# Ncert Solutions For Class 10 Maths Ex 14.1

## Ncert Solutions For Class 10 Maths Chapter 14 Ex 14.1

Question 1:

A group of pupil has conducted a survey as a as the part of their college program, They collected examined data’s of plants from 20 houses and nearby areas. Find the mean of the collected plants from the given houses.

 No. of Plants Taken from 20 houses 0-2 2-4 4-6 6-8 8-10 10-12 12-14 Given number of houses 2 1 5 1 2 6 3

Name the method you use for finding the mean and why?

Solution:

 Class Interval fi xi fi xi 0-2 2 1 1 2-4 1 3 6 4-6 5 5 5 6-8 1 7 35 8-10 2 9 54 10-12 6 11 22 12-14 3 13 40 Sum fi=20 Sum fixi=163

Mean can be calculated as follows:

x¯=ΣfixiΣfi=16320=8.15$\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{163}{20}=8.15$

In this issues the fiandxi$f_{i} and x_{i}$ are small so the direct method is given to solve the problem.

Question2:

The distributed wages is given for 50 workers working in the factory.

 Daily wages given to workers (in Rs) 100-120 120-140 140-160 160-180 180-200 Number of wage workers in factory 14 12 6 8 10

Find the mean of workers of daily wages given by the workers.

Solution:

In this case, value of xi is quite large and hence we should select the assumed mean method.

 Class given Interval fi xi di = xi – a fi di 100-120 14 110 -40 -480 120-140 12 130 -20 -280 140-160 6 150 0 0 160-180 8 170 20 120 180-200 10 190 40 400 Sum fi=50 Sum fi di = -240

Now, mean of deviations of daily wagesis calculated as follows:

d¯=fidifi=24050$\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{-240}{50}$

x = d + a = 150+(-4.8)= 145.20

Question 3:

This distribution shows that the money daily pocket given to students in the given area. The mean pocket money is Rs. 18. Find the f which is the missing frequency.

 Daily money 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:

 Class interval fi xi fixi 11-13 7 12 84 13-15 6 14 84 15-17 13 16 144 17-19 9 18 234 19-21 f 20 20f 21-23 4 22 110 23-25 5 24 96 Sum fi=44+f Sum fixi= 752+20f

We have;

x¯=ΣfixiΣfi$\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}$ 18=752+20f44+fi$18= \frac{752+20f}{44+ f_{i}}$

18(44+18f) = 752+20f

792+18f = 752+20f

2f = 40

Missing frequency f= 20

Question4:

The number of 30 women were checked by one doctor and the distributed the heartbeat of them in following distribution table. Find the mean of heart beats of the given per minute for these thirty women, by choosing a suitable following method.

 Heartbeat per minute in given number 65-68 68-71 71-74 74-77 77-80 80-83 83-86 Number of 30 women 2 3 4 7 8 2 4

Solution:

 Class Interval fi xi di = xi – a fi di 65-68 2 66.5 -9 -18 68-71 3 69.5 -6 -24 71-74 4 72.5 -3 -9 74-77 7 75.5 0 0 77-80 8 78.5 3 21 80-83 2 81.5 6 24 83-86 4 84.5 9 18 Sum fi= 30 Sum fi di = 12

Now, mean can be calculated as follows:

d¯=fidifi=1230=0.4$\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{12}{30}=0.4$ x¯=d¯+a=0.4+75.5=75.9$\bar{x}=\bar{d}+a=0.4+75.5=75.9$

Question 5:

The oranges were packed and were sold by the fruit seller in a  market. The numbers of oranges are arranged in different boxes by the fruit seller. Through the number of oranges the number of oranges are distributed in following manner.

 No.of oranges 50-52 53-55 56-58 89-61 62-64 No. of boxes 15 110 135 115 25

In the distributed pattern the number of oranges. Find the mean. Choose the method to find the mean:

Solution:

 Class interval fi xi di=x-a fi di 50-52 110 54 -6 90 53-55 15 51 -3 -330 56-58 135 60 0 0 59-61 25 57 3 345 62-64 115 63 6 150 Sum fi= 400 Sum fi di=75

Mean can be calculated as follows:

d¯=ΣfidiΣfi=75400=0.1875$\bar{d}=\frac{\Sigma f_{i}d_{i}}{\Sigma f_{i}}=\frac{75}{400}=0.1875$ x¯=d¯+a=0.1875+57=57.1875$\bar{x}=\bar{d}+a=0.1875+57=57.1875$

In this case, there are wide variations in fi$f_{i}$ and hence assumed mean method is used.

Question 6:

The following distribution shows the food expenditure of the given households from the locality:

 Daily food expenditure (in Rs) 50-100 100-150 150-200 200-250 250-300 Number of given households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Solution:

 Class Interval fi xi di= xi – a ui = di/h fiui 50-100 4 125 -100 -2 -8 100-150 5 175 -50 -1 -5 150-200 12 225 0 0 0 200-250 2 275 50 1 2 250-300 2 325 100 2 4 Sum fi = 25 Sum fiui = -7

Mean can be calculated as follows:

x¯=a+fiuifi×h$\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h$ =225+725×50=225$=225+\frac{-7}{25}\times 50 =225$

Question 7:

To find concentration in sulphur dioxide (in parts per million, i.e. ppm),The parts of data are collected by different areas from cities are given in distributed table:

 Concentration in SO2 (in ppm) Frequency of F 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

Give the mean of the sulphur dioxide collected from the given cities.

Solution:

 Class Interval fi xi fixi 0.00-0.04 9 0.02 0.08 0.04-0.08 4 0.06 0.54 0.08-0.12 9 0.10 0.90 0.12-0.16 2 0.14 0.28 0.16-0.20 2 0.18 0.72 0.20-0.24 4 0.22 0.44 Sum fi=30 Sum fixi=2.96

Mean can be calculated as follows:

x¯=ΣfixiΣfi=2.9630=0.099ppm$\bar{x}=\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{2.96}{30}=0.099 ppm$

Question 8:

The absentee record of 40 students is handled by the teacher for one complete term. Find the mean of absent days of the given students:

 Number of following days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 Number of following students 10 11 4 7 3 4 1

Solution:

 Class Interval fi xi fixi 0-6 11 3 33 6-10 10 8 80 10-14 4 12 84 14-20 7 17 68 20-28 3 24 96 28-38 4 33 99 38-40 1 39 39 Sum fi = 40 Sum fixi= 499

Mean can be calculated as follows:

x¯=fixifi$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$= 49940$\frac{499}{40}$=12.4

Question 9:

The data of the literacy digit from the 35 cities is given in the following table. Find the mean of the 35 literacy cities.

 Literacy rate from 35cities (in %) 40-45 50-55 60-65 70-75 80-85 Cities in values 10 3 8 11 3

Solution:

 Class Interval fi xi di=xi-a ui=di/h fiui 40-45 10 50 -20 -2 -6 50-55 3 60 -10 -1 -10 60-65 8 70 0 0 0 70-75 11 80 10 1 8 80-85 3 90 20 2 6 Sum fi=35 Sum fiui= -2

Mean can be calculated as follows:

x¯=a+ΣfiuiΣfi×h$\bar{x}=a+\frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}\times h$ =70+235×10=69.42$= 70+\frac{-2}{35}\times 10=69.42$