Vedanta Excel in Mathematics Book 8 TG Final (2079)

Evexdacnetal in

MATHEMATICS

TEACHERS' MANUAL

8Book

Authors
Hukum Pd. Dahal
Tara Bahadur Magar

vedanta

Vedanta Publication (P) Ltd.

Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

Evexdacnetal in

MATHEMATICS

TEACHERS' MANUAL

8Book

Authors
Hukum Pd. Dahal
Tara Bahadur Magar

All rights reserved. No part of this publication may
be reproduced, copied or transmitted in any way,
without the prior written permission of the publisher.

First Edition: B.S. 2079 (2022 A. D.)

Published by:
Vedanta Publication (P) Ltd.

Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

Preface

This “Teachers’ Manual of Vedanta EXCEL in MATHEMATICS BOOK-9” is
prepared for teachers to aiming at assistance in pedagogical teaching learning
activities. Its special focus intends to fulfillment the motto of text books EXCEL
in MATHEMATICS approved by the Government of Nepal, Ministry of Education,
CDC, Sanothimi, Bhaktapur.

EXCEL in MATHEMATICS has incorporated the applied constructivism which
focuses on collaborative learning so that the learners actively participate in the
learning process and construct the new knowledge. The project works given at
the end of each chapter provides the ideas to connect mathematics to the real
life situations. Similarly, the text book contains enough exercises for uplifting
critical thinking and creation as per the optimum goal of Bloom’s Taxonomy.
The objective questions at the end of each area of subject content strengthen the
students’ knowledge level.

This manual helps the teachers to have the chapter-wise learning competencies,
learning outcomes and level-wise learning objectives. Also, it helps the teachers
in selecting the effective instructional materials, adopting the productive teaching
activities, solving the creative problems and getting more extra objective and
subjective questions which can be more useful for the summative assessments.

Grateful thanks are due to all Mathematics Teachers throughout the country who
encouraged and provided the feedback to us in order to prepare the new series.

Last but not least, any constructive comments, suggestions and criticisms from the
teachers for the further improvements of the manual will be highly appreciated.

Authors

Contents Page No.
5
Topics 12
1. Set 15
2. Number System in Different Bases 17
3. Integers 23
4. Real Number System 30
5. Ratio, Proportion and Unitary Method 37
6. Percentage and Simple Interest 41
7. Profit and Loss 47
8. Algebraic Expressions 51
9. Indices 57
10. Factorisation, H.C.F. and L.C.M. 61
11. Rational Expression 80
12. Equation, Inequality and Graph 87
13. Transformation 95
14. Geometry – Angles 109
15. Geometry – Triangles 117
16. Geometry - Quadrilateral and Regular Polygon 122
17. Geometry – Construction 132
18. Coordinates 137
19. Circle 152
20. Area and Volume 155
21. Bearing and Scale Drawing
22. Statistics

Unit Set

1

Allocated teaching periods: 7

Competency
- To find the operations (union, intersection, difference and complement) of sets
and solve the verbal problems based on set operations.

Learning Outcomes
- To find the difference and complement of sets and solving verbal problems.

Level-wise learning objectives

S.N. Levels Objectives

1. Knowledge (K) - To define a set
- To tell the types of sets
2. Understanding (U) - To state the relations between sets
- To write the set operations for the portion as shaded
3. Application (A)
4. High Ability (HA) in the Venn-diagram

- To list the elements of the set operations as asked
from the Venn-diagram.

- To find the operations of the sets and show in the
Venn-diagram by shading

- To write the set notation with special terminologies
- To write the word problems based on cardinality

relations in set notations

- To find the set operations of three overlapping sets
and show them in the Venn-diagrams.

- To verify the identities based on set operations
- To solve the verbal problems on operations (union,

intersection, difference and complement) of sets by
using Venn-diagram or formulae.

- To link various real life/ contemporary problems
with sets and solve in groups.

Required Teaching Materials/ Resources
Colourful chart-paper with definitions, Venn-diagrams and formulae, scissors, cello
tape, different coloured markers, highlighter, different models of Venn-diagrams, ICT
tools (GeoGebra) etc.

Pre-knowledge: Check the pre-knowledge on cardinality of sets, relation of sets,
operations of sets and Venn-diagrams

5 Vedanta Excel in Mathematics Teachers' Manual - 8

Teaching Activity on Relationship between Sets

1. Make a group discussion on the definition of sets, set notation, types of sets and

cardinality of sets, relations of sets by using Venn-diagrams.

2. Make the group of students and make them present a short explanation about following

topics in the classroom.

(a) Group A: Membership and non-memberships of sets

(b) Group B: Methods of describing the sets

(c) Group C: Types of sets

(d) Group D: Types of relationships of sets

3. Introduce the Venn-diagrams. German mathematician George Cantor (1844-1918)

developed the ideas of set as mathematical theory in nineteenth century. John Euler

Venn developed the method of representing the sets by closed figures called Venn

diagrams

4. Guide the students to solve the problems based on the relations between the sets

5. Give the project work to make pair of sets in the following cases

(i) Equal sets (ii) Equivalent sets

(iii) Overlapping sets (iv) Disjoint sets

6. Give the contextual examples to make the lesson understandable.

The given figure represents a set of students who take F B
part in the game on the occasion of annual day of a Dorje Maya Ram
school. Answer the following questions by studying the
figure. Sita Riya
(a) How many subsets can be formed from the set F? Hari Aasha

(b) How many members are there in the set B? Chhiring Priyanka
(c) How many members are there in the sets B and F in total?

(d) Who are the common members of both sets B and F?

(e) Who are not the members of any sets B and F?

(f) How many members are there in the given Venn diagram all together?

7. After these discussions note that the set that contains all the members of the given

figure is called universal set. It is denoted by U. All the members of F and B are also the

members of U. So, F and B are proper subsets of universal set U.



Teaching Activity on Operations of Sets
A. Union of Sets

1. With various examples and using models of Venn-diagrams, make a discussion on

union of two/three sets.

2. Give priority for group works and project works.
3. Make a conclusion that the union of two sets A and B is a set which contains all

the elements of A and B without repeating. The union of sets A and B is denoted by

mathematical symbol A∪B. Thus, A∪B = {x: x∈ A or x ∈ B}.
4. Note that (i) A∪ B = A or B or both
5. Discuss about the union of sets under the following cases.
(i) When A⊂B, then A∪B = B. (ii) When B⊂A, then A∪B = A.

B. Intersection of Sets
1. To warm up, ask the definition of the intersection of the sets.
2. With various real life examples and using models of Venn-diagrams, make a discussion

Vedanta Excel in Mathematics Teachers' Manual - 8 6

on union of two/three sets.

3. Make a conclusion that the intersection of two sets A and B is a set which contains all

the common elements of both sets. The intersection of two sets A and B is denoted by

A∩B. Thus, A∩B = {x: x∈ A and x ∈ B}.

4. Ask the students to tell the set notation for the shared part in the Venn-diagram.

5. Give some example on solving the problems based on the intersection of sets.

6. Focus on group works and project works.

7. Discuss about intersection of sets under the following cases.

(i) If A⊂B, then A∩B = A. (ii) If B⊂A, then A∩B = B.

(iii) A∩U = A (iv) When A and B are disjoint sets, A∩U = φ

C. Difference of Sets

1. With various real life examples and using models of Venn-diagrams, make a

discussion on union of two/three sets.

2. Under discussion, define the difference of two sets A and B as the set of the elements

which are contained in A but not in B. It is denoted by A - B.

3. Ask the students to tell the set notation for the shaded region in the Venn-diagrams.

In set builder notation, A – B = {x: x∈ A but x ∉B}.

4. With discussion, draw out the ideas in the following cases.

(i) A – B: only in A but not in B (ii) B – A: only in B but not in A

5. Make the groups of students and give the group work.

Group A: If A = {1, 2, 3, 4, 5, 6, 7} and B = {2, 4, 6, 8, 10}, find A – B and B – A.

Show them in Venn-diagrams.

Group B: If P = {a, b, c, d, e} and Q = {2, 4, 6, 8, 10}, find P – Q and Q – P. Show

them in Venn-diagrams.

Group C: If P = {x: x is a prime number, x<10} and Q = {y: y < 6, y ∈ N}, find P –

Q and show it in a Venn-diagram.

Group D: If M = {x: x is a multiple of 3, x<20} and N = {y: y is a factor of 12}, find

M – N and show it in a Venn-diagram.

6. Discuss upon the group works and similar problems from the exercise.

D. Complement of a Set
1. Make the groups of students and give the group work.
Group A: List the names of the group members who like banana. Then, find the
friends who don’t like banana at all.
Group B: List the names of the group members who like dancing and singing
separately. Then, find the friends who don’t like both.
Group C: List the names of the group members who have visited to Manakamana and
Lumbini. Then, find the friends who have not visited both the places.
Group D: List the names of the group members who can speak Newari and Tamang
languages separately. Then, find the friends who speak neither Newari
nor Tamang language.

7 Vedanta Excel in Mathematics Teachers' Manual - 8

2. With more real life examples and use of Venn-diagram, make the students clear about
complement of a set.

3. After discussion, define the complement of a set A as “the set of elements which are
not contained in set A but contained in universal set U”. The complement of set A is

denoted by A or A’ or Ac.

4. In set builder notation, A∪B = {x: x∈ U and x ∉ B}.
5. With discussion, solved some problems related to complement of the set.
6. Make the group of students and give the problems to solve in group and let them

present.
Information: A, B and C are the subsets of a universal set U. If U = {1, 2, … , 20}, A =

{1, 2, 3, 4, 5, 6, 7}, B = {2, 4, 6, 8, 10, 12} and C = {3, 6, 9, 12, 15, 18}
Group A: Find A ∪ B and show it in Venn-diagram.
Group B: Find A ∩ B and show it in Venn-diagram.
Group C: Find A ∪ B ∪ C and show it in Venn-diagram.
Group D: Find A ∩ B ∩ C and show it in Venn-diagram.
Group E: Find A – B and show it in Venn-diagram.
7. Note that A = A

Teaching Activity on Cardinality Relations of Sets

1. Discuss about the cardinality of sets, list the following formula by using Venn-diagrams

along with examples. Give the work to the students to write the formulae in chart paper

after discussion and paste the best one in math corner of the classroom or math lab.

Case- I: When A and B are disjoint sets. A BU
(i) n (A∩B) = 0

(ii) n (A∪B) = n (A) + n (B)

Case- II: When A and B overlapping sets

(i) n (A∪ B) = n (A) + n (B) – n (A∩B)

(ii) n (only A) = no (A) = n (A) – n (A∩B)
(iii) n (only B) = no (B) = n (A) – n (A∩B)
(iv) n (only one) or n (exactly one) = no (A) + no (B)
(v) n (A∪B) = no (A) + no (B) + n (A∩B)
(vi) n(A ∪ B) = n(U) – n(A∪B)

(vii) n (U) = n (A) + n (B) – n (A∩B) + n(A ∪ B)

2. Explain the useful terminologies in solving verbal problems.

(i) No. of people who like at least one fruits / either apple or banana= n (A∪B)

(ii) No. of people who like both apple and banana / who like apple as well as banana

= n (A∩B)

(iii) No. of people who like only apple = no (A) and only banana = no (B)
(iv) No. of people who like only one fruit = no (A) + no (B)
(v) No. of people who don’t like both the fruits / like neither apple nor banana = n(A ∪ B)

Vedanta Excel in Mathematics Teachers' Manual - 8 8

Note:
1. If the information are given in percentage, consider that n (U) = 100 or x.
2. If each people participates in at least one activity or like at least one item then remember

that n (U) = n (A∪B) or n(A ∪ B) = 0

Solution of selected questions from Excel in Mathematics -Book 8

1. A, B and C are the subsets of a universal set U. If U = {1, 2, … , 15}, A =

{2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12, 15} and C = {1, 2, 3, 4, 5, 6, 7}, find the elements of

the set (A – C) ∪ B and illustrate it in Venn-diagram.
Solution:

Here, U = {1, 2, … , 15}, A = {2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12, 15} and C = {1, 2, 3, 4, 5, 6,

7}

Now, A – C = {8, 10, 12} ∴A – C = {1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 14, 15}

Again, (A – C) ∪ B = {1, 2, 3, 4, 5, 6, 7, 9, 11, 12, 13,14, 15}

Illustrating (A – C) ∪ B in Venn-diagram, A 9B U
8 12 14

10 2 6 15
4 3

11 1 7
13 5
C

2. P and Q are subsets of a universal set U. If U = {x: x is a whole number, x≤9}, P = {y:

y is a multiple of 2} and Q = {z: z is a factor of 12}, verify that:

(i) P ∪ Q = P ∩ Q (ii) P ∩ Q = P ∪ Q
Solution:

Here, U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, P = {2, 4, 6, 8} and Q = {1, 2, 3, 4, 6, 12}

∴ P = {0, 1, 3, 5, 7, 9} and Q = {0, 5, 7, 8, 9}
Now,

P ∪ Q = {0, 1, 3, 5, 7, 9} … (i)

P ∩ Q = U – {2, 4, 6} = {0, 1, 3, 5, 7, 8, 9} … (ii)
From (i) and (ii), we get

P ∪ Q = P ∩ Q Proved
Again,

P ∩ Q = {0, 5, 7, 9} … (iii)

P ∪ Q = U – {1, 2, 3, 4, 6, 8, 12} = {0, 5, 7, 9} … (iv)
From (iii) and (iv), we get

P ∩ Q = P ∪ Q Proved

3. Among 72 students, 24 are selected to participate in music only, 26 are selected to
participate in dance only and none of them participate in both the activities. The rest
of the students are selected in other activities.

9 Vedanta Excel in Mathematics Teachers' Manual - 8

(i) How many students are participating in music or dance?

(ii) How many students are participating in other than music or dance?

(iii) Illustrate this information in Venn-diagram.

Solution:

Let M and D denote the sets of students who are participating in music and dance

respectively.

Then, n (U) = 72, n (M) = 24, n (C) = 26

(i) n (M∪D) = n (M) + n (D) = 24 + 26 = 50

Hence, 50 students are participating in music or dance.

n(M ∪ C ) = n(U) – n(M ∪ C) = 60 – 50 = 10

Hence, 10 students are participating in other than music or dance.

(ii) Venn-diagram, U
D
M

24 26

10

3. In a survey of 250 people, 180 like farming, 120 like civil service and every people

liked at least one occupation. Draw a Venn-diagram and find the number of people

who like farming as well as civil service. Also, find how many people like only civil

service?

Solution:

Let F and C denote the sets of people who like farming and civil service respectively. Then, n

(U) = 250, n (F) = 180, , n(C) = 120 and n( F ∪ C ) = 0

Let n (F∩C) = x

Now, drawing a Venn-diagram to show the above information,

From Venn-diagram, n (U) = (180-x) + x + (120-x) F U
or, 250 = 300 – x ∴x = 50 C
Hence, 50 people like farming as well as civil service. 120–x
x180–x

Again, no (C) = 120 – x = 120 – 50 = 80
Hence, 80 people like only civil service.

4. In a school’s Annual Day, 60 students involved in Science Exhibition, 75 involved in

Mathematics Exhibition, 20 involved in both Exhibitions and the rest of 285 students

involved in many other activities but not in Exhibitions.

(i) Illustrate the above information in Venn-diagram.

(ii) How many students took part in the School’s Annual Day?

(iii) How many students involved in only one exhibition?

Solution:

Let S and M denote the sets of students who involved Science and Mathematics Exhibitions

respectively.

Then, n (S) = 60, n (M) = 75 and n (S∩M) = 20 and n( S ∪ M ) = 285

Vedanta Excel in Mathematics Teachers' Manual - 8 10

(i) Illustrating the above information in the Venn-diagram U
n (U) = n (S) + n (M) – n (S∩M) + n( S ∪ M ) = 285 SM
40 20 55
(ii) = 60 + 75 – 20 + 285 = 400

Hence, 400 students took part in School’s Annual Day. 285

no(S) = n(S) - n (S∩M) = 60 – 20 = 40 and no(M) = n(M) - n (S∩M) = 75 – 20
= 55 ∴ no(S) + no(M) = 40 + 55 = 95
Hence, 95 students involved in only one exhibition

Extra Questions

1. If M – N = {1, 2}, N – M = {5, 6} and M∩N= {3, 4}, find M ∪ N by showing in Venn-

diagram, Ans: {1,2,3,4,5,6}

2. If U = {1, 2, 3, …, 10}, A = {1, 2, 3, 4, 5}and B = {1, 3, 5, 7, 9}, verify that: A ∪ B = A ∩ B
3. If A = {a, b, c}, B = {c, d, e}and C = {a, c, e}, verify that: A – (B ∪ C) = (A – B) ∩ (A – C)

4. If U = {1, 2, …, 9}, A = {2, 4, 5, 8}, B = {2, 3, 5, 6} and C = {3, 4, 5, 7},

(i) Draw a Venn-diagram to show the relation of U, A, B and C.

(ii) Find A ∪ B ∪ C Ans: {1, 9}
5. If U = {1, 2, …, 15}, A = {2, 3, 5, 7, 11, 13}, B = {1, 3, 5, 7, 9} and C = {1, 2, 3, 4, 5, 6},

(i) Draw a Venn-diagram to show the relation of U, A, B and C.

(ii) Find A ∩ B ∩ C Ans: {1,2,4,6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

6. Out of 100 students participated in an examination, 60 passed in Math, 50 passed in

Science and 20 passed in both subjects.

(i) Draw a Venn-diagram to show the above information.

(ii) How many students failed in both the subjects? [Ans:10]

7. In a survey of 200 youths who love playing games in mobiles, it was found that 120 enjoy

playing PUBG, 100 enjoy playing Ludo and 40 do not play both the games.

(i) Represent the above information in a Venn-diagram.

(ii) Find the number of youths who enjoy playing both the games. [Ans:60]

(iii) How many youths enjoy playing only Ludo? Find. [Ans:40]

8. In a class of 50 students, it was found that 25 liked to sing and 35 liked to dance. If every

students liked at least one of the two activities:

(i) Represent the above information in a Venn-diagram.

(ii) How many students liked both activities? [Ans:5]

9. Out of 80 students of class VIII, 50 students like to go hiking, 45 like to go picnic and each

student like to go at least one of them.

(i) How many students like to go for both? [Ans: 15]

(ii) Show the above information in the Venn-diagram.

11 Vedanta Excel in Mathematics Teachers' Manual - 8

Unit Number System in Different Bases

2

Allocated teaching periods: 3
Competency

- Conversion of Binary and Quinary numbers into Denary numbers
Learning Outcomes

- To convert denary numbers in to binary and quinary number and vice versa
Level-wise learning objectives

S.N. Levels Objectives
1. Knowledge (K)
2. Understanding (U) - To tell the digits used in decimal number system
Application (A) - To tell the digits used in binary number system
3. - To name the number system in which the digits only 0,1,2,3

and 4 are used

- To convert decimal number into binary number and vice
versa.

- To convert decimal number into quinary number and vice
versa.

- To convert the binary numbers into quinary number system
- To convert the quinary numbers into binary number system

4. High Ability (HA) - To compare the numbers in different bases

Required Teaching Materials/ Resources
Base ten blocks, Colourful chart-paper with the number systems in different bases, switch on
off system, different coloured markers, highlighter etc.

Pre-knowledge: Decimal number system, binary and quinary number system

Teaching Activities
1. Make a discussion about decimal number system and its expanded form
2. Introduce binary number system
3. Make the numbers of blocks in the groups of 2 and explain about the expanded form of

binary number.
4. Show the conversion process of denary to binary or binary to denary system in the chart

paper with examples and give some group works.
5. Show the conversion process of denary to quinary or quinary to denary system in chart

paper examples and give some group works.

Vedanta Excel in Mathematics Teachers' Manual - 8 12

Solution of selected questions from Excel in Mathematics -Book 8

1. Convert 11102 into quinary number.
Solution:

Here, the given binary number is 11102
Now,

Converting binary number into decimal number, we get

1A1g1a0in2 ,= 1×23 + 1×22 +1×21 +0×20 = 8 + 4 + 2 + 0 = 1410

Converting 1410 in to quinary number, we get

5 14 Remainder

52 4

02

∴1410 = 245
Hence, 11102= 1410 = 245

2. Which one is greater between 11110112 and 1435? Also, find their difference in decimal
system.

Solution:
Here, 1111012 = 1×26 + 1×25 + 1×24 +1×23 +0×22 +1×21 +1×20

= 64+ 32 + 16 + 8 + 0 + 2 + 1= 123
Also, 1435 = 1×52 + 4×51 +3×50 = 25 + 20 + 3 = 48
Since, 123>48 So, 1111012 > 1435
Again, difference = 123 – 48 = 75

3. tTnhuhemreabenessrw.tuRedraem1n1itl0sa1Rg1ao0mt2.tihWlaeh,aoSnhgswoatsehtrhw1ea1ct1o0ar1nr1ed2c,tASarhnaasbswihewwrea?rtFegionatdstkhbeeydactnaoslcwcuoelnravt1ie1or1nt1.2021125ainnd to binary
Arabi got

Solution:

Here, the given quinary number is 2215
Now,

Converting the quinary number into decimal number, we get

2215 = 2×52 + 2×51 +1×50 = 50 + 10 + 1 = 61
Again,

Converting 6110 in to binary number, we get
2 61 Remainder

2 30 1

2 15 0

27 1

23 1

21 1

01

∴2215= 1111012
Hence, Shashwat got the correct answer.

13 Vedanta Excel in Mathematics Teachers' Manual - 8

Extra Questions

1. Which one is smaller between 1100112 and 1235? Also, find their difference in decimal
system. [Ans: 13]

2. Find the difference between 1010102 and 1305 in denary system. [Ans: 2]

3. Convert the sum of 100112 and 8110 in to quinary number system. [Ans: 4005]
4. For what value of p, the numbers 100p012 and p225 are equal? Find it. [Ans: 1]

Vedanta Excel in Mathematics Teachers' Manual - 8 14

Unit Integer

3


Allocated teaching periods: 4

Competency
- Simplifying the integers with mixed operations and brackets.

Learning Outcomes
- To simplify the integers using BODMAS rule.

Level-wise learning objectives

S.N. Levels Objectives
1. Knowledge (K)
2. Understanding (U) - To tell the additive inverse of any integer.
Application (A) - To identify the identity element of addition of integers.
3. - To identify the identity element of multiplication of

integers.
- To recall BODMAS rule.

- To perform simple simplification of integers.
- To verify commutative law of addition or multiplication of

integers.
- To verify the associative law of addition or multiplication of

integers.

- To simplify the integers with operations inside the brackets
- To make the expression for the given statement and

simplify them.

4. High Ability (HA) - To create the problems involving operational signs and
brackets then solve

Required Teaching Materials/ Resources
Number line, graph board/copy, coloured markers, highlighter, ICT tools etc.
Pre-knowledge: Set of natural /whole numbers
Teaching Activities

1. Recall the set of natural and whole numbers
2. Ask the students to find the differences between pair of whole numbers then discuss

about the results.
3. Draw the numbers line and show the integers in it.
4. Explain with examples about the laws of addition and multiplication of the integers
5. Discuss about the identity element of addition and multiplication of integers.
6. Recall the simplification of the integers using BODMAS rule.
7. Divide the students in to groups and give some problems to be solved in the group.

15 Vedanta Excel in Mathematics Teachers' Manual - 8

Solution of selected questions from Excel in Mathematics -Book 8
1. Simplify: 300÷25 [56÷7 {24 - 2(6 - 5 –9)}]

Solution:
300÷25 [56÷7 {24 - 2(6 - 5 –9)}]
=300÷25 [56÷7 {24 - 2(6 - –4)}]
=300÷25 [56÷7 {24 - 2(6 +4)}]
=300÷25 [56÷7 {24 - 2(10)}]
=300÷25 [56÷7 {24 - 20)}]
=300÷25 [56÷7 {4)}]
=300÷25 [56÷28}]
=300÷25 [2]
=300÷50
=6

2. Let’s simplify by making mathematical expression.
‘The sum of 48 and itself, its half and half of half is added to 18’
Solution:
{48 + 48 + (48÷2) + (48÷2) ÷2}+18
= {96+24 + (24÷2)} + 18
= {96+24+ (24÷2)} +18
= {120+ 12} + 18
=132 + 18
=150

3. Mr. Hamal had Rs 4400. He purchased 6 kg of rice at Rs 75 per kg, 2 packets of oil at
Rs 125 per packet and he gave Rs 3300 to his wife. If he divided the remaining sum
between his son and daughter equally, find the share of each of them.
Solution:
[4400 – {(6×75) + (2×125) + 3300}]÷2
= [4400 – {450+ 250 + 3300}]÷2
= [4400 – 4000]÷2
= [400] ÷2
= 200
Hence, the share of each daughter and son is Rs 200.

Extra Questions

1. Simplify: 48÷3 [10 - 100÷2 {21 - 2(5+ 4 – 11)}]
[Ans: 2]

2. Out of 80 students of class 8 with two sections A and B, 1 girl from section A and

3 boys from section B are absent. Then 23 boys in each section are found present.

Also, equal numbers of girls are present in each section, what is the total number

of girls in section A? [16]

Vedanta Excel in Mathematics Teachers' Manual - 8 16

Unit Real Number System

4

Allocated teaching periods: 9

Competency

- Defining real number system and distinguish rational and irrational numbers

- Expressing number in scientific notation

- Simplifying the problems on surds
Learning Outcomes

- To distinguish rational and irrational numbers

- To change the non-terminating recurring decimals into fractions

- To write the number in scientific notation

- To rationalize the denominator and simplify the problems on surds
Level-wise learning objectives

S.N. Levels - Objectives
1. Knowledge (K) -
- To define rational number
- To give examples of rational and irrational number
- To tell the definition of real number system
- To express the numbers in to scientific notation
To find the sum of difference of like surds
- To tell the rationalizing factor of the given surd

- To write the difference between the rational and irrational
2. Understanding (U) numbers
To show the relationship between the set of numbers to
- the real number system
- To perform basic operations in scientific numbers
- To perform operations on surds
To rationalize the denominator
-
To write the non-terminating recurring numbers in to
3. Application (A) - fraction.
- To simplify the problems using scientific notations
To simplify the surds by rationalizing the denominator

- To discover the irrational numbers using Pythagoras

4. High Ability (HA) theorem in number line

- To solve the contextual problems on surds

Required Teaching Materials/ Resources
Chart of relationship between the set of numbers to the real number system, Graph board/
copy, compass, scale, highlighter, ICT tools etc.

17 Vedanta Excel in Mathematics Teachers' Manual - 8

Pre-knowledge: Set of natural /whole numbers and integers

Teaching Activity-1

1. Recall the set of natural, whole numbers and integers
2. Make a discussion to realize the existence of another set of numbers.
For example in 2 + 4 =6, is 6 an integer, 2 – 4 = – 2, is – 2 an integer,

4 – 2 = 2, is 2 an integer, 4 × 2 = 8, is 8 an integer, 4 ÷ 2 = 2, is 2 an integer,
1 1
2÷4 = 2 , is 2 an integer?

3. Define rational number with examples.
4. Divide the students into groups and give some rational numbers to change in to decimal
numbers. Then make them see the result very carefully and ask to list as terminating

decimal numbers or non-terminating decimal numbers.

5. Make the following conclusions: ba,

6. The numbers which can be expressed in the form where ‘a’ and ‘b’ are integers and
b ≠ 0, are called rational numbers.

Rational numbers can be expressed as terminating or non-terminating but recurring decimal

numbers.

7. Recall Pythagoras theorem, and make the unit square and ask the students to find the
diagonal. Similarly, use graph board and draw the number line and using compass find the

numbers on the number line. Then discuss about the irrational numbers.

8. Make some numbers and ask to list whether they are rational numbers or not.
9. After discussion, show the chart of set of real numbers and discuss about it.
10. Discuss about the problems given ion the exercise.

Teaching Activity-2
1. Give some examples representing the larger or very smaller numbers. Then discuss about

the way of expressing the larger or smaller number in short0cut form as the integral
powers of 10.
2. With examples, discuss about the rules of expressing the general numbers in to scientific
notation.

Rule 1: The whole number part in scientific notation of a number should be of one
digit number. For example,

59400000000 = 5.94 × 1010 (But 59.4 × 109 is incorrect scientific notation)
0.0000000826 = 8.26 × 10–8 (But 82.6 × 10–9 is incorrect scientific notation)

Rule 2: In the case of a large number, after separating whole number part and
decimal part by using decimal point, count the number of digits after the decimal
point and express as the product of the same number of power of 10. For example:

Insert decimal point here.

3275000000 = 3.275 × 109 There are 9 digits after the decimal
point. So the power of 10 will be 109.
123456789

9 digits

Rule 3: In the case of a very small number, after separating whole number part and
decimal part by using decimal point, count the number of digits before the decimal

Vedanta Excel in Mathematics Teachers' Manual - 8 18

point and express as the product of the same number of negative power of 10. For

example:

0.0000000653 = 6.53 × 10–8 Decimal point is shifted after eight
number of digits to make a one digit whole
12345678 number. So, the power of 10 will be 10–8.

8 digits

3. With discussion, give the class work from exercise
4. Make the groups of students and provide proper guidelines to solve the problems

Teaching Activity-3
1. With examples, make discussion on the surds.
2. List out the following points under discussion of the surds.

In a surd (n a ) , ‘n’ is called order and it’s a natural number, ‘a’ is called the radical and it
should always be a rational number.

The surds are irrational numbers.
3. Give the surds and ask the students to collect the surds of equal order and same radicand.

Then, define the like surds with more examples.
4. Make a discussion about the addition and subtraction of like surds with example. Also,

ask the students for the sum and difference between the some like surds.
5. Discuss about rules of the multiplication and division of surds.
6. Make the groups of students and ask the problems on addition, subtraction, multiplication

and division of the surds Discuss about how the irrational number can be expressed as the
rational number.
7. Define rationalization or conjugation with examples.
8. Solve some problems on rationalizing the denominators and give same type of the questions.
9. Focus on group works and support the groups with proper instructions.

Solution of selected questions from Excel in Mathematics -Book 8
1. Convert the following non-terminating recurring decimals in to fractions.
(a) 0.•3 (b) 0.•2 (c) 0.•2 •4 (d) 1.•5 •7
(e) 0.•3 2•4 (f) 1.•4 3•2 (g) 0.8•3 (h) 0.2•6

Solution:
(a) Let x = 0.•3 = 0.33 …… …equation (i)

Multiplying equation (i) by 10, we get
10x = 3.33 …… …equation (ii)

Subtracting equation (i) from equation (ii), we get
9x = 3
3 1
∴ x = 9 = 3

Hence, 0.•3 = 1 …… …equation (i)
(b) Let x = 0.•2 = 30.22

Multiplying equation (i) by 10, we get
10x = 2.22 …… …equation (ii)

Subtracting equation (i) from equation (ii), we get
9x = 2
2
∴ x = 9

19 Vedanta Excel in Mathematics Teachers' Manual - 8

Hence, 0.•2 •4==92 0.2424 …… …equation (i)
(c) Let x = 0.•2 …equation (ii)

Multiplying equation (i) by 100, we get …equation (i)
100x = 24.2424 …… …equation (ii)

Subtracting equation (i) from equation (ii), we get …equation (i)
99x = 24 …equation (ii)

∴x =0.•229•494 = 24 …equation (i)
Hence, = 99 …equation (ii)
Let x = 1.•5 •7 1.5757
(d) …… … (A)
…equation (i)
Multiplying equation (i) by 100, we get …equation (ii)
100x = 157.5757 ……

Subtracting equation (i) from equation (ii), we get
99x = 156
156 19579
∴ x = 99 =

LHeetnxce=, 10..•5•3•72•4==159079.324324 ……
(e)

Multiplying equation (i) by 1000, we get
1000x = 324.324324 ……

Subtracting equation (i) from equation (ii), we get
999x = 324

∴ x = 324 = 108
999 333

Hence, 0.•3 2•4 = 108 ……
(f) Let x = 1.•4 3•2 = 333

1.432432

Multiplying equation (i) by 1000, we get
1000x = 1432.432432 ……

Subtracting equation (i) from equation (ii), we get

999x = 1431

∴ x = 1431 = 1943992
999
LHeetnxce=, 10..•483•3•2==0.189433932933
……
(g)
Multiplying (A) by 10, we get
10x = 8.3333 ……
Multiplying (A) by 100, we get
100x = 83.3333 ……

Subtracting equation (i) from equation (ii), we get
90x = 75

∴x = 75
Hence, 0.8•3 = 90
5
6

Vedanta Excel in Mathematics Teachers' Manual - 8 20

(h) Let x = 0.2•6 = 0.26666 …… … (A)
Multiplying (A) by 10, we get …equation (i)
10x = 2.6666 …… …equation (ii)
Multiplying (A) by 100, we get
100x = 26.6666 ……

Subtracting equation (i) from equation (ii), we get
90x = 24

∴ x = 24 = 4
90 15
0.2•6 4
Hence, = 15

2. Simplify: 6.7 × 10–11 × 2.5 × 1020 × 3.6 × 1015
(3.35 × 106)2

Solution:

Here, 6.7 × 10–11 × 2.5 × 1020 × 3.6 × 1015
(3.35 × 106)2

= (6.7 × 2.5 × 3.6) × 10–11 + 20 + 15
(3.35)2 × 1012

= 60.3 × 1024
11.2225 × 1012

= 5.37 × 1012

3. Simplify: 9.6 × 106 – 7.2 × 105
2.4 × 103

Solution:

Here, 9.6 × 106 – 7.2 × 105
2.4 × 103

= 9.6 × 106 – 0.72 × 106
2.4 × 103

= (9.6 – 7.2) × 106 = 8.88 × 106
2.4 × 103 2.4 × 103

= 3.7 × 103

4. Sound travels 3.43 × 102 metres in 1 second. Calculate the distance travelled by the sound
in 3.6 × 103 seconds.

Solution:

Here,

Distance travelled by the sound in 1 second = 3.43 × 102 metres

Now,

The distance travelled by the sound in 3.6 × 103 second = (3.6 × 103)(3.43 × 102)m

= 12.348 × 105m

= 1.2348 × 106m

5. How many petrol tanks of capacity 1.2 × 104 litres each are required to empty 4.8 × 105
litres of petrol?

Solution:

Here,

Quantity of petrol = 4.8 × 105 litres, capacity of a tank = 1.2 × 104 litres

21 Vedanta Excel in Mathematics Teachers' Manual - 8

Now, 4.8 × 105
1.2 × 104
The required number of tanks = = 4 × 10 = 40

6. Simplify: 15 – 30 – 18 + 20
Solution: 5 45 27

Here, 15 – 30 – 18 + 20
5 45 27

= 15 – 3 30 × 5– 3 18 × 3 + 2×2×5
5 ×3 ×3

= 15 – 30 – 18 +2 5
5 35 33

= 15 – 10 – 6 +2 5
5 5 3

= 15 × 5 – 10 × 5 – 6 × 3+ 2 5
5 55 53 3

= 75 – 10 5 – 63 +2 5
5 5 3

= 53 –2 5 –2 3 +2 5
5

= 3 – 2 3 = – 3

7. Find the perimeter of a triangle whose sides are 75 cm, 48 cm and 27 cm.
Solution:
Here,
The perimeter of the triangle = 75 cm + 48 cm + 27 cm
= 5 3 cm + 4 3 cm + 3 3 cm = 12 3 cm

Extra Questions
1. In a competitive examination, a question was asked as to evaluate 0.•5 + 0.•6 + 0.•7 . Anil
got the answer 1.•8 , Bishnu got the answer 0.1•8 and Chandani got the answer 2. Who got

the correct answer? Find it. [Ans: Chandani]

2. Simplify: (6.3 × 10–8) × (1.5 × 1013) [Ans: 43.2]
(2.5 × 10–3) × (3.5 × 109)

3. A rectangular field is 300 m long and 108 m wide. [Ans: 180m2, 32 2 m]
Find its (i) area (ii) perimeter

Vedanta Excel in Mathematics Teachers' Manual - 8 22

Unit Ratio, Proportion and Unitary Method

5

Allocated teaching periods: 9

Competency

- Solving daily life problems based on ratio and proportion
Learning Outcomes

- To find the ratio of quantities of same kind.

- To solve the problems related to ratio and proportion

- To solve the simple problems using unitary method
Level-wise learning objectives

S.N. Levels Objectives

1. Knowledge (K) - To define ratio
- To identify the antecedent and consequent of any ratio
- To recall means and extremes of the proportion
- To tell the types of proportion
- To define unitary method

- To find the ratio of two quantities
- To find the compound ratio
- To compare the ratios
2. Understanding (U) - To find the proportional asked in the problem
- To solve the simple problem using unitary method

- To solve the problems related to ratio

- To identify the direct or indirect proportion and solve the

3. Application (A) problems

- To solve the problems using unitary method

- To find the ratio of length and breadth of various

4. High Ability (HA) objects like mathematics book, room, table, bed etc after

measuring.

- To solve the contextual problems based on ratio and

proportion

Required Teaching Materials/ Resources
Water and juice, some amount of money, scale, highlighter, ICT tools etc.

Pre-knowledge: Quantities of same kind, comparison between numbers

Teaching Activity for Ratio
1. Take different notes of money and share between two students (i) notes with same value

(ii) notes with different values and ask to compare them by (a) subtraction and (b) division.
For example:
(i) ratio 2 five rupee notes and a five rupee note
(ii) ratio 3 ten rupee notes and a fifty rupee note

23 Vedanta Excel in Mathematics Teachers' Manual - 8

2. Make discussion on finding the ratios of weights, heights, distances etc. separately.
3. Ask to define ratio and discuss about the parts of ratio (antecedent and consequent) with

examples.
4. Discuss on compound ratios with examples.
5. Divide the students into the proper groups and give to solve the problems from exercise.
6. Give some project work on ratio.

Solution of selected questions from Excel in Mathematics -Book 8
1. The cost of book is two times more than the cost of box and the cost of box is two times
more than the cost of a pen. If the cost of these three items is Rs 350, find the cost of each

item.

Solution: 2 2 2 4
1 1 2 2
Here, the ratio of cost of the book and the box = 2:1 = = × =

The ratio of the box and the pen = 2:1 = 2
1

The cost of the book, box and the pen = 4:2:1

Let, the cost of the book = Rs 4x, the cost of the box = Rx 2x and the cost of the pen = Rs x.

Now, 4x + 2x + x = Rs 350 or, 7x = Rs 350 ∴x = Rs 50

Hence, the cost of the book = 4x =4× Rs50 = Rs 200

The cost of the box = 2x =2× Rs50 = Rs 100 and the cost of the pen = Rs x = Rs 50.

2. The monthly income of A is double than that of B, the monthly income of B is treble than
that of C. if the total income of three persons is Rs 80,000, find the monthly income of

each of person.
Solution:
2 2 3 6
Here, the ratio of monthly income of A and B = 2:1= 1 = 1 × 3 = 3

The monthly income of B and C = 3:1 = 3
1

Let, the monthly income of A, B and C are 6x, 3x and x.

Now, 6x + 3x + x = Rs 80,000 or, 10x = Rs 80,000 ∴x = Rs 8,000

Hence, the monthly income of A= 6x = 6×Rs 8,000 = Rs 48,000

The monthly income of B = 3x= 3×Rs 8,000 = Rs 24,000 and the monthly income of C

= x = Rs 8,000.

3. Last year 60 students of a school appeared SEE. Among them, 8 students secured grade

C, 4 students secured grade D and the rest of them secured grades A and B in 3: 5 ration.

(i) Find the number of students who secured grades A and B.

(ii) Find the ratio of the number of students who secured the grades A, B, C and D.

Solution:

Let, the number of students who secured grade A be 3x and grade B be 5x.

Now, the number of students who secured grades A and B = 60 – 8 – 4 = 48

or, 3x + 5x = 48 or, 8x = 48 ∴x = 6
(i) No. of students who secured grade A = 3x = 3×6 = 18 and the no. of students who
secured grade B = 5x = 5×6 = 30
(ii) The ratio of students who secured the grades A, B, C and D = 18:30:8:6 = 9:15:4:3
4. Four businessmen invested a sum of Rs 4,50,000 in the ratio 1:2:3:4 to start a new
business.
(i) Find the sum invested by each of them.
(ii) At the end of the year, if they gained Rs1,80,000 divide the profit among them

in the ratio of their shares.

Vedanta Excel in Mathematics Teachers' Manual - 8 24

Solution:
(i) Let, the sum invested by four businessmen A, B, C and D be x, 2x, 3x and 4x

respectively.
Then, x + 2x + 3x + 4x = Rs 4,50,000 or, 10x = Rs 4,50,000 ∴x = Rs 4,500
∴ The sum invested by A = x = Rs 4,500
The sum invested by B = 2x = 2×Rs 4,500 = Rs. 9,000
The sum invested by C = 3x = 3×Rs 4,500 = Rs. 13,500
The sum invested by D = 4x = 4×Rs 4,500 = Rs. 18,000
(ii) Again, profit = Rs 1,80,000 and the ratio of sharing profit is 1:2:3:4
Now, x + 2x + 3x + 4x = Rs 1,80,000 or, 10x = Rs 1,80,000∴x = Rs 1800
∴ The profit received by A = x = Rs 1,800
The profit received by B = 2x = 2×Rs 1,800= Rs. 3,600
The profit received by C = 3x = 3× Rs 1,800= Rs. 54,000
The profit received by D = 4x = 4× Rs 1,800= Rs. 72,000

5. Two numbers are in the ratio 4:5. When 6 is added to to each term, their ratio becomes
5:6. Find the number.

Solution:

Let, the first number be 4x and the second number be 5x
4x + 6 5
According to the question, 5x + 6 = 6 or, 25x + 30 = 24x + 36 ∴x = 6

Hence, the first number = 4x = 4×6= 2 4 and second number = 5x = 5×6=30

6. The ratio of the present ages of a mother and her daughter is 15:4. Four years ago, the
ratio of their ages was 13:2. Find the present ages of mother and the daughter.

Solution:

Let, the present age of mother be 15x years and that of her daughter be 4x years.

4 years ago, age of mother = (15x – 4) years and age of daughter = (4x – 4) years
15x – 4 13
According to the question, 4x – 4 = 2 or, 52x – 52 = 30x – 8 or, 22x = 44 ∴x = 2

Hence, the present age of mother = 15x = 15×2= 30 years and the present age of daughter =
4x = 4×2=8 years.

7. A rectangular field is 16 m broad and the ratio of the length and breadth is 3:2.

(i) Find the length of the field (ii) Find the perimeter of the field.

(iii) Find the area of the field.

Solution:

(i) Let, the length of the field (l) = x m.

According to the question, x = 3 or, 2x = 48 ∴x = 24
16 2
Hence the length of the field is 24 m .

(ii) The perimeter of the field (P) = 2 (l + b) = 2 (24 + 16) = 80 m

(iii) Area of the field (A) = l × b = 24 m × 16 m = 384 m2

25 Vedanta Excel in Mathematics Teachers' Manual - 8

Teaching Activity for Proportion

1. Mix juice and water in certain ratio and ask about the taste of the mixture.
For example:

S.N. Glass-A Glass-B

(i) 20 ml of juice and 50 ml of water 40 ml of juice and 50 ml of water

(ii) 50 ml of juice and 100 ml of water 75 ml of juice and 150 ml of water

Ask to compare the taste of the mixture.

2. Recall the comparison of two ratios

3. With examples, discuss about the proportion.

4. Discuss about the means and extremes of the proportion

5. Make a discussion on direct and indirect ratios with proper examples

6. Give hints and encourage students to solve the problems given in the exercise.

7. Give some group work/project work.

Solution of selected questions from Excel in Mathematics -Book 8

1. The bus fare of 8 passengers is Rs 640. If the fare is decreased by Rs 5 per passenger, how
many passengers can travel for Rs 900.

Solution:

Let x passengers can travel for Rs 900.

Here, the bus fare of 8 passengers = Rs 640 - 8×Rs 5 = Rs 600

The ratio of number of passengers = 8: x and ratio of bus fare = 600:900

Since, the number of passengers and the bus fare are in direct proportion.

So, 8: x = 600: 900

or, 8 = 600 or, 2x = 24 ∴x = 12
x 900
Hence, 12 passengers can travel for Rs 900.

2. If 8 workers can complete a piece of work in 18 days, how many workers are required to
complete the work in 12 days?

Solution:

Let, x number of workers are required to complete the work in 12 days.

The ratio of number of workers = 8: x

The ratio of the number of working days = 18: 12

Since, the number of workers and the working days are in inverse proportion.

So, 8: x = 12:18 or, 8 = 12 or, 2x = 24 ∴x = 12
x 18
Hence, 12 workers are required to complete the work in 12 days.

3. A bus completes its journey in 14 hours with an average speed of 40 km per hour. How
long does the bus take to complete the journey if its average speed is increased to 56 km
per hour?

Solution:
Let, the bus take x hour to complete the journey with average speed 56 km per hour.
The ratio of time taken = 14: x and ratio of speeds = 40: 56
Since, the time taken to complete the journey and speed are in inverse proportion.

Vedanta Excel in Mathematics Teachers' Manual - 8 26

So, 14: x = 56:40 or, 14 = 56 or, 7x = 70 ∴x = 10
x 40

Hence, the bus takes 10 hours to complete the journey at the average speed of 56 km per hour.

4. 15 workers were employed to complete a piece of work in 36 days. How many more
workers should be added to complete the work in 30 days?

Solution:

Let, x number of workers should be added to complete the work in 30 days.

The ratio of the number of workers = 15: (15 + x)

The ratio of the number of working days = 36: 30

Since, the number of workers and the working days are in inverse proportion.
15 30 15 5
So, 15: (15+x) = 30:36 or, 15 + x = 36 or, 15 + x = 6

or, 75 + 5x = 90 or 5x = 15 ∴x = 3

Hence, 3 workers should be added to complete the work in 30 days.

5. 20 students in a hostel have provisions for 60 days. If 10 more students are admitted to
the hostel, for how many days would the provisions be enough?

Solution:

Let, the provision would be enough for x days when 10 more students are admitted.

The ratio of the number of students = 20: (20 + 10) = 20: 30

The ratio of the working days = 60: x

Since, the number of students and the days for provision are in inverse proportion.
20 x
So, 20: 30 = x: 60 or, 30 = 60 or, 3x = 120 ∴x = 40

Hence, provisions would be enough for 40 days if 10 more students are admitted.

6. A garrison of 60 men had provision for 90 days. If 15 men left the garrison, how long
would the provisions last?

Solution:

Let, the provision would last in x days when 15 men left the garrison.

The ratio of number of men = =60: (60 – 15) = 60: 45

The ratio of the number of days = 90: x

Since, the number of men and the days for provision are in inverse proportion.
60 x
So, 60: 45 = x: 90 or, 45 = 90 or, 3x = 360 ∴x = 120

Hence, provisions would be enough for 120 days if 15 men left the garrison.

Teaching Activity for Unitary Method

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 copies is Rs is Rs 500, find the cost of 7 such copies.
Group-C: A motorcycle takes 20 minutes to cover the distance of 5 km, how long does it take

to cover a distance of 3 km?
Group-D: 6 workers can complete a piece of work in 15 days. In how many days, would10

workers complete the same work?
3. Solve the problems from exercise with discussion.
4. Focus on group or project work.

Solution of selected questions from Excel in Mathematics -Book 8

27 Vedanta Excel in Mathematics Teachers' Manual - 8

1. 4 machines of a factory can finish the required amount of production in 6 days. If
one machine had machinery defect and stop functioning, in how many days would
the remaining machines finish the same quantity of production at the same rate of
production?

Solution:

Here, number of machines = 4 – 1 = 3

4 machines can finish the required amount of production in 6 days.

1 machine can finish the same amount of production in 4×6 days = 24 days.
24
∴3 machines can finish the same amount of production in 3 = 8 days.

2. 12 workers were employed to complete the construction of a building in 60 days. How
many additional numbers of workers should be employed to complete the construction
in 45 days?

Solution:

In 60 days, 12 workers can construct of a building.

In 1 day, 60×12 = 720 workers can construct the building
720
In 45 day, 45 = 16 workers can construct the building

Hence, the additional number of workers = 16 – 12 = 4.

3. 120 students of hostel have food enough for 60 days. If 30 more students join the hostel
after 15 days, how long will the remaining food last?

Solution:

Here, remaining number of days =60 – 15 = 45

Total number of students with 30 more students = 120 + 30 = 150

Now,

120 students have food for 45 days

1 student has food enough for 120×45 = 5400 days
5400
150 students have the food enough for 150 = 36 days.

Hence, the remaining provisions will last in 36 days.

4. A garrison of 110 people had provisions for 30 days. If 22 people leave the garrison after
10 days, how long does the remaining provision last?

Solution:

Here,

After 10 days, remaining number of days =30 – 10 = 20

Remaining number of people = 110 – 22 = 88

Now,

110 people have provision for 20 days

1 person has provision for 110× 20 = 2200 days
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25 days. days.
last in 25

5. A computer can finish downloading an application file in 4 minutes at the rate of 600 MB
per minute.

(i) Find the size of the application file?
(ii) How long does it take to download the file when the download rate increases to

800 MB per minute?
Solution:
Here,

Vedanta Excel in Mathematics Teachers' Manual - 8 28

(i) In 1 minute, 600 MB of the file can finish downloading
In 4 minutes, 4×600MB = 2400 MB of file can download.

Hence, the size of the application file is 2400 MB.
(ii) 800 MB of the file can finish downloading in 1 minute.
1
1 MB of the file can finish downloading in 800 minutes.

2400 MB of the file can finish downloading in 1 × 2400 = 3 minutes.
Hence, it takes 3 minutes to download the file. 800

6. The size of movie file in YouTube is 1.8 gigabyte (GB) and the rate of download of the file
is 900 megabyte (MB) per minute.

(i) How long does it take to download the file?
(ii) Find the rate of download of the file per second.
Solution:

Here,
(i) 900 MB of the file can download in 1 minute.
1
1 MB of file can download in 900 minutes.

1.8 GB = 1800 MB of file download in 1 × 1800 = 2 minutes.
900

Hence, it takes 2 minutes to download the YouTube file of 1.8 GB.
(ii) In 1 minute, 900 MB of file can download.
In 60 seconds, 900 MB of the file can download.
900
In 1 second, 60 = 15 MB of the file can finish downloading.

Hence, rate of downloading the file is 15 MB per second.

Extra Questions
1. Mrs. KC exchanged Rs 5,600 in to the number of 10 rupee, 20 rupee and 50 rupee notes
in the ratio of 5:4:3 at a bank for Dashain Tika, find the number of each type of rupee
notes. [Ans: 100, 80, 60]
2. Three businessmen A, B and C invested a sum of Rs 7,00,000 such that the ratio of
investment of A and B is 2:3 and that of B and C is 4:5.
(i) Find the sum invested by each of them.
(ii) At the end of the year, if they gained Rs 1,15,500, divide the profit among them in the
ratio of their shares.
[Ans: (i) Rs 1,60,000; Rs 2,40,000; Rs 3,00,000
(ii) Rs 26,400; Rs 39,600;Rs 49,500]

3. The ratios of milk and water in glass-A is 1:2, in glass-B is 3:5 and in glass-C is 1:3. If
the mixtures of all these glasses are poured in a bigger bicker, find the ratio of milk and
water in the bicker. [Ans: 23:49]
4. In grocery, the cost of 6 kg of sugar is Rs 444.
(i) What is the cost of 13 kg of sugar?
(ii) How many kilograms of sugar can be bought for Rs 666?
[Ans: (i) Rs 962, (ii) 9 kg]
5. 36 workers can construct a wall in 30 days working 6 hours in a day. How many workers
are required to construct the same wall in 18 days working 9 hours in a day?
[Ans: 40 workers]

29 Vedanta Excel in Mathematics Teachers' Manual - 8

Unit Percentage and Simple Interest

6

Allocated teaching periods: 7
Competency

- Solving daily life problems based on percentage
- Solving contextual problems related to simple interest.

Learning Outcomes

- To solve the problems related to percentage
- To solve the problems related to simple interest
Level-wise learning objectives

S.N. Levels Objectives
1.
- To define percentage
2.
3. - To tell the rule to express decimal or fraction in to

Knowledge (K) percentage.

- To tell the rule to convert percentage in to fraction and

decimal

- To tell the formula of calculating simple interest or time or

rate or amount.

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

in percentage.

- To find the compound ratio

Understanding (U) - To calculate the simple interest when principal, time and

rate are given.

- To find the rate of interest when the principal, time and

simple interest are given.

- To find the time duration when the sum, rate and simple

interest are given.

Application (A) - To solve the problems related to percentage
- To solve the verbal problems based on simple interest.

- To solve the contextual problems based on percentage and

4. High Ability (HA) simple interest

Required Teaching Materials/ Resources
Blocks of hundreds, process of expressing decimals or fractions in to percentage and vice versa
in chart paper, definitions of principal, interest and amount in chart paper, list of formulae etc

Pre-knowledge: Percentage, simple interest

Teaching Activity for Percentage 30

Vedanta Excel in Mathematics Teachers' Manual - 8

1. Divide the students into the groups and discuss upon the following questions.
(i) Out of hundred marks in an exam, Rahul obtained 80 marks, what percentage of marks

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

of students secured A+ grade?
(iii) In a basket out of 50 guavas, 10 guavas were found damage and not fit for sale, what

percentage of the guavas were damage?
(iv) There are 24 boys and 16 girls in a class, find the percentage of boys and the girls.
2. Discuss about the meaning the percentage with real life problems.
3. Make a discussion about the process of expressing the fractions or decimals in to the

percentages.
4. With related examples, make a discussion about the process of expressing the percentage in

to the fractions or decimals.
5. Focus on structured problem solving method.
6. Focus on guided discovery method to encourage students for learning.
7. Support students to solve the problems given in the exercise
8. Focus on group or individual work.
9. Give some project work on percentage.

Solution of selected questions from Excel in Mathematics -Book 8

1. The monthly salary of a Mr. Nepali was Rs 18,500 before last year and it was increased
by 10% last year. Again, it was increased by 20% this year.

(i) How much was his salary last year?
(ii) How much is he drawing this year?
Solution:
(i) Last year, his salary was Rs 18,500 + 10% of Rs 18,500 = Rs 20,350
(ii) This year, his salary is Rs 20,350 + 20% of Rs 20,350 = Rs 24,420

2. In 2018 A.D., the number of tourists who visited Nepal was 12,60,000. In 2019, it was
decreased by 10% and in 2020, it was increased by 40%. How many tourists visited Nepal
in 2020?

Solution:
Here,
No. of tourists visited in 2019 = 12,60,000 – 10% of 12,60,000 = 11,34,000
No. of tourists visited in 2020 = 11,34,000 +40% of 11,34,000 = 15,87,600


3. Ganesh spends 60% of his income every month and saves Rs 11,120 in a month. Find his

monthly income and expenditure.
Solution:
Let, his monthly income be Rs x.
Here, his percentage of saving = 100% - 60% = 40% and monthly saving = Rs 11,120
Now, 40% of x = Rs 11,120 or, 0.4x = Rs 11,120 ∴x = Rs 27,800

31 Vedanta Excel in Mathematics Teachers' Manual - 8

Hence, his monthly income is Rs 27,800
Again, monthly expenditure = Rs 27,800 – Rs 11,120 = Rs 16,680

4. In an examination, an examinee needs 40% marks to pass the examination in grade C. If

Krishna got 350 marks and unable to get grade C by 10 marks, find the full marks of the

examination.

Solution:

Let, the full of the examination be x.

Marks required to get grade C = 350 + 10 = 360

Now, 40% of x = 360 or 0.4x = 360 ∴x = 900

Hence, the full mark of the examination was 900.

5. Two years ago, when a Publication House increased the monthly salary of its employees

by 10%, an employee started to draw Rs 23,100 per month. Again, the Publication House

increased the salary by 20% last year. Calculate the increased amount of monthly salary

of the employee in two years.

Solution:

Let, the monthly salary of an employee of a Publication House before 2 years be Rs x.

Then, x + 10% of x = Rs 23,100 or, 1.1x = 23,100 ∴x = 21,000

Hence, the monthly salary of the employee before 2 years was Rs 21,000

Again,

Last year, the monthly salary of the employee = Rs 23,100 + 20% of Rs 23,100 = Rs 27,720

Now, the increased monthly salary in 2 years = Rs 27,720 – Rs 21,000 = Rs 6,720

Teaching Activity for Simple Interest

1. Give real life examples and ask the definition of principal, interest and amount.

2. Ask the formula of finding the simple interest

Recall unitary method and derive that the interest on a sum Rs P at the rate of R% per
laisstRosu=t thPe×1foT0l0×loRwing
annum for T years formulae I × 100
Under discussion,

(i) I= P×T×R (ii) P = I× ×10R0 (iii) T = P × R (iv) R= I × 100
100 T P×T

(v) A = P + I (vi) P = 1A00×+10T0R


3. Divide the students in to groups and give classwork.

4. Give hints and encourage students to solve the problems given in the exercise.

5. Prepare some problems to develop students’ critical thinking.

6. Focus on the students for being problem maker and the problem solver themselves.

7. Give some group work/project work.

Solution of selected questions from Excel in Mathematics -Book 8

Vedanta Excel in Mathematics Teachers' Manual - 8 32

1. A businessman deposited Rs 60,000 in a Commercial Bank at 4.5% p.a. How long should
he deposit the sum to receive an amount of Rs 61,350?

Solution:

Principal (P) = Rs 60,000; Rate of interest (R) = 4.5%; Amount (A) = Rs 61,350; Time (T) =?

Now, interest (I) = A – P = Rs 61,350 – Rs 60,000 = Rs 1,350

Again, I × 100 Rs 1,350 × 100
TR Rs 60,000 × 4.5
T = = = 0.5 years = 12 × 0.5 months = 6 months

Hence, the required time duration is 6 months.

2. What sum of money amounts to Rs 5,454 at 8% p.a. in 30 months?

Solution:

Here, Amount (A) = Rs 5,454

Rate of interest (R) = 8% p.a.
30
Time (T) = 30 months = 12 years = 2.5 years

Principal (P) =? Rs 5454 × 100
A × 100 100 + 2.5 × 8
Again, P = 100 + TR =

Hence, the required sum is Rs 4,545.

3. Pradeep lent a sum to Santosh at the rate of 12 1 % p.a. After 4 years, Santosh paid him,
2
Rs 7,500 and clear the debt, what sum did Pradeep lenmd to Santosh?

Solution: 1
2
Here, Amount (A) = Rs 7,000 Rate of interest (R) = 12 % p.a. = 12.5% p.a.

Time (T) = 4 years Principal (P) =? 7,50,000
A × 100 Rs 7,500 × 100 = Rs 150
Again, 100 + TR = 100 + 12.5 = Rs 5,000

Hence, Pradeep lent Rs 5,000 to Santosh.

4. Mr. Gurung borrowed a loan of Rs 25,000 from Mr. Pandey at 12% p.a. At the end of 3

years, if he agreed to pay Rs 26,300 with a goat to clear his debt. Find the cost of the goat.

Solution:

Principal (P) = Rs 25,000; 12 Rate of interest (R) = 12% p.a.; Time (T) = 3 years
P×T×R Rs 15,000 × 3×
Now, I= 100 = 100 = Rs 9,000

∴Amount (A) = P + I = Rs 25,000 + Rs 9,000 = Rs 34,000

Again,

Let the cost of the goat be Rs x.

Then, according to question

Rs 26,300 + x = Rs 34,000 ∴x = Rs 34,000 – Rs 26,300 = Rs 7,700

Hence, the cost of the goat is Rs 7,700.

5. Sajina deposited Rs 20,000 at the rate of 8% p.a. in her saving account. After 2 years,

she withdrew Rs 5,000 and the total interest of 2 years. How long should she keep the

33 Vedanta Excel in Mathematics Teachers' Manual - 8

remaining amount to get total interest of Rs 6,800 from the beginning?

Solution:

Principal (P) = Rs 20,000; Rate of interest (R) = 8% p.a.; Time (T) = 2 years
P×T×R Rs 20,000 × 2 × 8
Now, I= 100 = 100 = Rs 3,200

∴Amount (A) = P + I = Rs 20,000 + Rs 3,200 = Rs 23,200

At the end of 2 years, the amount of withdrawal amount = Rs 5,000 + Rs 3,200 = Rs 8,200

Again,

Remaining sum (P1) = Rs 23,200 – Rs 8,200 = Rs 15,000
Total interest from the beginning = Rs 6,800

Interest for the remaining sum (I1) = Rs 6,800 – Rs 3,200 = Rs 3,600
Rate of interest (R1) = 8% p.a.
Time (T1) =?
We have,
I1 × 100 Rs 3,600 × 100
T1 = P1 × R1 = Rs 15,000 × 8 =3 years

Hence, the remaining amount should be deposited for next is 3 years.

5. A man took a loan of Rs 12,000 with simple interest for as many years as the rate of

interest per year. If he paid Rs 3,000 as interest at the end of loan period, what was the

rate of interest?

Solution:

Principal (P) = Rs 12,000; interest (I) = Rs 3,000

Let the rate of interest (R) be x % p.a. then the loan period (T) = x years
P×T×R Rs 12,000 × x × x
Now, I= 100 or, Rs 3000 = 100 or, 25 = x2 or, x = 25 = 5

Hence, the rate of interest was 5% pa.a

Extra Questions

1. Ram and Hari have rectangular plots. If Ram’s plot is 25% longer and 20% narrower than

that of Hari’s plot. Whose plot covers more area? Find it. [Ans: Equal]

2. Last week, the rate of cost of vegetable was increased by 20% and this week it has been

decreased by 20%. Find the percentage changes in the rate of cost of the vegetable during

two weeks. [Ans: Decreased by 4%]

3. Binisha 25% taller than Kajol. By what percent is Kajol shorter than Binisha? Find it.

[Ans: 20%]

4. Ranjita secored 20 marks out of 25 full marks in a Unit Test of mathematics. Bidur secured

72% marks in the same test.

(i) Who did better in the test? Find it.

(ii) Who secored more marks and by how much percent? Find it.

[Ans: (i) Ranjita (ii) Ranjita secored 10% more marks]

5. A money lender took a loan of Rs 40,000 from a bank at the rate of 6% per annum for 3 years

and immediately he deposited the same sum in a cooperative at the rate of 9% per annum

for the same duration. How much did he gain in this transaction? [Rs 3,600 gain]

Vedanta Excel in Mathematics Teachers' Manual - 8 34

6. Mr. Sharma borrowed a sum of money at the rate of 10% p.a. for 2 years 6 months and

immediately he lent it to Mrs. Gurung at the rate of 18% p.a. for the same duration of time.

If he gained Rs 1,100, find the sum he borrowed. [Ans: Rs 5,500]

7. The interest on Rs 4,050 in 5 years is Rs 2,025, what will be the interest on Rs 5000 in 8

years at the same rate of interest? [Ans: Rs 4,000]

8. Umesh deposited Rs 50,000 at the rate of 4% p.a. in saving account. After 5 years, he

withdrew Rs 10,000 and the total interst of 5 years. How long should he keep the remaining

amount to get total interest Rs 14,800 from the beginning?

[Ans: 3 years]

9. Find the simple interest on Rs 4500 for 5 years at the rate of 14% per annum. In how many

years, will Rs 5000 produces the same interest at the rate of 12% per annum?

[Ans: Rs 3150, 5 years and 3 months]

10. What amount of interest will be received when Rs 5,000 is deposited in a bank for 2 years

at the rate of 5% p.a., and 5% income tax on interest is taken by the bank? Find it.

[Ans: 760]

11. If a sum of money doubles itself in 5 years at a certain rate. In how many years will be treble

itself at the same rate of interest? [Ans: 10 years]



35 Vedanta Excel in Mathematics Teachers' Manual - 8

Unit Profit and Loss

7



Allocated teaching periods: 7

Competency
- To solve the daily life problems on profit and loss by using fundamental rules of
profit/loss and formulae.

Learning Outcomes
- To collect the real life problems on profit and loss and solve them.

Level-wise learning objectives

S.N. Level Objectives
1. Knowledge (K)
- To define cost price, selling price
2. Understanding (U) - To tell the relation of C.P., S.P. and profit % or loss%
- To define marked price
3. Application (A) - To tell the formula of finding the discount
4. High Ability (HA) - To define VAT

- To find the profit/ loss amount
- To calculate the profit and profit
- To find the discount/ VAT amount
- To calculate the rates of discount and VAT

- To solve the verbal problems on profit and loss
- To solve the verbal problems on discount and VAT

- To mathematize the daily life problems related to profit
and loss and solve them.

- To link various real life/ contemporary problems with
discount and VAT

Required Teaching Materials/ Resources
Colourful chart-paper with definitions and formulae, bills, VAT bills, audio/video clips related
to profit and loss, projector etc.
Pre-knowledge: cost price, selling price, profit and loss

Teaching Activities for Profit and Loss
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, profit

percentage and loss percentage.
3. Explain the following formulae with examples

Vedanta Excel in Mathematics Teachers' Manual - 8 36

(i) Profit amount = S.P. – C.P.

(ii) Loss amount = C.P. – S. P.

(iii) Profit amount = profit % of C.P.

(iv) Loss amount = loss % of C.P.

(v) S.P. = C.P. + P% of C.P.

(vi) S.P. = Cp.ePr.c–enLt%agoef=C.pPC.r.ofPi.t × 100% or, S. P. – C. P. × 100%
(vii) Profit C. P.

(viii) Loss percent = loss × 100% or, C. P. – S. P. × 100%
C. P. C. P.
4. Focus on group / project work

5. Give more priority in collaboration learning.

6. Work as the facilitator and encourage the learners.

Solution of selected questions from Excel in Mathematics -Book 8

1. A trader bought 1000 glass tumblers at Rs 8 each. 100 glass tumblers were broken and he/

she sold the rest at Rs 10 each. Find his/her gain or loss percent.

Solution:

Here, C.P. of 1000 glass tumbler = 1000 ×Rs 8 = Rs 8000

No. of broken tumblers = 100

Remaining number of tumblers = 1000 – 100 = 900

S.P. of 900 glass tumblers = 900 ×Rs 10 = Rs 9000

Since, S.P. > C.P., he/she made a profit

Profit amount = S.P. – C.P. = Rs 9000 – Rs 8000 = Rs 1000
Profit amount Rs 1000
Now, profit percent = Cost price (C.P.) × 100% = Rs 8000 × 100% = 12.5%

Hence, his/her profit percent is 12.5%.

2. A book-seller bought 7 story books for Rs 525 and sold 4 books for Rs 360. If he/she sold

the remaining books at the same rate of cost price, find his/her profit or loss percent.

Solution:

Here, C.P. of 7 story –books = Rs 525 ∴C.P. of 1 story-book= Rs 75

S.P. of 4 books = Rs 360 and S.P. of remaining 3 books = 3×Rs 75 = Rs 225

S.P. of 7 books = Rs 360 + Rs 225 = Rs 585

Since, C.P. > S.P., he/she made a loss

Loss amount = C.P. – S.P. = Rs 585 – Rs 525 = Rs 60
Loss amount Rs 60
Now, loss percent = Cost price (C.P.) × 100% = Rs 525 × 100% = 11.43%

Hence, his/her loss percent is 11.43%.

3. Shiva bought a second hand motorcycle for Rs 1,53,200. He spent Rs 1,200 to repair it and
sold it to Bishnu at 8% loss. How much did Bishnu pay for it?

Solution:
Here, C.P. of motorcycle with repairing = Rs 1,53,200 + Rs 1,200 = Rs 1,54,400
Loss percent = 8%

37 Vedanta Excel in Mathematics Teachers' Manual - 8

Now, S.P. = C.P.- Loss % of C.P. = Rs1,54,400- 8% of Rs 1,54,400 = Rs 1,42,048
Hence, Bishnu paid Rs 1,42,048.

4. A shopkeeper sold a T-shirt for Rs 665 at a loss of 5%. At what price did he/she buy the
T-shirt?

Solution:
Let the S.P. of T-shirt = Rs 665 loss percent = 5%
Let C.P. of the T-shirt be Rs x.
Then, loss amount = L% of C.P. = 5% of x = Rs 0.05x
Now, S.P. = C.P. – Loss or, 665 = x – 0.05x or, 665 = 0.95x ∴x = 700
Hence, he/she bought the T-shirt fort Rs 700.

5. A dealer bought a laptop for Rs 45,000 and sold it to a retailer at 12% profit. The retailer
sold it to a customer at 25% profit. How much did the customer pay for the laptop?

Solution:
For dealer:
C.P. of a laptop = Rs 45,000 and profit percentage = 12%
∴S.P. = C.P. + P% of C.P. = Rs 45,000 + 12% of Rs 45,000 = Rs 50,400
For retailer:
C.P. of a laptop = S.P. of dealer = 50,400 and profit percentage = 25%
∴S.P. = C.P. + P% of C.P. = Rs 50,400 + 25% of Rs 50,400 = Rs 63,000
Since, S.P. of retailer = C. P. of customer = Rs 63,000
Hence, the customer paid Rs 63,000 for the laptop.

6. Sunayana bought a mobile phone and sold to Pratik at 8% loss. Pratik sold it to Debashis
for Rs 11,040 and made 20% profit.

(i) Find the cost price of mobile to Pratik.
(ii) Find the cost price of mobile to Sunayana.
Solution:
(i) Let the cost price (C.P.) of the mobile for Pratik be Rs x.
Then, S.P. = C.P. + P% of C.P.
or, Rs 11,040 = x + 20% of x
or, 11,040 = 1.2x ∴x = 9200
∴C.P. of the mobile for Pratik = Rs 9,200
(ii) Again, C.P. of Pratik = S.P. of Sunayana = Rs 9,200
Let the cost price (C.P.) of the mobile for Sunayana be Rs y.
Then, S.P. = C.P. – L% of C.P.
or, Rs 9,200 = y – 8% of y
or, 9,200 = 0.92y ∴y = 10000
∴C.P. of the mobile for Sunayana = Rs 10,000

Vedanta Excel in Mathematics Teachers' Manual - 8 38

Teaching Activities for Discount and VAT

1. Recall cost price (C.P.) and selling price (S.P.) of an article.
2. With examples, discuss on marked price (M.P.) and discount.
3. Explain discount as the amount of reduction in the marked price of an article.
4. Paste/show the different types of taxes in the colourful chart paper and explain with

appropriate examples.

5. Clarify the VAT as tax levied on purchase of goods or service
6. List the following formulae after discussion
(i) Discount amount = M.P. – S. P.

(ii) Discount amount = Discount % of M.P.
Discount amount
Rate of discount = S.P. × 100%

S.P. = M.P. – Discount amount

S.P. = M.P. – Discount% of M.P. = M.P. (1 – Discount %)

S.P. = M.P. (1 – D1%)(1 – D2%) when two successive discount rates D1% and D2 % are given.
VAT amount = S.P. including VAT – S. P. excluding VAT

VAT amount = VAT% of .P.

Rate of VAT = VAT amount × 100%
S.P.

S.P. with VAT = S.P. + VAT amount

S.P. with VAT = S.P. + VAT% of S.P. = S.P. (1 + VAT %)

Solution of selected questions from Excel in Mathematics -Book 8

1. Kamala marked the price of a cosmetic item as Rs 400. She offered her customers a
discount of 20% and made a loss of Rs 30, what was the actual cost of the item to her?

Solution:
Marked price of the cosmetics (M.P.) = Rs 400 and discount percent = 20%
Now,
S.P. after discount = M.P. – D% of M.P. = Rs 400 – 20% of Rs 400 = Rs 320
Also, loss amount = Rs 30
Again, C.P. = S.P. + Loss amount = Rs 320 + Rs 30 = Rs 350
Hence, the actual price of the cosmetics is Rs 350.

2. The price of a rice cooker is marked as Rs 2,500. If 40% discount is allowed and then 13%

VAT is charged, find the selling price of the cooker with VAT.

Solution:

Marked price of the cooker (M.P.) = Rs 2,500 and discount percent = 40%

Now,

S.P. after discount = M.P. – D% of M.P. = Rs 2500 – 40% of Rs 2500 = Rs 1500

Again, S.P. with 13% VAT = S.P. + VAT% of S.P.

= Rs 1500 + 13% of Rs 1500 = Rs 1695

Hence, the selling price of the cooker with VAT is Rs 1,695.

39 Vedanta Excel in Mathematics Teachers' Manual - 8

3. A dealer marks the price of his/her mobiles 40% above the costy price and allowed 20%
discount. If his/her purchase price of a mobile is Rs 7,500, how much should a customer
pay for it with 13% VAT?

Solution:
Cost price (C.P.) of a mobile = Rs 7,500
Now, marked price of the cosmetics (M.P.) = C.P. + 40% of C.P.

= Rs 7,500 + 40% of Rs 7,500
= Rs 10,500
Also, S.P. after discount = M.P. – D% of M.P. = Rs 10,500 – 20% of Rs 10,500

= Rs 8,400
Again, S.P. with 13% VAT = S.P. + VAT% of S.P.

= Rs 8400 + 13% of Rs 8400 = Rs 9492
Hence, the selling price of the cooker with VAT is Rs 9,492.

4. A gift house allowed 20% discount on the marked price of a doll, 13% VAT was levied on
it. If the doll was sold at Rs 1,808, what was its marked price?

Solution:
Let M.P. of the doll be Rs x.
Now, S.P. after discount = M.P. – D% of M.P. = x – 20% of x = Rs 0.8x
Again, S.P. with 13% VAT = S.P. + VAT% of S.P.
or, Rs 1,808 = 0.8x + 13% of 0.8x or, 1,808 = 0.904x ∴x = 2000
Hence, the marked price of the doll was Rs 2,000.

Extra Questions

1. Subin marked the price of a watch at Rs 1,800. He sold it after allowing 15% discount and

made a profit of Rs 250. What was the cost price of the watch? Ans: Rs 1,280]

2. On the occasion of Dashain and Tihar, a shopping mall allows same percentage of discount

in every electronic item. If the selling price of a heater which is marked at Rs 2,000 is

Rs 1700 without VAT, what is the selling price of a refrigerator marked at Rs 30,000 with

13% VAT? [Ans: Rs 28,815]

Vedanta Excel in Mathematics Teachers' Manual - 8 40

Unit Algebraic Expressions

8

Allocated teaching periods: 8

Competency

- To solve problems using the squares or cubes of binomials

Learning Outcomes

- To find the degree of the polynomial

- To evaluate the polynomials

- To derive or explore the formulae of (a ± b)2, a2 ± b2, (a ± b)3 or a3 ± b3

Level-wise learning objectives

S.N. Level Objectives

- To define polynomial

- To identify the polynomial from the group of expressions

- To tell the degree of the polynomial

1. Knowledge (K) - To recall the formulae of (a ± b)2

- To tell the formulae of a2 ± b2

- To tell the formulae of (a ± b)3 or a3 ± b3
- To find the value of polynomials

- Two expand the squares of polynomials by using formulae.

2. Understanding (U) - To expand and simplify the cubes of polynomials by using

formulae.

- To factorize the polynomial of the form a2 – b2
- To simplify by using the formulae.

- To evaluate the value of sum of squares or square of sum of

3. Application (A) binomials using formulae

- To evaluate the value of sum of cubes or cube of sum of

binomials using formulae

- To link various real life/ contemporary problems based on

4. High Ability (HA) the sum of squares or cubes of binomials

- To link various real life problems based on the sum of

cubes or cube of binomials

Required Teaching Materials/ Resources
Square shaped card-boards, rectangular pieces of chart papers, formulae in colourful chart
paper, blocks etc.
Pre-knowledge: square and cubes of the algebraic terms, value of expressions
Teaching Activities for Polynomials and degree of polynomials and evaluation of polynomials
1. Recall on constant, variable, algebraic terms, algebraic expressions, types of expressions as
(i) 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.
Ram obtains always same marks in mathematical tests implies that his marks is

constant over the tests.

41 Vedanta Excel in Mathematics Teachers' Manual - 8

(ii) A letter which represents many values is considered as variable. For example, 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. If y represents the average temperature of a place which varies day

to day, then y is a variable.

(iii) 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.

(iv) The algebraic term or the combination of terms is an algebraic expression. For example,
4ab, x + 2y, a2 – a + 5 etc. are algebraic expressions.

2. Write the expressions with the numbers of unlike terms. Ask to count the number of unlike
terms and identify the expressions as monomial, binomial, trinomial or multinomial etc.

3. Ask about the coefficient, base and degree of the algebraic terms.
4. Discuss with lots of examples about the powers of variables in each term of the expressions.

Ask the students to separate the expressions in which the powers of variables in each terms

are whole numbers (not negatives, not decimals, not fractions). Then define the polynomial

as the expression in which the powers of variables are whole numbers.

5. Take a monomial polynomial like 5x3. Here x is multiplied 3 times. So the degree of 5x3 is
3. In 3xy4, the sum of powers of x and y = 1 + 4 = 5. So, the degree is 5.

6. Take a polynomial like x3y2 – 3x2y + 5xy. Here, the sum of powers of variables x and y in
the first term is 3 + 2 = 5, in the second term is 2 + 1 = 3, in the third term is 1 + 1 = 2.

The highest sum of powers of variables is 5. So the degree is 5. Make clear with sufficient

number of examples.

7. Discuss about the value of polynomials with examples.
8. If possible, use algebraic apps for evaluation of polynomials.
9. Focus on group / project work
10. Give more priority in collaboration learning.
11. Work as the facilitator and encourage the learners.

Solution of selected questions from Excel in Mathematics -Book 8

(a) Why is x2 a polynomial but 2 is not a polynomial?
Solution: 2 x2

IpBnoultyx2in2no, mtx2h2ie,aolp.row2xe–r2 of variable x is 2 which is a whole number. So x2 ,is a polynomial. 2
, the power of variable x is 2 x2
-2 which is not a whole number. So, is not a

(b) Why is 2x not a polynomial but 2x is a polynomial?
Solution:
In 2x = (2x)1/2x1/2, the power of variable x is 1/2 which is not a whole number. So, 2x is not a
polynomial.
But, in 2x, the power of variable x is 1 which is a whole number. Hence, 2x is a polynomial.

Vedanta Excel in Mathematics Teachers' Manual - 8 42

Teaching Activities for Squares or Cubes of Binomials
1. Recall about binomial.
2. Show the expansion of (a + b)2 with the following activities.

(i) Take a few rectangular sheet of paper (photocopy paper). Then fold them to get the
square sheet of paper as shown in the diagrams.

Again fold the square sheet of paper as shown in the diagram.




a1

a a2 1.a a+1

1 1.a 12
a+1

(ii) Use GeoGebra tool to visualise it (if possible).



(iii) Expand (a + b) 2 by multiplying.
(iv) Put different numbers to verify the result.
(v) To master on it, repeat the result/formulae by replacing ‘a’ and ‘b’ by other alphabets

or numbers.

3. Present the expansion of (a – b) 2 by the following activities.

(i) Use paper folding method to show the expansion of (a – b) 2.

D a–b bC

(a – b)2 b(a – b) a–b
a

b(a – b) b2 b

A a–b b B

(ii) Use GeoGebra tool to visualise it (if possible).

(iii) Expand (a – b) 2 by multiplying.

(iv) Put different numbers to verify the result.

(v) To master on it, repeat the result/formulae by replacing ‘a’ and ‘b’ by other alphabets

or numbers.

4. Show the factorization of a2 – b2 with the following activities.
(i) Use paper folding method to show the expansion of a2 – b2

43 Vedanta Excel in Mathematics Teachers' Manual - 8

Da CD C Da b
C

a–b (a + b) (a – b)
a–b
ab a2 – b2
b EF A a+b B

A B AB Area of ABCD
Area of ABCD Area of ABCDEF

= a2 = a2 – b2 a2 – b2 = (a + b) (a – b)

(ii) Use GeoGebra tool to visualise it (if possible).



(iii) Put different numbers to verify the result.

(iv) To master on it, repeat the result/formulae by replacing a and b by other

alphabets or numbers.

5. From the above discussion, list out the following relations

(i) (a + b)2 = a2 + 2ab + b2

(ii) (a – b)2 = a2 – 2ab + b2

(iii) a2 + b2 = (a + b)2 – 2ab or (a – b)2 + 2ab

(iv) a2 – b2 = (a + b) (a – b)

(v) a2 + b2 = (a + b)2 – 2ab or (a – b)2 + 2ab

6. Show the expansions of (a + b)3 by following activities.

(i) Cut a cubical shop into the following pieces.

aa a a+b
b
ab b b a+b a+b
a a bb

a3 + 3ab2 + 3ba2 + b3 = (a + b)3

(ii) Use GeoGebra tool to visualise it (if possible).

(iii) Put different numbers to verify the result.

(iv) Repeat the result/formulae by replacing ‘a’ and ‘b’ by other alphabets or numbers on

mastering it.

7. Make discussion in the following formulae or relations.

(i) (a + b)3 = a3 + 3a2b + 3ab2+ b3

(ii) (a – b)3 = a3–3a2b + 3ab2– b3

(iii) a3 + b3 = (a + b)( a2 –ab +b2) or, (a + b)3 – 3ab (a + b)

(iv) a3 – b3 = (a – b)( a2 + ab +b2) or, (a – b)3 + 3ab (a – b)

8. Focus on group / project works to solve the problems.

9. Work as the facilitator and encourage the learners.

Vedanta Excel in Mathematics Teachers' Manual - 8 44

Solution of selected questions from Excel in Mathematics -Book 8

1. Express 9p2 + 12pq + 4q2 as perfect square.
Solution:

9p2 + 12pq + 4q2 = (3p)2 + 2.3p.2q + (2q)2 = (3p+2q)2

2. Express 8a3 – 36a2 + 54a – 27 as perfect cube.
Solution:

8a3 – 36a2 + 54a – 27 = (2a)3 – 3 . (2a)2. 3 + 3. (2a).32 – (3)3 = (2a – 3)3

3. If a + 1 = 3, find the values of (i) a2 + 1 (ii) a – 1 2
Solution: a a2 a

Here, a + 1 = 3
a

(i) We have, a2 + b2 = (a + b)2 – 2ab

∴ a2 + 1 = a + 1 2 – 2 × a × 1 = 32 – 2 = 9 – 2 = 7
a2 a a

(ii) We have, (a – b) 2 = a2 – 2ab + b2
1 2 = 1 1 1
∴ a – a a2 – 2 × a × a + a2 = a2 + a2 – 2 = 32 – 2=7

4. If m2 – 1 = 6, find the values of (i) m2 + 1 (ii) m + 1 2
m m2 m
Solution:
m2 – 1 m2 1 1
Here, m = 6 or, m – m = 6 ∴m– m = 6

(i) We have, a2 + b2 = (a – b)2 + 2ab

∴ m2 + 1 = m + 1 2+ 2×m × 1 = 62 + 2 = 36 + 2 = 38
m2 m m

(ii) We have, (a + b) 2 = a2 + 2ab + b2

∴ m + 1 2 = m2 + 2 × m × 1 + 1 = m2 + 1 + 2 = 38 + 2 = 40
m m m2 m2

5. a + 1 = 4, find the value of a3 + 1
a a3

Solution: 1
a
Here, a + = 4

We have, a3 + b3 = (a + b)3 – 3ab (a + b)

∴ a3 + 1 = a + 1 3 – 3 × a ×a1 a + 1 = 43 – 3 × 4 = 64 – 12 = 52
a3 a a

6. If m – 1 = 3, find the value of m3 – 1
m m3

Solution:

Here, m – 1 = 3
m
We have, a3 – b3 = (a - b)3 + 3ab (a – b)

∴ m3 + 1 = m– 1 3+3×m× 1 m – 1 = 33 + 3 × 3 = 27 + 9 = 36
m3 m m m

45 Vedanta Excel in Mathematics Teachers' Manual - 8

7. If a + b = 2, find the values of a3 + b3 + 6ab
Solution:

Here, a + b = 2

We have, a3 + b3 = (a + b)3 – 3ab (a + b)

∴ a3 + b3 + 6ab = (a + b)3 – 3ab (a + b) + 6ab
= 23 – 3ab (2) + 6ab

= 8 -6ab + 6ab

= 8



Extra Questions

1. If a + 1 = 2, show that a2 + 1 = a3 + 1
a a2 a3
1
2. If x2 – 3x + 1 = 0, find the values of, find the values of x2 + x2 [Ans: 7]

3. If y2 – 5y – 1 = 0, find the values of y2 + 1 [Ans: 27]
y2

4. If a2 + 1 = 6, find the values of a6 + 1. [Ans: 198]
a a3
1 1
5. If x + x = 3, show that x3 + x3 .

6. If p2 + 1 = 7, find the positive value of p3 + 1 . [Ans: 18]
p2 p3



Vedanta Excel in Mathematics Teachers' Manual - 8 46

Unit Indices

9

Allocated teaching periods: 4

Competency
- Simplifying the expressions using the laws of indices.
Learning Outcomes
- To simplify the expressions by using the laws of indices
- To find the use of indices in daily life activities.
Level-wise learning objectives

S.N. Levels Objectives

- To define index
1. Knowledge (K) - To recall the product and quotient laws of indices

- To tell the law of zero index
- To write the power law of index

- To evaluate the simple expressions by using laws of indices
2. Understanding (U) - To simplify the simple given expressions by using laws of

indices

- To simplify the given rational expressions (involving roots

3. Application (A) as well) by applying the laws of indices

- To prepare the report about the laws of indices
4. High Ability (HA) - To contextualize the laws of indices

Required Teaching Materials/ Resources
Squared card-board, cubical box, chart papers with laws of indices, scissors, ruler, glue-stick
and computer/projector if possible
Pre-knowledge: Base, coefficient, power/index

Teaching Activities

1. Show a squared card-board of 10 cm or 15 cm or of any measures. Then ask the area of
square.

2. Show a squared card-board of x units and ask the following questions. x cm
(i) What is the area of the square? x cm
(ii) Why is x times x equal to x2 by observing the nature of powers?

3. By presenting a cube of side length x units, make a discussion under the following questions.

(i) What is area of each face of the cube?

(ii) What is the total surface of the cube?

(iii) What is the volume of the cube? x units

47 Vedanta Excel in Mathematics Teachers' Manual - 8

4. Discuss upon the following questions.
The cost of 1 kg of apple is Rs 125. Find the cost of 5 kg of apples by using the product law

of indices. For, 5 × 125 = 51 × 53 = 51+3 = 54 = 625

Divide 81 copies are equally among 9 students by using the quotient law of indices. For,
81 34
9 = 32 = 34–2 = 32 = 9

5. Recall the following laws of indices by presenting in chart paper with proper examples

Pradeep sir, yo table class 9 ko TG ma xa hai



S.N. Name of laws Rules Examples

(i) Product law am × an = am + n 23×25 =23+5 = 28

am ÷ an = am – n when m > n 36÷32=36–2= 34=81
1 32÷36= 1 = 1 = 1
(ii) Quotient law am ÷ an = an – m when m < n
(iii) Power law 36–2 34 81
(am)n = am × n, (ab)m = ambm, (24)2=24×2=28
am
a m = bm
b

(iv) Law of negative a– m = 1 or am = 1 2–3= 1 , 2 –4 = 3 4
index am a–m 23 3 2

(v) Law of zero a0= 1 where a ≠ 0 20=1, 50=1, etc.
index

(vi) Root law of index n am = m 3 52 2

an = 53

6. Give classwork from the textbook from the classroom section.

7. Discuss, give the way of solving the various problems and involve the students in solving
the problems from exercise

8. Under given condition, prove the expressions and give the same type problems to the
students and tell them to prove in the class.

9. Call the students randomly to solve the problems on the board in order to make them
confident to solve the problems

Solution of selected questions from Excel in Mathematics -Book 8

1. If 3x = 1, what is the value of x?
Solution:
Here, 3x = 1 or, 3x = 30 x = 0. Hence, the required value of x is 0.
2. What is the value of 1a+b?
Solution:
Since, any power of 1 is equal to 1. So, the value of 1a+b is 1.

Vedanta Excel in Mathematics Teachers' Manual - 8 48

3. For what power of (a+2) the value of a 3 2 is 3?
Solution: +

Here, 3 3 3 or, (a + 2) = 1 = (a + b) 0
a + 2 = 3 or, a + 2= 1
Hence, the power of (a + 2) is 0.

3

4. Evaluate: 274

Solution:

3 3 4 272 = 3 27 = 3

Here, 274 = 272 = 3

5. Simplify: 35 × 255 × 225
93 × 1254
Solution:

Here, 35 × 255 × 225 = 35 × (52)5 × (3×5)2 = 35 × 510 × 32 × 52 = 3 × 55+2–610+2–12

93 × 1254 (32)3 × (53)4 36 × 512
= 3 × 50 = 3 × 1 = 3

6. Simplify: xp – q + 1 × xq – r + 1 × xr – p + 1
Solution: x3

Here, xp – q + 1 × xq – r + 1 × xr – p + 1 = xp – q + 1 + q – r + 1 + r – p + 1 = x3 = x0= 1
x3 x3 x3

7. Simplify: xa + b × xc + d
Solution: xc + b xd + a

xa + b xc + d
Here, xc + b × xd + a = x ×xa+b–c–b c+d–d–a = xa–c × xc–a = xa–c+c–a = x0 = 1

8. Simplify: xa–b × xb–c × xc–a
Solution: xb–c ×
a–b b–c c–a a–b + b–c + c–a a–b+b–c+c–a = x0 = 1
Here, xa–b × 2 2
xc–a = x 2 x2 x2 =x 2 =x 2

9. Simplify: 2x + 3 + 2x + 2
Solution: 2x + 2 – 2x + 1

Here, 2x + 3 + 2x + 2 = 2x × 23 + 2x × 22­ = 2x(8 + 4) = 12 = 6
2x + 2 – 2x + 1 2x × 22 – 2x × 21 2x(4 – 2) 2

10. If a=1, b=2 and c = -3, find the value of ab–cbc–aca–b
Solution:
Here, a=1, b= 2 and c = -3

49 Vedanta Excel in Mathematics Teachers' Manual - 8

Now, ab–cbc–aca–b = 12–(–3) × 2–3–1 × (–3)1–2 = 15 × 2–4 × (–3)–1 = 1 × 1 × ( 1 ) = – 1
24 –3 48

11. If x=2, y=4, m=-1 and n = 3, find the value of xm + n × yn – m
xm – n × yn + m
Solution:

Here, x=2, y=4, m=-1 and n = 3

Now, xm + n × yn – m = 2–1+3 × 43–(–1) = 22 × 44 = 22 × (22)4 = 22 × 28 = 210 = 1024
xm – n × yn + m 2–1–3 × 43+(–1) 2–4 × 42 2–4 × (22)2 2–4 × 24 20

Extra Questions

1. If a = -3 and b = 2, find the value of ab [Ans: 9]

2. If m + n = m – n, find the value of mn [Ans: 1]

3. If a ≠ 0, find the values of (8a)o and 8ao. [Ans: 1, 8]

4. If a = -1, b = 2 and c = 3, find the value of (a + b)c + (b + c)a + (c + a)b [Ans: 5 1 ]
5

5. Simplify: 52077 – 52076 [Ans:1]
20 × 52075

6. Simplify: (xa – b)a2 + ab + b2 × (xb – c)b2 + bc + c2 × (xc – a)c2 + ca + a2 [Ans:1]




Vedanta Excel in Mathematics Teachers' Manual - 8 50