Vedanta Excel in Mathematics Book 8 TG Final (2079)

Let the length of the room (l) x=cxmc∴mh. T=hex3nc, m2×breadth (b) = length (l) = x cm ∴b = x cm and
3× height (h) = length (l) = 2

Now,

Volume (V) = l × b × h

or, 4500 = x × x × x
2 3
or, 27000 = x3

∴ x = 30 (l) = x = 30 cm, breadth (b) = x cm = 30 cm = 15 cm and height
Thus, length of the cartoon 2 2

(h) = x cm = 30 cm = 10 cm.
3 3

Practice Questions

1. The ratio of length, breadth and height of a cuboid is 5:4:3. If its volume is 480m3, find

the area of its base. [Ans: 80 m2 ]

2. The volume of a cubical box is 1331 cm3; find its total surface area. [Ans: 726 m2 ]

3. The width of box is two-third of its length and height is one third of its length. If the

volume of the box is 48cm3, find the area of its base. [Ans: 24 m2 ]

4. Find the volume of the given solid. [Ans: 36 cm3]

151 Vedanta Excel in Mathematics Teachers' Manual - 8

Unit Bearing and Scale Drawing

21

Allocated teaching periods: 4

Competency

- Finding the bearing of one place with reference to another place

- Solving the problems related to scale drawing

Learning Outcomes

- To find the bearing of one place with reference to another place

- To solve the problems related to scale drawing

Level-wise learning objectives

S.N. LEVELS OBJECTIVES
1. Knowledge (K)
- To define bearing
2. Understanding (U) - To tell the bearing of various directions
- To recall the scale drawing

- To find the bearing of a point A from B when the bearing of
B from A is given.

- To find the actual distance with reference to the given scale
drawing

Required Teaching Materials/ Resources
Compass, map, measuring tap etc
Pre-knowledge: compass bearing

Teaching Activities
1. Give more real life examples representing the bearing.
2. Draw the directions and discuss about the followings.
(i) Directions of compass bearing
(ii) Base line for finding the bearing
(iii) Bearing
(iv) Expression of bearing with 3 digits
3. Support the students to solve the problems related to bearing
4. Give real life examples of enlargement and mapping scale factor
5. Make a discussion on scale drawing

Vedanta Excel in Mathematics Teachers' Manual - 8 152

Solution of selected questions from Excel in Mathematics -Book 8

1. The bearing of A from B is given in the figure. Find the bearing of B from A.

Solution: N N1
Here, the bearing of A from B = 0400

Now, ∠BAN1 = 1800 - ∠NBA [∵ NB // N1A and co-interior angles] 040° A
B
= 1800 - 0400 = 1400

Hence, the baring of B from A = 3600 - 1400 = 2200

2. The bearing of A from B is given in the figure. Find the bearing of B from A.

Solution: N1

Here, the bearing of A from B = 2850 N

Now, ∠NBA = 3600 – 2850 = 750 A
∠BAN1 = 1800 - ∠NBA [∵ NB // N1A and co-interior angles] B 285°
= 1800 - 750 = 1050

Hence, the baring of B from A = 1050

Short-cut tricks:

1. If the bearing of A from B is x0 and x0 less than 1800 then the bearing of B from A
will be x0 + 1800

2. If the bearing of A from B is x0 and x0 greater than 1800 then the bearing of B from A
will be x0 - 1800.

3. If the bearing of Kathmandu from Pokhara is 0950, what is the bearing of N N1
Pokhara from Kathmandu? 095°
PK
Solution:
N1
Here, the bearing of Kathmandu from Pokhara = 0950 075° C
A
Now, ∠PKN1 = 1800 - ∠NPK [∵ NP // N1K and co-interior angles] 310°

= 1800 - 950 = 0850

Hence, the baring of Pokhara from Kathmandu = 3600 - 0850 = 2750

4. An aeroplane was flying in the bearing of 0750 from Kathmandu. NB
After flying over certain distance it changed its direction and flew
in the bearing of 3100. By what angle did the aeroplane change the 075°
direction? K

Solution:

Here, the bearing of a place A from Kathmandu (K) = 0750

Bearing of a place B from A = 3100
Now, ∠N1AC = 0750 [∵ NK // N1A and corresponding angles]
Hence, the aeroplane changed the direction by = 3100 - 0750 = 2350

153 Vedanta Excel in Mathematics Teachers' Manual - 8

5. The scale drawing distance between Kathmandu and Biratnagar is 4.7 cm. If the scale
factor is 1: 10000000, find the actual distance between these two places.

Solution:
Here, the scale1:10000000 means 1 cm represents 10000000 cm

= 10000000 km = 100 km
100 × 1000

∴4.7 cm represents 4.7 × 100 km = 470 km

Practice Questions

N

1. What is the compass bearing of SE? [Ans: 1350] T E
S SE
2. What is the bearing direction of A from B
from the given figure? [Ans: 2250] N A

3. Sketch to show the bearing of 0800. 45°
4. Draw a diagram to show N450E. B

Vedanta Excel in Mathematics Teachers' Manual - 8 154

Unit Statistics

22

Allocated teaching periods: 9

Competency
- Representing the given data in pie-chart
- Solving the problems based on mean, median and quartiles
-
Learning Outcomes
- To construct the frequency distribution table for the collected data
- To construct the pie-chart for the collected data and solve the related problems
- To find the mean, median, mode and the quartiles of the ungrouped data.
- To collect the real (primary and secondary) data and analyse the data using the

appropriate statistical measure.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES
1. Knowledge (K)
- To define statistics
2. Understanding (U) - To tell the meaning of frequency
3. Application (A) - To define pie-chart
- To recall the formula of finding the mean, median and
4. High Ability (HA)
quartiles on ungrouped data

- To answer the questions from line graph
- To construct the line to represent the given groped data
- To find the mean, median of individual series

- To construct a cumulative frequency table for the given
data

- To represent the data in pie-chart
- To calculate the mean, median and quartiles of discrete

series

- To collect the real (primary and secondary) data and
analyse the data using the appropriate statistical measure

Required Teaching Materials/ Resources
Colourful chart-paper, colourful markers, chart paper with required formulae, graph-board,
graph paper, highlighter etc.

Pre-knowledge: Frequency, mean, median etc.

155 Vedanta Excel in Mathematics Teachers' Manual - 8

Teaching Activities

1. Explain about statistics

2. Discuss on data and types of data with real life examples

3. With example, discuss about the frequency tables

4. Explain with examples about the line graph, its importance and procedures of constructing

histogram

5. Give a project work as “Collect the marks obtained by your friends in the recently conducted

mathematics examination. Group the marks into the class interval of 10 and show them in

a cumulative frequency distribution table.”

6. Ask about the pie-chart, its importance and procedures of its construction.

7. Divide the students and tell to show the data in pie-charts from the questions given in

exercise and present in class

8. With appropriate examples, discuss on the central tendencies and their calculations

9. With discussion, list out the following formulae

(i) Mean åx
n
(a) For individual data; Mean (x) =

(b) For discrete data; Mean (x) = å fx
n

(c) If the mean of a group of n1 items = x1 and the mean of a group of n2 items = x2

then the mean of total of (n1 +n2) items is given by (x) = x = n1  x1 + n2 x2
n1 + n2
(ii) Median th

For individual and discrete data; position of median = N + 1 term
2
(iii) Quartiles
th
N+1
For individual and discrete data; position of Q1 = 4 term

N+1 th
4
For individual and discrete data; position of Q3 = 3 term

(iv) Mode
For individual and discrete data; mode = item having highest frequency
(v) Range = largest item – smallest item

Solution of selected questions from Excel in Mathematics -Book 8

1. Mrs. Rai’s monthly expenditure is Rs 10,800. The Food Rent
diagram on the right is a pie chart showing her 150° 70°
expenditure on different headings. Work out how much
was spent under each heading. 30° Miscellaneous

Solution: 60° 50°
Here, total monthly expenditure = Rs 10,800 Study Transportation

Vedanta Excel in Mathematics Teachers' Manual - 8 156

Headings Expenditures
Rent
Rs 10,800 × 70 = Rs 2100
Food 360
Study
Transportation Rs 10,800 × 150 = Rs 4500
Miscellaneous 360

Rs 10,800 × 60 = Rs 1800
360

Rs 10,800 × 50 = Rs 1500
360

Rs 10,800 × 30 = Rs 900
360

2. The pie chart given alongside shows the votes secured by three BA
candidates A, B and C in an election. If A secured 5,760 votes, 120°

i) how many votes did C secure? C
140°
ii) who secured the least number of votes? How many votes did

he secure?

Solution:

Here, 5760
120
(i) For A; 1200 represents 5,760 votes. i.e., 10 represents = 48 votes.

∴1400 represents 48× 140 = 6,720 votes

Hence, C secured 6, 720 votes.

(ii) Central angle of the sector of B = 3600 – (1200 + 1400) = 1000

No. of votes secured by B = 100× 48 = 4800

Thus, B secured the least number of votes.

3. The given pie chart shows the composition of different materials Cotton
in a type of cloth in percent. 90° Nylon
54°
i) Calculate the percentage of each material found in the cloth.
ii) Calculate the weight of each material contained by a bundle 144° 72°

of 50 kg of cloth. Polyester Others

Solution:

Here, the weight of material contained by a bundle of cloth = 50 kg

Materials Percentage Weight

Nylon 54 × 100% = 15% 15% of 50 kg = 7.5 kg
Cotton 360 25% of 50 kg = 12.5 kg
Polyester 40% of 50 kg = 20 kg
90 × 100% = 25%
360

144 × 100% = 40%
360

157 Vedanta Excel in Mathematics Teachers' Manual - 8

Others 72 × 100% = 20% 20% of 50 kg = 10 kg
360



4. The average age of 5 students is 9 years. Out of them the ages of 4 students are 5, 7, 8

and 15 years. What is the age of remaining student?

Solution:

Let the age of reaming student be x years.

Then, sum of ages (Σx) = 5 + 7 + 8 + 15 + x = 35 + x

Total no. of students = 5

Average age (x) = 9 years 35 + x
HWeenhcaev, eth, e(xag) e=ofånrexmoarin, 9in=g stud5ent
or, 35 + x = 45 ∴x = 10
is 10 years.

5. The average weight of 30 girls in class 8 is 42 kg and that of 20 boys is 45 kg. Find the

average weight of the students of the class.

Solution: x1 )= 42
45 kg
Here, number of girls (n1) = 30, average weight of girls ( kg
Number of boys (n2) = 30, average weight of boys (x2 )=

Combined average weight (x) =?

We have, (x) = n1  x1 + n2 x2 = 30 × 42 + 20 × 45 = 2160 = 43.2
n1 + n2 30 + 20 50

Hence, the average weight of the students of the class is 43.2 kg.

6. The average height of x number of girls and 15 boys is 123 cm. If the average height of

boys is 125 cm and that of girls is 120 cm, find the number of girls.

Solution:
HNCoeumrmeb,binenuremodfbabevoreyrosafg(gnei2rh)lse=i(gn1h15)t ,=(axvxe),r=aavg1ee2rha3geceigmhh,et ixogfh=bt?ooyf sgi(rxl2s)(=x1
) = 120 cm
125 cm

We have, (x) = n1  x1 + n2 x2 = x × 120 + 15 × 125
n1 + n2 x + 15

or, 123x + 1845 = 120x + 1875 or, 3x = 30 ∴x = 10
Hence, the number of girls is 10.

7. If the mean of a series having Σfx = 100 – k and N = k – 4 is 15, find the value of k.

Solution:

Here, Σfx = 100 – k, N = k – 4 and mean ( x ) = 15, k =?

We have, (x) = å fx
or, 15 n

= (x) = 100 – k
k–4

or, 15k – 60 = 100 – k

Vedanta Excel in Mathematics Teachers' Manual - 8 158

or, 16k = 160

∴ k = 10

He nce, the value of k is 10.

8. Find the mean from the given table.

Marks obtained 15 25 35 45 55

No. of students 7 8 12 7 6

Solution:
Computation of mean:

Marks obtained (x) No. of students (f) f×x

15 7 105

25 8 200

35 12 420

45 7 315

55 6 330

N = 40 Σfx = 1370

We have, (x) = å fx = 1370 = 34.25
n 40

Hence, the average marks is 34.25

9. The table below gives the daily earning of 110 workers in a textile mill.

Daily earning 200-300 300-400 400-500 500-600 600-700

No. of workers 40 20 15 25 10

Find the daily average earning.
Solution:
Computation of mean:

Daily earning No. of workers (f) Mid-value (m) f×m

200 – 300 40 250 10000

300 – 400 20 350 7000

400 – 500 15 450 6750

500 – 600 25 550 13750

600 - 700 10 650 6500

N = 110 fx = 44000

We have, (x) = å fx = 44000 = 400
n 110

Hence, the average daily earning is Rs400.

159 Vedanta Excel in Mathematics Teachers' Manual - 8

10. Find the median marks from the data given below.

Marks 24 36 50 65 78

No. of students 24 12 11 6
HeSreo,lution: c.f.
No. of students (f)
Marks

24 2 2

36 4 6

50 12 18

65 11 29

78 6 35

N = 35

Now, N+1 th
Position of median 2
= item

= N + 1 th
2
item = 18th item

Since, the c.f. just greater than or equal to 18 is 18 and its corresponding value is 50.
Hence, the required median is 50.

Practice Questions

1. Find the arithmetic mean of the following data.

Marks obtained 10 20 30 40 50 60
4
No. of students 4263 2
[Ans: 33.33]

2. Find the median from the given data. Wages (in Rs.) 500 600 700 800 900

No. of workers 6 9 10 15 8

3. Present the given data in a pie-chart. [Ans: Rs 700]

Division Distinction 1st Division 2nd Division 3rd Division
56 32 16
No. of students 40

4. The present data of world’s consumption of energy is given below

Coal – 25% Natural gas – 20% Hydro – 10%

Oil – 35% Nuclear – 5% Others – 5%

Represent the above data in a pie-chart.

Vedanta Excel in Mathematics Teachers' Manual - 8 160