Evexdacnetal in
MATHEMATICS
TEACHERS' MANUAL
6Book
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
6Book
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
Topics Page No.
1. Set 5
2. Operations on Whole Numbers 11
3. Operations on Composite Numbers 17
4. Integers 27
5. Fraction and Decimals 30
6. Unitary Method 52
7. Percentage 56
8. Profit and Loss 62
9. Algebra 67
10. Equation, Inequalities and Graphs 80
11. Coordinates 89
12. Geometry - Point and Line 93
13. Geometry - Angles 97
14. Geometry - Triangles and Polygons 107
15. Geometry - Construction 115
16. Perimeter, Area and Volume 122
17. Symmetrical Figures, Design of Polygons and Tesselations 138
18. Statistics 140
Unit Set
1
Allocated teaching periods: 7
Competencies
- To describe the set in three different methods
- To identify the type of sets
- To describe the relations (equal, equivalent, overlapping and disjoint) between the
sets
Level-wise learning objectives
S.N. Levels Objectives
- To define a set
1. Knowledge (K) - To identify the member of a set
- To identify the type of set
- To state the relations between sets
- To write the cardinal number of set
- To describe the sets in listing method
2. Understanding (U) - To describe the sets in descriptive method
- To describe the sets in set-builder form
- To make possible subsets of the given set
- To express the given set in to descriptive, listing, and set-
3. Application (A) builder methods
- To describe the relations (equal, equivalent, overlapping
and disjoint) between the sets
- To make the list of the common elements of the given sets
and show in the diagrams.
4. High Ability (HA) - To link various real life/ contemporary problems with sets
and solve
Required Teaching Materials/ Resources
Collections of well-defined objects, models of sets, chart-paper, scissors, colourful markers,
charts of alphabets, different types of sets in chart etc.
Pre-knowledge: collection, group etc.
Teaching Activities
Activity 1 Set and membership of set
1. Show the following collection of well-defined objects in real/
chart-paper and ask the questions in groups
(a) What types of animals are there?
There is a cow. There is a dog.
There is a hen. There is a rabbit.
5 Vedanta Excel in Mathematics Teachers' Manual - 6
There is a sheep
These all are domestic animals.
This is a collection of domestic animals.
(b) What numbers are there inside the boundary?
1 is an odd number. 3 is an odd number.
5 is an odd number. 7 is an odd number.
9 is an odd number.
These all are odd numbers less than 10.
This is a collection of odd numbers less than 10.
Similarly, make the interactive class by discussing with the following collections
From above examples, discuss about the well-defined collection of objects.
Give more examples of not well-defined collection of objects. For example
(a) Collection of tall teachers (It’s not defined how tall they are)
(b) Collection of beautiful girls (It cannot be determined what type of beauty)
(c) Collection of tasty fruits. (Tasty fruits varies as per individual’s interest)
(d) Collection of long rivers of Nepal (how long are they?) etc.
(e) Define the set as the well-defined collection of objects.
Activity 2 Methods of writing the members of a set
1. With examples, describe a set by the following methods
(a) Diagrammatic method
With discussion, present the members/elements of the sets inside the circular
or oval shape. For example:
(b) Description method
Make a discussion in groups and describe the common property of members of the set
inside the braces.
For example: E = {even numbers less than 10}.
Give more examples to clear it out and give the questions to write in description method.
(c) Listing method
Give an example of a set and ask the elements of the set. List the elements of the set by
separating with commas and enclose within braces.
For example: E = {2, 4, 6, 8}. Give more examples to clear it out and give the questions to
write in listing method.
(d) Set-builder method
Use the variable like x, y, z ... to represent the common property (or properties) of the
members of set.
For example: E = {2, 4, 6, 8} → E = {x: x is an even number less than 10}
Give more examples to clear it out.
2. Write the whole numbers up to 20 and ask to write the following sets in listing and set-
builder methods.
(a) C = {composite numbers} (b) P = {prime numbers}
(c) E = {even numbers} (d) O = {odd numbers}
(e) S = {square numbers} (f) C = {cube numbers}
Vedanta Excel in Mathematics Teachers' Manual - 6 6
(g) F = {factors of 20} (h) M = {multiples of 3}
3. Discuss upon the questions given in exercise.
4. Give the project work to select any four objects which are available in the kitchen or
in classroom and express the set in diagrammatic, descriptive, listing and set-builder
methods.
Solution of selected questions from EXERCISE 1.1
1. Let’s tick ( √) to the well-defined collections.
a) A collection of tasty fruits.
b) A collection of fruits. [√]
c) A collection of long rivers of Nepal.
d) A collection of rivers of Nepal. [√]
2. Let’s tell and write the members of these sets in listing method.
a) A= {prime numbers between 10 and 20} Solution: A = {11, 13, 17, 19}
b) B = {letters of word ‘elephant’} Solution: B = {e, l, p, h, a, n, t}
c) C = {x: x is an odd number less than 5} Solution: C = {1, 3}
d) D = {y: y is a factor of 12} Solution: D= {1, 2, 3, 4, 6, 12}
3. Let’s answer the following questions.
a) What is a set? Give an example of a set.
Solution:
The collection of well-defined objects is called a set. For example, A = {even numbers
less than 10}
b) Is the collection of nice Nepali songs a set? Why?
Solution:
We cannot determine what types/genres of songs are nice or melodious. So, the collection
of nice Nepali songs is not a set.
c) How do we write a set in set-builder method?
Solution:
In set-builder method, the members of a set are represented by a variable. The common
property (properties) of members of the set is described by the variable inside the
braces. For example: A = {1, 2, 3, 4} → A = {x: x is a natural number less than 5} or,
A = {x: x< 5, x∈N}
4. Let’s describe the following sets in listing method.
a) W= {whole numbers less than 6} Solution: W = {0, 1, 2, 3, 4, 5}
b) M = {the first five multiples of 4} Solution: M = {4, 8, 12, 16}
c) A = {the letters of word ‘MATHEMATICS’} Solution: A = {M, A, T, H, E, I, C, S}
d) P = {x: x is a prime number, x< 20} Solution: P = {2, 3, 5, 7, 11, 13, 17, 19}
e) B = {y: y is a composite number, y< 10} Solution: B = {4, 6, 8, 9}
5. Let’s express the following sets in description method.
a) A= {क, ख, ग, घ, ङ} Solution: A = {the first five letters of Devanagari alphabets}
b) N = {1, 2, 3, 4, 5, 6, 7, 8, 9} Solution: N = {natural numbers less than 10}
7 Vedanta Excel in Mathematics Teachers' Manual - 6
c) P = {2, 3, 5, 7} Solution: P = {prime numbers less than 10}
d) V = {a, e, i, o, u} Solution: V = {vowels of English alphabets}
e) F = {1, 2, 7, 14} Solution: F = {factors of 14}
f) D = {4, 8, 12, 16} Solution: D= {the first four multiples of 4}
6. Let’s express the following sets in set-builder method.
a) W = {0, 1, 2, 3, 4} Solution: W= {x: x is a whole number, x<5}
b) S = {1, 4, 9, 16, 25} Solution: S = {x: x is perfect square number, x< 26}
c) O = {1, 3, 5, 7, 9} Solution: O = {x: x is an odd number, x< 10}
d) A = {a, b, c, d, e} Solution: A= {x: x is the first five English alphabets}
e) F = {1, 3, 5, 15} Solution: F = {x: x is a factor of 15}
f) M = {3, 6, 9, 12, 15, 18} Solution: M = {x: x is a multiple of 3, x< 20}
Note: Please! Focus on project word to consolidate the knowledge and skills.
Activity 3 Types of sets
1. With examples, discuss about the cardinality of sets.
(a) If A = {a, e, i, o, u} then n (A) = 5
(b) If S = {1, 4, 9} then n (S) = 3
2. Make the groups of students and ask to find the cardinality of the sets by listing the
elements as given below
Group A: N = {natural number less than 1}
Group B: H = {highest pick in the world}
Group C: S = {square numbers up to 100}
Group D: P = {prime numbers}
3. Under discussion, classify the sets as follows
(a) N = { }. The set N has no element. i.e., n (N) = 0. Define it as an empty set
(b) H = {Mt. Everest}. The set H contains only one element. Define it as singleton set
(c) S = {1, 4, 9, 16, 25, 36, 49, 645, 81, 100}. This set contains finite number of elements
and consider it as finite set.
(d) P = {2, 3, 5, 7, 11, ... }. This set contains infinite number of elements. Define it as
infinite set.
4. After discussion with more examples, make the students define the empty, singleton, finite
and infinite sets.
5. Give the questions on sets and ask to identify the type of the sets by listing on the elements.
Activity 4 Relationship between sets
1. Ask the students to compare the elements of the sets as given below. Then discuss about
equal sets.
(a) A = {a, e, i, o, u} and B = {u, i, a, o, e}
(b) P = {2, 3, 5, 7} and Q = {3, 2, 7, 5}
With discussion, conclude as A = B and P = Q
2. Ask the students to observe the number of elements of the sets as given below. Then
discuss about equivalent sets.
(a) A = {a, e, i, o, u} and B = {2, 4, 6, 8, 10}
Vedanta Excel in Mathematics Teachers' Manual - 6 8
(b) M = {5, 10, 15, 20, 25, 30} and F = {1, 2, 3, 4, 6, 12}
With discussion, conclude as A ~ B and M ~ N.
3. Show two sets having some element common to both in diagram.
From diagram, A = {1, 2, 3, 4} and B = {2, 3, 5, 7}
Discuss about the overlapping sets.
4. Similarly, give some pair of sets to write in listing method. Tell to observe the elements
common to both the sets or not. And identify the sets are whether they are overlapping or
not.
5. Make the students define the overlapping sets themselves.
6. Show the examples of sets having no any common elements as
shown in the diagram. Make the students define the disjoint
sets themselves
7. Divides the students into groups and make them give examples of overlapping and
disjoint sets.
8. Give a project work to conduct a survey among at least 10 classmates as given below.
Make separate sets the following sets
(a) Set of classmates who like apple
(b) Set of classmates who like banana
(c) Set of classmates who like mango
(d) Set of classmates who like apple and banana both
(e) Set of classmates who like apple and mango both
(f) Set of classmates who like banana and mango both
Draw the diagram and show the name of classmates in the diagram
9. Discuss upon the questions given in the exercise.
Solution of selected questions from EXERCISE 1.2
1. Let’s list the elements of these sets and write whether they are empty (null), singleton,
finite or infinite sets.
a) A = {prime number between 5 and 7} Solution: A = { } . It is a null set.
b) B = {multiples of 2 less than 20}
Solution: B = {2, 4, 6, 8, 10, 12, 14, 16, 18}. It is a finite set.
c) C = {multiples of 2 greater than 20}
Solution: C = {22, 24, 26, ...}. It is an infinite set.
d) D = {Square numbers between 10 and 20}
Solution: D = {16}. It is a singleton set.
2. Let’s list the elements and write the cardinal numbers of these sets.
9 Vedanta Excel in Mathematics Teachers' Manual - 6
a) A = {natural numbers less than 10} → A = {1, 2, 3, 4, 5, 6, 7, 8, 9} ∴n (A) = 9
b) B = {factors of 18} → B = {1, 2, 3, 6, 9, 18} ∴n (B) = 6
c) C = {letters of the word ‘apple’} → C = {a, p, l, e} ∴n (C) = 4
d) D = {x: x is a square number, x< 75} → D = {1, 4, 9, 16, 25, 36, 49, 64} ∴n (D) = 8
3. Let’s list the elements and write with reasons whether the following pairs of sets are
equal or equivalent.
(a) A = {whole number less than 5} and B = {1, 2, 3, 4, 5}
(b) P = {x: x is a prime number, x< 10} and Q = {7, 5, 3, 2}
Solution
(a) A = {whole numbers less than 5} = {0, 1, 2, 3, 4} ∴n (A) = 5
B = {1, 2, 3, 4, 5} ∴n (B) = 5
Since, the sets A and B do not have exactly the same elements but they have equal
numbers of elements. So, they are equivalent sets.
(b) P = {x: x is a prime number, x< 10} = {2, 3, 5, 7} and Q = {7, 5, 3, 2}
Since, the sets P and Q have exactly the same elements. So, they are equal set.
4. Let’s list the elements and write with reasons whether the following pairs of sets are
overlapping or disjoint.
(c) A = {factors of 12} and B = {factors of 18}
(d) P = {first five multiples of 4} and Q = {first five multiples of 7}
Solution
(e) A = {factors of 12} = {1, 2, 3, 4, 6, 12} and B = {factors of 18} = {1, 2, 3, 6, 9, 18}
Since, the elements 1, 2, 3 and 6 are common to both the sets A and B. So, the sets A and
B are overlapping.
(f) P = {first five multiple of 4} = {4, 8, 12, 16} and Q = {first five multiple of 7} = {7, 14,
21, 28}.
Since, there is no element common to both the sets P and Q. So, the sets P and Q are
disjoint.
Note: Please! Focus on Project Work to validate the knowledge and skills.
Extra Questions
1. Express the set A = {the prime number less than 6} in listing and set builder methods.
2. Describe the set N = {1, 2, 3, 4, 5, 6, 7, 8, 9} in description and set-builder methods.
3. Re-write the set M = {x: x is a multiple of 6, x<30} in description and listing methods.
4. Write all the possible subset of the set {s, i, x}
5. State with reason whether the sets A = {x: x is a composite number, x<10} and B = {y: y
is a natural number, y<5}
(a) Equal or equivalent (b) Overlapping or disjoint
Vedanta Excel in Mathematics Teachers' Manual - 6 10
Unit Operations on Whole Numbers
2
Allocated teaching periods: 6
Competencies
- To tell and write the face value and place value of the digit in the given number
- To write the number name of the numbers up to kharab/billion
- To simplify the problems with mixed operations and brackets
Level-wise learning objectives
S.N. Levels Objectives
- To identify the face/place value of digit in a number
1. Knowledge (K) - To state the fundamental operations
- To tell the steps of BODMAS rule
- To write the number name in Nepali and International
2. Understanding (U) numeration systems
- To write the relation between the places in Nepali and
Internal numeration system
- To perform simplification
- To solve the contextual problems based on place values
3. Application (A) - To simplify the problems with mixed operations and
brackets
4. High Ability (HA) - To make the mathematical expression for the verbal problem
and simplify using BODMAS rule
Required Teaching Materials/ Resources
Scale, scissors, card-board paper, number line, markers, place-value table (Nepali and
international systems), audio-visual tape, internet etc
Pre-knowledge: number system (Nepali and international), simplification
Teaching Activities
Activity 1 Natural and whole numbers
1. Ask about basics of the Hindu-Arabic numeration (base ten or decimal) system.
2. Discuss upon the following question/s.
(a) What types of numbers are used in counting the objects?
(b) Are the numbers 1, 2, 3, 4 etc. counting numbers?
With real life examples, explain about the set of natural numbers
11 Vedanta Excel in Mathematics Teachers' Manual - 6
3. Under discussion, conclude the following facts
(i) The counting numbers are called natural numbers.
(ii) The set of natural numbers is denoted by N. Thus, N = {1, 2, 3, 4, 5, ...}
(iii) The least natural number is 1 and the greatest natural number is infinite
4. To make the students understand about whole numbers, ask the questions as given below.
(i) Suppose you have 2 apples. You ate 1 and gave 1 to your friend, how many apples are
left with you?
(ii) If all the students passed the exam, how many of them were fail in the exam?
(iii) How much is left when 10 is subtracted from 10?
During discussion with the above questions, make the students introduce ‘none’ or
‘zero’
5. Use the number line and conclude the following facts under discussion
(i) The counting numbers including zero are called whole numbers.
(ii) The set of whole numbers is denoted by W. Thus, W = {0, 1, 2, 3, 4, 5, ...}
(iii) The least whole number is 0 and the greatest natural number is infinite.
Activity 2 Place, place value and face value
1. Take the four digit numbers. Use the blocks of ones, tens, hundreds and thousands,
discuss about the face value and place value of each digit.
Similarly, show the places, face value and place value of each digit of a large number in the
chart paper.
For example
Numeral Places Face values Place values
5823974
Ones 4 4×1 = 4
Tens 7 7×10 = 70
Hundreds 9 9×100 = 900
Thousands 3 3×1000 = 3000
Ten thousands 2 2×10000 = 20000
Lakhs 8 8×100000 = 800000
Ten lakhs 5 5×1000000 = 5000000
Similarly, give some numerals and get the students engaged in finding the places, face
and pale values.
2. Present the following periods and places of both the systems in chart paper and give the
group work to prepare the same table in the colourful chart paper and paste the in the
math corner.
(a) Periods and places in Nepali system
Periods Kharabs Arabs Crores Lakhs Thousands Units Ones
Places Ten kharabs Ten Arabs Ten Crores Ten Lakhs Ten Thousands Hundreds Tens
kharabs arabs crores lakhs thousands
Vedanta Excel in Mathematics Teachers' Manual - 6 12
(b) Periods and places in International system
Periods Billions Billions Hundred Millions Millions Hundred Thousands Units Ones
Places Hundred Ten millions Ten thousands Ten Thousands Hundreds Tens
billions billions
millions thousands
3. Discuss on the use of commas in both Nepali and international systems.
4. Take a numeral and write its number names in both Nepali and international systems.
5. Recall the expanded form of the numbers.
6. Ask the smallest and the greatest numbers of (i) certain digits and (ii) the numbers
formed by the given digits
7. Make the students complete the Classwork Section from exercise as classwork and
check their performance.
8. Focus on group work and project work.
Solution selected problems from EXERCISE 2.1
1. The estimated budget of a hydro-electric project in Nepal is twenty-eight billions five
hundred forty-two million rupees.
(i) Express the budget in figure.
(ii) Re-write the value of the budget in Nepali numeration system.
Solution:
Here,
(i) The estimated budget of a hydro-electric project is Rs 28, 542, 000,000
(ii) In Nepali numeration system, Rs 28,54,20,00,000: Twenty-eight arab fifty-four crore and
twenty lakh rupees.
2. a) How many lakhs are there in 4 million?
Solution:
Here, 4 millions = 4,000,000 = 40,00,000 = 40 lakhs
b) How many crores are there in 20 million?
Solution:
Here, 20 million = 20,000,000 = 2,00,00,000 = 2 crores
c) How many arabs are there in 30 billion?
Solution:
Here, 30 billion = 30,000,000,000 = 30,00,00,00,000 = 30 arabs
d) How many millions are there in 5 crores?
Solution:
Here, 5 crores = 5, 00,00,000 = 50,000,000 = 50 million
e) How many billions are there in 7 arab?
Solution:
Here, 7 arab = 7,00,00,00,000 = 7,000,000,000 = 7 billion
3. Find the sum and difference between the place values of two sevens in the number
6784037525.
Solution
Here,
The place value of 7 in thousand’s place = 7,000
13 Vedanta Excel in Mathematics Teachers' Manual - 6
The place value of 7 in ten crore’s place = 70,00,00,000
Now,
700000000 700000000
+7000 - 7000
Sum = 700007000 Difference = 699993000
4. What is the sum and difference between the numbers 473 and the number obtained by
reversing the digits at ones and hundreds places?
Solution
Here,
The given number = 473
The number obtained by reversing the digits at one and hundred places = 374
Now,
473 473
+374 - 374
Sum = 847 Difference = 99
5. By how much is 6345987 smaller than 1 crore?
Solution
Here,
1 crore = 1,00,00,000 and the given number = 63,45,987
Now,
10000000
- 6345987
Difference = 3654013
Hence, 6345987 smaller than 1 crore by 3654013
Activity 3 Order of operations and use of BODMAS rule
1. For making students interested towards mathematics, give the classroom activities on
fundamental operations to fill in the blanks in the book with tricks, quiz and puzzles.
2. Make a discussion on order of operations to review the facts those learnt in the previous
classes. For, ask individually or in group the rule of DMAS.
3. For more activities, give the classwork to fill up the blanks given in page number 27.
4. Make the groups of students. Give the verbal problems on order of operations. Tell them
to write in mathematical expressions and then simplify.
Group A: 5 is added to the product of 2 and 3.
Group B: 3 is subtracted from the sum of 7 and 4.
Group C: product of 2 and 9 is divided by 6.
Group D: 10 is added to the quotient of 24 divided by 8
5. Discuss about the brackets ( ), { } and [ ] which are used in simplification.
6. With proper examples, explain about BODMAS rule where
B stands for brackets ( ), { } and [ ]
Vedanta Excel in Mathematics Teachers' Manual - 6 14
O stands for ‘Of’ (times/direct multiplication)
D stands for ‘Division: ÷’
M stands for ‘Multiplication: ×’
A stands for ‘Addition: +’
S stands for ‘Subtraction: - ’
7. Make the groups of students. Give the verbal problems on order of operations with
brackets. Tell them to write in mathematical expressions and then simplify.
Solution selected problems from EXERCISE 2.2
1. Let’s make mathematical expressions and simplify.
a) 5 is subtracted from the sum of 8 and 7 Solution: 8 + 7 – 5 = 15 – 5 = 10
b) 7 is added to the product of 5 and 6. Solution: 5×6 + 7 = 30 + 7 = 37
c) 9 is subtracted from the product of 7 and 3. Solution: 7×3 – 9 = 21 – 9 = 12
d) 8 is added to the quotient of 36 divided by 4. Solution: 36÷4 + 8 = 9 + 8 = 17
e) The product of 7 and the quotient of 40 divided by 5. Solution: 7×40÷5 = 7×8 = 56
2. a) Sunayana bought 6 pencils at Rs 8 each and she gave a Rs 50 note to the shopkeeper
for the payment. How much change did the shopkeeper return her?
Solution:
Here, total cost for 6 pencils = Rs 8×6
So, the amount returned by shopkeeper = Rs 50 - Rs 8×6 = Rs 50 – Rs 48 = Rs 2
b) The cost of 1 kg of rice is Rs 90 and 1 kg of sugar is Rs 80. Find the total cost of 5 kg
of rice and 2 kg of sugar.
Solution:
Here, total cost for 5 kg of rice = Rs 90×5 and total cost for 2 kg of sugar = Rs 80×2
So, the total cost of 5 kg of rice and 2 kg of sugar = Rs 90×5 + Rs 80×2
= Rs 450 + Rs 160
= Rs 610
c) On a day, there were 27 students present in class six. 16 of them were girls and the
rest were boys. If only 3 boys were absent on that day, find the number of boys in
class six.
Solution:
The number of boys who were present on that day = 27 – 16
The number of boys who were absent on that day = 3
∴Total number of boys in class six = 27 – 16 + 3 = 30 – 16 = 14
Solution selected problems from EXERCISE 2.3
1. The sum of 60 and 30 is divided by 10 and the difference of quotient and 3 is divided
by 2.
Solution
The expression = {(60+30)÷10 – 3}÷2
= {90÷10 – 3}÷2 = {9 – 3}÷2 = 6÷2 = 3
15 Vedanta Excel in Mathematics Teachers' Manual - 6
2. The distance between Dakshe’s house and his school is 7 km. How many kilometres
does he travel in 6 days?
Solution
The distance that he travels in 1 day = 2×7 km
The expression = 6(2×7) = 6×14 = 84 km
Hence, he travels 84 km in 6 days.
3. Mrs. Chamling earns Rs 6000 in a week. She spends Rs 250 for food and Rs 75 for
transportation everyday. How much money does she save in a week?
Solution
The expression = 6000 – 7(250 + 75)
= 6000 – 7(325) = 6000 – 2275 = 3725
Hence, she saves Rs 3725 in a week.
4. There are two baskets each of them containing 18 apples. 10 apples are rotten and not
fit for eating. You add 4 more apples and then divided among 6 friends. How many
apples would each get?
Solution
The expression = {(2×18 – 10) + 4}÷6
= {(36 – 10) + 4}÷6
= {26 + 4}÷6 = 30÷6 = 5
Hence, each would get 5 apples.
Extra questions
1. Find the sum and difference between greatest and least numbers of 6 digits.
[Ans:1099999, 899999]
2. Find the sum and difference between greatest and least numbers of 7 digits formed by 2,
9, 0, 4, 1, 6, 5. [Ans:10678779, 8629631]
3. By how much is 55 billion smaller than 55 kharab?
[Ans:545000000000]
4. What will be the answer for 2 plus 2 divided by 2? [Ans: 3]
5. Mr Pradeep types 5 pages of word file in every half of an hour. He types only 4 hours
everyday. How many pages of file does he have in a week if 10 pages of the entire file is
mistakenly deleted permanently? [Ans: 270 pages]
6. The sum of 30 and 18 is divided by 8 and the difference of quotient and 2 is divided by
4. [Ans: 1]
7. Simplify: 88÷11[160÷5{3+(10 – 7 + 2)}] [Ans:2]
Vedanta Excel in Mathematics Teachers' Manual - 6 16
Unit Properties of on Whole Numbers
3
Allocated teaching periods: 13
Competencies
- To separate prime and composite numbers
- To test of divisibility of the numbers up to 11
- To prime factorize the number
- To find the H.C.F. and solve the real life problems related to H.C.F.
- To find the L.C.M. and solve the real life problems related to L.C.M.
- To solve the simple problems related to square and square root of numbers
- To solve the simple problems related to cube and cube root of numbers
- To use the nth term rules to find out the other terms of sequences
- To find out the rules for the nth terms of the sequences
Level-wise learning objectives
S.N. Levels Objectives
1. Knowledge (K)
2. Understanding (U) - To define prime and composite numbers
- To define factors and multiples of numbers
3. Application (A) - To define H.C.F.
4. High Ability (HA)
- To factorize the number by both the factor tree and division
methods.
- To test the divisibility of numbers
- To find the H.C.F. and L.C.M.
- To find the square and square roots
- To solve the problems based on H.C.F. and L.C.M.
- To solve the problems related to square and square root
- To link various real life/ contemporary problems with
H.C.F and solve
Required Teaching Materials/ Resources
Scale, scissors, card-board paper, blocks, tiles, multiplication table chart, markers, audio-visual
tape etc.
Pre-knowledge: prime/composite numbers, factor and multiples
Teaching Activities
Activity 1 Prime and composite numbers
1. Give the tiles and tell to form the rectangular arrangements as possible as shown below.
(a) 1: ← Only one possible rectangular arrangement
17 Vedanta Excel in Mathematics Teachers' Manual - 6
(b) 2: ← Two possible rectangular arrangements
(c) 3: ← Two possible rectangular arrangements
(d) 4: ← Three possible rectangular arrangements
(e) 5: ← Two possible rectangular arrangements
← Three possible rectangular arrangements
(f) 6:
(g) 12: ← Three possible rectangular arrangements
Discuss about the different number of tiles and number of rectangular arrangements.
The number of tiles from which only two rectangular arrangements can be formed,
are prime numbers. From the above activity, make sure that 2, 3, and 5 are prime
numbers. At last, conclude the prime numbers as those numbers which are divisible
by one and itself.
(h) Again, on the basis of above example, ask the students to identify the composite
numbers and make the students define composite numbers.
2. Use the Sieve of Eratosthenes method for finding prime numbers from 1 to 100. This
method is very simple and everyone can understand easily.
For this method, follow the procedures
(a) First, write all the numbers from 1 to 100 as shown in the table.
(b) Cross “1” as it is not a prime number.
(c) Now leave number “2 “as it is a prime number
because it is exactly divisible by 1 and itself. Then
cross all other multiples of 2 starting with 4.
Similarly leave 3, 5 and 7 then cross their all multiples. The remaining numbers
are called prime numbers. Not only up to 100 Eratosthenes method can be applied
to any extent
Vedanta Excel in Mathematics Teachers' Manual - 6 18
Activity 2 Test of divisibility
1. Recall even numbers and with sufficient examples make sure that the even numbers are
exactly divisible by 2.
2. Take a number and find the sum of the digits used in the number. Then test whether the
sum is exactly divisible by 3 or not. Also, divide the number by 3. Give sufficient number
of examples so that the students themselves could make a conclusion that if the sum of the
digits is divisible by 3, the given number is exactly divisible by 3 otherwise not.
3. With examples, verify that the given number is exactly divisible by 4 if the number formed
by last two digits of the given number is divisible by 4 or it is 00.
4. If the digit at ones’ place is 0 or 5, then the number is exactly divisible by 5. Discuss with
examples.
5. Since, 6 = 2×3. So, all the numbers divisible by both 2 and 3 are also divisible by 6.
6. If the difference of two times the digit at ones’ place and number formed by rest of the
digits of the given number is divisible by 7 or equal to 0 then the given number is divisible
by 7. For example: In 42, two times of 2 is 4 and 4 – 4 = 0. So 42 is divisible by 7; in 203,
two times of 3 is 6 and 20 – 6 = 14 which is divisible by 7. So, 203 is divisible by 7. In
3661, two times of 1 is 2 and 366 – 2 = 364. Also, in 364, two times of 4 is 8 and 36 – 8 =
28 which is divisible by 7. So, 3661 is exactly divisible by 7.
7. As divisibility test of 4; the given number is exactly divisible by 8 if the number formed by
last three digits of the given number is divisible by 4 or it is 000.
8. As divisibility test of 3; if the sum of the digits is divisible by 9, the given number is exactly
divisible by 9 otherwise not.
9. If the digit at one’s place is 0 then the given number is exactly divisible by 10.
10. If the difference of the sum of the digits at alternate places is either 0 or divisible by 11,
then the given number is divisible by 11.
Solution selected problems from EXERCISE 3.1
1. Let’s use the divisibility test and write which of the following numbers are exactly
divisible by 7.
a) 525 b) 798 c) 836 d) 3696 e) 5706
Solution:
Here,
a) In 525; two times of 5 is 10 and 52 – 10 = 42, which is divisible by 7. So, 525 is exactly
divisible by 7.
b) In 798; two times of 8 is 16 and 79 – 16 = 63, which is divisible by 7. So, 798 is exactly
divisible by 7.
c) In 836; two times of 6 is 12 and 83 – 12 = 71, which is not divisible by 7. So, 836 is not
exactly divisible by 7.
d) In 3696; two times of 6 is 12 and 369 – 12 = 357. Also, for 357; two times of 7 is 14 and 35
– 14 = 21; which is divisible by 7. So, 3696 is exactly divisible by 7.
e) In 5706; two times of 6 is 12 and 570 – 12 = 558. Also, for 558; two times of 8 is 16 and
55– 16 = 39; which is not divisible by 7. So, 5706 is not exactly divisible by 7.
19 Vedanta Excel in Mathematics Teachers' Manual - 6
2. Let’s apply the rule of test of divisibility and write which of the following numbers are
exactly divisible by 11.
a) 6358 b) 2845 c) 71852 d) 258995 e) 999666
Solution:
Here,
a) In 6358; 6+5 = 11, 3 + 8 = 11 and 11 – 11 = 0. So, 6358 is exactly divisible by 11.
b) In 2845; 2 + 4= 6, 8 + 5 = 13 and 13 – 6 = 7, which is not divisible by 11. So,2845 is not
exactly divisible by 11.
c) In 71852; 7 + 8 + 2 = 17, 1 + 5 = 6 and 17 – 6 = 11, which is divisible by 11. So,71852 is
exactly divisible by 11.
d) In 258995; 2 + 8 + 9 = 19, 5 + 9 + 5 = 19 and 19 – 19 = 0. So,258995 is exactly divisible
by 11.
e) In 999666; 9 + 9 + 6 = 24, 9 + 6 + 6 = 21 and 24 – 21 = 3, which is not divisible by 11.
So,999666 is not exactly divisible by 11.
3. Write all possible factors of these pair of numbers. Circle the common factors and select
the highest common factor.
a) 8 and 12 b) 12 and 18 c) 16 and 24 d) 28 and 42
Solution:
Here,
a) F8 = {1, 2, 4 , 8} and F12 = {1, 2, 3, 4 , 6, 12} Common factors लाई Circle लाउने
So, Highest common factor = 4
b) F12 = {1, 2, 3, 4, 6 , 12} and F18 = {1, 2, 3, 6 , 9, 18} Common factors लाई Circle लाउने
So, Highest common factor = 6
c) F16 = {1, 2, 4, 8 , 16} and F24 = {1, 2, 3, 4, 6, 8 , 12, 24} Common factors लाई Circle लाउने
So, Highest common factor = 8
d) F28 = {1, 2, 4, 7, 14 , 28} and F42 = {1, 2, 3, 6, 7,14 ,21, 42} Common factors लाई Circle लाउने
So, Highest common factor = 14
4. Let’s find the digit at blank space of the following numbers so that the number formed is
divisible by 11.
a) 27… 3 b) 1... 574 c) 939… d) 10… 57
Solution:
Here,
a) For the number 27… 3
Let x be the required digit. Then, sum of digits in the odd places = 2 + x
The sum of digits in the even places = 7 + 3 = 10
Now, 10 – (2 + x) = 0 [Since, the difference should be 0 or multiple of 11]
or, 8 – x = 0
or, x = 8
Hence, required digit is 8.
Vedanta Excel in Mathematics Teachers' Manual - 6 20
b) For the number 1 … 574
Let x be the required digit. Then, sum of digits in the odd places = 1 + 5 + 4 = 10
The sum of digits in the even places = x + 7
Now, 10 – (x + 7) = 0 [Since, the difference should be 0 or multiple of 11]
or, 10 – x – 7 = 0
or, x = 3
Hence, required digit is 3.
c) For the number 939 …
Let x be the required digit. Then, sum of digits in the odd places = 9 + 9 = 18
The sum of digits in the even places = 3 + x
Now,
18 – (3 + x) = 0 [Since, the difference should be 0 or multiple of 11]
or, 18 – 3 - x = 0
or, x = 15 which is not a digit
Also, 18 – (3 + x) = 11 [Since, the difference should be 0 or multiple of 11]
or, 18 – 3 - x = 11
or, 15 – x = 11
or, 15 – 11 = x
or, x = 4
Hence, required digit is 4.
d) For the number 10 … 57
Let x be the required digit. Then, sum of digits in the odd places = 1 + x + 7 = 8 + x
The sum of digits in the even places = 0 + 5 = 5
Now, (8 + x) – 5 = 0 [Since, the difference should be 0 or multiple of 11]
or, 3 + x = 0
or, x = - 3 which is impossible
Also, (8 + x) – 5 = 11 [Since, the difference should be 0 or multiple of 11]
or, 3 + x = 11
or, x = 8
Hence, required digit is 8.
Activity 3 Highest Common Divisor (H.C.F)
1. Recall factors of numbers.
2. Take a pair of numbers and factorize them. List out the common factors and select the
greatest common factor.
3. Make a discussion on H.C.F. and process of find the H.C.F.
4. Give real life examples based on H.C.F.
21 Vedanta Excel in Mathematics Teachers' Manual - 6
Solution selected problems from EXERCISE 3.2
1. Find the greatest number that divides 36 and 60 without leaving a remainder.
Solution:
Here, the required greatest number is the H.C.F. of 36 and 60.
2 36 2 60
2 18 2 30
39 3 15
3 5
36 = 2×2×3×3
60 = 2×2×3×5
Thus, H.C.F. = 2 × 2 × 3 = 12
Hence, the required greatest number is 12.
2. On the occasion of Bishwant’s birthday, he distributed 24 snickers and 36 cadburies
equally to his friends. Find the greatest number of his friends. How many snickers and
cadburies did everyone get?
Solution:
Here, the required greatest number of friends is the H.C.F. of 24 and 36.
2 24 2 36
2 12 2 18
26 39
33
24 = 2×2×2×3
36 = 2×2×3×3
Thus, H.C.F. = 2 × 2 × 3 = 12
Hence, the required greatest number of friends is 12.
Again; 24 ÷ 12 = 2. So, each friend got 2 snickers
36÷ 12 = 3. So, each friend got 3 cadburies.
3. A container can hold 40 l of milk and another container can hold 60 l of milk. What is the
maximum capacity of a jar that can be used to measure the milk of both containers?
Solution:
Here, the required maximum capacity of jar is the H.C.F. of 40 l and 60 l.
2 40 2 60
2 20 2 30
2 10 3 15
55
40 = 2×2×2×5
60 = 2×2×3×5
Thus, H.C.F. = 2 × 2 × 5 = 20
Hence, the required maximum capacity of the jar is 20 l which can be used to measure the milk
of both the containers.
Note: Please! Focus on Project Work
Vedanta Excel in Mathematics Teachers' Manual - 6 22
Activity 4 Lowest Common Multiples (L.C.M.)
1. Recall multiples of numbers.
2. Take a pair of numbers and write their first 10 multiples. List out the common multiples
and select the lowest common multiple.
3. Make a discussion on L.C.M. and process of find the L.C.M.
4. Give real life examples based on L.C.M.
Solution selected problems from EXERCISE 3.3
1. Find the least number which is exactly divisible by 18 and 24.
Solution:
Here, the required least number is the L.C.M. of 18 and 24.
2 18, 24
3 9, 12
3, 4
∴L.C.M. = 2 × 3 × 3 × 4 = 72
Hence, the required least number is 72.
2. Two measuring tapes are 20 m and 30 m long. Find the shortest length that can be measured
exactly by one of these tapes.
Solution:
Here, the required shortest length is the L.C.M. of 20m and 30 m.
2 20, 30
5 10, 15
2, 3
∴L.C.M. = 2 × 5 × 2 × 3 = 60
Hence, the required shortest length is 60 m.
3. On the occasion of Tihar, Deepshikha uses the decorative lights at her house. The red light
flickers in every 4 seconds and blue light flickers in every 6 seconds. If she switches on
the lights together, after what time will both coloured lights flicker together again?
Solution:
Here, the required least interval of time is the L.C.M. of 4 seconds and 6 seconds.
2 4, 6
2, 3
∴L.C.M. = 2 × 2 × 3 = 12
Hence, both the lights flicker together after 12 seconds.
4. Two bells ring at the interval of 30 and 40 minutes respectively. Find the least time interval
after which they ring at a time. If they ring at 9:00 a.m. at once, at what time will they ring
together again?
Solution:
23 Vedanta Excel in Mathematics Teachers' Manual - 6
Here, the required least time interval is the L.C.M. of 30 and 40 minutes.
2 30, 40
5 15, 20
3, 4
∴L.C.M. = 2 × 5 × 3 × 4 = 120
Hence, they ring at a time after120 minutes i.e., 2 hours.
If they ring at 9:00 a.m. at once, they will ring together again at 9:00 a.m. + 2 hours = 11:00 a.m.
5. Two TV movies starts at 8:00 p.m. One channel airs commercial breaks in every 12
minutes and the other channel airs commercials breaks in every 15 minutes. At what
time will both the channels start commercial breaks at the same time?
Solution:
Here, the required least time interval is the L.C.M. of 12 and 15 minutes.
3 12, 15
4, 5
∴L.C.M. = 3 × 4 × 5 = 60
Hence, they ring at a time after 60 minutes i.e., 1 hour.
Both TV starts commercial breaks together at 8:00 pm + 1 hr = 9:00 pm.
6. A spider has 8 legs and an ant has 6 legs. There are a group of spiders and a group of ants.
If both groups have equal number of legs, find the least number of spiders and ants in
each group.
Solution:
Here, the required least number of legs in each group is the L.C.M. of 8 and 6.
2 8, 6
4, 3
∴L.C.M. = 2 × 4 × 3 = 24
So, the least number of legs in each group is 24.
Again; 24÷8 = 3. So, there are 3 spiders. 24÷6 = 4. So, there are 4 ants.
Note: Please! Focus on Project Work
Activity 5 Perfect square and Square root
1. Arrange the students in a square form so that each row and column has 2/2 students, 3/3
students and so on. Then, discuss about the number of students required.
2. Using the multiplication table, ask the students about the product of 2×2, 3×3, 4×4, 5×5,
6×6, … so on.
By using square patterns of dots, discuss about the perfect square numbers.
3. Make a discussion, about the square roots as one of the two identical factors of the number.
For example, 4 is the square of 2 and 2 is the square root of 4, 9 is the square of 3 and 3 is
the square root of 9, 16 is the square of 4 and 4 is the square root of 16, 25 is the square of
5 and 5 is the square root of 25, …, 100 is the square of 10 and 10 is the square root of 100
etc.
4. Introduce the sign of square root.
Vedanta Excel in Mathematics Teachers' Manual - 6 24
Activity 6 Cube numbers and Cube roots
1. Recall the formula to find the volume of a cube,
2. Give more examples of cube numbers,
1×1×1 = 13 = 1→1 is the cube of 1 and cube root of 1 is 1, 2×2×2 = 23 = 8 → 8 is the cube
of 2 and 2 is the cube root of 8, 3×3×3 = 33 = 27 → 27 is the cube of 3 and 3 is the cube
root of 27 and so on.
3. Introduce the sign of cube root.
Solution selected problems from EXERCISE 3.4
1. 25/25 plants are planted along the length and breadth of a square garden. How many
plants are there in the garden?
Solution:
Here, total number of plants in the garden = 252 = 25×25 = 625
2. There are 45 students in class VI. If every student donates the same amount of money as
their number to flood victims, how much money do they donate altogether?
Solution:
Here, number of students = 45, amount of money donated by each student = Rs 45
Now, total amount of money donated by students = Rs 452 = Rs 45×45 = Rs 2025
3. In an afforestation programme, every student planted as many plants as their number. If
they plated 576 plants altogether, how many students took part in the programme?
Solution:
Here, number of students is the square root of 576.
2 576
2 288
2 144
2 72
2 36
2 18
3 9
3
∴576 = 2×2×2×2×2×2×3×3 = 22 ×22 ×22 ×32
∴ 576 = 2 × 2 × 2 × 3 = 24
Hence, 24 students took part in the programme.
25 Vedanta Excel in Mathematics Teachers' Manual - 6
Extra questions
1. Test whether the number 112 is exactly divisible by 7 or not. [Ans: Yes]
2. Test whether the number 1331 is exactly divisible by 11 or not. [Ans: Yes]
3. On the occasion of Ojaswi’s birthday, she distributed 36 snickers and 54 cadburies equally
to his friends. Find the greatest number of his friends. How many snickers and cadburies
did everyone get? [Ans: 18, 2, 3]
4. A group of youths want to 45 sweaters and 60 trousers to a certain number of children
equally. Find the greatest number of children. Also, find the share of each item among
them. [Ans: 15, 3, 4]
5. Two bells ring at the interval of 20 and 30 minutes respectively. Find the least time interval
after which they ring at a time. If they ring at 11:00 a.m. at once, at what time will they ring
together again? [Ans: 12:00 noon]
6. All the students of a school are assembling in 25 rows and 25 columns in the play-ground.
Find the total number of students. [Ans: 625]
7. The length of a square garden is 30 ft; find its area. [Ans: 900 sq. ft]
Vedanta Excel in Mathematics Teachers' Manual - 6 26
Unit Integers
4
Allocated teaching periods: 4
Competencies
- To write the integers
- To compare the integers
- To perform operations on integers
Level-wise learning objectives
S.N. Levels Objectives
1. Knowledge (K) - To define the set of integers
- To compare the integers
2. Understanding (U) - To perform the operations (addition, subtraction,
multiplication and division) of integers
3. Application (A) - To solve the problems based on integers
4. High Ability (HA) - To link various real life/ contemporary problems with
integers
Required Teaching Materials/ Resources
Number line, graph board, ICT tools etc.
Pre-knowledge: natural/whole numbers
Teaching Activities
Activity 1 Integers
1. Recall the set of natural numbers, set of whole numbers.
2. Give some real life examples based on positive and negative numbers (integers).
3. Show the integers on the number line.
4. Give more examples of operations on integers using number line.
5. List out the signs rules of multiplication or division of integers.
Solution selected problems from EXERCISE 4.1
1. Let’s tell and write the correct answers in the blank spaces.
a) The positive integers lie towards right from 0 (zero) in a number line.
b) The negative integers lie towards left from 0 (zero) in a number line.
27 Vedanta Excel in Mathematics Teachers' Manual - 6
c) Integers between -3 and +3 are -2, -1, 0, 1 and 2.
d) Integers greater than – 2 and less than +1 are -1, and 1.
e) The ascending order of the integers 2, -3, 0, -1, 4 is -3, -1, 0, 2, 4
f) The descending order of the integers 0, -2, 3, -1, 2, -3, 1 is 3, 2, 1, 0, -1, -2, -3
2. Let’s tell and write True or False for the following statements.
a) 0 is less than every positive integer. True
b) 0 is s positive integer. False
c) 0 is the smallest integer. False
d) -1 is the greatest negative integer. True
e) The opposite of -4 is 4. True
f) The opposite of +6 is +9. False
g) -5 is greater than 0. False
h) The sum of an integer and its opposite is always zero. True
i) The sum of two negative integers is positive. False
j) The product of two negative integers is negative. False
3. Mrs. Yadav walks 18 km due North and then 12 km due South. Find her position with
respect to her starting point.
Solution:
Here, 18 km – 12 km = 6 km North from her starting point.
4. Mr. Chamling travels 20 km due East and then 25 km due West. Find his position from
the starting point.
Solution:
Here, 25 km – 20 km = 6 km North from her starting point.
5. Sunayana walks 8 km due East, then turns rounds and goes 4 km due West of starting
point. Then, she again turns back and returns to the starting point. What is the total
distance travelled by her?
Solution:
Here, total distance travelled by Sunayana = 8 km + (8km + 4km) + 4km = 24 km.
Vedanta Excel in Mathematics Teachers' Manual - 6 28
6. The temperature of a body first rises by 20° C and then falls by 24° C. Find the final
temperature of the body, if its initial temperature is 10° C.
Solution:
Final temperature = 10°+ (+ 20°) + (– 24°) = 30° – 24°= 6° C.
7. The temperature of a body first rise by 25° C and then falls by 30° C. Find the final
temperature of the body, if its initial temperature is (i) 8°C. (ii) 2°C (iii) 0°C. (iv) – 3° C
Solution:
(i) Final temperature = 8°+ (+ 25°) + (– 30°) = 33° – 30°= 3° C.
(ii) Final temperature = 2°+ (+ 25°) + (– 30°) = 33° – 30°= 3° C.
(iii) Final temperature = 0°+ (+ 25°) + (– 30°) = 25° – 30°= -5° C.
(iv) Final temperature = -3°+ (+ 25°) + (– 30°) = 25° – 33°= -8° C.
8. The temperature of a body first falls by 15° C and then rises by 20°C. Find the final
temperature of the body, if its initial temperature was:
(i) 28° C (ii) 12° C (iii) 5° C. (iv) – 2°C
Solution:
(i) Final temperature = 28°+ (-15°) + (20°) = 48° – 15°= 13° C.
(ii) Final temperature = 12°+ (-15°) + (20°) = 32° – 15°= 17° C.
(iii) Final temperature = 5°+ (-15°) + (20°) = 25° – 15°= 10° C.
(iv) Final temperature = -2°+ (-15°) + (20°) = 20° – 17°= 3° C.
29 Vedanta Excel in Mathematics Teachers' Manual - 6
Unit Fraction and Decimals
5
Allocated teaching periods: 20
Competencies
- To compare the functions
- To perform addition, subtraction, multiplication and division of fractions
- To convert decimal into fraction and vice-versa.
- To perform addition, subtraction, multiplication and division of decimal numbers.
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To define function
1. Knowledge (K) - To identify proper, improper fraction and mixed number
- To tell the fractions equivalent to the given fraction.
- To tell the fraction for a decimal
- To find the equivalent fractions of the given fraction
2. Understanding (U) - To find the sum or difference of the fractions
- To multiply or divide the fractions
- To perform addition, subtraction, multiplication and
division of decimal numbers.
- To round off decimal numbers
- To solve the word problems based on comparison of fractions
- To solve the problem related to fundamental operations on
3. Application (A) fractions
- To solve the problems related to decimal numbers.
- To prepare the models to show the equivalent fractions using
4. Higher Ability (HA) rectangular / circular card paper and share in the class.
Required Teaching Materials/ Resources
Rectangular or circular cardboard paper, fraction charts, scissors, blocks, ICT tools like Geo-
Gebra etc.
Pre-knowledge: Proper, improper fractions and mixed number.
Teaching Activities
Activity 1 Equivalent fractions
1. Recall proper, improper fractions and
mixed numbers.
2. (i) Take some rectangular chard papers.
Divide it into equal parts and colour one
part and ask the fraction to represent the
shaded portion.
Vedanta Excel in Mathematics Teachers' Manual - 6 30
(ii) Take another rectangular chart paper as (i) and with the help of a ruler, draw a
horizontal straight line that divides the rectangular into four equal parts and ask the
fraction to represent the shaded portion.
(iii) Similarly, continue the same process and collect the fractions representing for the
shaded portions as 1 , 2 , 3 etc.
2 4 6
3. (i) Take a few numbers of rectangular chart paper
and divide it into three equal parts, colour two of
the three parts and express into fraction.
(ii) Take a rectangular chart as coloured in (i) and
divide it into 6 equal parts as shown in the figure
alongside and express into fraction.
(iii) Take a rectangular chart as coloured in (i) and divide it into 12 equal parts as shown
in the figure alongside and express into fraction.
(iv) Collect the fractions and discuss upon the equivalent fractions.
4. Discuss about the process of finding the equivalent fractions.
5. Take a pair of equivalent fractions and investigate the idea on testing equivalent fractions
oeqr unaolt..SFiomr ielxaarlmy,ptlaek,e13aannodth62erapftaeirr cross multiplication, 1×6 = 6 and 2×3 = 6, which are
of equivalent fractions and find the product after cross
multiplication. Take a pair of fractions which are not equivalent and find the product after
cross multiplication. With such examples, make a conclusion.
Activity 2 Comparison of like and unlike fractions
1. With examples, discuss about the like and unlike fractions.
2. With examples, discuss about the like and unlike fractions and discuss upon the method
of conversion of unlike fractions into like by taking L.C.M. of denominators.
To convert unlike fractions, we follow the following process.
- Find the L.C.M. of denominators.
- Divide L.C.M. by each denominator
- Multiply the numerator and denominator both by the quotient.
31 Vedanta Excel in Mathematics Teachers' Manual - 6
3. Show a few examples of like fractions as shown in the figure below and ask to compare
them.
(a) b)
125 < 175 < 1153 2 < 7 < 13
15 15 15
Note:
(i) To compare like fractions, we compare the numerators. The greater fraction has larger
numerator and the smaller fraction has smaller numerator.
(ii) To compare unlike fractions, first, we should convert unlike fractions into like and compare
the numerators of like fractions.
(iii) Short-cut trick to compare unlike fractions:
Find the cross-products of numerators and denominators then we compare the products.
The fractions along which the product is greater is larger.
3 78.
For example, in the fractions 5 and The product of 3 and 8 = 24 and the product of 5 and
7 = 35. Since, 24<35. Hence, 3 < 7
5 8
Solution of selected questions from EXERCISE 5.1
1. a) In class - VI of a school, there are 15 girls and 20 boys. 9 students are wearing
spectacles, 11 girls are wearing ribbons and 7 boys are wearing caps.
(i) What fraction of the students are girls and boys?
(ii) What fraction of the students are wearing spectacles?
(iii) What fractions of the girls are wearing ribbons and boys are wearing cap?
Solution: 15 20
35 35
(i) The fraction of girls = and the fraction of boys=
(ii) The fraction of students who are wearing spectacles = 9
35
11
(iii) The fraction of girls who are wearing ribbons = 15
The fraction of boys who are wearing cap = 7
20
b) Mrs. Maharjan bought one dozen of bananas and ate 5 of them.
(i) what fraction of the bananas was eaten?
(ii) What fraction of the bananas was left?
Vedanta Excel in Mathematics Teachers' Manual - 6 32
Solution: 5
12
(i) The fraction of bananas which was eaten =
(ii) The fraction of bananas which was left = 7
12
2. Find the first three fractions equivalent to the following fractions by multiplication
process.
Solution:
(i) 21 × 2 = 2 , 1 × 3 = 3 , 1 × 4 = 4 .
× 2 4 2 × 3 6 2 × 4 8
So, 2 , 3 and 4 are the equivalent fractions of 1 .
4 6 8 2
(ii) 23 ×× 2 = 4 , 2 × 3 = 6 , 2 ×4 = 182 .
2 6 3 × 3 9 3 ×4
So, 4 , 6 and 8 are the equivalent fractions of 2 .
6 9 12 3
(iii) 43 ×× 2 = 6 , 3 × 3 = 9 , 3 ×4 = 12 .
2 8 4 × 3 12 4 ×4 16
So, 6 , 9 and 12 are the equivalent fractions of 3 .
8 12 16 4
(iv) 98 ×× 2 = 16 , 8 × 3 = 24 , 8 ×4 = 32 .
2 18 9 × 3 27 9 ×4 36
So, 16 , 24 and 4 are the equivalent fractions of 8 .
18 27 8 9
3. Find the first three fractions equivalent to the following fractions by division process.
Solution:
(i) 162÷÷22 = 3 , 6÷3 = 2 , 6÷6 = 1 .
6 12 ÷ 3 4 12 ÷ 6 2
So, 3 , 2 and 1 are the equivalent fractions of 6 .
6 4 2 12
(ii) 186÷÷22 = 4 , 6÷4 = 2 , 8÷8 = 1 .
8 16 ÷ 4 4 16 ÷ 8 2
So, 4 , 2 and 1 are the equivalent fractions of 8 .
8 4 2 16
(iii) 1284 ÷÷ 2 = 9 , 18 ÷ 3 = 6 , 18 ÷ 6 = 3 .
2 12 24 ÷ 3 8 24 ÷ 6 4
So, 9 , 6 and 3 are the equivalent fractions of 18 .
12 8 4 24
4. Test whether the following pairs of fractions are equivalent.
Solution:
(i) The product of 1 and 6 = 6 and the product of 3 and 2 = 6. Since,1×6 = 3×2.
So, 1 and 2 are equivalent fractions.
3 6
33 Vedanta Excel in Mathematics Teachers' Manual - 6
(ii) The product of 2 and 7 = 14 and the product of 5 and 3 = 16. Since,2×7 ≠ 5×3. So,
2 and 3 are not equivalent fractions.
5 7
5. Convert the following pairs of fractions into like fractions.
Solution:
(i) The L.C.M. of denominators 2 and 3 is 6.
Now, 1 = 1 × 3 = 3 and 2 = 2 × 2 = 4 . So, 3 and 3 are like fractions.
2 2 3 6 3 3 2 6 6 6
(ii) The L.C.M. of denominators 3 and 6 is 6.
Now, 2 = 2 × 2 = 4 and 5 . So, 4 and 5 are like fractions.
3 3 2 6 6 6 6
6. Arrange the following fractions in ascending order.
Solution:
(i) The L.C.M. of denominators 2, 4 and 8 is 8
Now, 1 = 1 × 4 = 4 and 3 = 3 × 2 = 6 and 5 .
2 2 4 8 4 4 2 8 8
Since, 4<5<6. So, 4 < 5 < 86. i.e. 12, < 5 < 3
8 8 8 4
(ii) The L.C.M. of denominators 3, 6 and 9 is 18
Now, 2 = 2 × 6 = 12 , 5 = 5 × 3 = 15 and 7 = 7 × 2 = 14
3 3 6 18 6 6 3 18 9 9 2 18
Since, 12<14<15. So, . i.e., 12 < 14 < 15 . i.e., 2 < 7 < 5
18 18 18 3 9 6
7. Father spends 3 parts of his monthly income on food and 1 parts in your education. On
4 8
which heading does he spend more money?
Solution:
Here, expenditure on food = 3 and the expenditure on education = 18.
4
The L.C.M. of denominators 4 and 8 is 8
Now, expenditure >on81f.oHoden=ce43, = 3 × 2 = 6 and expenditure on education =18
he 4 2 8
6
Since, 6>1. So, 8 spends more on food.
8. Sunita travelled 3 parts of distance by bus and 5 parts by taxi when she was going to
7 14
airport. By which vehicle did she travel more?
Solution:
Here, distance travelled by bus = 3 and the distance travelled by taxi = 5 .
7 14
The L.C.M. of denominators 7 and 14 is 14
Now, distance travelled by bus = 3 = 3 × 2 = 6 and distance travelled by taxi = 5
7 7 2 14 14
6 5
Since, 6>5. So, 14 > 14 . Hence, Sunita travelled more distance by bus.
9. Dhaniya can do 2 parts of a work in 1 day and Bhuniya can do 3 parts of the work in 1
3 4
day. Who can do more work in 1 day?
Vedanta Excel in Mathematics Teachers' Manual - 6 34
Solution:
Here,
In 1 day, Dhaniya can do 2 parts of a work and Bhuniya can do 3 parts of the work
3 4
The L.C.M. of denominators 3 and 4 is 12
Now, 2 = 2 × 4 = 8 and 3 = 3 × 3 = 9 Since, 9>8. So, 9 > 8 .
3 3 4 12 4 4 3 12 12 12
Hence, Bhuniya can do more work in 1 day.
10. A man gave 2 parts of his property to his wife, 3 parts to his son and 1 parts to his
5 10 6
daughter. To whom did he give more property? Find।
Solution:
Here,
The L.C.M. of denominators 5, 10 and 6 is 30.
Now, 2 = 2 × 6 =So31, 0213,20 3 = 1>3035×0 3 = 9 and 1 = 1 × 5 = 5
5 5 6 10 9 3 30 6 6 5 30
> 30
Since, 12>9>5.
Hence, the man gave more property to his wife.
Activity 2 Addition of fractions
1. Recall proper, improper fractions and mixed numbers with examples.
2. Recall like and unlike fractions with examples.
3. Take an example: For, take a bread and divide it into 5 parts. Ram eats 1 part of the breads
and Sita eats 2 parts of it. Make a discussion with students.
(i) What fraction of bread is eaten by Ram?
(ii) What fraction of the bread is eaten by Sita?
(iii) What fraction of bread is eaten by by them?
4. If possible, show the addition of like fractions by using Geo-Gebra or live worksheets.
5. Recall equivalent fractions and the process of converting unlike fractions into like. Then,
make a discussion on addition of unlike fractions.
(i)
(ii)
35 Vedanta Excel in Mathematics Teachers' Manual - 6
6. Encourage the students to add the fractions from exercise given in the textbook.
Activity 3 Subtraction of fractions
1. Recall proper, improper fractions and mixed numbers with examples.
2. With examples, discuss upon the subtraction of like fractions.
For example: 4 – 1 = 3
5 5 5
3. If possible, show the addition of like fractions by using Geo-Gebra or live worksheets.
4. Recall equivalent fractions and the process of converting unlike fractions into like. Then,
make a discussion on subtraction of unlike fractions.
5. With discussion, give classwork to fill up the sum and difference of fractions from the
worksheets given in the textbook.
6. Encourage the students to add the fractions from exercise given in the textbook.
Solution of selected questions from EXERCISE 5.3
1. Samriddhi bought a pencil for Rs 545 and an eraser for Rs 354 . What is the total cost of
both the articles? Find.
Solution:
Here, total cost of pencil and eraser = Rs (545 + 3170)
= 29 + 37 = 29 × 2 + 37 × 1
5 10 10
= 58 + 37 = 95 = 19 = 921
10 10 2
Hence, the total cost of the pencil and eraser i6s65Rksm921b. y bus to reach his uncle’s home. What
2. Kamal travelled 341 km by tempo and
distance did he travel altogether?
Solution:
Here, total distance travelled by tempo and bus = (314 + 656) km
= 13 + 41 = 13 × 3 + 41 × 2
4 6 12
= 39 + 82 = 121 = 10112
Hence, he travelled 10112 km altogether. 12 12
3. The weight of Manisha is 2512 kg and that of her school bag with stationery is 241 kg. What
will be the total weight while she is carrying her school bag?
Solution:
Here, total weight of Manisha while carrying bag==52(125+12 49+=2415) 1kg×
2 + 9 × 1
4
= 102 + 9 = 111 = 2734
4 4
Vedanta Excel in Mathematics Teachers' Manual - 6 36
Hence, total weight of Manisha while carrying bag is 2743 kg. new school’s dress. If 143 m of
4. Lakhan needs 341 m of cloth for a pants and a shirt of his
cloth is required for the pants, how much cloth is required for a shirt?
Solution:
Here, cloth required for a shirt = cloth required for a pants and a shirt – cloth required for a
pants = 314 – 134 = 13 – 7 = 6 = 3 = 112
4 4 4 2
Hence, Lakhan needs 121 m for a shirt.
5. The weight of empty gas cylinder is 1654 kg and that of the cylinder filled with gas is 31
filled in the cylinder?
7 kg. What is the weight of the gas
15
Solution:
Here, weight of gas filled in cylinder = 31175 – 1654 = 472 – 84
15 5
= 472 × 1– 84 × 3 = 220 = 44 = 1432
Hence, the weight is 1432 kg. 15 15 3
of empty cylinder
6. A man gives 3 parts of the property to his son, 5 parts to his daughter and 1 parts to a
8 12 6
charity. What parts of the property are left with him?
Solution:
Here, parts of property given to son, daughter and charity = 3 + 5 + 1
8 12 6
= 3 × 3 + 5 ×2 + 1 × 4
24
= 9 + 10 + 4 = 23
24 24
23 1
Hence, the property left with him =1 – 24 = 24
7. The distance between two places is 15 km. A man travelled 841 km by a taxi 581 km by a
bus and the remaining distance he walked. How many kilometre did he walk?
Solution:
Here, distance travelled by bus and taxi = 841 + 518 = 33 + 41
4 8
= 33 × 2 + 41 × 1 = 66 + 41 = 107
8 8 8
37 Vedanta Excel in Mathematics Teachers' Manual - 6
Hence, remaining distance he walked =15 – 107 = 120 – 107 = 13 = 185 km.
8 8 8
Activity 4 Multiplication of fractions
(i) Multiplication of a fraction by a whole number
1. Recall multiplication as the recurring addition of numbers.
2. Give the proper examples of multiplication of a fraction by a whole number.
a)
b)
3. Note that product of a whole number and a fraction is equal to “product of whole
number and numerator divided by denominator.”
For example: 4 × 2 = 8 = 171, 3 × 1 = 83, 5 × 2 = 10 = 2 etc.
7 7 8 15 15 3
(ii) Multiplication of a fraction by another fraction
1. With examples, discuss on investigating the idea about the multiplication of fraction
by another fraction.
Example:
a) Half of a half circle: → 1 × 1 = 1× 1 = 1
2 2 2× 2 4
b) Two thirds of a quarter rectangle: → 2 × 1 = 2 × 1 = 2
3 4 3 × 4 12
2. Note that product of a fraction and another fraction is equal to “product of numerators
divided by denominators.”
Activity 5 Division of fractions
(i) Division of a whole number by a fraction
1. Describe about the reciprocal of
fractions with examples.
2. Ask a few examples of real life
problem, for example, how many group of halves are in 6 whole apples?
3. Draw the models of division of a whole number by a fraction and discuss about these
questions.
a) How many halves are there in 1 rectangle?
Vedanta Excel in Mathematics Teachers' Manual - 6 38
b) How many halves are there in 2 rectangles?
c) How many thirds are there in 3 rectangles?
4. Under discussion, write the answers of above questions as follows.
1
a) 1 ÷ 2 = 2 → 1 × 2 = 2
b) 2 ÷ 1 = 4 → 2 × 2 = 4
2
1
c) 3 ÷ 2 = 9 → 3 × 3 = 9
5. Note that division of a whole number by a fraction is equal to “product of whole
number and the reciprocal of the fraction.”
(ii) Division of a fraction by a whole number
1. Let the model of fraction and divide it by a whole number.
(i) Take a half of circle and divide it into two
equal parts. Discuss about the result so
obtained. Obviously, 1 ÷ 2 = 1
2 4
(ii) Take one third of a square. Divide into 5
parts. Discuss upon the resulting fraction.
2. Discussing on the related examples,
make a conclusion of dividing a fraction
by a whole number.
3. Draw the models of division of a whole
number
(iii) Division of a fraction by a fraction
1. Let the model of fraction and divide it by a fraction.
(i) Discuss on the question- “How many groups of quarters are there in one half?”
→a half = 1
2
→1 quarter= 1
4
→2 quarters →There are 2 quarters in 1 half.
Solution of selected questions from EXERCISE 5.4
1. Let’s complete multiplication from the given shaded diagrams
Solution:
a) = 2 × 1 = 1
2
39 Vedanta Excel in Mathematics Teachers' Manual - 6
b) = 3 × 12==332×=231=21 2
c)
d) = 5 × 3 = 15 = 334
4 4
2. Let’s complete multiplication from the given shaded diagrams.
Solution:
a) = 1 of 1 = 1 × 1 = 1
b) 3 2 3 2 6
= 1 of 1 = 1 × 1 = 1
4 3 4 3 12
3. Let’s simplify
Solution:
a) 21 = 1 ÷ 2 = 1 × 3 = 3 = 112
3 2 2
3
b) 84 8 3 3 121
= 4 ÷ 3 = 4 × 8 = 2 =
3
2
c) 43 2 2 1 1
= 3 ÷ 4 = 3 × 4 = 6
5
d) 160 5 10 5 18 3 112
= 6 ÷ 18 = 6 ÷ 10 = 2 =
18
4. Let’s simplify
Solution:
a) 1 ÷ 3 of 2 = 1 ÷ 3 × 2 = 1 ÷ 10 = 1 × 10 = 5 = 132
2 4 5 2 4 × 5 2 3 2 3 3
b) 2 ÷ 5 of 3 = 2 ÷ 5×3 = 2 ÷ 1 = 2 × 4 = 8 = 232
3 6 10 3 6 × 10 3 4 3 1 3
c) 3 of 2 ÷ 8 = 3 × 2 × 21 = 3 ÷ 21 = 9
4 7 21 4 × 7 8 14 8 16
d) 1 of 227 ÷ 3 = 1 of 16 ÷ 35= 1 × 16 × 5 = 4 ÷ 5 = 20
4 5 4 7 4×7 3 7 3 21
5. a) If the cost of 1 kg of rice is Rs 8521 , find the cost of 10 kg of rice.
Solution:
Here, the cost of 1 kg of rice is Rs 8512 = Rs 171
2
The cost of 10 kg of rice = 10 × Rs 171 = Rs 855
2
Hence, the cost of 10 kg of rice is Rs 855.
b) A car can travel 134 km in 1 minute. How many kilometres does it travel in 12 minutes
Vedanta Excel in Mathematics Teachers' Manual - 6 40
with the same speed?
Solution:
Here, in 1 minute, a car can travel 134 km.
In 12 minutes, the car can travel 12×134 km = 12× 7 km = 21 km
4
Hence, in 12 minutes, the car can travel 21 km.
6. a) Teacher cut one-third part of a chart paper and she gave it to her student to colour
half of this part. What fraction of the whole chart paper did the student colour?
Solution:
Here, the fraction of coloured paper = 1 of 1 = 1 × 1 = 1
2 3 2 3 6
b) Mother gives half of a whole bread to brother. Brother divides this half bread into three
equal parts and he gives one part to his sister. (i) What fraction of the whole bread
does sister get? (ii) What fraction of the whole bread is left with brother?
Solution:
Here, (i) The fraction of bread given to sister = 1 of 1 = 1 × 1 = 1
3 2 3 2 6
1 5
(ii) The fraction of bread left with brother= 1 – 6 = 6
7. a) Anamol can do 1 part of a work in 1 day. How much work does he do in 3 days?
Solution: 9
Here, in 1 Mhdearysc,.aAMnnadahoma3orjl×acna91nc=adno31d61opp16aarptrtaoorftftohafewawoworkork.r.k in 1 day. She worked for 2 days and left. The
In 3 days,
b)
remaining parts of the work is done by Mrs. Rai.
(i) How much work did Mrs. Maharjan do in 2 days?
(ii) How much work was left to do?
(iii) How much work was done by Mrs. Rai?
Solution:
(i) In 1 day, Mrs. Maharjan can do 1 part of a work.
6
In 2 days, she can do 2 × 1 = 1 part of the work.
6 3
(ii) Remaining part of work = 1 – 1 = 2
3 3
(iii) The part of work done by Mrs. Rai = remaining part of the work = 2
4. a) How many jars each of capacity 212 litres are 3
of milk?
needed to empty a vessel of 50 litres
Solution:
Here, required number of jars = 50÷212 =50÷52 = 50 × 1 = 20
2
b) A rope is 40 m long. How many 313 metre pieces can be cut from the rope?
Solution:
Here, required number of pieces = 40÷331 = 40 ÷130 =40 × 3 = 12
10
41 Vedanta Excel in Mathematics Teachers' Manual - 6
5. a) In the Polymerise Chain Reaction (PCR) test among 7,290 suspected people 3 of
10
them were found having Corona positive. Find the number of people who have
Corona positive.
Solution:
Here, the number of people who have Corona positive = 3 of 7,290 = 2,187
10
b) Mr. Shakya earns Rs 18,000 in a month. He spends 5 parts of his earning to run his
family. 9
(i) How much money does he spend to run his family?
(ii) How much money does he save in a month?
Solution:
(i) The monthly expense to run family = 5 of Rs 18,000= Rs 10,000
9
(ii) Monthly saving = Rs 18,000 – Rs 10,000 = Rs 8,000
c) Father earns Rs 20,000 in a month. He spends 1 part of her earning on food and 2
parts on the education of his children. 4 5
(i) How much money does he spend on food?
(ii) How much money does he spend on education?
(iii) What is his total expenditure in a month?
(iv) How much money does he save in a month?
Solution:
(i) The monthly expense on feodoudca=tio41no=f R25s 20,000= Rs 5,000
(ii) The monthly expense on of Rs 20,000= Rs 8,000
(iii) Total monthly expenses = Rs 5,000 + Rs 8,000 = Rs 13,000
(iv) Total monthly saving = Rs 20,000 – Rs 13,000 = Rs 7,000
d) The distance from Dhanghadi to Mahendranagar is 60 km. Mr Chaudhari travelled
1 part of the distance by motorbike, 3 parts of the distance by taxi and the remaining
3 5
distance he walked. How many kilomertres did he walk?
Solution:
The distance from Dhanghadhi to Mahendranagar = 60 km
Now, the distance travelled by motorbike = 1 of 60 km = 20 km
3 36 km
The distance travelled by taxi= 3 of 60 km =
5
So, the distance travelled by walking = 60 – (20 + 36) = 60 – 56 = 4 km.
6. Divide: 2 2 4 4 1 7 7 141
3 3 9 9 4 8 8
a) 6 ÷ and ÷ 6 b) 24 ÷ and ÷ 24 c) 1 ÷ 1 and 1 ÷ .
Look at the quotients in every pair of division. What do you notice? Discuss with
your friends.
d) Perform the operations as given in each pair of numbers.
(i) 412 – 3 and 4 1 ÷ 3 (ii) 531 – 4 and 5 1 ÷ 4 (iii) 614 – 5 and 641 ÷ 5.
2 3
Are the difference and quotient of numbers in each pair equal? Find.
Solution:
Vedanta Excel in Mathematics Teachers' Manual - 6 42
a) 126441÷÷÷32941=78=6=2×445×23÷49=1=895 a5=n4da54n23×d÷491856÷==243223=×an9416d×==219141 = 1 = 15 ÷ 5 = 15 ÷ 4 = 3
b) 7 ÷54114 8 4 8 5 2
c) 8
From the above pair of division, we observed that the quotient becomes reciprocal to each
other if we interchange the dividend and divisor.
d) (i) 412 –3= 9 –3 = 9–6 = 3 = 121 and 412 ÷3= 9 × 1 = 3 = 121
2 2 2 2 3 2
(ii) 513 – 4 = 16 – 4 = 16 – 12 = 4 = 1 1 and 531 ÷ 4 = 16 × 1 = 4 = 131
3 3 3 3 3 4 3
(iii) 641 – 5 = 25 – 5 = 25 – 20 = 5 = 1 1 and 641 ÷ 5 = 25 × 1 = 5 = 114
4 4 4 4 4 5 4
Yes, we observed that the difference and quotients are equal.
7. a) Out of 30 students in a class, two-third were successful to get A+ grade in Maths
test. Among these successful students, one-quarter got A+ grade even in science
also.
(i) What fraction of the successful students got A+ grade in science?
(ii) How many students were successful to get A+ grade in science?
(iii) How many students got other than A+ grade in these two subjects?
Solution:
Here,
(i) The fraction of the successful students who got A+ grade Science = 1 of 2 =61
4 3
(ii) The number of students who were successful to get A+ grade Science= 1 of 30 =5
(iii) The number of students who 6
2 T
got A+ grade in Maths = 3 of 30 =20
M
The number of students who got other than A+ grade in these two S
subjects=30 – 20 = 10 15 5
(For clarity, look at the diagram given alongside)
10
b) Out of 40 students of class VI, three-fifth of the students participated in various games on
sport day. Among the participants, 1 took part in high jump.
4
(i) What fraction of the students took part in high jump?
(ii) How many students took part in high jump?
(iii) How many students did not take part in any games?
Solution:
Here,
(i) The fraction of tsshttuueddseetunndttssewwnthhsoowtthoooookktoppoaakrrttpiiannrthviaingrhihoiujugshmgjpaumm=eps23=0=o1453f of 3 = 3
(ii) The number of 40 5 20
(iii) The number of of
=6
40 =24
So, number of students who did not take part in any games = 40 – 24 = 16.
43 Vedanta Excel in Mathematics Teachers' Manual - 6
Activity 6 Decimals
1. Give a few examples of the fractions like whose denominators are 10 or the power of 10. For
eMxaamkepaled,i1s7c0u, s1s10i3o0n, o1n2007c0onevtce.rsSiuocnhoffrfarcatcitoinons sarheacvainllgeddetnhoemdeincaimtoarls
2. decimals. fractions. of 10 as the
10 or power
a) Conversion of Fractions into decimals
(i) Fractions with denominators 10 or power of 10
Present the following models and discuss about the fraction and decimals which are
represented by the shaded number of grids to the whole.
Discuss about the process of converting the fractions with denominators 10 or power
of 10.
The decimal point is shifted as many number of digits to the left as there are zeros in
power of 10. For example: 2 = 0,2, 67 = 0.67, 3 = 0.003 etc.
10 100 1000
(ii) Fractions having denominators any other natural numbers
In this case, the numerator is directly divided by the denominator to get decimal
number. For example: 1 = 1 × 5 = 5 = 0.5 or directly divide 1 by 2 and write 0.5,
2 2 × 5 10
25
8 = 3.125 etc.
b) Conversion of decimals into fraction
Take a few number of fractions and count the number of digits after the decimal point
under discussion.
(i) If there is only one digit after the decimal point, remove the decimal and write 10 in
the denominator of the required fraction. For example: 0.3 = 3
10
(ii) If there are two digits after the decimal point, remove the decimal and write 100 in
the denominator of the required fraction. For example: 0.47 = 47
100
(iii) If there are three digits after the decimal point, remove the decimal and write 1000 in
the denominator of the required fraction. For example: 0.239 = 239 nd so on.
1000
3. Recall the places of digits in a whole number. Similarly, take a few examples of fractions.
3 24
In 10 = 0.3, 3 is in tenth place, in 100 = 0.24, 2 is in tenth place and 4 is in hundredths
place. Thus, tenths, hundredths, thousandths, etc. are the places of decimal numbers.
Vedanta Excel in Mathematics Teachers' Manual - 6 44
Activity 7 Addition and Subtraction of Decimals
1. With the following models, discuss about the following questions.
Model 1:
(i) The decimal represented by green grids is ….
(ii) The decimal represented by blue grids is ….
(iii) The decimal represented by coloured grids is ….
Model 2:
2. Discuss about the calculation process of addition and subtraction of decimals.
(i) Arrange the decimal numbers in such a way that the digits at the same places should
lie in the same column.
(ii) To make the equal decimal places, write as many zeros as required at the end of the
decimal numbers.
(iii) Perform the addition and subtraction operations
3. Focus on project work and students’ activities as given in textbooks.
Solution of selected questions from EXERCISE 5.6
1. a) Find the difference between 8.72 and 0.872.
Solution:
Here, the difference between 8.72 and 0.872 is 8.72 – 0.872
= 8.720 – 0.872 = 7.848
b) What should be added to 64.35 to make 100?
Solution:
Here, the required number to be added = 100 – 64.35
= 100.00 – 64.35
= 35.65
c) What should be subtracted from 55.55 to get 5.555?
Solution:
Here, the required number to be subtracted = 55.55 – 5.555
= 55.550 – 5.555
= 49.995
2. a) On your birthday, you bought a cake for Rs 250.25, and a packet of sweets for
Rs 145.90. If you gave Rs 500 rupee note to the shopkeeper, what change did the
shopkeeper return to you?
Solution:
Here, the cost of a cake and a packet of sweets =Rs 250.25 + Rs 145.90
= Rs 396.15
So, change = Rs 500 .00 – Rs 396.15 = Rs 103.85
b) The initial temperature of a body is 40.5°C. It’s temperature first rises by 35.7°C and
then falls by 29.8°C. Now, what is the temperature of the body?
Solution:
Here, temperature of body =40.5°C + 35.7°C - 29.8°C
= 75.2°C - 29.8°C
= 45.4°C
45 Vedanta Excel in Mathematics Teachers' Manual - 6
c) The distance between Itahari and Damak is 42 km. Mr. Dhamala travelled 18.325
km by a bus, 15.675 km by a taxi and the remaining distance by a motorbike. How
many kilometres did he travel by motorbike?
Solution:
Here, the distance between Itahari and Damak = 42 km
Distance travelled by bus and taxi = 18.325 km + 15.675 km = 34 km
So, the distance travelled by motorbike = 42 km – 34 km = 8 km.
d) A dairy man has 50.5l of milk. He has to deliver 30.75l and 32.5l of milk in two
marriage ceremonies in the same time. How much more milk should he manage?
Solution:
Here, the quantity of milk to be delivered = 30.75 l + 32.5 l
=30.75 l + 32.50 l
= 62.25 l
Required quantity of milk to be managed = 62.25 l - 50.50 l = 11.75 l
3. a) Find the sum of place values of 3’s and subtract it from the place value of 2 in the
numeral 0.23436.
Solution:
Here, in the number 0.23436; the place values of 3’s are 0.03 and 0.0003 and the place
value of 2 is 0.2
Now, sum of 0.03 and 0.0003 = 0.03+0.0003 = 0.0303
Again, difference between 0.2 and 0.0303 = 0.2000 – 0.0303 = 0.1697
b) In the numeral 6.89098, subtract the sum of place values of 9’s from the sum of
place values of 8’s.
Solution:
Here, in the number 6.89098; the place values of 9’s are 0.09 and 0.0009 and the place
value of 8’s are 0.8 and 0.00008
Now, sum of 0.09 and 0.0009 = 0.09+0.0009 = 0.0909
Sum of 0.8 and 0.00008 = 0.8+0.00008 = 0.80008
Again, difference between 0.80008 and 0.0909 = 0.80008 – 0.0909 = 0.70918
4. a) From a sack of a quintal of potatoes, a shopkeeper sold 25.65 kg of potatoes yesterday
and 20.48 kg this morning. How many kilograms of potatoes are left to sell?
Solution:
Here, 1 quintal of potatoes = 100 kg
Total sold potatoes = 25.65 kg + 20.48 kg = 46.13 kg
So, quantity of potatoes left to sell = 100 kg – 46.13 kg = 100.00 – 46.13 = 53.87 kg
c) The distance between Sita’s house and her school is a kilometer. On the way to the
school, her friend Gita’s house is 0.256 km away from her house and their friend
Rita house is 0.375 km far from Gita’s house. How far is the Rita’s house from
school?
Solution:
Here, the distance between Sita’s house and her school is a kilometre = 1 km
The distance between Sita and Gita’s house =0.256 km
The distance between Gita and Rita’s house =0.375 km
Now, the distance between Sita and Rita’s house =0.256 km + 0.375 km = 0.631km
So, the distance between Rita’s house and the school =1.000 km – 0.631 km =0.369 km
Vedanta Excel in Mathematics Teachers' Manual - 6 46
Activity 8 Multiplication and Division of Decimals
1. Multiplication of decimal numbers by whole numbers
Represent each cube by 0.1 then make a sense of
multiplication of decimals by whole numbers.
For example,
2 times of 0.3 = 2×0.3 = 0.3 + 0.3= 0.6
3 times of 0.4 = 3×0.4 =0.4+0.4+0.4 = 1.2
2. Multiplication of decimals by 10 or powers of 10
(i) Represent each cube by 0.1 then make conform that there are 10 cubes in 1. So, there
is 1 block of 10 cubes. Thus, 10 ×0.1 = 1
Similarly, 10×0.3=3 (blocks of 10 cubes)
(ii) Note that while multiplying a decimal number by 10 or powers of 10, the decimal
point is shifted as many digits to the right as there are zero in 10 or powers of 10.
For example: 0.6789 × 10 = 6.789, 0.6789 × 100 = 67.89, 0.6789 × 100 = 678.9 etc.
3. Multiplication of decimal numbers by decimal numbers
Discuss upon the following examples.
0.2×0.3 = 0.2 of 0.3 = 6-hundredths = 0.06
More precisely, 0.2×0.3 = 2 × 3 = 6 = 0.06
10 10 100
Similarly, 0.5×0.2 = 5 × 2 = 1 = 0.1
10 10 10
Thus, when a decimal number is multiplied by decimal
number (or whole number), the product has as many
decimal places as there are total number of decimal places in multiplicand and multiplier.
4. Division of decimal numbers by whole numbers
a) Represent each cube by 0.1 and take 6 cubes and divide them equally among 2 students
and discuss about 0.6÷2. Obviously, its 0.3.
b) Make a discussion on the following questions.
(i) How many 0.1s are there in 1?
(ii) How many 0.1s are there in 2?
(iii) How many 0.2s are there in 1?
47 Vedanta Excel in Mathematics Teachers' Manual - 6
(iv) How many 0.4s are there in 2?
5. Division of decimals by 10 or powers of 10
Give a few examples and note that when a decimal number is divided by 10, 100, or 1000,
we should shift the decimal point as many number of digits to the left as there are zeros in
10, 100, or 1000.
6. Division of decimals by decimals
(i) With the following model
discuss on 1.2 divided by 0.3.
(ii) Note that when a decimal
number is divided by another
decimal number, at first, remove the decimal point from the divisor by multiplying
a suitable number 10 or 100 or 1000 to both numerator and denominator and then
divide as usual process.
(iii) Note that if both the dividend and divisor have equal number of digits after decimal
points then remove the decimal points and divide as usual.
Solution of selected questions from EXERCISE 5.8
1. Simplify. c) (0.2 + 0.6) ÷ 2
a) 2 × (0.4 + 0.3) b) 3 × (0.9 – 0.4)
d) (3 × 0.8) ÷ (2 × 0.2) e) 2.36 – 1.28 ÷ 1.6 f) 1.3 × 0.4 + 2.4 × 0.3 – 1.8 × 0.2
Solution:
a) 2 × (0.4 + 0.3) = 2 × 0.7 = 1.4
b) 3 × (0.9 – 0.4) = 3 × 0.5 = 1.5
c) (0.2 + 0.6) ÷ 2 = 0.8 ÷ 2 = 0.4
d) (3 × 0.8) ÷ (2 × 0.2) = 2.4 ÷ 0.4 = 6
e) 2.36 – 1.28 ÷ 1.6 = 2.36 – 1.28 ÷ 1.60 = 2.36 – 0.8 = 2.36 – 0.80 = 1.56
f) 1.3 × 0.4 + 2.4 × 0.3 – 1.8 × 0.2 = 0.52 + 0.72 – 0.36 = 1.24 – 0.36 = 0.88
2. a) If the cost of a pen is Rs. 15.25, find the cost of 9 pens.
Solution:
Here, the cost of 1 pen = Rs 15.25
So, the cost of 9 pens = 9×Rs 15.25 = Rs 137.25
b) 1.35 m of cloth is required for 1 shirt. What will be the total length of cloth required
for 6 shirts of the same size?
Solution:
Here, the length of cloth required for 1 shirt = 1.35 m
Hence, the length of cloth required for 6 shirts = 6×1.35 m = 8.10 m
3. a) If the cost of one dozen of exercise books is Rs. 225.72, find the cost of 1 exercise book.
Solution:
Vedanta Excel in Mathematics Teachers' Manual - 6 48
Here, the cost of 1 dozen of exercise books = Rs 225.72
Hence, the cost of 1 dozen of exercise books = Rs 225.72÷12 = Rs 18.81
b) The length of a rope is 11.55 m. If it is cut into 7 equal pieces, find the length of each
piece of rope.
Solution:
Here, the length of a rope = 11.55 m
Hence, the length of a piece of rope = 11.55m÷7 = 1.65m
4. a) Mother recorded the fuel consumption of her car as:
City: 21.4 km per litre Highway: 23.5 km per litre
The fuel tank of car holds 40.2 litre of petrol.
(i) How long could she drive on a full tank of petrol in the city?
(ii) How long could she drive on a full tank of petrol in the highway?
Solution:
Here,
The consumption of fuel in City = 21.4 km per litre
The consumption of fuel in Highway = 23.5 km per litre
The petrol in the fuel tank of car =40.2 litre
(i) Distance travelled in city with the fuel = 40.2×21.4 km=860.28 km
Hence, she could drive 860.28 km on a full tank of petrol in the city.
(ii) Distance travelled in highway with the fuel = 40.2×23.5 km=944.70 km
Hence, she could drive 860.28 km on a full tank of petrol in the city.
b) According to money exchange rate announced by Nepal Rastra Bank on 2nd November
2020, the buying and selling rates of 1 Australian dollar are: Buying rate: Rs 77.96
Selling rate: Rs 78.30
(i) How much Nepalese rupees do you need to buy 25.5 Australian dollar?
(ii) How much Nepalese rupees do you get while selling 80 Australian dollars?
Solution:
Here,
Buying rate of 1 Australian dollar = Rs 77.96
Selling rate of 1 Australian dollar = Rs 78.30
(i) Nepalese rupees required to buy 25.5 AUD = 25.5× Rs 77.96 = Rs 1,987.98
(ii) Nepalese rupees required to sell 80 AUD = 80× Rs 78.30 = Rs 6,264
5. a) In a supermarket a 1.4 kg bag of apples costs Rs 196.70.
(i) What is the actual cost of 1 kg of apples?
(ii) If you spend Rs 491.75 for apples, what would be the weight of the apples?
Solution:
Here,
The cost of 1.4 kg of apples = Rs 196.70
(i) The cost of 1 kg of apples = Rs 196.70÷1.4 = Rs 140.50
(ii) Rs 140.5 is the cost of 1 kg of apples
49 Vedanta Excel in Mathematics Teachers' Manual - 6
Re 1 is the cost of 1÷140.5 kg of apples.
So, Rs 491.75 is the cost of 491.75 ÷140.5 = 3.5 kg of apples.
So, 3.5 kg of apples can be bought for Rs 491.75
b) Last week, Mr. Lama worked 38.5 hours and earned Rs 5563.25.
(i) How much money did he earn per hour?
(ii) How many hours would he work to earn Rs 6574.75?
Solution:
Here,
The earning of 38.5 hours = Rs 5563.25
(i) The earning of 1 hour = Rs 5563.25÷38.5 = Rs 144.50
Rs 144.50 is the earning of 1 hour.
(ii) Re 1 is the cost of 1÷144.5 hours.
So, Rs 6574.75 is the earning of 6574.75 ÷144.5 = 45.5 hours.
So, Mr. Lama should work for 45.5 hours to earn Rs 6574.75
6. a) Anamol, Shandeep and Bishal got different answer for their problem:
12 × (4.8 ÷ 0.3) – 3.5 × 3.64. Anamol’s answer was 39.12, Shandeep’s answer was
179.26 and Bishal’s answer was 64.6.
(i) Simplify and then identify which student had correct answer.
(ii) Find the mistake done by other two students and explain.
Solution:
Here,
(i) 12 × (4.8 ÷ 0.3) – 3.5 × 3.64
= 12 × 16 – 3.5 × 3.64
= 192– 12.74
= 179.26
So, Shandeep’s answer was correct.
(ii) Anamol’s answer was 39.12 which is incorrect.
His process could be:
12 × (4.8 ÷ 0.3) – 3.5 × 3.64
= 12 × 16 – 3.5 × 3.64
= 12 × 16 – 12.74
= 12 × 3.26 (he made mistake here) = 39.12
He went wrong way to the order of operations in simplification.
Also, Bishal’s answer was 64.6 which is incorrect.
His process could be:
12 × (4.8 ÷ 0.3) – 3.5 × 3.64
= 12 × 16 – 3.5 × 3.64
= 192 – 127.4 (he made mistake here)
= 64.6
He made mistake in multiplication of decimals.
Vedanta Excel in Mathematics Teachers' Manual - 6 50
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