data. Interpret the two measures.

 Number of students per teacher Number of states/UT 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:

Modal class = 30-35, l = 30, a1=10,a0=9,a2=3$a_{1} = 10, a_{0} = 9, a_{2} = 3$ and h = 5

Mode=l+(a1a02a1a0a2)×h$Mode = l +\left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

=30(2092×1093)×5$= 30 – \left ( \frac{20 – 9}{2 \times 10 – 9 – 3 } \right ) \times 5$

=3018×5=30.625$= 30 – \frac{1}{8} \times 5 = 30.625$

Calculating mean:

 Number of students per teacher ai zi di= zi –b ui = ui/h aiui 15 – 20 3 17.5 -15 -3 -9 20 – 25 8 22.5 -10 -2 -16 25 – 30 9 27.5 -5 -1 -9 30 – 35 10 32.5 0 0 0 35 – 40 3 37.5 5 1 3 40 – 45 0 42.5 10 2 0 45 – 50 0 47.5 15 3 0 50 – 55 2 52.5 20 4 8 Sum ai = 35 sum aiui= -23

p¯=aixiai×h$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}\times h$

=30.52335×5$= 30.5 – \frac{23}{35}\times 5$

=30.5237=29.22$= 30.5 – \frac{23}{7} = 29.22$

Therefore, from the mode we know that the maximum number of states has 30 – 35 students per teacher, and from the mean we know that the average ratio of students per teacher is 29.22

Question 5:

Provided table shows the total runs scored by some of the top batsman of the world in one-day cricket matches.

 Runs Scored Number of Batsman 3000 – 4000 4 4000 – 5000 18 5000 – 6000 9 6000 – 7000 7 7000 – 8000 6 8000 – 9000 3 9000 – 10000 1 10000 – 11000 1

Calculate the mode of the given information.

Solution:

Modal class = 4000-5000, l = 4000, a1=18,a0=4,a2=9$a_{1} = 18, a_{0} = 4, a_{2} = 9$ and h = 1000

Mode=l+(a1a02a1a0a2)×h$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$ =4000+(1842×1849)×1000$= 4000 + \left ( \frac{18 – 4}{2 \times 18 – 4 -9} \right )\times 1000$ =4000+(1423)×1000=4608.70$= 4000 + \left ( \frac{14}{23} \right )\times 1000 = 4608.70$

Question 6:

A complete observation of the number of cars on the roads for 100 periods of 3 minutes are summarized in the following table. Calculate the mode of the data.

 Number of cars 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 7 14 13 12 20 11 15 8

Solution:

Modal class = 40 – 50, l = 40, a1=20,a0=12,a2=11$a_{1} = 20, a_{0} = 12, a_{2} = 11$ and h = 10

Mode=l+(a1a02a1a0a2)×h$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

=40+(20122×201211)×10$= 40 + \left ( \frac{20 – 12}{2 \times 20 – 12 – 11} \right )\times 10$

=40+(817)×10=44.70$= 40 + \left ( \frac{8}{17} \right )\times 10 = 44.70$