Vedanta Excel in Mathematics TG Book - 6 Final

b) Evaluate by showing all steps: {2.5 × (4.3 – 3.4) ÷ (1.5)2} + 0.1
Solution:
Here,
{2.5 × (4.3 – 3.4) ÷ (1.5)2} + 0.1
= {2.5 × (4.3 – 3.4) ÷ 2.25} + 0.1
= {2.5 × (0.9) ÷ 2.25} + 0.1
= {2.25÷ 2.25} + 0.1
= 1 + 0.1
= 1.1

Activity 9 Rounding off decimal numbers (Approximation)

1. Recall round off a whole number to the nearest tens, hundreds, thousands, etc.
2. Similarly, with proper examples, round off a decimal number to the nearest whole number,

tenths, hundredths, thousandths, etc.
3. Note that
(i) If the digit which is to be rounded off is less than 5, its place is considered zero and the

digit at the higher place remains unchanged.
(ii) If the digit which is to be rounded off is 5 or greater than 5, it is considered as zero and 1

is added to the digit at the higher place.

Solution of selected questions from EXERCISE 5.9

1. Round off to 1 decimal place (d.p.):

Solution:

a) 3.43 ≈ 3.4 b) 6.47 ≈ 6.5 c) 17.142 ≈ 17.1

d) 25.258 ≈ 25.3 e) 0.386≈ 0.4

2. Round off to 2 decimal places (d.p).

Solution:

a) 3.416≈3.42 b) 8.321≈8.32 c) 21.9832≈21.98

d) 0.1647≈0.16 e) 0.5273≈0.53

3. Round off to 3, 2 and 1 decimal places.

Solution:

a) 4.2134≈4.213 4.213≈4.21 4.21≈4.2

b) 10.6579≈10.658 10.658≈10.66 10.66≈10.7

c) 8.10925≈8.109 8.109≈8.11 8.11≈8.1

d) 0.39768≈0.398 0.398≈0.40 0.40≈0.4

e) 0.095487≈0.095 0.095≈0.10 0.10≈0.1

51 Vedanta Excel in Mathematics Teachers' Manual - 6

Unit Unitary Method

6


Allocated teaching periods: 5

Competency
- Solving daily life problems using unitary method

Level-wise learning objectives

S.N. Levels Objectives

1. Knowledge (K) - To define unitary method
- To tell the unit value

2. Understanding (U) - To solve the simple problem using unitary method

3. Application (A) - To solve the problems using unitary method

4. High Ability (HA) - To solve the contextual problems based on unitary method

Required Teaching Materials/ Resources
Copies, pens, scale, highlighter, ICT tools etc.
Pre-knowledge: Quantities of same kind, comparison between numbers
Teaching Activity
1. Discuss about the unit value with examples.
2. Make the groups of students and give the works as follows
Group-A: If the cost of 2 kg of apples is Rs 200, what is the cost of 5 kg of apples?
Group-B: If the cost of a dozen of bananas is Rs 60, find the cost of 3 bananas.
Group-C: A car takes 10 minutes to cover the distance of 5 km, how long does it take to

cover a distance of 1 km?
3. Focus on group or project work.
4. Solve the problems from exercise with discussion.

Solution of selected questions from EXERCISE 6.1

1. If the cost of 1 dozen of pencils is Rs 96, find the cost of 18 pencils.
Solution:
Here, 1 dozen of pencils = 12 pencils
Now, the cost of 1 dozen of pencils = Rs 96
or, the cost of 12 pencils = Rs 96
or, the cost of 1 pencil= Rs 96÷12 = Rs 8
∴ The cost of 18 pencils = 18×Rs 8 = Rs 144

2. The wages of a labour for a week is Rs 3,500, find the wages of the labour for 12 days.
Solution:

Vedanta Excel in Mathematics Teachers' Manual - 6 52

Here, 1 week = 7 days
Now, the wages of a labour for a week = Rs 3,500
or, the wages of a labour for 7 days = Rs 3,500
or, the wages of a labour for 1 day = Rs 3500÷7 = Rs 500
∴ The wages of a labour for 12 days = 12×Rs 500 = Rs 6,000

3. If the cost of 1 quintal of onions is Rs 4,000, what is the cost of 15 kg of onions?
Solution:
Here, 1 quintal of onions = 100 kg of onions
Now, the cost of 1 quintal of onions = Rs 4000
or, the cost of 100 kg of onions = Rs 4000
or, the cost of 1 kg of onion= Rs 4000÷100 = Rs 40
∴ The cost of 15 kg of onions = 15×Rs 40 = Rs 600

4. If the cost of carpeting 1 sq. m of a floor is Rs 120, what is the cost of carpeting 96 sq. m
of floor?

Solution:
Here, the cost of carpeting 1 sq. m of floor = Rs 120
∴ The cost of carpeting 96 sq. m of floor = 96×Rs 120 = Rs 11,520

5. The speed of a bus is 40 km per hour. How much distance does it cover in (i) 45 minutes

(ii) 3 hours (iii) 2 hours 30 minutes?

Solution:

Here, 1 hour = 60 minutes

(i) In 1 hour i.e., 60 minutes, the bus can cover a distance of 40 km

or, In 1 minute, the bus can cover a distance of 40 km = 2 km
60 3

∴In 45 minutes, the bus can cover a distance of 45× 2 km = 30 km
3
(ii) In 1 hour, the bus can cover a distance of 40 km

∴In 3 hours, the bus can cover a distance of 3×40= 120 km

(iii) 2 hours 30 minutes = 2 hours + 0.5 hours = 2.5 hours

In 1 hour, the bus can cover a distance of 40 km

∴In 2.5 hours, the bus can cover a distance of 2.5×40= 100 km

6. A car can cover a distance of 240 km in 4 hours. (i) Find the speed of the car in km per

hour. (ii) How many kilometre does it cover with the same speed in 7 hours?

Solution:

Here, a car can cover a distance of 240 km in 4 hours.

(i) Speed of the car = 240 km/hr = 60 km/hr
4
(ii) In 1 hour, the bus can cover a distance of 40 km

In 7 hours, the car can cover a distance of 7×60 km = 420 km

53 Vedanta Excel in Mathematics Teachers' Manual - 6

7. The cost of 6 kg of oranges is Rs 540. How many kilograms of oranges can be

purchased for Rs 720?

Solution:

Here, Rs 540 is the cost of 6 kg of oranges.

Re 1 is the cost of o5f64702=0×910910 kg of oranges.
Rs 720 is the cost = 8 kg of oranges.

Hence, 8 kg of oranges can be purchased for Rs 720.

8. The cost of 5 story books is Rs 800. How many books can be bought for Rs 1,440?

Solution:

Here, Rs 800 is the cost of 5 story books.

Re 1 is the cost of 5 = 1 story books.
Rs 1440 is 800 160
1
the cost of 1440× 160 = 9 story books.

Hence, 9 story books can be bought for Rs 1440.

9. If 3/4 of the distance between two places is 30 km, how much is 2/5 of the distance of

these places?

Solution:

H∴eWreh, o43lep(a1r)tsdoisfttahnecedi=stan33c0ek=m3=0 km 4 km = 40 km
30× 3

4

Again, parts of the distance = of 40 km = × 40 km = 16 km

Hence, the required distance is 16 km.

10. The cost of carpeting 2/3 parts of a floor is Rs 6,000. What is the cost of carpeting 3/5

parts of the floor?

Solution:

Here, the cost of carpeting 2 parts of the floor = Rs 6000
∴The 3
6000 3
cost of carpeting the whole (1) floor =Rs 2 = Rs 6000× 2 = Rs 9000

3 3 3 3
5 5 5
Again, the cost of carpeting parts of the floor = of Rs 9000 = × Rs 9000 = Rs 5400

Hence, the required cost for carpeting is Rs 5,400.

11. A pipe can fill a water tank of capacity 7,500 l completely in 4 hours. How much

water does it fill in 3 hours?

Solution:

Here, in 4 hours, a pipe can fill 7,500 litres of water.

In 1 hour, the pipe can fill 7500 litre = 1875 litre of water.
4

∴ In 3 hours, the pipe can fill 3×1875 litre = 5,625 litre of water.

Hence, the pipe can fill 5,625 litres of water in 3 hours.

Vedanta Excel in Mathematics Teachers' Manual - 6 54

12. A tap can fill 6,000 l of water in 3 hours. How long does it take to fill a water tank of

capacity 4,000 l completely?

Solution:

Here, 6,000 litre of water can be filled by a tap in 3 hours.

4,0010liltirtreeosfowf wataetrecracnanbebfeilflielldedbybythtehteatpapinin60430,000h0×ou6r0s30. 0 = 2 hours.
∴

Hence, the pipe can fill 4,000 litres of water in 2 hours.

13. A computer can download a 200 megabytes (MB) of a game file in 20 seconds. How
many megabytes does it download in 1 second? What is the download speed of the
internet?

Solution:
Here, in 20 seconds, a game file of 200 MB can be downloaded.
In 1 seconds, the game file of 200 MB÷20 = 10 MB can be downloaded.
Hence, the speed of internet is 10 MBps

14. A computer can download a 2.5 GB of a movie file in 50 seconds. Find the download
speed of the internet in megabytes (MB) per second.

Solution:
Here, 1 GB = 1000 MB ∴2.5 GB = 2.5×1000 MB = 2500 MB
In 50 seconds, a movie file of 2500 MB can be downloaded.
In 1 seconds, the movie file of 2500 MB÷50 = 50 MB can be downloaded.
Hence, the speed of internet is 50 MBps.

Extra Questions

1. The cost of 15 pens is Rs 300, What is the cost of 7 such pens?
2. If a student pays Rs. 24,000 as annual fees to the school. Find his/her fee of 5 months.
3. The cost of 6 kg of sugar is Rs 444. How many kilograms of sugar can be bought for

Rs 666?

4. A car running with uniform speed covers a distance of 120 km in 3 hours. How much
distance will the car cover in 4 hours running with the same speed?

5. A bike can travel 450 km consuming 15 litres of petrol. How much petrol will it consume
while travelling through a distance of 300 km?

6. A pipe can fill a water tank of capacity 10,000 l completely in 5 hours. How much water
does it fill in 2 hours?

55 Vedanta Excel in Mathematics Teachers' Manual - 6

Unit Percentage

7

Allocated teaching periods: 6

Competencies
- To convert fraction and decimals into percentage.
- To solve the problems related to percentage.

Level-wise learning objectives

S.N. Levels Objectives

1. Knowledge (K) - To define percentage
- To tell the rule to express decimal or fraction in to
2. Understanding (U)
3. Application (A) percentage.
4. High Ability (HA) - To tell the rule to convert percentage in to fraction and

decimal

- To express the part of whole quantity in fraction and then
in percentage.

- To find the value of given percentage of quantity

- To solve the problems related to percentage

- To solve the contextual problems based on percentage

Required Teaching Materials/ Resources
Blocks of hundreds, process of expressing decimals or fractions in to percentage and vice
versa in chart paper etc.
Pre-knowledge: Fraction, decimal

Teaching Activities
Activity 1 Percentage
1. Take a block of hundred, colour a few grids and discuss about the fraction of coloured
parts.
2. Make a discussion on meaning of percentage with various examples.
3. Discuss upon the following questions.
(i) Have you observed the battery percentage on mobile phones or laptop?

(ii) Have you ever seen the discount percentage in items in stores?
(iii) Out of hundred marks in an exam, Shashwat obtained 70 marks,

what percentage of marks did he obtain?
(iv) Out of 100 students, 65 students secured A+ grade in mathematics,

what percentage of students secured A+ grade?
(v) In a basket out of 50 guavas, 10 guavas were found damage and not

fit for sale, what percentage of the guavas were damage?
(vi) There are 24 boys and 16 girls in a class, find the percentage of boys and the girls.
4. Make a discussion about the process of expressing the fractions or decimals in to the

percentages.

Vedanta Excel in Mathematics Teachers' Manual - 6 56

5. With related examples, make a discussion about the process of expressing the percentage
in to the fractions or decimals.

6. Focus on group or individual work.
7. Give some project work on percentage.

Solution of selected questions from EXERCISE 7.1
1. Express the parts of whole as the percentage of the whole.
a) 5 as a percent of 10 b) 6 as a percent of 15
c) Rs 40 as a percent of Rs 250 d) 60 marks as a percent of 75 marks
e) 9 months as a percentage of a year f) 250 gram as a percentage of 1 kg

Solution: 5
10
a) 5 as a percent of 10 = × 100% = 50%

b) 6 as a percent of 15 = 5 × 100% = 50%
10

c) Rs 40 as a percent of Rs 250 = 40 × 100% = 16%
250

d) 60 marks as a percent of 75 marks = 60 × 100% = 80%
75

e) 9 months as a percentage of a year = 9 × 100% = 75%
12

f) 250 gram as a percentage of 1 kg = 250 × 100% = 25%
1000

2. H52 oowf the number of students of a school are girls?
(i) many percentage of the students are girls?
(ii) How many percentage of students are boys?
Solution:
2
(i) Percentage of girls = 5 × 100% = 40%

(ii) Percentage of boys = 100% - 40% = 60%

3. There are 360 students in a school. 198 of them are boys.
(i) Find the percentage of the boys.

(ii) Find the percentage of the girls.

Solution: 198
360
(i) Percentage of boys= × 100% = 55%

(ii) Percentage of girls = 100% - 55% = 45%

5. In an educational excursion, the students of class 6 of a school visited to an animal-farm
and recorded the list of animals alongside. Express each animal as percentage of the total
animals.

Solution:
Here, total number of animals = 24 + 16 + 40 = 80

Percentage of cow = 24 × 100% = 30%
80
16
Percentage of buffalo = 80 × 100% = 20%

Percentage of goat = 40 × 100% = 50%
80

57 Vedanta Excel in Mathematics Teachers' Manual - 6

6. Look at the monthly progress
report of Dipti Shah and solve
the given problems.

a) Express her marks of every
subject in percent.

b) In which subject did she
show the better performance?

c) Find the percent of her total marks out of total full marks.

Solution:

a) Here, 34
40
Percentage of obtained marks in English = × 100% = 85%

Percentage of obtained marks in Nepali = 15 × 100% = 75%
20

Percentage of obtained marks in Maths = 45 × 100% = 90%
50

Percentage of obtained marks in Science = 21 × 100% = 84%
25

b) She showed better performance in Maths.
c) Total full marks = 40 + 20 + 50 + 25 = 135
Total obtained marks = 34 + 15 + 45 + 21 = 115
115
Percentage of total obtained marks out of total full marks = 135 × 100% = 85.16%

Solution of selected questions from EXERCISE 7.2

1. Find the value of 5% of Rs 300.
Solution:
5
Here, 5% of Rs 300 = 100 × Rs 300 = Rs 15

2. Find the whole quantity whose 75% of saving is Rs, 2,700.
Solution:

Let, required quantity (saving amount) be Rs x.

Then, 75% of x = Rs 2700
or, 17050 × x = 2700

or, 75x = 270000

or, x = 3600

Hence, the required saving is Rs. 3600.

3. Out of 40 students of grade VI, 80% of them are attending their virtual classes in zoom
regularly. Find the number of students who are not attending the virtual classes.

Solution:

Here,

The number of students who are attending the virtual classes = 80% of 40
80
= 100 × 40 = 32

Hence, the number of students who are not attending the virtual classes = 40 – 32 = 8


4. Mrs. Bhatta earns Rs 5,400 and spends 65% every week.
(i) How much does she spend every week?

Vedanta Excel in Mathematics Teachers' Manual - 6 58

(ii) How much does she save every week?
(iii) How much does she save in 1 year (52 weeks)?
Solution:
65
(i) Weekly expenditure = 65% of Rs 5,400 = 100 × Rs 5400 = Rs 3,510

(ii) Weekly saving = Rs 5400 – Rs 3510 = Rs 1,890
(iii) Her saving in 1 year (52 weeks) = 52×Rs 1890 = Rs 98,280

5. A man purchased a piece of land for Rs 4,50,000. He sold it at a loss of 5%.
(i) Find his loss amount.
(ii) At what price did he sell the land

Solution:
Here,
5
(i) Loss amount = 5% of Rs 4,50,000 = 100 × Rs 450000 = Rs 22,500

(ii) Selling price (S.P.) = C.P. – Loss = Rs 4,50,000 – Rs 22,500 = Rs 4, 27,500

6. Pragya got 21 marks out of 25 full marks in a spelling test. Pooja got 80% marks in the
same test. Who got the more mark? By what percent was the marks more than the lower
marks?

Solution:

Pragya’s obtained marks = 21

Pooja obtained marks = 80% of 25 = 80 × 25 = 20
100

Pragya obtained more marks by 21 – 20 = 1.

Percentage of difference = 1 × 100% = 5%
20

Hence, Pragya obtained 5% more than Pooja.

7. Are the numbers 10 and 10% same? By what percent is the number smaller than the
greater one?

Solution: 10
100
We know, 10% = = 0.1

So, the numbers 10 and 10% are not same.

Also, difference = 10 – 0.1 = 9.9
9.9
Percentage of difference = 10 × 100% = 99%

Hence, 10% i.e., 0.1 is smaller than 10 by 99%.

8. In a mock test of mathematics, the maximum marks is 80. Bodhraj gets 85% marks and
cannot get A+ grade by at least 4 marks. Find:
a) least marks for A+ grade

b) least percent for A+ grade.

Solution: 85
100
Marks obtained by Bodhraj = 85% of 80 = × 80= 68

a) The least marks for A+ grade = 4 marks more than the obtained marks of Bodhraj
=68 + 4 = 72

b) The least percent for A+ grade = 72 × 100% = 90%
80

59 Vedanta Excel in Mathematics Teachers' Manual - 6

9. Out of 50 students in a class, 60% are boys. If 50% of boys and 40% of girls wear spectacles,
find:
a) number of boys and girls.
b) number of boys and girls who wear spectacles.
c) number of students who wear spectacles.
d) percentage of students who wear spectacles in all.

Solution: 60
100
a) The number of boys = 60% of 50 = × 50 = 30

The number of girls = 50 – 30 = 20

b) The number of boys who wear spectacles = 50% of 30 = 50 × 30 = 15
100
40
The number of girls who wear spectacles = 40% of 20 = 100 × 20 = 8

c) The number of students who wear spectacles = 15 + 8 = 23

d) The percentage of students who wear spectacles in all = 23 × 100% = 46%
50

10. Shreyasha spends 40% of her income and saves Rs 18,000 in a month. Sahayata spends
60% of her income and saves Rs 12,800 in a month.

a) Who has more monthly income? Find by calculation.
b) Who spends more money and by how much?
Solution:
a) Let, the monthly income of Shreyasha be Rs x.

Expenditure = 40%. ∴Saving = 100% - 40% = 60%

Now, 60% of x = Rs 18000

or, 60 × x= 18000
100

or, 60x = 1800000

or, x = 30,000

Thus, the monthly income of Shreyasha is Rs 30,000.
Also,
Let, the monthly income of Sahayata be Rs y.

Expenditure = 60%. ∴Saving = 100% - 60% = 40%

Now, 40% of y = Rs 12800

or, 40 × y= 12800
100

or, 40x = 1280000

or, x = 32,000

Thus, the monthly income of Sahayata is Rs 32,000.

Difference = 32000 – 30000 = 2000

The monthly income of Sahayata is more than Shreyasha’s income by Rs 2,000.
b) Expenditure of Shreyasha = Rs 30,000 – Rs 18,000 = Rs 12,000
Expenditure of Sahayata = Rs 32,000 – Rs 12,800 = Rs 19,200
Difference = Rs 19,200 – Rs 12,000 = Rs 7,200

Hence, Sahayata spends more than Shreyasha by Rs 7,200.

Vedanta Excel in Mathematics Teachers' Manual - 6 60

Extra Questions

1. Convert 0.125 into percentage.
[Ans:12.5%]

2. Express 6 hours of a day in to percentage.
[Ans:25%]

3. There are 16 girls and 24 boys in a class. Find the percentage of (i) boys (ii) girls
[Ans: 40%, 60%]

4. There are 300 students in a school. If 60% of them are boys, find the number of girls.
[Ans: 120]

5. In a farmer’s animal firm 45% of the animals are cows and the rest are buffalos. If there
are 110 cows, find (i) the total number of animals. (ii) the number of buffallos.
[Ans: (i) 200 (ii) 90]

61 Vedanta Excel in Mathematics Teachers' Manual - 6

Unit Profit and Loss

8

Allocated teaching periods: 5

Competency
- To solve the simple problems based on profit and

Level-wise learning objectives

S.N. Level Objectives

- To define cost price, selling price
1. Knowledge (K) - To tell the relation of C.P., S.P. and profit or loss

- To define marked price

2. Understanding (U) - To find the profit/ loss amount
- To find the discount

3. Application (A) - To solve the verbal problems on profit and loss
4. High Ability (HA) - To prepare the project work and present in the class.

Required Teaching Materials/ Resources
Colourful chart-paper with definitions and formulae, bills, audio/video clips related to profit
and loss, projector etc.
Pre-knowledge: cost price, selling price, profit and loss
Teaching Activities
1. With some articles like watch, mobile, books, copies, bags etc. discuss upon cost price,

selling price, profit and loss.
2. Divide the class into 5/6 groups and ask the formulae of profit amount, loss amount
3. Explain the following formulae with examples under discussion.
(i) Profit amount = S.P. – C.P.
(ii) Loss amount = C.P. – S. P.
4. Involve the students in play acting as a shopkeeper and customer.
5. Discuss upon the marked price and discount amount.
6. Under discussion, list out the following formulae.
(i) S.P. = M.P. – Discount amount
(ii) Discount amount =M.P. – S.P.
7. Focus on group / project work
8. Give more priority in collaboration learning.
9. Work as the facilitator and encourage the learners.

Vedanta Excel in Mathematics Teachers' Manual - 6 62

Solution of selected questions from EXERCISE 8.1

1. Mr. Yadav buys a calculator for Rs 375 and sells it for Rs 450. Find his profit.

Solution:

Here, C.P. of a calculator = Rs 375

S.P. of the calculator = Rs 450

Now, profit amount = S.P. – C.P.

= Rs 450 – Rs 375

= Rs 75

Hence, Mr. Yadav makes a profit of Rs 75.

2. Mrs. Shakya bought a saree for Rs 1,650 and sold it for Rs 1,485. Find her loss.

Solution:

Here, C.P. of a saree= Rs 1,650

S.P. of the saree = Rs 1,485

Since, C.P. is more than S.P. So, there is a loss.

Now, loss amount = C.P. – S.P.

= Rs 1650 – Rs 1485

= Rs 165

Hence, Mrs. Shakya bears a loss of Rs 165.

3. A shopkeeper sold a camera for Rs 2,550 and made a loss of Rs 480. At what price

did he/she purchase the camera?

Solution:

Here, S.P. of a camera= Rs 2,550

Loss amount = Rs 480

C.P. of the camera =?

Now, C.P. = S.P. + Loss

= Rs 2550 + Rs 480

= Rs 3030

Hence, the camera was bought at Rs 3030.

4. Mr. Sharma bought a bicycle for Rs 5,995 and sold at a profit of Rs 1,001. At what

price did he sell the bicycle?

Solution:

Here, C.P. of a bicycle= Rs 5,995

Profit amount = Rs 1,001

S.P. of the bicycle =?

Now, S.P. = C.P. + Profit

= Rs 5995 + Rs 1001

= Rs 6996

Hence, Mr. Sharma sold the bicycle for Rs 6,996.

63 Vedanta Excel in Mathematics Teachers' Manual - 6

5. A stationer sold 1 dozen of pens at the rate of Rs 25 each and gained Rs 120 altogether.

How much was the cost price of the pens?

Solution:

Here, S.P. of a pen = Rs 25

S.P. of 1 dozen of pens = S.P. of 12 pens = 12×Rs 25 = Rs 300

Profit amount = Rs 120

C.P. of the pens=?

Now, C.P. = S.P. – Profit

= Rs 300 – Rs 120

= Rs 180

Hence, the cost price of the pens was Rs 180.

6. A fruit seller purchased 10 kg of apples at the rate of Rs 75 per kg and sold at the

total loss of Rs 50. How much was the selling price of the apples?

Solution:

Here, C.P. of 1 kg of apples = Rs 75

C.P. of 10 kg of apples = 10×Rs 75 = Rs 750

Loss amount = Rs 50

S.P. of the apples=?

Now, S.P. = C.P. – Loss

= Rs 750 – Rs 50

= Rs 700

Hence, the selling price of the apples was Rs 700.

7. A merchant bought a quintal of potatoes for Rs 4,000. He sold all potatoes at Rs 390

per 10 kg. Find his profit or loss.

Solution:

Here, C.P. of 1 quintal of potatoes = C.P. of 100 kg of potatoes = Rs 4,000

S.P. of 10 kg of potatoes = Rs 390

S.P. of 1 kg of potatoes = Rs 390 ÷ 10 = Rs 39

S.P. of 100 kg of potatoes = 100 ×Rs 39 = Rs 3900

Since, C.P. is more than the S.P. So, there is a loss.

Now, Loss amount = C.P. –S.P.

= Rs 4,000 – Rs 3,900

= Rs 100

8. Mr. Poudel buys 60 bananas at Rs 80 per score and sells all the bananas at Rs 60 per

dozen. What is his profit or loss? (1 score = 20)

Solution:

Here, C.P. of 1 score of bananas = C.P. of 20 bananas = Rs 80

C.P. of 1 banana = Rs 80÷20 = Rs 4

C.P. of 60 bananas = 60 ×Rs 4 = Rs 240

S.P. of 1 dozen of bananas = S.P. of 12 bananas = Rs 60

S.P. of 1 banana = Rs 60÷12 = Rs 5

Vedanta Excel in Mathematics Teachers' Manual - 6 64

S.P. of 60 bananas = 60 ×Rs 5 = Rs 300

Since, S.P. is more than the C.P. So, there is a profit.

Now, profit amount = S.P. – C.P.

= Rs 300 – Rs 240

= Rs 60

So, Mr. Poudel makes a profit of Rs 60.

9. A shopkeeper buys a table fan for Rs 1,200 and expenses Rs 40 on its repairs. If he
sells the fan for Rs 1,550, find his profit.

Solution:

Here, C.P. of a table fan = Rs 1,200

C.P. of the fan with repairing = Rs 1200 + Rs 40 = Rs 1240

S.P. of the fan = Rs 1,550

Now, profit amount = S.P. – C.P.

= Rs 1550 – Rs 1240

= Rs 310

Hence, the shopkeeper makes a profit of Rs 310.

10. Mr. Thapa bought a gross of copies for Rs 2,880 and made a profit of 288, find the
selling price of each copy? (1 gross = 12 dozens)
Solution:
Here,1 gross of copies = 12 dozens of copies = 12×12 copies = 144 copies
C.P. of 1 gross of copies = C.P. of 144 copies = Rs 2,880
Now, S.P. of 1 gross of copies = C.P. + Profit = Rs 2,880 + Rs 288
or, S.P. of 144 copies = Rs 3168
or, S.P. of 1 copy = Rs 3,168÷144 = Rs 22
Hence, the selling price of each copy was Rs 22.

11. A retailer bought 6 watches for Rs 3,330 and sold them at a gain of Rs 111 from each
watch. At what price did he sell each watch?
Solution:

Here, C.P. of 6 watches = Rs 3,330 ∴C.P. of 1 watch = Rs 3,330 ÷6 = Rs 555
Profit on each watch = Rs 111

Now, S.P. of each watch = C.P. + Profit = Rs 555 + Rs 111 = Rs 666
Hence, the retailer sold each watch for Rs 666.

65 Vedanta Excel in Mathematics Teachers' Manual - 6

Extra Questions

1. Mr. Shakya bought a carpet for Rs 5,000 and sold it for Rs 5,500. Find his profit
or loss. [ A n s :

Profit=Rs 500]

2. A shopkeeper bought a refrigerator for Rs 24,000 and sold it for Rs. 28,000. Find his

profit or loss. [Ans: Profit=Rs

4,000]

3. Mr Ravi bought a washing machine for Rs. 30,000 and sold it for Rs. 27,000. Find
his profit or loss. [ A n s :

Loss=Rs 3,000]

4. Shila bought 20 rice bags for Rs. 500 per bag. Due to spoilage, she sold all the rice bags

for Rs. 8,500. Find her gain or loss. [Ans: Loss=Rs 1,500]

5. Mahesh has a laptop worth Rs 60,000 and he wants to sell it with a profit of Rs 2,000. At

what price should he sell his laptop? [Ans: Rs 62,000]

6. Janak sold a motorbike for Rs 2,50,000 and made a profit of Rs 20,000. Find his cost
price. [Ans: Rs 2,30,000]

7. A merchant bought a quintal of onions for Rs 5,000. He sold all onions at Rs 540 per

10 kg. Find his profit or loss. [Ans: Profit=Rs 400]

8. Nirmal sold his television for Rs. 25,000 at a loss of Rs. 3,500. Find the cost price of the
television. [Ans: Rs 28,500]

9. Pramod bought 100 bunches of methi greens for Rs 500. In a sudden downpour, 20 of

the bunches on his handcart got spoilt. He sold the rest at the rate of Rs 8 each. Find

his profit or loss. [Ans: Profit: Rs 140]

10. From a wholesaler, Pradip bought 300 eggs for Rs 2,200 and spent Rs 200 on transport.

50 eggs fell down and broke. He sold the rest at Rs 10 each. Find his profit or loss.

[Ans: Profit = Rs 100]

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Unit Algebra

9

Allocated teaching periods: 20
Competencies
- To classify the algebraic expressions on the basis of the number of terms.
- To identify the base, coefficient and exponent of the algebraic terms.
- To evaluate the algebraic expressions.
- To perform the operations of the algebraic expressions.
Level-wise learning objectives

S.N. Level Objectives

- To identify constant and variable.

1. Knowledge (K) - To define algebraic term and expressions.

- To identify the type of expressions.

- To tell the base, coefficient and exponent of the term.

- To find the value of expressions
- To write the base, coefficient and exponent of the term.
2. Understanding (U) - To add and subtract the like terms.
- To multiply and divide the monomials.

- To classify the algebraic expressions.
3. Application (A) - To evaluate the algebraic expressions.

- To solve the problems based on fundamental operations of
algebraic expressions.

4. High Ability (HA) - To link various real life/ contemporary problems based on
the algebraic expressions.

Required Teaching Materials/ Resources
Algebraic models, tiles, blocks, sheet of paper etc.
Pre-knowledge: like and unlike terms
Teaching Activities

Activity 1 Constant and variables
1. Discuss upon the following questions.

(i) How many head do you have?
(ii) How many eyes/hands do you have?
(iii) How many fingers are there in your hands?
(iv) How many days are there in a week?
(v) How many months are there in a year?

67 Vedanta Excel in Mathematics Teachers' Manual - 6

(vi) How many provinces/districts are there in Nepal?
(vii) What is the highest mountain in the world?
2. Also, discuss upon the following questions.
(i) What does the number 1 represent? Does it represent two, three, four etc.?
(ii) What is the sum of 4 and 6? Is it 8 or 9 or 10 or 11 or any number?
3. List out the following statements.
(i) The numbers 1, 2, 3, 4, 5, …etc. always represent the fixed numbers. These

numbers are called constants.
(ii) A number or letter which represents the fixed value is considered as constant. For

example, x represents the sum of 2 and 3. That is x = 5.
(iii) Ram obtains always same marks in mathematical tests implies that his marks is

constant over the tests.
4. Discuss upon the following questions.
(i) Are the heights of students a class same?
(ii) Are the ages of teachers of a school same?
(iii) Are the marks obtained by the students in an exam same?
5. The height/ ages/ weights of even a person cannot be same throughout the life.
6. A letter which represents many values is considered as variable. For example,
(i) x represents the prime number less than 10. Here, the value of x may be 2 or 3 or

5 or 7. Thus x is a variable.
(ii) If y represents the average temperature of a place which varies day to day, then y

is a variable.
(iii) If x represents the heights of the students of a class, then x is a variable.
7. Ask such various examples to the students and given some questions on constants and

variables.

Activity 2 Algebraic terms and expressions
1. A single constant number or variable or product of constant and variable/s or product

of variables, quotient when a variable is divided by a constant or quotient when a
constant is divided by a variable or quotient when a variable is divided by another
variable is considered as the algebraic terms. For example, 3, x, 5x, 8ab, x2, x/y etc. are
algebraic terms.
2. The algebraic term or the combination of terms is an algebraic expression. For example,
4ab, x + 2y, a2 – a + 5 etc. are algebraic expressions.
3. Write a few expressions with the unlike terms. Ask the students to count the number of
terms.
4. With various examples, make a discussion of types of expressions.

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Types of expressions Meaning Definitions Examples
Monomial
The prefix ‘mono’ means ‘one’, An algebraic 2, 5x, xyz,
Binomial expression 8ab, x/y etc.
‘alone’ or ‘single’. Examples: containing only
Trinomial one term is called
monorail, monotonous, monomial.
Multinomial
monocotyledon, monomial

etc.

The prefix ‘bi’ means ‘two’ An algebraic a+5, x- y, xy
or ‘twice’. Examples: bicycle, expression +8, 3p – 4q
binoculars, binary, binomial containing two etc.
etc. terms is called
binomial.

The prefix ‘tri’ means ‘three’ An algebraic 2x+3y-5,
or ‘thrice’. Examples: triangle, expression xy+y + 1 etc.
triple, trinomial etc. containing three
terms is called
trinomial.

The prefix ‘multi’ means An algebraic a+3b,
expression x+y+z,
‘many’. Examples: containing two 2a-3b+c-8
or more terms is etc.
multiplication, multimedia, called multinomial.

multinomial etc.

Activity 3 Coefficient, base and exponent of algebraic terms
1. Take and discuss with students upon such examples.
(i) 3×3 = 32, x×x×x = x3, y×y×y×y = y4 etc.
(ii) x3 + x3 = 2x3. In 2x3, 2 is called coefficient, x is

base and 3 is exponent.

(iii) The exponent x is 3 means x is multiplied three
times.

2. The power is the product of repeated multiplication of
the same base. For example, in 52, 5 is base, 2 is exponent and 52

itself is a power.

3. Give some terms to the students and ask to find their base,
coefficient and exponent.

4. Also, make a discussion about the numerical and literal coefficient of the terms. For
example: in 7x, 7 is a numerical coefficient. In ax3, a is the literal coefficient of x3.

Solution of selected questions from EXERCISE 9.1

1. Rewrite the following statements in algebraic expressions.

(a) Product of x and y is added to z. Solution: xy + z

(b) Three times the sum of x and y is increased by 5. Solution: 3(x + y) + 5

69 Vedanta Excel in Mathematics Teachers' Manual - 6

(c) Two times the difference of p and q is decreased by 4. Solution: 2 (p – q) – 4

(d) Product of p and q is subtracted from r. Solution: r - pq

(e) Five times the product of a and b is increased by x. Solution: 5ab + x

(f) The sum of x and y is divided by 2 and decreased by 7. Solution: (x + y)/2 – 7

2. a) The present age of Anamol is x years.

(i) How old was he 2 years before?
(ii) How old will he be after 2 years?
(iii) If his father is four times older than him, how old is his father?
Solution:

Here, the present age of Anamol is x years.

(i) Before 2 years, he was (x – 2) years old.
(ii) After 2 years, he will be (x + 2) years old.
(iii) The present age of his father is 4x years

b) The breadth of a room is b metre. If its length is 5 metres longer than its breadth,
represent the length of the room by an expression.

Solution:
Here, the breadth of the room = b m. Then, length of the room = (b + 5) m.

c) The marks obtained by A in maths is x. The marks obtained by B is double than that
of A and marks obtained by C is double than that of B. Represent the marks obtained
by B and C by expressions.

Solution:
Here, the mark obtained by A = x. The marks obtained by B = 2x and the marks obtained by
C = 2 (2x) = 4x.

3. Rewrite the following formulae in algebraic expressions.

a) The perimeter of a triangle is sum of its three sides a, b and c of the triangle. What is

the formula of perimeter of the triangle?

Solution: Here, the perimeter of triangle = The sum of three sides a, b and c

= a + b + c.

b) The area of a triangle is half of the product of base (b) and height (h). What is the

formula of area of the triangle?

Solution: Here, the area of triangle = half of the product of base (b) and height (h)

= 1 b × h
2
c) The perimeter of a square is four times of its length (l). What is the perimeter of the

square?

Solution: Here, the perimeter of square = Four times of its length (l) = 4l

d) The area of a square is the square of its length (l). What is the area of the square?

Solution: Here, the area of square = Square of length (l)= l 2

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e) The perimeter of a rectangle is two times the sum of its length (l) and breadth (b).

What is the perimeter of the rectangle?

Solution: Here, the perimeter of rectangle = two times the sum of its length (l) and breadth

(b)= 2 (l + b)

f) The area of a rectangle is the product of its length (l) and breadth (b). What is the area

of the rectangle?

Solution: Here, the area of rectangle = the product of length (l) and breadth (b)

= l ×b



Solution of selected questions from EXERCISE 9.2

1. If x = 2 and y = 3 and z = 4, evaluate the following expressions.

a) x + y + z b) 2x + 3y – z c) 3(x – y + z) d) 5xy
f) y2 + z2 – x2
e) x2 + y2 g) x + 2y h) (x – y+ z)2
Solution: z y

Here, x = 2, y = 3 and z = 4

Now,

a) x+ y + z = 2 + 3 + 4 = 9

b) 2x + 3y – z = 2×2 + 3×3 – 4 = 4 + 9 – 4 = 9

c) 3 (x – y + z) = 3 (2 – 3 + 4) = 2 (6 – 3) = 2×3 = 6

d) 5xy = 5×2×3 = 30

e) x2 + y2 = 22 + 32 = 4 + 9 = 13

f) y2 + z2 – x2 = 32 + 42 – 22 = 9 + 16 – 4 = 21

g) x + 2y = 2+2 ×3 =2 + 6 = 3948==32
h) (x –zy + z)2 = (24 – 3 + 4)2 4 32 =
= 3
y 3
2. If x = 5 cm, find the length of the

following line segment.

Solution:

Here,

The length of AD = AB + BC + CD

= (x + 1) cm + x cm + (x – 1) cm

= (5 +1) cm + 5 cm + (5 – 1) cm

= 6 cm + 5 cm + 4 cm

= 15 cm

3. Write the algebraic expressions to represent the perimeters

of the following figure. If x = 2, y = 3 and z = 5, find the

perimeters of the figures.

Solution:

Here,

The perimeter of the given figure = 2x cm + y cm + z cm + (y +1) cm + (z + 2) cm

= (2x + 2y + 2z + 3) cm

If x = 2, y = 3 and z = 5, actual perimeter = (2×2 + 2×3 + 2×5 + 3) cm = 18 cm

71 Vedanta Excel in Mathematics Teachers' Manual - 6

4. x is a variable on the set A = {1, 2, 3}, that is, x can be replaced by 1, 2 and 3.
Evaluate the expressions (i) x + 5 (ii) 2x – 1.
Solution:
(i) When x = 1 then x + 5 = 1 + 5 = 6
When x = 2 then x + 5 = 2 + 5 = 7
When x = 3 then x + 5 = 3 + 5 = 8
(ii) When x = 1 then 2x – 1 = 2×1 – 1 = 2 – 1 = 1
When x = 2 then 2x – 1 = 2×2 – 1 = 4 – 1 = 3
When x = 3 then 2x – 1 = 2×3 – 1 = 6 – 1 = 5
5. If y = 2x + 1 and x is a variable on the set B = {2, 4, 6}, find the possible values of y.
Solution:
Here, y = 2x + 1 and replacement set = {2, 4, 6)
When x = 2 then = 2x + 1 = 2×2 +1 = 4 + 1 = 5
When x = 4 then = 2x + 1 = 2×4 +1 = 8 + 1 = 9
When x = 6 then = 2x + 1 = 2×6 +1 = 12 + 1 = 13
6. If x = 2y, express 2x + 5y in terms of y and evaluate the expression when y = 3.
Solution:
Here, x = 2y
Now, 2x + 5y = 2 (2y) + 5y = 4y + 5y = 9y
When y = 3 then 2x + 5y = 9y = 9×3 = 27
7. If x = 2a + 1, show that 3x – 6a + 7 = 10
Solution:
Here, x = 2a +1
Now, 3x – 6a + 7 = 3 (2a + 1) – 6a + 7 = 6a + 3 – 6a + 7 = 10 Proved.

Activity 4 Addition and Subtraction of algebraic expressions

1. With examples, make a discussion on like and unlike terms.
2. 3 books and 4 books are like (same) type of things. Similarly, x and 2x are like terms. 3y

and 4y are like terms. The algebraic terms having the same base and equal power are

called like terms. For examples: 2x, 7x, 9x, etc. are like terms because they have same base

x and equal power 1. 5a2, 9a2, 12a2, etc. are like terms because they have the same base a

and equal power 2.

3. The algebraic terms which have different bases or different powers are called unlike terms
4. Explain the addition and subtraction of like terms.

Solution of selected questions from EXERCISE 9.4

1. Add: 8abc – 5ab + 3a and 2a + 7ab – 4abc

Solution:

Here, 8abc – 5ab + 3a+ 2a + 7ab – 4abc = 8ab – 4abc – 5ab + 7ab + 3a + 2a

= 4abc + 2ab + 5a

2. Subtract: xyz – 4xy + 5 from 5xyz + xy + 3yz – 4

Solution:

Vedanta Excel in Mathematics Teachers' Manual - 6 72

Here, 5xyz + xy + 3yz – 4 – (xyz – 4xy + 5) = 5xyz + xy + 3yz – 4 – xyz + 4xy – 5

= 4xyz + 5xy + 3yz – 9

3. What should be added to 2x – 3 to get 5x + 7?

Solution:

Here, the required expression to be added is (5x + 7) – (2x – 3)

= 5x + 7 – 2x + 3

= 3x + 10

4. What should be subtracted from 5m – 3n to get 2m + 5n?

Solution:

Here, the required expression to be subtracted is (5m – 3n) – (2m + 5n)

= 5m – 3n – 2m – 5n

= 3m – 8n

5. To what expression 2a – 3b + 1 must be added to get 4a + 7b – 3?

Solution:

Here, the required expression to be added is (4a + 7b – 3) – (2a – 3b + 1)

= 4a + 7b – 3– 2a + 3b – 1

= 2a + 10b – 4

6. From what expression x2 + 5y2 – 3xy must be subtracted to get 2x2 – y2+ 4xy?

Solution:

Here, the required expression to be subtracted is 2x2 – y2+ 4xy +( x2 + 5y2 – 3xy)

= 3x2 + 4y2 + xy

7. If x = 2m – n and y = m + n, show that x – y = m – 2n.

Solution:

Here, x = 2m – n and y = m + n

Now, L.H.S. = x – y

= 2m – n – (m + n)

= 2m – n – m –n

= m – 2n = R.H.S. Proved

8. If x2 = 2a2 – b2, and y2 = 2b2 – c2 and z2 = 2c2 – a2, show that: x2 + y2 + z2 = a2 + b2 + c2

Solution:

Here, x2 = 2a2 – b2, and y2 = 2b2 – c2 and z2 = 2c2 – a2

Now, L.H.S. = x2 + y2 + z2

= 2a2 – b2+ 2b2 – c2 + 2c2 – a2

= a2 + b2 + c2 = R.H.S. Proved

73 Vedanta Excel in Mathematics Teachers' Manual - 6

9. The sides of a triangle are x + 4, 2x – 3 and 3x + 1, find its perimeter.

Solution:

Here, the sides of a triangle are x + 4, 2x – 3 and 3x + 1
Now, perimeter = x + 4 + 2x – 3 + 3x + 1

= 6x + 2

Activity 5 Multiplication of algebraic expressions

1. Recall the addition and subtraction of like terms.

2. Recall the exponent, base and coefficient of algebraic term.

3. Make a discussion on the following activities.

a) Take a square sheet of paper and measure the length of side of the paper and ask

about the perimeter and area of the paper.

b) Similarly, suppose the length of side of square by x and ask about the perimeter and

area of the paper.

c) Take the cube and denote its every edge by x unit and discuss about its volume.

4. Under the discussion about the indices, list out the points.

a) While multiplying two powers of same base, the exponents are added.

For example: (i) x × x =x1+1 = x2, (ii) x2 × x3 = x2+3 = x5 (iii) y5 × y = y6,

(iv) ab2 × a3 b = a1+3b2+1 = a4b3 etc.

b) While multiplying the algebraic terms, the coefficients of the terms are multiplied

and the exponents of the same bases are added.

For example: (i) 2x ×3x = 6x2 (ii) 5ab ×2a2 = 10a3b (iii) a (b +c) = ab + ac

(iv) (x + y) (a + b) = x(a + b) + y(a + b) = ax + bx + ay + by etc.

5. Discuss upon the worksheets on multiplication.

a) The area of rectangle = length (l) ×breadth (b). Find the area of the following rectangles.

(i) (ii) (iii)



Area (A)= …………. Area (A) = ………… Area (A) = ………

b) The volume of cuboid (V) = length (l) × breadth (b) × height (h). Let’s find the

volume of following cuboids.

(i) (ii)

Volume (V) = …………… Volume (V) = ………….

6. Focus on group work and project works to identify the connection of multiplication in real
life situation.

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Solution of selected questions from EXERCISE 9.5

1. Simplify.

a) x ×x×x2 b) 2a2 × 3a×5a c) 3x×2y×2x ×y

d) p2 × 5q × 2p × 3q2 e) 3b3 × c2 × 2b2×4c3 f) xyz×2xy × yz × 5zx

g) (– 3ab)×(– 2bc) × (– abc) h) (–5qr) × (– 2pqr) × pq i) 6x2y× (– xy2) ×xyz× (– 3yz2 )

Solution:

a) x ×x×x2 = x1+1+2 = x4

b) 2a2 × 3a×5a = 30a2+1+1 = 30a4

c) 3x×2y×2x ×y = 12x2y2

d) p2 × 5q × 2p × 3q2 = 30p3q3

e) 3b3 × c2 × 2b2×4c3 = 24b5c5

f) xyz×2xy × yz × 5zx = 10x3y3z3

g) (– 3ab) × (– 2bc) × (– abc) = (+6ab2c) × (– abc) =-6a2b3c2

h) (–5qr) × (– 2pqr) × pq = 10p2q3r2

i) 6x2y× (– xy2) ×xyz× (– 3yz2 )=(-6x3y3) × (– 3xy2z3 )=18x4y5z3

2. If p = 5x and q = 3x – 1, show that 2pq = 30x2 – 10x.
Solution:
Here, p = 5x and q = 3x – 1
Now, L.H.S. = 2pq = 2×5x×(3x – 1)= 10x (3x – 1) = 30x2 – 10x = R.H.S. Proved
3. If m = n = 2a and a = 3, find the value of mn .
Solution:
Here, m = n = 2a and a = 3
Now, mn = 2a ×2a = 2a
When a = 3, then 2a = 2 × 3 = 6
4. If p = 2x, q = 3x and x = 3, find the value of 6pq
Solution:
Here, p = 2x, q = 3x and x = 3
Now, 6pq = 6 × 2x × 3x = 6x
When x = 3, then 6x = 6 × 3 = 6
5. Simplify. x (x + 1) + 2(x + 1)
Solution:
Here, x (x + 1) + 2(x + 1) = x2 + x + 2x + 2 = x2 + 3x + 2

Solution of selected questions from EXERCISE 9.6

1. Multiply: (5a – 2b) (5a + 2b)

Solution:

Here, (5a – 2b) (5a + 2b) = 5a (5a + 2b) – 2b (5a + 2b)

= 25a2 + 10ab – 10ab – 4b2

= 25a2– 4b2

2. Multiply: (4a – 5b) (3a + 2b – 7)

Solution:

75 Vedanta Excel in Mathematics Teachers' Manual - 6

Here, (4a – 5b) (3a + 2b – 7) = 4a (3a + 2b – 7) – 5b (3a + 2b – 7)

= 12a2 + 8ab – 28a – 15ab – 10b2 + 35b

=12a2 –-28a – 7ab + 35b – 10b2

3. If a = (2x – 3) and b = (2x + 3), show that ab + 9 = x2
4
Solution:

Here, a = (2x – 3) and b = (2x + 3)

Now, ab = (2x – 3) (2x + 3)

= 2x (2x + 3) – 3 (2x + 3)

= 4x2 + 6x – 6x – 9

= 4x2 – 9

L.H.S. = ab + 9 = 4x2 –9 + 9 = 4x2 = x2 = R.H.S. Proved
4 4 4

Activity 6 Division of algebraic expressions

1. Recall multiplication of algebraic terms.
2. Under discussion about the quotient law of indices, divide a monomial by another

monomial.
The process of dividing an algebraic term by another term:
(i) Divide the coefficient of dividend by the coefficient of divisor.
(ii) Subtract the exponent of the base of divisor from the exponent of the same base of

dividend.

Example: Divide a) 10x2 by 2x b) 15a3b4 by 5ab2

Solution: 10x2
2x
a) 10x2 ÷ 2x = 10x2 ÷ 2x = 10x2 In division of the same
2x base we should subtract
150 × x × x lower exponent from
= 2×x = 10 x2 – 1 higher exponent of the
2 same base.
= 5x = 5x

b) 15a3b4 ÷ 5ab2 = 15a3b4
5ab2

= 135 × a ×a × a × b ×b × b ×b 15a3b4 ÷ 5ab2 = 15a3b4
5× a × b ×b 5ab2

15a3–1 b4 – 2
5

=3×a×a×b×b =

= 3a2b2 = 3a2b2

3. Under discussion, divide polynomials by monomials by using the following process.
The process of dividing polynomials by monomials

(i) Divide each term of dividend by the divisor.

(ii) Subtract the exponent of the base of divisor from the exponent of the same base of
dividend.

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Example: Divide (a) 10x2 – 15x by 5x (b) 14a3 b2 – 8a2 b3 by – 2ab

(b) 14a3b2 – 8a2b3 by – 2ab

Example: Divide (a) 10x2 – 15x by 5x

Solution:

a) (10x2 – 15x) ÷ 5x

= 120x2 – 135x
5x 5x
11
= 2x2 – 1 – 3

= 2x –

4. Give examples of division of polynomials by binomials by using the following process.
The process of dividing polynomials by binomials
1. Arrange the terms of divisor and dividend in descending (or ascending) order of exponents

of common bases.
2. Divide the term with highest exponent of the polynomial by the term with highest exponent

of the binomial. Write the result above the division line.
3. Multiply this result by the divisor, and subtract the resulting binomial from the polynomial.
4. Divide the term with highest exponent of the remaining polynomial by the highest term

with highest exponent of the binomial.
5. Repeat this process until the remaining polynomial has lower exponent than the binomial.

Example: Divide (x2 + 5x + 6) by (x + 3)
Solution:

Divide x2 by x x2 ÷ x = x (In quotient)
Multiply the divisor (x + 3) by the quotient x
x(x + 3) = x2 + 3x
Subtract the product from the dividend.

x2 + 5x + 6
± x2 ± 3x

2x + 6
Again, divide the first term of the remainder by the
first term of the divisor. Continue the process till the
remainder is not divisible by divisor.

77 Vedanta Excel in Mathematics Teachers' Manual - 6

Solution of selected questions from EXERCISE 9.6

1. Divide: (20b4 c6 – 50b4 c4) ÷ (–10b4 c4)

Solution:

Here, (20b4 c6 – 50b4 c4) ÷ (–10b4 c4) = 20b4c6 – 50b4c4 = 20b4c6 – 50b4c4 =-2c2+5
–10b4c4 –10b4c4 –10b4c4
2. Find the quotient: (m2 + 7m + 12) ÷ (m + 4)

Solution:

m + 4) m2 + 7m + 12 (m+3

± m2 4m

3m + 12

± 3m ± 12

0

Hence, quotient = m + 3

3. Find the quotient: (b2 + 2b – 15) ÷ (b – 3)

Solution:

b – 3) b2 + 2b – 15 (b+5

± b2 ∓ 3b

5b – 15

± 5b ∓15

0

Hence, quotient = b + 5

4. Find the quotient: (x2 – 3x – 10) ÷ (x – 5)

Solution:

x - 5) x2 – 3x – 10 (x+2

± x2 ∓ 5x

2x – 10

± 2x∓10

0

Hence, quotient = x + 2

5. The product of two algebraic terms is 6a3 b2. If one of the terms is 3ab, find the other term.

Solution:

Here, another term = 6a3b2 = 2a2b
3ab

6. The area of a rectangle is x2 + 4x + 3 sq. units and its length is (x + 1) units. Find its

breadth.

Solution:

Here, breadth = Area ÷length = (x2 + 4x + 3) ÷ (x +1)

x + 1) x2 + 4x +3 (x+3

± x2 ± x

3x + 3

± 3x ± 3

0

Hence, the required breadth of the rectangle is (x + 3) units.

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7. The area of a rectangle is x2 + 3x – 10 square units. If its breadth is (x – 2) units, find its

length.

Solution:

Here, length = Area ÷breadth = (x2 + 3x – 10) ÷ (x – 2)

x - 2) x2 + 3x – 10 (x+5

± x2 ∓ 2x

5x – 10

± 5x∓10

0

Hence, the required length of the rectangle is (x + 5) units.

8. If a = 3x2 y, b = 4xy2 and c = 2xy, find the value of ab/c.

Solution:

Here, a = 3x2 y, b = 4xy2 and c = 2xy

Now, ab = 3x2y×4xy2 = 12x3y3 = 6x2y2
c 2xy 2xy

9. If x = 12p3 q4, y = 6p2 q3 and z = 2p2 q2, show that x/y + y/z = 2pq + 3q.

Solution:

Here, x = 12p3 q4, y = 6p2 q3 and z = 2p2 q2

Now, L.H.S. = x + y = 12p3q4 + 6p2q3 = 2pq + 3q = R.H.S. Proved
y z 6p2q3 2p2q2

Extra Questions

1. Define binomial with an example. [Ans: 20]
2. If a = 2, b = 3 and c =4, find the value of ab + bc.

3. What should be subtracted from 5m – 3n to get 2m + 5n? [Ans: 8n -3m]

4. Find the perimeter of a triangle in which the length of sides are (4x + 5) cm, (10– 3x) cm and

(x – 4) cm. [Ans: (2x + 11) cm]

5. Multiply: (-3bc) ×(-8abc) ×ca [Ans: 24a2b2c3]

6. If p = 3x + 4 and q = 3x – 4, show that: 9x2 – pq = 16.

7. If x2 = 2a2 – b2, and y2 = 2b2 – c2 and z2 = 2c2 – a2, show that: x2 + y2 + z2 = a2 + b2 + c2

8. Divide (42p5q3 – 24p3q2) by (-6p2q2) [Ans: 4p – 7p3q]

9. Find the quotient: (x2 + 3x + 2) ÷ (x + 2) [Ans: (x + 1)]

10. The area of a rectangle is x2 + 7x +10 square units. If its breadth is (x + 2) units, find its

length. [Ans: (x + 5) units]

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Unit Equation, Inequality and Graph

10

Allocated teaching periods: 10
Competencies

- To solve linear equations of single variable.
- To solve the verbal problems related to linear equation of single variable.
- To represent the inequality on a number line.
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define open statement

1. Knowledge (K) - To define equation

- To make the equation for statements.

- To write down the inequality from the number line.

2. Understanding (U) - To solve linear equation of single variable.
- To represent the inequality on number line.

3. Application (A) - To solve linear equation of single variable.
- To make the equation and solve.

4. High Ability (HA) - To solve the real life problems based on equation.

Required Teaching Materials/ Resources

Chart papers, beam balance, graph board, graph copy, ICT tools (if possible) etc.

Pre-knowledge: algebraic expressions

Teaching Activities

Activity 1 Equation

1. Make a discussion on mathematical statements (true, false, open) with proper examples.

2. Take a beam balance and present equality of quantities.

3. By using the beam balance, show the following facts of solving equations.

Fact 1: When an equal number is added to two equal quantities the sum will be equal. For

example, If x = 2, then x + 3 = 2 + 3

Fact 2: When an equal number is subtracted from two equal quantities the difference will

be equal. For example, If x = 8, then x – 5 = 8 – 5

Fact 3: When two equal quantities are multiplied by an equal number, the product will be

equal. for example, If x = 4, then 3×x = 3×4

Fact 4: When two equal quantities are divided by an equal number, the quotient will be

equal. For example, If x = 24, then x÷8 = 24÷8

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Solution of selected questions from EXERCISE 10.2

1. Solve the equations:

(i) 5x – 2=8
6

Solution: Here, 5x –2 = 8 or, 5x = 10 or, 5x = 60 ∴x = 12
6 6

(ii) 3a – 1 = 321
4

Solution: Here, 3a – 1 = 312 or, 3a – 4 = 7 or, 6a – 8 =28 or, 6a = 36 ∴a = 6
4 4 2

(iii) 49x + 2 = 423

Solution: Here, 4x + 2 = 432 or, 4x + 18 = 14 or, 12x + 54 = 126 or, 12x = 72 ∴x= 6
9 9 3

2. Solve the equations:

Solution:

Here,

(i) 3x – 2 = 2x + 5
4 3

or, 3 (3x – 2) = 4 (2x + 5)

or, 9x – 6 = 8x + 20

or, 9x – 8x = 20 + 6

∴ x = 26

(ii) x + x = 5
2 3

Here, (i) 4x – 3x = 2
5 10
3x + 2x
or, 6 = 5 Solution:

or, 5x = 30 Here, 4x – 3x = 2
or, 85x – 31x0 = 2
∴ x = 6

10
or, 5x = 20
(i) or, 5x = 30
∴ x = 4
∴ x = 6

3. From these balance, let’s make the equation and solve it.
Solution:
Here, 4y = 2y + 6
or, 4y – 2y = 6
or, 2y = 6
or, y = 3

4. Make an equation for the perimeter of each square. Solve it to find the value of x. Also
find the actual length of each square.

81 Vedanta Excel in Mathematics Teachers' Manual - 6

Solution:
a) The perimeter of the square = 4×length of side (l)
or, 16 = 4x
or, x = 16/4 = 4

Hence, the actual length of square is x cm i.e., 4 cm.
b) The perimeter of the square = 4×length of side (l)
or, 24 = 4 (3x)
or, 24 = 12 x
or, x = 2 and 3x = 3×2 = 6

Hence, the actual length of square is 18 cm.
c) The perimeter of the square = 4×length of side (l)
or, 40 = 4 (x +2)
or, 40 = 4x + 8
or, 40 – 8 = 4x
or, 32 = 4x
or, x = 8 and (x + 2) = (8+2) = 10

Hence, the actual length of square is 10 cm.

5. Make an equation for the area of each rectangle. Solve it to find the value of x. Also
find the actual length and breadth of each rectangle.

Solution:
a) Here, length of rectangle (l) = (x + 2) cm and breadth (b) = 3 cm

Now, the area of the rectangle = l × b
or, 18 = (x +2) × 3
or, 18 = 3x + 6
or, 18 – 6 = 3x
or, 12 = 3x
or, x = 4
Hence, the actual length (l) = (x + 2) cm = (4 + 2) cm = 6 cm
b) Here, length of rectangle (l) = (3x – 1) cm and breadth (b) = 4 cm

Now, the area of the rectangle = l × b
or, 20 = (3x – 1) × 4

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or, 20 = 12x – 4
or, 20 + 4 = 12x
or, 24 = 12x
or, x = 2
Hence, the actual length (l) = (3x – 1) cm = (3×2 – 1) cm = 5 cm
c) Here, length of rectangle (l) = 9 cm and breadth (b) = (5x -6) cm

Now, the area of the rectangle = l × b
or, 36 = 9 × (5x – 6)
or, 36 = 45x – 54
or, 36+ 54 = 45x
or, 90= 45x
or, x = 2
Hence, the actual breadth (b) = (5x – 6) cm = (5×2 – 6) cm = 4 cm

6. Make an equation for the perimeter of each rectangle. Solve it to find the value of x.
Also find the actual length and breadth of each rectangle.

Solution:
a) Here, length of rectangle (l) = 3x cm and breadth (b) = 2x cm

Now, the area of the rectangle = 2 (l +b)
or, 20 = 2(3x + 2x)
or, 20 = 2 (5x)
or, 20 = 10x
or, x = 2
Also, 3x cm = 3×2 cm = 6 cm and 2x cm = 2×2 cm = 4 cm
Hence, the actual length (l) is 10 cm and breadth (b) is 4 cm.
b) Here, length of rectangle (l) = 2x cm and breadth (b) = (x + 2) cm

Now, the area of the rectangle = 2 (l +b)
or, 22 = 2(2x + x + 2)
or, 22 = 2 (3x + 2)
or, 22 = 6x + 4
or, 18 = 6x
or, x = 3
Also, 2x cm = 2×3 cm = 6 cm and (x+ 2) cm = (2+3) cm = 5 cm
Hence, the actual length (l) is 6 cm and breadth (b) is 5 cm.

c) Here, length of rectangle (l) = (x + 6) cm and breadth (b) = (x + 2) cm
Now, the area of the rectangle = 2 (l +b)

or, 36 = 2(x +6 + x + 2)

83 Vedanta Excel in Mathematics Teachers' Manual - 6

or, 36 = 2 (2x + 8)
or, 36 = 4x + 16
or, 20 = 4x
or, x = 5
Also, (x +6) cm = (5 + 6) cm = 11 cm and (x+ 2) cm = (5+2) cm = 7 cm
Hence, the actual length (l) is 11 cm and breadth (b) is 7 cm.

Solution of selected questions from EXERCISE 10.3

1. a) The sum of two numbers is 12. If one of them is 7, find the other number.
Solution:
Let, the other number be x.
Now, x + 7 = 12
or, x = 12 – 7
or, x = 5
Hence, the required number is 5.

b) The difference of two numbers is 6. If the greater one is 17, find the smaller number.
Solution:
Let, the smaller number be x.
Now, 17 – x = 6
or, 17 – 6 = x
or, x = 11
Hence, the required smaller number is 11.

c) The difference of two numbers is 9. If the smaller one is 6, find the greater number.
Solution:
Let, the greater number be x.
Now, x – 6 = 9
or, x = 9 + 6
or, x = 15
Hence, the required greater number is 15.

d) The quotient of dividing a number by 4 is 5. Find the number.

Solution:

Let, the number be x.
x
Now, 4 =5

or, x = 20

Hence, the required number is 20.

2. a) Bishwant thought of a number. He doubled it and added 7. If he got 15, find the
number he thought of.
Solution:
Let, the number that Bishwant thought be x.
Now, 2x + 7 = 15

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or, 2x = 15 – 7
or, 2x = 8
or, x = 4
Hence, the required number is 4.

b) Sunayana thought of a number. She trebled it and subtracted 4 from it. If she got 5,
find the number she thought of.
Solution:
Let, the number that Sunayana thought be x.
Now, 3x – 4 = 5
or, 3x = 5 + 4
or, 3x = 9
or, x = 3
Hence, the required number is 3.

3. a) The cost of 10 pencils is Rs 15 more than the cost of such 7 pencils. What is the cost of
a pencil?

Solution:
Let the cost of a banana be Rs x.
Now, 10x = 7x + Rs 15
or, 10x – 7x = 15
or, 3x = 15
or, x = 5
Thus the cost of a banana is Rs 5.

b) The cost of a dozen copies is Rs 50 more than that of 10 copies of same kind. What is
the cost of a copy?

Solution:
Let the cost of a copy be Rs x.
Then, cost of a dozen of copies = cost of 12 copies = Rs 12x.
Now, 12x = 10x + Rs 50
or, 12x – 10x = 50
or, 2x = 50
or, x = 25
Thus the cost of a copy is Rs 25.

4. a) If the sum of two consecutive odd numbers is 16, find the numbers.
Solution:
Let the smaller odd number be x.
Then, the consecutive odd number for x is x + 2.
Now, x + (x + 2) = 16
or, 2x + 2 = 16
or, 2x = 16 – 2
or, 2x = 14
or, x = 7 and x + 2 = 7 + 2 = 9
Hence, the required consecutive odd numbers are 7 and 9.

85 Vedanta Excel in Mathematics Teachers' Manual - 6

b) If the sum of two consecutive even numbers is 22, find the numbers.
Solution:
Let the smaller even number be x.
Then, the consecutive even number for x is x + 2.
Now, x + (x + 2) = 22
or, 2x + 2 = 22
or, 2x = 22 – 2
or, 2x = 20
or, x = 10 and x + 2 = 10 + 2 = 12
Hence, the required consecutive even numbers are 10 and 12.

5. a) A sum is Rs x. If its 10% is Rs 40, find the sum.

Solution:

Here, the sum = Rs x

or, 10% of x = Rs 40

or, 10 × x = Rs 40
100
or, x= Rs 400

Hence, the required sum is Rs 400.

b) A sum is Rs x. If 2 parts of the sum is Rs 50, find the sum.
3
Solution:

Here, the sum = Rs x

or, 2 × x = Rs 50
3
or, 2x = 150

or, x= Rs 75

Hence, the required sum is Rs 75.

6. The cost price (C.P.) of an article is Rs x. If 10% profit on C.P. amounts to Rs 250, find the

C.P. of the article.

Solution:

Here, the cost price (C.P.) of an article = Rs x

Now, the profit amount = Rs 250

or, 10% of x = Rs 250

or, 10 × x = Rs 250
100
or, x= Rs 2500

Hence, the cost price (C.P.) of an article is Rs 2,500.

7. The perimeter of a rectangle is 42 cm. If its length is double than that of its breadth, find
the length and breadth of the rectangle.
Solution:
Let, the breadth of the rectangle (b) = x cm. Then, its length (l) = 2×breadth = 2x cm
Now, the perimeter (P) = 42 cm

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or, 2 (l + b) = 42 cm
or, 2 (2x + x) = 42
or, 2 (3x) = 42
or, 6x = 42
or, x = 7
Hence, the length of the rectangle (l) = 2x cm = 2×7 cm = 14 cm and breadth (b) = x cm
= 7 cm.
8. a) If the sum of two numbers is 10 and the difference is 6, find the numbers.
Solution:

Let, the smaller number be x then the greater number = (x + 6)
Now, x + (x + 6) = 10
or, 2x + 6 = 10
or, 2x = 10 – 6
or, 2x = 4
or, x = 2
Hence, the smaller number = x = 2 and greater number = x + 6 = 2 + 6 = 8.

b) A number is greater than another number by 3. If their sum is 33, find the numbers.
Solution:

Let, the smaller number be x then the greater number = (x + 3)
Now, x + (x + 3) = 33
or, 2x + 3 = 33
or, 2x = 33 – 3
or, 2x = 30
or, x = 15
Hence, the smaller number = x = 15 and the greater number = x + 3 = 15 + 3 = 18.

c) A number is less than another number by 7. If their sum is 37, find the numbers.
Solution:

Let, the greater number be x then the smaller number = (x – 7)
Now, (x – 7) + x = 37
or, 2x – 7 = 37
or, 2x = 37 + 7
or, 2x = 44
or, x = 22
Hence, the greater number = x = 22 and the smaller number = x – 7 = 22 – 7 = 15.

9. a) Father is 30 years older than his son. If the sum of their ages is 50 years, find their
ages.
Solution:
Let, the age of son be x years. Then, the age of the father = (x + 30) years
Now, x + (x + 30) = 50

87 Vedanta Excel in Mathematics Teachers' Manual - 6

or, 2x + 30 = 50
or, 2x = 50 – 30
or, 2x = 20
or, x = 10 and (x + 30) = 10 + 30 = 40
Hence, the age of son is 10 years and the age of the father is 40 years.

b) Daughter is 25 years younger than her mother. If the sum of their ages is 35 years, find
their ages.

Solution:
Let, the age of mother be x years. Then, the age of the daughter = (x – 25) years
Now, (x – 25) + x = 35
or, 2x – 25 = 35
or, 2x = 35 + 25
or, 2x = 60
or, x = 30 and (x – 25) = 30 – 5 = 25
Hence, the age of mother is 30 years and the age of the daughter is 5 years.

10. a) The total cost of a pencil and an eraser is Rs 15. If the pencil is more expensive than
eraser by Rs 7, find their costs.

Solution:
Let, the cost of an eraser be Rs x. Then, the cost of a pencil = Rs. (x + 7)
Now, (x + 7) + x = Rs 15
or, 2x + 7 = 15
or, 2x = 15 – 7
or, 2x = 8
or, x = 4 and (x + 7) = 4 + 7 = Rs 11
Hence, the cost of an eraser is Rs 4 and the cost of a pencil is Rs 11.

b) There are 32 students in a class. If the number of girls are 6 more than the number of
boys, find the numbers.
Solution:
Let, the number of boys be x. Then, the number of girls = (x + 6)
Now, (x + 6) + x = 32
or, 2x + 6 = 32
or, 2x = 32 – 6
or, 2x = 26
or, x = 13 and (x + 6) = 13 + 6 = 19
Hence, the number of boys is 13 and the number of girls is 19.

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Unit Coordinates

11

Allocated teaching periods: 4

Competencies
- To find the coordinates of points

- To plot the points on the graph paper

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To tell the x-coordinate and y-coordinate of point
- To state the origin

- To tell the coordinates of point from the graph
2. Understanding (U) - To write a point on x-axis and y-axis

3. Application (A) - To write the coordinates of vertices of figure from the graph
- To plot the points and join them in order to form a figure

- To find the coordinates of mid-point of a segment joining
the two points

4. Higher Ability (HA) - To mark the points in the first, second, third and fourth
quadrants and write their coordinates.

- To draw a polygon (triangle, rectangle) in a graph paper and
write down the vertices.

Required Teaching Materials/ Resources
Graph board/ graph chart /graph paper, ruler, ICT tools like GeoGebra etc.
Pre-knowledge: Number lines

Teaching Activities

Activity 1 Coordinates

1. Make a short discussion on ordered pairs with examples.
(i) Husband and wife of relatives: (father, mother), (uncle, aunt) etc.
(ii) Countries and their capital cities: (Nepal, Kathmandu), (India, New Delhi) etc.
(iii) Numbers and their squares: (1, 1), (2, 4), (3, 9), (4, 16) etc.

89 Vedanta Excel in Mathematics Teachers' Manual - 6

2. By using graph board/chart and make a short discussion on the coordinate system: origin,
axes and quadrants.

3. Show the points and their coordinates in Geo-Gebra.
4. Mark some points on the graph chart/graph board and

ask the coordinates of them points.
5. List out the following points under discussion.
(i) The coordinates of origin O is (0, 0)
(ii) X-coordinate is called abscissa.
(iii) Y-coordinate is called ordinate
(iv) The coordinates of a point on x-axis is (x, 0)
(v) The coordinates of a point on y-axis is (0, y)

Solution of selected questions from EXERCISE 11.1

1. Plot the following points and join them in alphabetical order by using ruler.
Name the figures so formed.
(a) M (3, 1), A (3, 4), T (–2, 4), H (–2, 1)
(b) N (– 2, 2), I (0, 5), C (2, 2) E (0, – 7)
Solution:
(a) Here, the given points are M (3, 1), A (3, 4), T (–2, 1) and H (–2, 4)
Plotting these points on a graph paper and joining them in alphabetical order.

From graph, MATH is a rectangle.
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(b) Here, the given points are N (– 2, 2), I (0, 5), C (2, 2) and E (0, – 7)
Plotting these points on a graph paper and joining them in alphabetical order.
From graph, NICE is a kite.

2. Plot the points A (1, 2) and B (5, 2) on a graph paper and join them. Mark the
mid-point M of line segment AB and find its coordinates.
Solution:
Here, the given points are A (1, 2) and B (5, 2).
Measuring the length of segment AB with the help of ruler and marking its mid-
point.
From the graph, we observed that the coordinates of mid-point M is (3, 2).


3. Draw a line segment AB with ends A (–1, 3) and B (2, 3). Produce AB to the point
C so that AB = BC. Then, write the coordinates of the point C.
Solution:
91 Vedanta Excel in Mathematics Teachers' Manual - 6

Here, the given points are A (-1, 3) and B (2, 3).
Plotting the points on the graph and join them with the help of a ruler.
Measuring the length of AB and producing it to C where the length of BC is equal
to AB.
From the graph, we observed that the coordinates of point C is (5, 3).

Extra Questions

1. Write the coordinates of vertices of triangle ABC from the graph given aside.
2. Plot the points P (1, 1), Q (3, 1), R (3, 3) and S (1, 3) on a graph paper. Join them in

alphabetical order by using ruler. Also, name the figures so formed.
3. Plot the points P (2, 3) and Q (6, 5) on a squared paper and join them. Mark the mid-point

M on the line segment PQ and write its coordinates.
4. Draw a rectangle in a graph paper. Write the coordinates of the vertices of the rectangle.



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Unit Geometry: Point and Line

12

Allocated teaching periods: 4

Competencies

- To draw parallel line segments
- To draw perpendicular line segments
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To define point
- To define line
- To identify the parallel or perpendicular line segments

- To identify the parallelism and perpendicular relationship

2. Understanding (U) of line segments

- To write down the names of vertices of polygon

3. Application (A) - To find the distances between two given line segments and

write down the parallel relation between them

4. High Ability (HA) - To make a project work on parallel and perpendicular line
segments

Required Teaching Materials/ Resources
Ruler, straight sticks, box, maps, different types of objects representing the parallel and the
perpendicular segments etc

Pre-knowledge: Point, line

Teaching Activities

Activity 1 Point, line, ray, line segment, plane
1. Mark the position of students of the class by drawing a sketch of classroom.
2. Take a map of Nepal or district and mark the position of places
3. Ask students to make a dot on the paper with a well-sharpened pencil.
4. Ask the definition of point/s with examples.
5. Take a robber-band and stretch it and discuss about the line.
6. Draw different types of lines and explain them.
7. List out the following points regarding line/s.
(i) An infinite number of lines can be drawn through a given point.
(ii) One and only one straight line can be drawn through two points.
(iii) An infinite number of curved can be drawn through two points.

93 Vedanta Excel in Mathematics Teachers' Manual - 6

8. Give the real life examples of rays like sun ray, ray of torch light, ray from projector etc.
9. Define rays with examples and notation to represent rays.
10. Take a box/book, give it to students to touch its surfaces and discuss about its plane

faces.
11. Give more examples of plane figures like triangle, square, rectangle etc.

Activity 2 Parallel and Perpendicular line segments

1. Give the sheet of paper to each student and ask to fold it and again. Then tell to unfold the
paper. Observe and discuss the lines made on the papre such as intersecting lines, parallel
lines and perpendicular lines etc.

2. Make a discussion on the opposite edges of white-board, table, book etc. and investigate the
ideas of parallel line segments.

3. Define parallel line segments with its notation.
4. Discuss on the distance between the parallel line segments.
5. Give real life examples which represent the intersecting lines such as cross paths, arms of

scissors, hands of watch etc.
6. Draw some intersecting line segments on the board. Give names of the segments and point

of intersection.
7. Give as many as examples of perpendicular line segments.
8. Ask the definition of the perpendicular lines to the students.

Solution of selected questions from EXERCISE 12.1

1. Let’s read the incomplete sentences given below and write ‘parallel’ or ‘perpendicular’ to
complete them.

a) The opposite edges of a ruler are parallel.
b) The arrangements of books in the bookshelf are parallel.
c) The X-axis and Y-axis are perpendicular.
d) The adjacent sides of a rectangle are parallel.
e) The diagonals of a square are perpendicular.
2. Let’s tell and write the answers as quickly as possible.
a) How many straight lines can be drawn from a point? Infinite
b) How many straight lines can be drawn through two points? Only one
c) How many curved lines can be drawn through two points? Infinite
d) Through how many points do two lines interest each other? None or one or infinite
e) What is the angle made by two perpendicular lines? 90o
f) What is the angle made by two parallel lines? 0o
g) If the perpendicular distances between two lines at any point are equal, the lines are

said to be parallel
h) Are the opposite edges of your desk parallel? Yes
i) Are the breadths of your books perpendicular to their lengths? Yes
j) How many plane surfaces are there in your exercise book? Six

Vedanta Excel in Mathematics Teachers' Manual - 6 94

3. a) If AB//CD and CD//EF, what is the relation
between AB and EF?

Solution:
Here, AB//CD and CD//EF
Hence; the relation between AB and EF is parallel.
i.e., AB//EF.

b) If PQ ⊥ AB and RS ⊥ AB, what is the relation
between PQ and RS?

Solution:
Here, PQ ⊥ AB and RS ⊥ AB
Hence; the relation between PQ and RS parallel.
i.e., PQ//RS.

c) If AB // CD, EF and GH are perpendicular lines on AB and
CD, what is the relation between EF and GH?

Solution:
Here, AB // CD, EF and GH are perpendicular lines on
AB and CD.
Hence; EF = GH.

d) If WX = YZ, what is the relation between AB and CD
Solution:
Here, WX = YZ, WX and YZ are perpendiculars to AB.
Hence; AB//CD.
4. Name the parallel and perpendicular line segments. Express them in geometrical notations.

Solution:
a) AD⊥BC, BE⊥AC and CF⊥AB
b) PQ//SR and PS//QR
c) DC//AB and CP⊥AB.

95 Vedanta Excel in Mathematics Teachers' Manual - 6

Extra Questions

1. Write down the relation between the line segments OP and MN by
using geometrical notation. [Ans: OP ⊥ MN]

2. Draw a line segment AB of length 5 cm. Take a point P outside the line segment AB. Draw
the perpendicular line segment from the point P on to the segment AB.

3. Write down the relation between the line segments OP and MN
by using geometrical notation.
[Ans: CD//EF]

4. Copy the table and write the measurements of the
perpendicular distances PQ, RS and TU between
line segments AB and CD. State whether the line
segments are parallel or not.

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Unit Geometry: Angles

13

Allocated teaching periods: 9
Competencies
- To measure angle by using protractor
- To draw angles by using protractor.
- To use the relationship of (i) linear pair (ii) vertically opposite angles.
- To use the relationship between the following pair of angles made by a transversal with the

parallel lines (i) alternate angles (ii) corresponding angles (iii) co-interior angles

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To identify the types of angles.
- To define adjacent angles, linear pair, vertically opposite
2. Understanding (U)
3. Application (A) angles, complementary and supplementary angles
4. High Ability (HA) - To write the sum of adjacent angles formed by two

intersecting lines
- To tell the relation between the vertically opposite angles
- To identify the alternate, corresponding and co-interior

angles made by the transversal with parallel lines.
- To state the relationships of alternate, corresponding and

co-interior angles

- To draw the given angle by using protractor.
- To classify the given angles in different types.
- To solve the simple problems based on right angle,

complete turn, linear pair, vertically opposite angles etc.

- To solve the problems based on linear pair, vertically
opposite angles etc.

- To find the unknown angles from the figure using the
relationships of alternate, corresponding and co-interior
angles

- To prepare the project works on the relationships of
vertically opposite angles, alternate, corresponding and
co-interior angles and present in the classroom.

Required Teaching Materials/ Resources
Ruler, protractor, strips, sticks, chart paper, scissors, threads, ICT tools etc.

Pre-knowledge: Angles, types of angles

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Teaching Activities

Activity 1 Measurement of angles

1. Make a discussion on meaning of angle. Draw the figure and describe

the arms, vertex and name of angle. In the figure aside, O is the

vertex, OA and OB are arms of the angle. Draw such more angles

with different names and ask the vertex and arms of the angles.

2. Make discussion about the angles made by the revolving line with a

fixed line.

3. Draw some angles on the board and measure the angles by using protractor or ICT tools

following the steps below.

(i) Place the centre of protractor on the vertex of the angle.

(ii) Line up one of the arms of the angle with the base line (zero line) of the protractor.

(iii) Count up the angles in degrees starting from 0° until the other arm meets the protractor.

(iv) Discuss about the use of inner and outer scale of protrctor.



4. Use protractor or set-squares, classify the angles into various types: acute, right, obtuse,
straight, reflex angle and complete turn.

5. Give some measurements of angles and ask the students to identify their types.
6. Under discussion, define the types of angles with examples.
7. Teacher can use work sheets on types of angles by visiting various websites too.

Solution of selected questions from EXERCISE 13.1
1. Let’s tell and write the correct answers in the blank spaces.

a) The quarter turn of a revolving line makes an angle of 90o.
b) The half turn of a revolving line makes an angle of 180o.
c) The third - quarter turn of a revolving line makes an angle of 270o.
d) The complete turn of a revolving line makes an angle of 360o.
e) The angle made by a circle is a complete turn.
f) The degree measurement of 1 right angle is 90o.
g) The degree measurement of 2 right angle is 180o.
h) If x° and 40° make a right angle, then x° = 50o.
i) If y° and 120° make a straight angle, then y° = 600.
j) If z° and 300° make an angle of complete turn, then z° = 60o.
2. Let’s look at the given watch. Tell and write in how many minutes will the revolving

minute hand turn through.

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a) a quarter turn in 15 minutes
b) a half turn in 30 minutes.
c) a third-quarter turn in 45 minutes.
d) a complete turn in 60 minutes.
3. A revolving minute hand of a watch subtend an angle of 360o in 60 minutes (1 hour).

Apply unitary method and calculate the following.
a) The angle subtended by the minute hand in 10 minutes.
Solution:
Here,
In 60 minutes, the minute hand of a watch subtends an angle of 360o.
In 1 minute, the minute hand of a watch subtends an angle of 360o÷60 = 6o
In 10 minute, the minute hand of the watch subtends an angle 6o×10 = 60o.

b) The angle subtended by the minute hand in 25 minutes.
Solution:
Here,
In 60 minutes, the minute hand of a watch subtends an angle of 360o.
In 1 minute, the minute hand of the watch subtends an angle of 360o÷60 = 6o
In 25 minute, the minute hand of a watch subtends an angle 6o×25 = 150o.

c) The time taken by the minute hand to turn through 90o.
Solution:
Here,
The minute hand of a watch subtends an angle of 360o in 60 minutes.
The minute hand of the watch subtends an angle of 1o in 60o÷360 =1/6 minutes.
The minute hand of the watch subtends an angle of 90o in 90×1/6 =15 minutes.

d) The time taken by the minute hand to turn through 210o.

Solution:

Here,

The minute hand of a watch subtends an angle of 360o in 60 minutes.

The minute hand of the watch subtends an angle of 1o in 60o÷360 =1/6 minutes.

The minute hand of the watch subtends an angle of 210o in 210×1/6 =35 minutes.

4. Calculate the size of unknown angles.

a) Solution:

Here, y o +40o = 90o [ The sum of parts of a right angle is 90o]

or, yo = 90o – 40o

∴ y = 50o

b) Solution:

Here, zo =25o+90o [By whole part axiom]

zo = 115o

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c) Solution: po +90o =120o [By whole part axiom]
Here, po = 120o - 90o
or, po = 30o
∴

d) Solution:

Here, 2xo +xo = 90o [ The sum of parts of a right angle is 90o]

or, 3xo = 90o

or, xo = 30o

So, xo = 30 o and 2xo = 2×30o = 60o

e) Solution:

Here, xo +3xo = 360o [ The sum of parts of a complete angle is 360o]

or, 4xo = 360o

or, xo = 90o

So, xo = 90 o and 3xo = 3×270o = 60o

f) Solution:

Here, yo +2yo = 150o [By whole part axiom]

or, 3yo = 150o

or, yo = 50o

So, yo = 50 o and 2yo = 2×50o = 100o

g) Solution:

Here, 2po +3po+150o = 360o [ The sum of parts of a complete angle is 360o]

or, 5po = 360 o – 150 o

or, 5po = 210o

or, po = 42o

So, 2po = 2×42o = 84o and 3po = 3×42o = 126 o

Activity 2 Different pairs of angles

1. With sufficient examples, describe about adjacent angles. Ask cross questions to identify
the adjacent angles.

2. Measure the pair adjacent angles in a straight lines and find their sum. Ask the students
to say the sum of adjacent angles in a straight line. Conclude that the sum of linear pair is

always 180o.

3. Draw a pair of intersecting line segments, show the pair of vertically opposite angles.
4. Note that the vertically opposite angles are always equal.
5. Take a pair of angles whose sum is a right angle (90o) then define them as complementary

angles.

6. Note that the complementary angles may or may not be adjacent.
7. Take a pair of angles whose sum is 180o then define them as supplementary angles.
8. Note that the supplementary angles may or may not be adjacent.

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