Mensuration Textbook Solution

Practice Set 6.1

1. The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.

Time (hrs.)	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10
No. of students	7	18	12	10	3

Solution

Class (Time in hours)	Class Mark xi	Frequency (Number of students) fi	Class mark × Frequency xifi
0 – 2	1	7	7
2 – 4	3	18	54
4 – 6	5	12	60
6 – 8	7	10	70
8 – 10	9	3	27
		[sum_{} f_i = 50]	[sum_{} x_i f_i = 218]

Mean = [frac{sum_{} x_i f_i}{sum_{} f_i}]

[= frac{218}{50}]

= 4.36 hours
Hence, the mean time spent by students for their studies is 4.36 hours.

2. In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by ‘assumed mean’ method.

Toll (Rupees)	300 – 400	400 – 500	500 – 600	600 – 700	700 – 800
No. of vehicles	80	110	120	70	40

Solution

Class (Toll in rupees)	Class Mark xi	di = xi − A	Frequency (Number of vehicles fi	Frequency × deviation fi × di
300 – 400	350	−200	80	−16000
400 – 500	450	−100	110	−11000
500 – 600	550 = A	0	120	0
600 – 700	650	100	70	7000
700 – 800	750	200	40	8000
			[sum f_i = 420_{}]	[sum_{} f_i d_i = – 12000]

Required Mean = [A + frac{sum_{} f_i d_i}{sum_{} f_i}]
[= 550 – frac{12000}{420}]
= 550 − 28.57
​= Rs 521.43  
Hence, the mean of toll is Rs 521.43.

3. A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.

Milk Sold (Litre)	1 – 2	2 – 3	3 – 4	4 – 5	5 – 6
No. of Customers	17	13	10	7	3

Solution:-

Class (Milk sold in litres)	Class Mark xi	Frequency (Number of customers) fi	Class mark × Frequency fixi
1 – 2	1.5	17	25.5
2 – 3	2.5	13	32.5
3 – 4	3.5	10	35
4 – 5	4.5	7	31.5
5 – 6	5.5	3	16.5
Total	–	`sumf_i` = 50	`sumf_i x_i` = 141

Mean = `(sumf_i x_i)/(sumf_i)`

= `141/50`

= 2.82 litres

Hence, the mean of the milk sold is 2.82 litres.

4. A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by ‘assumed mean’ method.

Production (Thousand rupees)	25 – 30	30 – 35	35 – 40	40 – 45	45 – 50
No. of Customers	20	25	15	10	10

Solution

Class (Production in Thousand rupees)	Class Mark xi	di = xi − A	Frequency (Number of farm owners) fi	Frequency × deviation fi × di
25 – 30	27.5	−10	20	−200
30 – 35	32.5	−5	25	−125
35- 40	37.5= A	0	15	0
40 – 45	42.5	5	10	50
45 – 50	47.5	10	10	100
			[sum f_i = 80_{}]	[sum_{} f_i d_i = – 175]

Required Mean = [A + frac{sum_{} f_i d_i}{sum_{} f_i}]
[37 . 5 – frac{175}{80}]
= 37.5 − 2.19
​= 35.31 thousand rupees
= Rs 35310
Hence, the mean production of oranges is Rs 35310.

5. A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of the funds by ‘step deviation’ method.

Fund (Rupees)	0 – 500	500 – 1000	1000 – 1500	1500 – 2000	2000 – 2500
No. of workers	35	28	32	15	10

Class (Production in Thousand rupees)	Class Mark xi	di = xi − A	[u_i = frac{d_i}{h}]	Frequency (Number of farm owners) fi	Frequency × deviation fi × ui
0 – 500	250	−1000	−2	35	−70
500 – 1000	750	−500	−1	28	−28
1000 – 1500	1250 = A	0	0	32	0
1500 – 2000	1750	500	1	15	15
2000 – 2500	2250	1000	2	10	20
				[sum f_i = 120_{}]	[sum_{} f_i u_i = – 63]

6. The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by ‘step deviation’ method.

Weekly wages (Rupees)	1000 – 2000	2000 – 3000	3000 – 4000	4000 – 5000
No. of workers	25	45	50	30

Solution:-

Class (Weekly wages rupees)	Class Mark (xi)	di = xi − A = xi − 2500	ui = `”d”_”i”/”g”` = `”d”_”i”/1000`	Frequency (Number of workers) (fi)	Frequency × deviation (fi × ui)
1000 – 2000	1500	−1000	−1	25	−25
2000 – 3000	2500 →A	0	0	45	0
3000 – 4000	3500	1000	1	50	50
4000 – 5000	4500	2000	2	30	60
				∑fi = 150	∑fiui = 85

The required mean can be represented mathematically as:
$$text{Required Mean} = frac{sum_{i=1}^n f_iu_i}{sum_{i=1}^n f_i}$$
where $f_i$ is the frequency of the $i$th observation and $u_i$ is the value of the $i$th observation.
Given that $f_1=20, u_1=2000$, $f_2=30, u_2=2500$, $f_3=50, u_3=3000$, we have:
$$text{Required Mean} = frac{(20times 2000)+(30times 2500)+(50times 3000)}{20+30+50} = frac{85000}{100} = 850$$
The mean can be represented mathematically as:
$$bar{X} = A + bar{u}g$$
where $A$ is the assumed mean, $bar{u}$ is the mean deviation from the assumed mean, and $g$ is the common factor by which all values are multiplied or divided to obtain the new set of values.
Given that $A=2500$, $bar{u}=0.57$, and $g=1000$, we have:
$$bar{X} = 2500 + 0.57times 1000 = 3070$$
Therefore, the mean of the weekly wages is Rs 3070.

Practice Set 6.2

1. The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.

Daily No. of hours	8 – 10	10 – 12	12 – 14	14 – 16
Number of workers	150	500	300	50

Solution

Class (Number of working hours)	Frequency (Number of workers) fi	Cumulaive frequency less than the upper limit
8 – 10	150	150
10 – 12 (Median Class)	500	650
12 – 14	300	950
14 – 16	50	1000
	[N = 1000]

From the above table, we get
L (Lower class limit of the median class) = 10
N (Sum of frequencies) = 1000
h (Class interval of the median class) = 2
f (Frequency of the median class) = 500
cf (Cumulative frequency of the class preceding the median class) = 150
Now, Median = [L + left( frac{frac{N}{2} – cf}{f} right) times h]
[= 10 + left( frac{frac{1000}{2} – 150}{500} right) times 2]
[ = 10 + 1 . 4]
[ = 11 . 4text{ hours }]

>Hence, the median of the number of hours they work is 11.4 hours

2. The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.

No. of Mangoes	50 – 100	100 – 150	150 – 200	200 – 250	250 – 300
No. of trees	33	30	90	80	17

Solution:-

Class (Number of working hours)	Frequency (Number of workers) fi	Cumulaive frequency  less than the upper limit
50 – 100	33	33
100 – 150	30	63
150 – 200 (Median Class)	90	153
200 – 250	80	233
250 – 300	17	250
	N = 250

From the above table, we get

L (Lower class limit of the median class) = 150

N (Sum of frequencies) = 250

h (Class interval of the median class) = 50

f (Frequency of the median class) = 90

cf (Cumulative frequency of the class preceding the median class) = 63

Now, Median = [L + left( frac{frac{N}{2} – cf}{f} right) times h]

[= 150 + left( frac{frac{250}{2} – 63}{90} right) times 50]

= 150 + 34.44

= 184.44 mangoes

= 184 mangoes

Hence, the median of data is 184 mangoes.

3. The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.

Average Speed of Vehicles(Km/hr)	60 – 64	64 – 69	70 – 74	75 – 79	79 – 84	84 – 89
No. of vehicles	10	34	55	85	10	6

Solution:-

Class (Number of working hours)	Continuous classes	Frequency (Number of workers)	Cumulaive frequency  less than the upper limit
60 – 64	59.5 – 64.5	10	10
64 – 69	64.5 – 69.5	34	44
70 – 74	69.5 – 74.5	55	99 → cf
75 – 79	74.5 – 79.5	85 → f	184
79 – 84	79.5 – 84.5	10	194
84 – 89	84.5 – 89.5	6	200
Total	–	N = 200	–

Here, total frequency = ∑fi = N = 200 

∴ `”N”/2 = 200/2 = 100`

Cumulative frequency which is just greater than (or equal) to 100 is 184. 

∴ The median class is 74.5 – 79.5.

Now, L = 74.5, f = 85, cf = 99, h = 5

Now, Median = [L + left( frac{frac{N}{2} – cf}{f} right) times h]

`= 74.5 + ((100 – 99)/85) xx 5`

= 74.5 + 0.059

= 74.559 ≈ 75

∴ The median of the given data is 75 km/hr (approx.).

4. The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.

No. of bulbs produced (Thousands)	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90	90 – 100
No. of factories	12	35	20	15	8	7	8

Solution:-

Class (Number of bulbs produced in thousands)	Frequency (Number of factories) fi	Cumulaive frequency less than the upper limit
30 – 40	12	12
40 – 50	35	47
50 – 60 (Median Class)	20	67
60 – 70	15	82
70 – 80	8	90
80 – 90	7	97
90 – 100	8	105
	N = 105

From the above table, we get
L (Lower class limit of the median class) = 50
N (Sum of frequencies) = 105
h (Class interval of the median class) = 10
f (Frequency of the median class) = 20
cf (Cumulative frequency of the class preceding the median class) = 47
Now, Median = [L + left( frac{frac{N}{2} – cf}{f} right) times h]
[= 50 + left( frac{frac{105}{2} – 47}{20} right) times 10]
= 50 + 2.75
= 52.75 thousand lamps
= 52750 lamps 
Hence, the median of the productions is 52750 lamps.

Practice Set 6.3

1. The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.

Content of fat (%)	2 – 3	3 – 4	4 – 5	5 – 6	6 – 7
Milk collected (Litre)	30	70	80	60	20

The maximum class frequency is 80.
The class corresponding to this frequency is 4 – 5.
So, the modal class is 4 – 5.
L (the lower limit of modal class) =  4
f₁ (frequency of the modal class) = 80 
f_o (frequency of the class preceding the modal class) = 70
f₂ (frequency of the class succeeding the modal class) = 60
h (class size) = 1 
Mode = [L + left( frac{f_1 – f_0}{2 f_1 – f_0 – f_2} right) times h]
[= 4 + left( frac{80 – 70}{2 times 80 – 70 – 60} right) times 1]

= 4 + 0.33
= 4.33
Hence, the modal fat content is 4.33 litres.