vedanta
EXCEL in
MATHEMATICS
9Book
Authors
Hukum Pd. Dahal
Tara Bahadur Magar
vedanta
Vedanta Publication (P) Ltd.
Vanasthali, Kathmandu, Nepal
+977-01-4382404, 01-4362082
[email protected]
www.vedantapublication.com.np
vedanta
EXCEL in
MATHEMATICS
9Book
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.
Published by:
Vedanta Publication (P) Ltd.
Vanasthali, Kathmandu, Nepal
+977-01-4382404, 01-4362082
[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 useful for the summative assessments.
Grateful thanks are due to all Mathematics Teachers throughout the country who
encouraged and provided the feedback to me 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.
1
Topics 5
1. Set
2. Profit and Loss 15
3. Commission and Taxation 23
4. Household Arithmetic 33
5. Mensuration 48
6. Algebraic Expressions 52
7. Indices 58
8. Simultaneous Linear Equations 62
9. Quadratic Equations 70
10. Ratio and Proportions 78
11. Geometry - Triangle 106
12. Geometry - Similarity 113
13. Geometry - Parallelogram 127
14. Geometry - Circle 140
15. Geometry - Construction 143
16. Trigonometry 148
17. Statistics 154
18. Probability
Unit Set
1
Competency Allocated teaching periods 8
- To find the relation between the sets and solve the related problems by demonstrating
the relations on the basis of the properties of the sets
Learning Outcomes
- To solve the verbal problems related to the cardinality of sets by using Venn diagrams
and solve the behavioural problems by using the properties of relations of sets.
Level-wise learning objectives
S.N. Levels Objectives
- To define a set
- To tell the types of sets
1. Knowledge (K) - To state the relations between sets
- To list the operations on sets
- To define cardinality of sets
- To tell the formulae involving two sets
- To find the cardinality of a set.
- To identify the set notation with special terminologies
2. Understanding (U) - To write the word problems based on cardinality
relations in set notations
-
To solve the verbal problems on operations (union,
3. Application (A) intersection, difference and complement) of sets by
- using Venn-diagram
To solve the verbal problems on operations of sets by
using formulae
- To relate the problem related to set with other areas of
4. High Ability (HA) learning like percentage, ratio and so on.
- To link various real life/ contemporary problems with
sets and solve
Required Teaching Materials/ Resources
Definitions and formulae in colourful chart-paper, scissors, cello tape, different coloured
markers, highlighter, models of Venn-diagrams, ICT tools etc.
Pre-knowledge: Check the Pre-knowledge on cardinality of sets, relation of sets, operations
of sets and Venn-diagram
Teaching Activities
1. Make a group discussion on the definition of sets, set notation, types of sets, cardinality
of sets and operations of sets by using Venn-diagram.
2. Ask individually the basic concepts from set as revision.
1 Vedanta Excel in Mathematics Teachers' Manual - 9
3. Make the group of students and give the questions on set operations from exercise.
4. Solve the problems and verify the relations from the exercise with discussion.
5. 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 is a subset of B
(i) n (AªB) = A BA U
(ii) n (A«B) = n (B)
(iii) n o (B) = n (B) – n (A)
Case- II: When A and B are disjoint sets.
(i) n (AªB) = 0 A BU
(ii) n (A«B) = n (A) + n (B)
Case- III: When A and B overlapping sets
(i) n (A« B) = n (A) + n (B) – n (AªB)
(ii) n (only A) = no (A) = n (A) – n (AªB)
(iii) n (only B) = no (B) = n (A) – n (AªB)
(iv) n (only one) or n (exactly one) = no (A) + no (B)
(v) n (A«B) = no (A) + no (B) + n (AªB)
(vi) n(A « B ) = n(U) – n (A « B)
(vii) (U) = n (A) + n (B) – n (AªB) + n(A « B )
6. 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 (AªB)
(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 )
(vi) No. of people who don’t like apple only = no (B) and no. of people who don’t like
banana only = no (A)
Note:
1. If the information are given in percentage, consider that n (U) = 100 or x.
2. If the data are given in fraction then suppose that n (U) = x.
3. If each people participates in at least one activity then n (U) = n (A«B) or n(A « B )= 0
Solution of selected questions from Vedanta Excel in Mathematics
1. In a survey of 900 students in a school, it was found that 600 students liked tea, 500
liked coffee and 125 did not like both drinks.
(i) Draw a Venn-diagram to illustrate the above information.
(ii) Find the number of students who like both drinks
(iii) Find the number of students who didn’t like tea only.
Solution:
Let T and C denote the sets of students who liked tea and coffee respectively.
Vedanta Excel in Mathematics Teachers' Manual - 9 2
Then, n (U) = 900, n (T) = 600, n (C) = 500 and n(T « C )= 125 U
Let the no. Of students who liked both the drinks be x. TC
(i) Drawing a Venn-diagram to show the above information.
600-x x 500 – x
(ii) From Venn-diagram,
n (U) = (600 – x) + x + (500 – x) + 125 or, 900 = 1225 – x = 325
Hence, 325 students liked both the drinks.
(iii) No. of students who didn’t like tea only i.e., n o (C) = 500 – x = 500 – 325 = 175
2. There are 900 students in a school. They are allowed to cast vote either only for A or
for B as their school prefect. 36 of them cast vote for both A and B, 483 cast vote for
A and 367 cast vote for B.
(i) How many students did not cast the vote?
(ii) Find the number of valid votes.
(iii) Show the information in Venn-diagram.
Solution:
Here, n (U) = 900, no(A) = 483, no(B) = 367 and n (Aª∩ B) = 36
Now, we have, n (A « B) = n (U) - n(A « B )= 900 – 36 = 864.
Again, n (A«B) = no (A) + no (B) + n (AªB) U
or, 864 = 483 + 367 + n (AªB) ? n (AªB) = 14 AB
Hence, 14 students didn’t cast the vote. 483 36 367
14
(i) The number of valid votes (ii) Illustration in Venn-diagram
= no (A) + no (B)
= 483 + 367 = 850
3. 54 students of class IX are taking part in sports or in music or in both activities. Out
of them 9 students are taking part in both activities. The ratio of the number of the
students who are taking part in sports to those who are taking part in music is 5:4.
(i) How many students are taking part in sport?
(ii) How many students are taking part in music only?
(iii) Illustrate this information in a Venn-diagram.
Solution:
Let S and M denote the sets of students who are participating in sports and music respectively.
Then, n (U) = n (S«M) = 54, n (S) = 5x, n (M) = 4x (say) and n (SªM) = 9
Now, n (S«M) = n (S) + n (M) – n (SªM) or, 54 = 5x + 4x – 9 ?x = 7
(i) The number of students who are taking part in sport n (S) = 5x = 35
(ii) The number of students who are taking part in music n (M) = 4x = 28
No. of students who are taking part in sport only, no (S) = n (S) - n (SªM) = 28 – 9 = 19
(iii) Illustration in Venn-diagram
U
SM
26 9 19
3 Vedanta Excel in Mathematics Teachers' Manual - 9
4. In a survey of some students, it was found that 60% of students were studying
commerce and 40% were studying science. If 40 students were studying both the
subjects and 10% did not study any of two subjects, by drawing a Venn-diagram,
(i) Find the total number of students.
(ii) Find the number of students who were studying science only.
Solution: C (60 – X)% U
Let C and S denote the sets of students who were studying commerce x% S
and science respectively. (40 – x)
Then n (U) = 100%, n (C) = 60%, n (S) = 40%, n(C « S)= 10% and
n (CªS) = 40 10%
Let n (CªS) = x%
Now, representing the above data in a Venn-diagram
Again,
From the Venn-diagram,
n (U) = n (C) + n (S) – n (CªS) + n(C « S)
or, 100% = (60 – x)% + (40 – x)% + x% + 10% ?x % = 10%
(i) Let the total number of students n (U) = x
Then n (CªS) =10% of x or, 40 = 0.1x ?x = 400
Hence, there were 400 students.
(ii) The number of students who were studying science only no(S) = (40-10)% of 400 =
30% of 400 = 120
Extra Questions
1. Out of 30 students of class IX, 15 students like to play volleyball, 20 students like to
play basketball and each student like to play at least one of the game.
(i) How many students like to play volleyball and basketball both?
(ii) Show the above information in the Venn-diagram. [Ans: 5]
2. Out of 77 districts of Nepal, 27 districts have shared their boarder with India, 15
districts have shared their boarder with China and 37 districts have not shared their
boarder with India and China both.
(i) How many districts have shared their boarder with China only?
(ii) How many districts have shared their boarder with India only?
(iii) Draw a Venn-diagram to show the above information. [Ans: (i) 13, (ii) 25]
3. In a group of 30 children, 13 favored apple only, 8 favored guava only and 3 favored
none of these fruits, by showing in Venn-diagram
(i) Find the ratio of number of children who like both and don’t like both the fruits.
(ii) What percent of the students like guava? [Ans: (i) 2:1, (ii) 60%]
4. In a survey of community, it was found that 50% of the people preferred yoga, 60%
preferred jugging and 10% preferred neither yoga nor jugging. If 200 people preferred
both yoga and jugging, by using a Venn-diagram find:
(i) How many people were participated in the survey?
(ii) How many people preferred only one of these? [Ans: (i)1000, (ii) 900]
Vedanta Excel in Mathematics Teachers' Manual - 9 4
Unit Profit and Loss
2
Competency Allocated teaching periods 6
- 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. LEVELS OBJECTIVES
- To define cost price, selling price
- To tell the relation of C.P., S.P. and profit % or loss%
1. Knowledge (K) - To define marked price
- To tell the formula of finding the discount
- To define VAT
- To find the profit/ loss amount
2. Understanding (U) - 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
3. Application (A) - To solve the verbal problems on discount and VAT
- To mathematize the daily life problems related to
4. High Ability (HA) 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 required 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
A. Profit and Loss
Teaching Activities
1. Discuss upon cost price, selling price, profit and loss of the articles like watch,
mobile, books, copies, bags etc.
2. Divide the class into 5/6 groups and ask the formulae of profit amount, loss amount,
profit percentage and loss percentage.
3. List out the following formulae with examples
(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.
5 Vedanta Excel in Mathematics Teachers' Manual - 9
(vi) S.P. = C.P. – L% of C.P. S. P. – C. P.
Profit C. P.
(vii) Profit percent = C.P. × 100% o × 100%
(viii) loss percent = Loss × 100% o C. P. – S. P. × 100%
C.P. C. P.
Solution of selected problems from Vedanta Excel in Mathematics
1. A grocer bought 20 kg of sugar at Rs 70 per kg. He sold 15 kg of sugar at Rs 75 per
kg and the remaining quantity at Rs 69 per kg. Find his profit or loss percent in the
transaction.
Solution:
Here, C.P. of 20 kg of sugar = 20 ×Rs 70 = Rs 1400
S.P. of 15 kg of sugar = 15 ×Rs 75 = Rs 1125
Remaining quantity of sugar = 20 kg – 15 kg = 5 kg
S.P. of 5 kg of sugar = 5 ×Rs 69 = Rs 345
S.P. of 20 kg of sugar = Rs 1125 + Rs 345 = Rs 1470
Since, S.P. > C.P., he made a profit
Profit amount = S.P. – C.P. = Rs 1470 – Rs 1400 = Rs 70
Profit amount Rs 70
Now, profit percent = Cost price (C.P.) × 100% = Rs 1400 × 100% = 5%
Hence, his profit percent is 5%.
2. Mrs. Rokaya bought 2 quintals of apples in Jumla for Rs 17000. She paid Rs 15 per
kg for the transportation from Jumla to Nepalgunj. 10 kg of apples is damage and
she sold the remaining quantity of apples at Rs 120 per kg. Calculate her profit or
loss percent in the total transaction.
Solution:
Here, C.P. of 2 quintals i.e., 200 kg of apples = Rs 17000
C.P. of the apples with the transportation cost = Rs 17000 + 200×Rs 15 = Rs 20000
Saleable quantity of apples = 200kg – 10 kg = 190 kg
S.P. of 190 kg of apples = 190 ×Rs 120 = Rs 22800
Since, S.P. > C.P., she made a profit
Profit amount = S.P. – C.P. = Rs 22800 – Rs 20000 = Rs 2800
Profit amount Rs 2800
Now, profit percent = Cost price (C.P.) × 100% = Rs 20000 × 100% = 14%
Hence, her profit percent is 14% in the total transaction.
3. A stationer sells 8 pencils for the cost of 10 pencils, find his gain percent.
Solution:
Let the C.P. of 1 pencil be Rs x.
Then, C.P. of 10 pencils = Rs 10x and C.P. of 8 pencils = Rs 8x
According to the question, S.P. of 8 pencils = C.P. of 10 pencils = Rs 10x
But C.P. of 8 pencils = Rs 8x. So, gain amount = Rs 10x – Rs 8x = Rs 2x
Gain amount Rs 2x
Now, gain percent = Cost price (C.P.) × 100% = Rs 10x × 100% = 20%
Hence, his gain percent is 20%.
4. A stationer bought 2000 exercise books. He distributed 200 exercise books to
the students from poor economical background as donation. He sold each of the
remaining exercise books at Rs 10 more than the cost price of each and gained 8%,
find the cost price of each exercise book.
Vedanta Excel in Mathematics Teachers' Manual - 9 6
Solution:
Let the C.P. of each exercise book be Rs x.
Then S.P. of each exercise book is Rs (x + 10)
Now, C.P. of 2000 exercise books = Rs 2000x
Saleable quantity of exercise books = 2000 – 200 = 1800
S.P. of 1800 exercise books = Rs 1800 (x + 10) = Rs (1800x + 18000)
Gain percent = 8%
Now, S.P. = C.P. + G% of C.P. or, 1800x + 18000 = 2000x + 8% of 2000x
or, 1800x + 18000 = 2160x or, 18000 = 360x ?x = Rs 50
Hence, the cost price of each exercise book is Rs 50.
5. Bikash purchased 10 pens. He sold 5 pens at 25% profit and the remaining 5 pen at
1632 % loss. If he received Rs 625 in total, find the cost price of each pen.
Solution:
Let the C.P. of each pen be Rs x. Then C.P. of 5 pens = Rs 5x and C.P. of 10 pens = Rs 10x
5oxf C+.P.2=5%5xof–51x623=%5xof+5x12=0505×x =×50R10s02×45x5x
S.P. of 5 pens = C.P. + P% of C.P. = 5x = Rs 25x
S.P. of remaining 5 pens = C.P. – L% –3 6
According to question, total S.P. = Rs 625
25x 25x 125x
or, 4 + 6 = Rs 625 or, 12 = Rs 625 ?x = 60
Hence, the cost price of each pen is Rs 60.
6. A dealer bought a pen-drive for Rs 500. He sold it at 10% loss. If he wanted to make
a profit of 12.5% without increasing the selling price, by how much should the cost
price of the pen-drive be reduced?
Solution:
Here, C.P. of a pen-drive = Rs 500 and loss percent = 10%
Now, S.P. of pen-drive = C.P. – L% of C.P. = Rs 500 – 10% of Rs 500 = Rs 450
Again, S.P. of pen-drive = Rs 450 and profit percent = 12.5%
We have, S.P. = C.P. + P% of C.P. or, Rs 450 = C.P. + 12.5% of C.P.
or, Rs 450 = 1.125 C.P. ?C.P. = Rs 400
Difference between cost price = Rs 500 – Rs 400 = Rs 100
Hence, the cost price of the pen-drive should be reduced by Rs 100 to make 12.5% profit.
7. A trader sold an article at 10% profit. If he sold it at 10% loss, it would yield Rs 140
less than the previous selling price; find the cost price of the article.
Solution:
Let the cost price of an article be Rs x.
Now, S.P. of article = C.P. + P% of C.P. = Rs x + 10% of Rs x = Rs 1.1x
According to the question, new S.P. of the article = Rs (1.1x – 140)
We have, S.P. = C.P. – L% of C.P.
or, 1.1x – 140 = x – 10% of x
or, 1.1x – 140 = 0.9x
or, 0.2x = 140
? x = 700
Hence, the cost price of the article was Rs 700.
7 Vedanta Excel in Mathematics Teachers' Manual - 9
8. Rajesh Das bought two calculators for Rs 1000. He sold one of them at 20% profit
and the other at 20% loss. If the selling prices of both the calculators are same,
find the cost price of each calculator. Also, calculate his gain or loss in the total
transaction.
Solution:
Let the cost price of the one calculator be Rs x then that of the another one is Rs (2000-x)
From the first calculator: C.P.1 = Rs x, profit percent = 20%
?S.P.1 = C.P.1 + P% of C.P.1 = Rs x + 20% of Rs x = Rs 1.2x
From the second calculator: C.P.2 = Rs (1000 – x), loss percent = 20%
?S.P.2 = C.P.2 - L% of C.P.1
= Rs (1000 – x) – 20% of Rs (1000 – x) = Rs (1000 – x – 200 + 0.2x) = Rs (800 – 0.8x)
According to the question, S.P.1 = S.P.2
or, 1.2x = 800 – 0.8x
or, 2x = 800
?x = Rs 400
Now, the cost price of the first calculator was Rs 400 and that of the other was
Rs (1000 – 400) = Rs 600
Again, S.P.1 = 1.2×Rs 400 = Rs 480
S.P. of two calculators = 2×Rs 480 = Rs 960 and C.P. = Rs 1000
Since, C.P. > S.P., there is a loss
Loss amount = C.P. – S.P. = Rs 1000 – Rs 960= Rs 40
Loss amount Rs 40
Now, loss percent = Cost price (C.P.) × 100% = Rs 1000 × 100% = 4%
Hence, his loss percent was 4% in the total transaction.
9. Ajaya bought a fan and a heater for Rs 4000. He sold the fan at a profit of 25% and
the heater at a loss of 5%. If he gained 7% on his total outlay, at what price did he
buy each item?
Solution:
Let the cost price of the fan be Rs x then that of the heater is Rs (4000-x)
From the fan: C.P.1 = Rs x, profit percent = 25%
?S.P.1 = C.P.1 + P% of C.P.1 = Rs x + 25% of Rs x = Rs 1.25x
From the heater: C.P.2 = Rs (4000 – x), loss percent = 5%
?S.P.2 = C.P.2 - L% of C.P.1
= Rs (4000 – x) – 5% of Rs (4000 – x)= Rs (4000 – x – 200 + 0.05x)= Rs (3800 – 0.95x)
Total C.P. = Rs 4000
Total S.P. = Rs (1.25x + 3800 – 0.95x)= Rs (3800 + 0.3x) and profit percent = 7%
S.P. = C.P. + P% of C.P.
or, 3800 + 0.3x = Rs4000 + 7% of 4000
or, 3800 + 0.3x = Rs4280
or, 0.3x = Rs480
?x = Rs 1600 and Rs (4000 – x) = Rs (4000 – 1600) = Rs 2400
Hence, the cost price of the fan was Rs 1600 and that of heater was Rs 2400.
Vedanta Excel in Mathematics Teachers' Manual - 9 8
10. A shopkeeper decided to make equal rate of profit in each fancy item. If he sold a
jacket costing Rs 4400 for Rs 5060, at what price did he sell the shoes which was
purchased for Rs 3420?
Solution:
Here,
The cost price (C.P.) of a jacket = Rs 4400 and selling price (S.P.) = Rs 5060
S. P. – C.P. Rs 5060 – Rs 4400
Now, profit percent = C. P. × 100% = Rs 4400 × 100% = 15%
Again, cost price of the shoes (C.P.) = Rs 3420 and profit percent = 15%
We have, S.P. = S.P. = C.P. + P% of C.P. = Rs 3420 + 15% of Rs 3420 =Rs 3933
Hence, he sold the shoes for Rs 3933.
11. A dishonest shopkeeper has two false balances. One balance weighs 10% more
while buying the goods and other weighs 10% less while selling the goods. Find the
gain percent just by weighing.
Solution:
Here,
While buying the goods, he weighs 10% more.
So, C.P. of goods of worth Rs 110 = Rs 100
While selling the goods, he weighs 10% less.
So, S.P. of goods of worth Rs 100 = Rs 110
S.P. of goods costing Re 1 = Rs 110
100
Since, he has the goods of worth Rs 110.
?S.P. of goods of worth Re 110 = Rs 110 × 110 = Rs 121
100
S. P. – C.P. Rs 121 – Rs 100
Now, profit percent = C. P. × 100% = Rs 100 × 100% = 21%
12. A crooked shopkeeper sells goods at the cost price. But his 1 kg weight weighs
900 g only. Find his gain percent.
Solution:
Let the C.P. of 1 g of goods be Rs x. Then, C.P. of 1 kg of goods = Rs 1000x and S.P. of 900 g
of goods = Rs 900x
According to the question,
S.P. of 900 g = C.P. of 1 kg of goods = Rs1000x
But C.P. of 900 g of goods = Rs 900x.
So, gain amount = Rs 1000x – Rs 900x = Rs 100x
Gain amount Rs 100x 1119%
Now, gain percent = Cost price (C.P.) × 100% = Rs 900x × 100% =
Hence, his gain percent is1191%. Rs 3000. He sold 1 of it with 10% loss. By how many
13 A grocer has some rice of worth 3
percent must the selling price be increased for making 10% profit on the outlay?
Solution:
Here, C.P. of certain quantity of rice = Rs 3000
Total S.P. for making 10% profit = C.P. + P% of C.P. = Rs 3000 + 10% of Rs 3000 = Rs 3300
1 1
C.P. of 3 of the rice = 3 of Rs 3000 = Rs 1000 and loss percent = 10%
9 Vedanta Excel in Mathematics Teachers' Manual - 9
?S.P. of 13of the rice = C.P. – L% of C.P. = Rs 1000 – 10% of Rs 1000 = Rs 900
othf ethreemreaminaiinngin23go23f
For making no any gain, S.P. of the rice = Rs 3000 – Rs 900 = Rs 2100
For making 10% gain, S.P. of the rice = Rs 3300 – Rs 900 = Rs 2400
Difference in S.P. = Rs 2400 – Rs 2100 = Rs 300 Rs 300 100% = 1427
Rs 2100 be increased
Now, increased percent in S.P. of remaining quantity = × 1472
Hence, the selling price of the remaining quantity of rice should by for
making 10% profit on the outlay.
Extra Questions
1. Mr. Lama bought a second hand bike for Rs 1,10,000 and immediately he spent Rs 5,000
to repair it. Then he sold it for Rs 1,26,500. Find his profit or loss percent.
[Ans: Profit percent = 10%]
2. A grocer sold 5 kg of wheat flour at Rs 55 per kg and gained 10%. If he sold all the flour
for Rs 260, what would be his gain or loss percent? [Ans:Profit 4%]
3. A man bought a hen and a duck for Rs 3,400. He sold the hen at 25% profit and then duck
at 10% loss. If he gained 10% on his total outlay. At what price did he buy the hen and
the duck each?
[Ans: Rs 2,000, Rs 1500]
4. Rajendra is a stationer. Once, he bought 2,000 books. Out of them he donated 200 books
to a school library. He sold the remaining books with 8% profit at the rate of Rs. 120 per
book. What will be the cost price of a book? Find it.
[Ans:Rs 100]
5. A shopkeeper sold a sewing machine for Rs 3,600 and made a loss of 10%. For what price
should he sell it to gain 10%? [Ans:Rs 4,400]
B. Marked price (M.P.) and Discount
Teaching Activities
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. Make clear 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
(iii) Rate of discount = M.P. × 100%
(iv) S.P. = M.P. – Discount amount
(v) S.P. = M.P. – Discount% of M.P. = M.P. (1 – Discount %)
(vi) S.P. = M.P. (1 – D1%)(1 – D2%) when two successive discount rates D1% and D2 %
are given.
(vii) VAT amount = S.P. including VAT – S. P. excluding VAT
(viii) VAT amount = VAT% of .P.
Vedanta Excel in Mathematics Teachers' Manual - 9 10
(ix) Rate of VAT = VAT amount × 100%
S.P.
(x) S.P. with VAT = S.P. + VAT amount
(xi) S.P. with VAT = S.P. + VAT% of S.P. = S.P. (1 + VAT %)
Solution of selected problems from Vedanta Excel in Mathematics
1. A retailer allows 15% discount on the marked price of an electric fan. If a customer
pays Rs 2,244 with 10% VAT, find the marked price of the fan.
Solution:
Let the marked price of the electric fan be Rs x
Then, S.P. after 15% discount = M.P. – D% of M.P. = x – 15% of x = Rs 0.85x
Again, S.P. with VAT = S.P. + VAT% of S.P.
or, 0.85x + 10% of 0.85x = Rs 2,244
or, 0.935x = Rs 2,244
? x = Rs 2,400
Hence, the marked price of the electric fan is Rs 2,400.
2. A tourist paid Rs 5,610 for a carved window made up of wood with discount of 15%
including 10% value added tax (VAT). How much does he get back while leaving
Nepal?
Solution:
Let the marked price of the carve window be Rs x
Then, S.P. after 15% discount = M.P. – D% of M.P. = x – 15% of x = Rs 0.85x
Also, S.P. with VAT = S.P. + VAT% of S.P.
or, Rs 5,610 = 0.85x + 10% of 0.85x
or, Rs 5,610 = 0.935x
? x = Rs 6,000
Again, S.P. = Rs 0.85x = Rs 0.85× 6000 =Rs 5,100
And VAT amount = 10% of Rs 5,100 = Rs 510
Hence, the tourist gets Rs 510 back while leaving Nepal.
3. The marked price of a cycle is Rs 5,500. After allowing certain percent of discount
with 10% VAT levied, the cycle is sold at Rs 5,445, find the discount percent.
Solution:
The marked price of a cycle = Rs 5,500, VAT percent = 10% and S.P. with VAT = Rs 5,445,
discount percent =?
Let S.P. after discount of the cycle be Rs x.
Then,S.P. with VAT = S.P. + VAT% of S.P.
or, Rs 5,445 = x + 10% of 0.85x
or, Rs 5,445 = 1.1x
x = Rs 4,950
Also, discount amount = M.P. – S.P. = Rs 5,500 – Rs 4,950 = Rs 550
Discount amount Rs 550
Again, discount percent = M.P. × 100% = Rs 5500 × 100% = 10%
Hence, the required discount percent is 10%.
4. The mobile price is tagged Rs 5,000. If a customer gets 12% discount and adding
certain percent VAT reaches as Rs 4,972, find out the VAT percent.
11 Vedanta Excel in Mathematics Teachers' Manual - 9
Solution:
The marked price of the mobile = Rs 5,000, discount percent = 12% and S.P. with VAT =
Rs 4,972, VAT percent =?
Now,
Then, S.P. after discount = M.P. – D% of M.P. = Rs 5,000 – 12% of Rs 5,000 = Rs 4,400
Also, VAT amount = S.P. with VAT – S.P. = Rs 4,972 – Rs 4,400 = Rs 572
VAT amount Rs 572
Again, VAT percent = S.P. × 100% = Rs 4400 × 100% = 13%
Hence, the required VAT percent is 13%.
5. After allowing 15% discount on the marked price of a camera, 15% VAT was levied
and sold it. If the difference between the selling price with VAT and selling price
after discount is Rs 1,122, find the marked price of the camera.
Solution:
Let the marked price of the camera be Rs x
Then, S.P. after 15% discount = M.P. – D% of M.P. = x – 15% of x = Rs 0.85x
Again, S.P. with 15% VAT = S.P. + VAT% of S.P. = Rs 0.85x + 15% of Rs 0.85x = 0.9775x
According to the question,
S.P. with VAT – S.P. after discount = Rs 1,122
or, 0.9775x – 0.85x = Rs 1,122
or, 0.1275x = Rs 1,122
? x = Rs 8,800
Hence, the marked price of the camera is Rs 8,800.
6. The marked price of an article is 25% above the cost price. When it is sold at a
discount of 15%, there is a gain of Rs 200. Find.
(i) The cost price of the article. (ii) The marked price of the article.
Solution:
Let the cost price (C.P.) of the article be Rs x
Then, M.P. of the article = C.P. + 25% of C.P. = Rs x + 25% of x = Rs 1.25x
Now, S.P. after 15% discount = M.P. – D% of M.P. =1.25x – 15% of 1.25 x = Rs 1.0625x
Again, profit amount = S.P. – C.P.
or, Rs 200 = 1.0625x – x
or, Rs 200 = 0.0625x
? x = Rs 3,200
And, M.P. = Rs 1.25x = Rs 1.25×3200 = Rs 4,000
Hence, the cost price of the article is Rs 3,200 and its marked price is Rs 4,000.
7. When a pen is sold at a discount of 15%, there is a gain of Rs 10. But if it is sold at
25% discount, there is a loss of Rs 2. Find the marked price of the pen.
Solution:
Let the marked price (M.P.) of the pen be Rs x.
According to the given first condition, discount percent = 15%, profit = Rs 10
We have, S.P. after 15% discount = M.P. – D% of M.P. = x – 15% of x = Rs 0.85x
Again, profit amount = S.P. – C.P.
or, Rs 10 = 0.85x – C.P.
? C.P. = Rs (0.85x – 10) … (i)
According to the given second condition, discount percent = 25%, loss = Rs 2
Vedanta Excel in Mathematics Teachers' Manual - 9 12
We have, S.P. after 25% discount = M.P. – D% of M.P. = x – 25% of x = Rs 0.75x
Again, loss amount = C.P. – S.P.
or, Rs 2 = Rs ( 0.85x – 10) – Rs 0.75x [From (i)]
or, Rs 12 = 0.1x
? x = Rs 120
Hence, the marked price of the pen is Rs 120.
8. A shopkeeper marked the price of an article a certain percent above the cost price
and he allowed 16% discount to make 5% profit. If a customer paid Rs 9,492 with
13% VAT to buy the article, by what percent is the marked price above the cost
price of the article?
Solution:
Let the marked price of the article be Rs x
Then, S.P. after 16% discount = M.P. – D% of M.P. = x – 16% of x = Rs 0.84x
Again, S.P. with VAT = S.P. + VAT% of S.P.
or, 0.84x + 13% of 0.84x = Rs 9,492
or, 0.9492x = Rs 9,492
? x = Rs 10,000
Hence, the marked price of the article is Rs 10,000.
Also, S.P. = Rs 0.84x = Rs 0.84×10,000 = Rs 8,400
Let C.P. of the article be Rs y
We have, S.P. = C.P. + P% of C.P.
or, Rs 8,400 = y + 5% of y
or, Rs 8,400 = 1.05y
? y = Rs 8,000
Again, difference between M.P. and C.P. = Rs 10,000 – Rs 8,000 = Rs 2,000
Rs 2,000
?The marked price of the article is above the cost price by Rs 8,000 × 100% = 25%
9. The marked price of an item is Rs 1,500 and 10% discount is given to make 20%
profit. By what percent is the discount to be increased to get only 12% profit?
Solution:
Here, the marked price (M.P.) of the item = Rs 1,500, discount = 10 % and
profit percent= 20%
Then, S.P. after discount = M.P. – D% of M.P. = Rs 1,500 – 10% of Rs 1,500 = Rs 1,350
Let the cost price (C.P.) of the item be Rs x.
We have, S.P. = C.P. + P% of C.P.
or, Rs 1,350 = x + 20% of x
or, Rs 1,350 =1.2x
? x = Rs 1,125
Thus, the cost price (C.P.) of the item is Rs 1,125
Again, to get only 12% profit
S.P. = C.P. + P% of C.P. = Rs 1,125 + 12% of Rs 1,125 = Rs 1,260
Discount amount = M.P. – S.P. = Rs 1,500 – Rs 1,260 = Rs 240
Discount amount Rs 240
Discount percent = M. P. × 100% = Rs 1500 × 100% = 16%
Hence, the discount should be increased by 16% - 10% = 6% to make only 12% profit.
13 Vedanta Excel in Mathematics Teachers' Manual - 9
10. The price of a watch is marked Rs 11,250. When it is sold allowing 20% discount,
20% profit is made. By what percent is the discount to be reduced to increase the
profit by 3%?
Solution:
Here, the marked price (M.P.) of the item = Rs 11,250, discount = 20 % and profit percent=
20%
Then, S.P. after discount = M.P. – D% of M.P. = Rs 11,250 – 20% of Rs 11,250 = Rs 9,000
Let the cost price (C.P.) of the item be Rs x.
We have, S.P. = C.P. + P% of C.P.
or, Rs 9,00 = x + 20% of x
or, Rs 9,000 =1.2x
? x = Rs 7,500
Thus, the cost price (C.P.) of the item is Rs 7,500
Again, to get only (20 + 3)% = 23% profit
S.P. = C.P. + P% of C.P. = Rs 7,500+ 23% of Rs 7,500 = Rs 9,225
Discount amount = M.P. – S.P. = Rs 11,250 – Rs 9,225= Rs 2.025
Discount percent = Discount amount × 100% = Rs 20,250 × 100% = 18%
M. P. Rs 11,250
Hence, the discount should be decreased by 20% - 18% = 2% to increase the profit by 3%.
11. The marked price of a digital watch is Rs 6,000. Allowing 10% discount and including
same percent of value added tax, the watch is sold. By how much percent is the VAT
amount less than discount amount?
Solution:
Here, the marked price (M.P.) of the digital watch = Rs 6,000, discount = 10 % and
VAT = 10%
Now, Discount amount = D% of M.P. = 10% of Rs 6,000 = Rs 600
Also, S.P. after discount = M.P. – D = Rs 6,000 – Rs 600 = Rs 5,400
Again, VAT amount = VAT% of S.P. = 10% of Rs 5,400 = Rs 540
Difference between discount and VAT amounts = Rs 600 – Rs 540 = Rs 60
Rs 60
VAT amount is less than discount amount by Rs 600 × 100% = 10%
Hence, the VAT amount is less than discount amount by 10%.
Extra Questions
1. What is the price of a bag costing Rs 2000 after allowing 15% discount? Find it.
[Ans: Rs 1700]
2. Mr. Ajay sold a watch for Rs 880 after allowing 20% discount, what was the marked
price of the watch? [Ans:Rs 1,100]
3. A shopkeeper in Nepalgunj fixed the price of a suitcase in such a way that the he could
gain 10% after allowing 10% discount on it. If the customer paid Rs 9900 for the suitcase,
find the marked price and the selling price of the suitcase. [Ans: Rs 11,000; Rs 9000]
4. What will be the price of a calculator costing Rs 600 with 13% value added tax (VAT)?
[Ans: Rs 678]
5. The marked price of a scooter is Rs 2,40,000. If the shopkeeper allows 15% discount and
levies 13% value added tax, how much should a customer have to buy the scooter? Find
it. [Ans: Rs 2,30,520]
6. A cycle was sold after allowing 20% discount on the marked price and levying 10% VAT.
If the customer got Rs 555 as the discount, how much VAT amount was levied on the
cost of the cycle? Find it. [Ans: Rs 222]
Vedanta Excel in Mathematics Teachers' Manual - 9 14
Unit Commission and Taxation
3
Competency Allocated teaching periods 6
- To solve the daily life problems on profit and loss by using fundamental rules of profit/
loss and formulae.
Competency
- To solve the daily life problems on by using fundamental rules of commercial
mathematics and formulae
Learning Outcomes
- To solve the daily life problems related to Commission, Tax and Dividend.
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To define commission
- To tell the formula of finding the commission
1. Knowledge (K) amount
- To define bonus
- To tell the formula of finding the dividend
- To define income tax
- To calculate the commission amount
2. Understanding (U) - To calculate the bonus amount received by each
employee
- To find the commission or total sales
3. Application (A) - To find the income tax.
- To solve the verbal problems related to dividend
- To compare the income taxes paid by an individual
4. High Ability (HA) and the married couple.
- To mathematize the daily life problems about the
shares of some companies or banks or business
solve them.
Required Teaching Materials/ Resources
Colourful chart-paper with definitions of buyer, seller, agent, commission and formulae,
bills, income tax rates published by IRD etc
Pre-knowledge: cost price, selling price, profit and loss, discount and VAT
A. Bonus
Teaching Activities
1. Start the classroom discussion with the following questions:
(i) Have you heard about bonus?
(ii) Have anyone of your family members received bonus yet?
(iii) What is Net profit?
(iv) How do you define bonus?
15 Vedanta Excel in Mathematics Teachers' Manual - 9
2. Explain bonus as the extra amount of money that is distributed to the employees by the
company for their goods performance from the certain percent of net profit of the fiscal
year.
3. Focus on group work to let the students formulate the concepts through examples.
4. Ask the formula of finding the bonus amount , bonus amount received by each employee,
bonus , explain the following formulae
(i) Bonus amount = Bonus % of net profit Total bonus amount
No. of employee
(ii) Bonus amount received by each employee =
(iii) Bonus percent = Bonus amount × 100%
Net profit
(vi) Annual income = annual salary + bonus
Solution of selected problems from Vedanta Excel in Mathematics
1. A publication housed announced to distribute 10% bonus equally to its 20 employees
from the net profit of Rs 18,36,000 at the end of a fiscal year, find the bonus received
by each employee.
Solution:
Here, net profit of the company = Rs 18,36,000, bonus percent = 10%, number of employees = 20
Now, bonus amount =10% of Rs 18,36,000 = Rs 1,83,600
Rs 1,83,600
Bonus amount received by each employee = 20 = Rs 9,180
Hence, each employee received Rs 9,180.
2. A garment factory announced 20% bonus to its 25 workers from the net profit at the
end of last fiscal year. If every worker received Rs 18,500, how much was the profit
of the factory?
Solution:
Here, bonus percent= 20%, number of employees = 25, bonus amount received by each
worker = Rs 18,500
Now, bonus amount =25× Rs 18,500 = Rs 4,62,500
Let net profit of the factory be Rs x
Then, bonus amount = bonus % of net profit
or, Rs 4,62,500 = 20% of x = 0.2x
?x = Rs 23,12,500
Hence, the profit of the factory is Rs 23,12,500.
3. A business company distributed bonus to its 24 employees from the net profit of
Rs 16,48,000. If every employee received Rs 8,240, what was the bonus percent?
Solution:
Here, net profit of the company =Rs 16,48,000, number of employees = 24, bonus amount
received by each worker = Rs 8,240, bonus percent = ?
Bonus amount =24× Rs 8,240 = Rs 1,97,760 Rs 1,97,760
Bonus amount Rs 16,48,000
Now, bonus percent = Net profit × 100% = × 100% = 12%
Hence, the required bonus percent was 12%.
4. A garment factory made a net profit of Rs 48,00,000 in the last year. The management
of the factory decided to distribute 18% bonus from the profit to its 25 employees.
(i) Find the bonus amount received by each employee.
(ii) By what percent should the bonus be increased so that each employee can
receive Rs 38,400?
(iii) What should be the profit of the company so that it can provide Rs 40,000 to
each at 15% bonus?
Vedanta Excel in Mathematics Teachers' Manual - 9 16
Solution:
Here, net profit of the company = Rs 48,00,000, bonus percent = 18%
number of employees = 25
(i) Bonus amount =18% of Rs 48,00,000 = Rs 8,64,000
Rs 8,64,000
Bonus amount received by each employee = 25 = Rs 34,500
Hence, each employee received Rs 34,500.
(ii) Net profit of the company =Rs 48,00,000, number of employees = 25, bonus amount
received by each employee= Rs 38,400, bonus percent = ?
Bonus amount =25× Rs 38,400 = Rs 9,60,000
Bonus amount Rs 9,60,000
Bonus percent = Net profit × 100% = Rs 48,00,000 × 100% = 20%
Hence, the bonus should be increased by 20% - 18% = 2% so that each employee can
receive Rs 38,400.
(iii) Bonus percent= 15%, number of employees = 25, bonus amount received by each
worker = Rs 40,000
Now, bonus amount =25× Rs 40,000 = Rs 10,00,000
Let net profit of the factory be Rs x
Then, bonus amount = bonus % of net profit
or, Rs 10,00,000 = 15% of x = 0.15x
? x = Rs 66,66,666.67
Hence, the profit of the factory is Rs 66,66,666.67
5. When a publication house increased its profit from 20% to 25%, the amount of profit
increased to Rs 52,08,000. If the company decided to distribute 60% bonus to its 30
employees equally from the increased amount of profit, how much bonus will each
employee receive?
Solution:
Let the yearly income of the publication house be Rs x
Then, 25% profit of yearly income = Rs 52,08,000
or, 0.25x = Rs 52,08,000
? x = Rs 2,08,32,000
Also, 20% profit of the yearly income = 20% of Rs 2,08,32,000 = Rs 41,66,400
Increased amount of profit = Rs 52,08,000 – Rs 41,66,400 = Rs 10,41,600
Now, bonus amount = 60% of Rs 10,41,600 = Rs 6,24,960
Rs 6,24,960
?Bonus amount received by each employee = 30 = Rs 20,832
Hence, each employee received Rs 20,832.
Extra Questions
1. A noodle factory announced to distribute 10% bonus equally to its 45 employees from
the net profit of Rs 34,65,000 at the end of a fiscal year, find the bonus received by each
employee. [Ans:Rs 7700]
2. A garment factory announced 20% bonus to its 70 workers from the net profit at the end
of last fiscal year. If every worker received Rs 15,000, how much was the profit of the
factory? [Ans:Rs 52,50,000]
3. The management of a supermarket decided to distribute a bonus to its 180 employee
from the net profit of Rs 3,00,00,000 at the end of a fiscal year. If every employee received
Rs 25,000, what was the bonus percent? [Ans:15%]
17 Vedanta Excel in Mathematics Teachers' Manual - 9
B. Commission
Teaching Activities
1. With story/ related examples on commission, explain the words sales, agent and
commission.
2. Define commission as the amount of money paid to the agent for performing the business
service such as buying and selling goods, property (land, building, car etc.) or collection
of money.
3. Under discussion, explain the following formulae
(i) Commission amount = commission % of total sales
Commission amount
(ii) Commission percent = Total sales × 100%
Monthly income = salary+ commission
Solution of selected problems from Vedanta Excel in Mathematics
1. A real estate company gives 5% commission on selling a piece of land for Rs 10,00,000
and 7% commission for the additional amount of selling price above the fixed price.
If the agent sold the land for Rs 12,99,000, how much commission did he/she receive
from the company?
Solution:
Here, the fixed selling price of the land = Rs 10,00,000
The selling price of the land = Rs 12,99,000
Now, the commission received by the agent = 5% of Rs 10,00,000 + 7% of (Rs 12,99,000 – Rs
10,00,0000) = Rs 50,000 + Rs20,930 = Rs70,930
Hence, the agent received the commission of Rs 70,930.
2. The monthly salary of a sales person of a subway restaurant is Rs 21,600 and an
additional incentive of 1.5% on the total monthly sale is provided as commission.
(i) Calculate his/her total income in a month if he/she makes a total sale of Rs
5,80,000 in the month.
(ii) What should be his/her total sale in the next month so that he/she can receive a
total income of Rs 31,350 in the month?
Solution:
Here, the monthly salary of a sales person = Rs 21,600
(i) The total sales of the month = Rs 5,80,000 and commission percent = 1.5%
Now, the commission received by the sales person = 1.5% of Rs 5,80,0000 = Rs 8,700
Hence, the income of the sales person in the month = salary + commission
= Rs 21,600 + Rs 8,700 = Rs 30,300
(ii) The income of the next month = Rs 31,350
? Commission amount received in the next month = Rs 31,350 – Rs 21,600 = Rs 9,750
Let the total sales of the next month be Rs x.
Then, the commission amount = commission % of total sales
or, Rs 9,750 = 1.5% of x
or, Rs 9,750 = 0.015x
? x = Rs 6,50,000
Hence, the total sales of the next month should be Rs 6,50,000 so that he/she can receive
a total income of Rs 31,350 in the month.
3. Mr. Bibek is an online salesperson in an online shopping store. His monthly salary
Rs 18,700 and 2% commission is given to him when the monthly sales is more than
5 lakh rupees. If the sales of the store in a month is Rs 7,20,000, calculate his total
income of the month.
Vedanta Excel in Mathematics Teachers' Manual - 9 18
Solution:
Here, the monthly salary of a salesperson = Rs 18,700
The fixed sales of the month = 5 lakh rupees = Rs 5,00,000
The total sales of the month = Rs 7,20,000
The sales eligible for commission = Rs 7,20,000 – Rs 5,00,000 = Rs 2,20,000
Now, the commission received by the sales person = 2% of Rs 2,20,0000 = Rs 4,400
Hence, the income of the sales person in the month = salary + commission
= Rs 18,700 + Rs 4,400 = Rs 23,100
4. Mrs. Nepali draws Rs 19,800 as her monthly salary in a wholesale cosmetic shop and
a certain commission is given as per the monthly sales. If the sales of the month is
Rs 12,00,000 and her total income of the month including commission is Rs 31,800,
find the rate of commission.
Solution:
Here, monthly salary = Rs 19,800 and the income of the month = Rs 31,800
?Commission amount received in the next month = Rs 31,800 – Rs 19,800 = Rs 12,000
The total sale of the month = Rs 12,00,0000
Commission amount Rs 12,000
Now, commission percent = Total sales × 100% = Rs 12,00,000 × 100% = 1%
Hence, the rate of the commission is 1%.
Extra Questions
1. Mr. Shrestha is a salesperson in a hardware shop. His monthly salary Rs 22,500 and
1.5% commission is given to him when the monthly sales is more than 10 lakh rupees.
If the sales of the hardware shop in a month is Rs 17,50,000, calculate his total income
of the month. [Ans: Rs 33,750]
2. Aravi draws Rs 15,000 as her monthly salary in a wholesale fancy shop and a certain
commission is given as per the monthly sales. If the sales of the month is Rs 10,00,000
and her total income of the month including commission is Rs 27,500, find the rate of
commission. [Ans:1.25%]
3. The monthly salary of Sanjay, a salesman of a departmental store, is Rs 25,000 and an
additional incentive of 1.5% on the total monthly sale is provided as commission.
(i) Find his total income in a month if he makes a total sale of Rs 5,55,000 in the month.
(ii) What should be his total sale in the next month so that he can receive a total income
of Rs 38,320 in the month? [Ans:Rs 33,250;Rs 8,88,000]
C. Income Tax
Teaching Activities
1. Ask about the yearly income of parents of the student and discuss upon the tax is to be
paid to the government.
2. Divide the students into groups and give them to study the printed form of the present
rates of income taxes fixed by Inland Revenue Department (IRD) and discuss upon the
following questions
(i) What is income tax?
(ii) Which authentic body is responsible to collect the tax?
(iii) Why should we pay tax to the government?
(iv) What do you mean by taxable income?
(v) Which incomes are entitles for tax rebate?
(vi) What is the rate of social security tax?
19 Vedanta Excel in Mathematics Teachers' Manual - 9
3. Present the income tax slabs for individual and married couple separately on a chart
paper.
4. Tell the students to write the important notes and formulae on the colourful chart paper
as project work
5. With some related examples, let the students identify the following formulae then
explain them one by one.
(i) Taxable income = Yearly income – tax free income
Income tax
(ii) Income tax = rate of tax of taxable income (iii) Tax rate = Taxable income × 100%
Solution of selected problems from Vedanta Excel in Mathematics
1. Mr Sunil Jha is a Secomdary level Mathematics Teacher in a community school. His
monthly salary is Rs 38,700 and one month’s salary as Dashain Bonus. 10% of his
salary is deducted to deposit in his provident fund. If his marital status is single,
calculate the annual income tax paid by him.
Solution:
Here, his monthly income after deducting the provident fund
= Rs 38,700 – 10% of Rs 38,700 = Rs 34,830
His annual income with Dashain bonus = 12Rs 34,830 + Rs 38,700 = Rs 4,56,660
Social security tax up to Rs 4,00,000 = 1% of Rs 4,00,660 = Rs 4,000
Taxable income = Rs 4,56,660 – Rs 4,00,000 = Rs 56,660
Income tax for Rs 56,660 = 10% of Rs 56,660 = Rs 5,666
Hence, the total annual income tax paid by him = Rs 4,000 + Rs 5,666 = Rs 9666
2. After deducting 10% provident fund a married person draws Rs 40,500 salary per
month and one month’s salary as festival bonus. He/she pays Rs 14,500 annually as
premium of his/her insurance. Calculate the annual income tax paid by the person.
Solution:
Here, his monthly income after deducting the provident fund = Rs 40,500
Let the monthly income of the person be Rs x.
Then, x – 10% of x = Rs 40,500 ?x = Rs 45000
His annual income with Dashain bonus = 12Rs 40,500 + Rs 45,000 = Rs 5,31,000
Taxable income after premium of insurance = Rs 5,31,000 – Rs 14,500 = Rs 5,16,500
Social security tax up to Rs 4,50,000 = 1% of Rs 4,50,660 = Rs 4,500
Taxable income for 10% tax = Rs 5,16,500 – Rs 4,50,000 = Rs 66,500
Income tax for Rs 66,500 = 10% of Rs 66,500 = Rs 6,650
Hence, the total annual income tax = Rs 4,500 + Rs 6,650 = Rs 11,150
3. Mr. Sayad Sharma an unmarried employee of a UN Project draws monthly salary
of Rs 51,000 after deducting 10% salary in his provident fund and 5% in citizen
investment trust. He also receives a Dashain Bonus of one month’s salary. He pays
Rs 22,000 annually as the premium of his life insurance. How much income tax does
he pay in a year?
Solution:
Here, his monthly income after deducting 10% provident fund and 5% citizen investment
trust = Rs 51,000
Let his monthly income before deducting the provident fund and citizen investment trust
be Rs x.
Vedanta Excel in Mathematics Teachers' Manual - 9 20
Then, x – (10 + 5) % of x = Rs 40,500 ?x = Rs 60,000
His annual income with Dashain bonus = 12Rs 51,000 + Rs 60,000 = Rs 6,72,000
Taxable income after premium of insurance = Rs 6,72,000 – Rs 22,000 = Rs 6,50,000
Social security tax up to Rs 4,00,000 = 1% of Rs 4,00,660 = Rs 4,000
Taxable income for 10% tax = Rs 5,00,000 – Rs 4,00,000 = Rs 1,00,000
Income tax for Rs 1,00,000 = 10% of Rs 1,00,000 = Rs 10,000
Taxable income for next 20% tax = Rs 6,50,000 – Rs 5,00,000 = Rs 1,50,000
Income tax for Rs 1,50,000 = 20% of Rs 1,50,000 = Rs 30,000
Hence, the total annual income tax = Rs 4,000 + Rs 10,000 + Rs 30,000 = Rs 44,000.
Extra Questions
1. The monthly income of an unmarried individual is Rs 45,000. If 1% social security tax is
charged up to Rs 4,00,000. Then 10% and 20% are charged for the next Rs1,00,000 and
Rs 2,00,000 respectively. Calculate the annual income tax paid by the individual.
[Ans: Rs 22,000]
2. Rumakanta Jha is a married professor. His monthly income is Rs 55,000. If 1% social
security tax is charged up to Rs 4,50,000; 10% tax for the income from Rs 4,50,000 to
Rs 5,50,000 and 20% tax from Rs 5,50,000 to Rs 7,50,0000 are to be paid. Calculate the
annual income tax paid by the individual. [Ans: Rs 36,500]
3. Mrs Pandey is a Branch Manager of a commercial bank. Here monthly salary is Rs 95,400
and 10% of her salary is deducted as provident fund. She pays Rs 24,520 as the premium
of her life insurance. If 1% social security tax is levied upon the income of Rs 4,50,000,
10%,20% and 30% taxes are levied upon the next incomes of Rs 1,00,000, Rs2,00,000
and up to Rs 12,50,000 respectively, how much income tax should she pay in a year?
[Ans: Rs 1,31,240]
D. Dividend
Teaching Activities
1. Recall the bonus.
2. Create a short story about dividend and tell in the class.
3. Define dividend as the certain amount distributed among the shareholders of a
corporation as per the number of shares from the net profit.
4. Ask the following questions during classroom discussion
(i) What is dividend?
(ii) Among whom the dividend is distributed?
(iii) Tell the difference between the bonus and the dividend.
5. Make a discussion upon the following formulae with examples.
(i) Dividend = Rate of dividend (in %) Net profit
(ii) Dividend = Value of dividend per share Total number of shares
(iii) Value of dividend for each share = Dividend amount
No. of shares
Dividend amount
(iv) Rate of dividend (in %) = No. of shares × 100%
21 Vedanta Excel in Mathematics Teachers' Manual - 9
Solution of selected problems from Vedanta Excel in Mathematics
1. Mrs Rai bought 250 shares out of 10,000 shares from a financial company. The
company earned a net profit of Rs 85,20,000 and declared 17% dividend to its
shareholders. Calculate the amount of dividend received by Mrs Rai.
Solution:
Here, dividend amount = 17% of Rs 85,20,000 = Rs 14,48,400
Dividend amount Rs 14,48,400
Value of dividend for each share = No. of shares = 10,000 = Rs 144.84
?Dividend for 250 shares = 250 Rs 144.84 = Rs 36,210
Hence, Mrs Rai received Rs 36,210 as dividend.
2. A Cable Car Company sold 3000 shares to the local people. The company earned
a profit of Rs 1,20,00,000 in a year and distributed a certain percent of profit as
dividend. If a shareholder who has bought 125 shares received Rs 1,10,000 dividend,
what percent of profit was distributed as dividend?
Solution:
Rs 1,10,000
Here, dividend amount distributed for 1 share = 125 = Rs 880
?Dividend amount distributed for 3000 shares = 3000Rs880 = Rs 26,40,000
Net profit = Rs 1,20,00,000 Rs 26,40,000
Dividend amount Rs 1,20,00,000
Now, rate of dividend = Net profit × 100% = × 100% = 22%
Hence, 22 of the profit was distributed as dividend.
3. Mr. Dhurmus bought 500 shares out of 10,000 shares sold by a commercial bank. The
bank earned some profit and distributed 14% of the net profit as the dividend in a
year. If Dhurmus received Rs 1,03,600 in the year, find the net profit of the bank.
Solution:
Here, dividend received for 500 shares = Rs 1,03,600
1,03,600
or, dividend received for 1 share = Rs 500 = Rs 207.20
?Dividend distributed for 10,000 shares = 10,000 Rs 207.20 = Rs 20,72,000
Let the net profit of a year of the commercial bank be x.
Then, dividend amount = Rate of divided Net profit
or, Rs 20,72,000 = 14% of x ?x = Rs 1,48,00,000
Hence, the net profit of the bank was Rs 1,48,00,000
Extra Questions
1. Mr Dahal bought 400 shares out of 20,000 shares from a Business Company. The company
earned a net profit of Rs 2,25,00,000 and declared 15% dividend to its shareholders.
Calculate the amount of dividend received by Mr Dahal. [Ans: Rs67,500]
2. Mr Jeevan bought 225 shares out of 10,000 shares from commercial bank. If the company
earned a profit of Rs 1,50,00,000 in a year and distributed a certain percent of profit
as dividend. If Mr Jeevan received Rs 54,000 dividend, what percent of profit was
distributed as dividend? [Ans:16%]
3. Rajesh bought 300 shares out of 5,000 shares sold by an insurance company. The
company earned some profit and distributed 20% of the net profit as the dividend in a
year. If Rajeshs received Rs 66,600 in the year, find the net profit of the bank.
[Ans:Rs 55,50,000]
Vedanta Excel in Mathematics Teachers' Manual - 9 22
Unit Household Arithmetic
4
Allocated teaching periods 10
Competency
- To solve the daily life problems by using the basic rules of household arithmetic and
financial arithmetic.
Learning Outcomes
- To solve the problems on household arithmetic such as electricity bills, telephone bills,
water bills, taxi meter reading (including discount and VAT)
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To define 1 unit of electricity
- To tell the formula of finding the consumed units
- To tell the formula of finding the tariff (Sub-total)
- To write the relation of total charge, tariff, TSC and
VAT
1. Knowledge (K) - To define 1 unit of water consumption
- To name the concerned authority of ministry of water
supply
- To name the concerned authority for implement the
rules and regulations regarding taximeter.
- To recall the minimum taxi fares during 6:00 am to
9:00 pm or 9:00 pm to 6:00 am.
- To find the electricity charge for simple problems
2. Understanding (U) - To calculate the telephone charge
- To calculate the taxi fare
- To find the electricity charge with rebate/fine
- To find the number of consumed units of electricity
3. Application (A) - To apply the rules and find the water charge
- To find the telephone charge with TSC and VAT
- To estimate the consumed unit of electricity of a
4. High Ability (HA) month and find the bill amount as per present rate of
electricity
- To prepare report on few ways of reducing unnecessary
use of electricity/telephone/water and use of ICT used
for payment of the bills
Required Teaching Materials/ Resources
Electricity bills, models of meter box, telephone bills, water bills, calculator, recharge card,
chart paper mentioning the taxi fare rates etc.
23 Vedanta Excel in Mathematics Teachers' Manual - 9
Pre-knowledge: discount, VAT
A. Electricity bills
Teaching Activities
1. Ask about the electric appliances used in the students’ houses.
2. With some samples/models of meter readings of two successive months, ask the following
questions:
(i) What is the reading of recent month?
(ii) What is the reading of preceding month?
(iii) What is the number of consumed units of electricity?
(iv) What do you mean by 1 unit of consumed electricity?
3. Present the recent rates, rules and regulations of electricity fixed by Nepal Electricity
Authority (NEA) on chart paper or power point presentation or though available website.
4. Encourage the students to find the electricity charge with electricity service charge and
rebate/fine.
Solution of selected problems from Vedanta Excel in Mathematics
1. The rate of electricity charge up to 20 units is Rs 3 per unit and Rs 7 per unit from 21
to 30 units. Find the charge of consumption of 28 units with Rs 50 service charge.
Solution:
Here, consumption of electricity = 28 units
Rate of charge up to 20 units = Rs 3 per unit
? Charge up to 20 units = 20 ×Rs 3 = Rs 60
The excessive number of units = 28 – 20 = 8 units
? Charge of excessive 8 units = 8 ×Rs 7 = Rs 56
Total charge of electricity with service charge = Rs 60 + Rs 56 + Rs 50 = RS 166
2. Electricity tariff rates and rebate/fine rules are given below
kWh 5 Ampere 15 Ampere 30 Ampere 60 Ampere
(Monthly)
Service Energy Service Energy Service Energy Service Energy
Units Charge Charge Charge Charge Charge Charge Charge Charge
per unit per unit per unit per unit
0 – 20 Rs30.00 Rs 3.00 Rs50.00 Rs4.00 Rs75.00 Rs5.00 Rs125.00 Rs6.00
The rules of rebate/fine
Meter Within 7 Within 8-22 Within 23-30 Within 31-40 Within 41-60
reading days days days days days
Rebate/fine 3% rebate - 5% fine 10% fine 25% fine
From the tables given above, workout the following problems
a) A household having 5A electricity meter consumed 18 units of electricity in one
month. Find the amount of payment made by the household within 7 days.
b) A household having a 15A meter consumed 16 units of electricity in one month. Find
the amount of payment made by the household on 25th day of meter reading.
c) A household having a 30A meter consumed 19 units of electricity in one month and
if the payment was made on 20th day of meter reading, find the amount of payment.
Vedanta Excel in Mathematics Teachers' Manual - 9 24
d) A household having a 60A meter consumed 20 units of electricity in one month
and the payment was made on 50th day of meter reading. Calculate the amount of
payment made by the house.
Solution:
(a) Here, capacity of meter box = 5A, consumption of electricity = 18 units
Rate of charge up to 20 units = Rs 3 per unit
Now, charge for 18 units = 18×Rs 3 = Rs 54
? Total charge of electricity with service charge = Rs 54 + Rs 30 = Rs 84
Since, the payment was made within 7 days of meter reading. So, 3% rebate was allowed.
Hence, the required payment was Rs 84 – 3% of Rs 84 = Rs 81.48
(b) Here, capacity of meter box = 15A, consumption of electricity = 16 units
Rate of charge up to 20 units = Rs 4 per unit
Now, charge for 16 units = 16×Rs 4 = Rs 64
? Total charge of electricity with service charge = Rs 64 + Rs 50 = Rs 114
Since, the payment was made on 25th day of meter reading. So, 5% fine was charged.
Hence, the required payment was Rs 114 + 5% of Rs114 = Rs 119.70
(c) Here, capacity of meter box = 30A, consumption of electricity = 19 units
Rate of charge up to 20 units = Rs5 per unit
Now, charge for 19 units = 19×Rs5 = Rs95
? Total charge of electricity with service charge = Rs95 + Rs75 = Rs170
Since, the payment was made on 20th day of meter reading. So, there was no rebate no fine.
Hence, the required payment was Rs170.
(d) Here, capacity of meter box = 60A, consumption of electricity = 20 units
Rate of charge up to 20 units = Rs 6 per unit
Now, charge for 20 units = 20 ×Rs 6 = Rs 120
? Total charge of electricity with service charge = Rs 120 + Rs 125 = Rs 245
Since, the payment was made on 50th day of meter reading. So, 25% fine was charged.
Hence, the required payment was Rs 245 + 25% of Rs 245 = Rs 306.25
3. Mr. Sharma has a 5A meter in his house. He uses 5 CFL bulbs of 15 watt each for
4 hours and an electric heater of 1200 watt for 1 hour every day. Find the cost of
payment of the bill of the month at the rate of Rs 3 per unit up to 20 units, Rs 7 per
unit from 21 to 30 units and Rs 8.50 from 31-50 units with Rs 75 service charge, if the
payment is made on 10th day of the meter reading.
Solution:
Here, consumption of electricity in 1 day = (5 × 15×4 + 1200×1) watts = 1500 watts
Consumption of electricity in 1 month = 30×1500 watts = 45000 watts = 45 kW
? Number of consumed units = 45 units because the time duration of use of electric
appliances were measured in hours.
Now,
Consumption block No. of units Rate of charge Electricity charge
0 – 20 20 – 0=20 Rs 3 20×Rs 3 = Rs 60
21 – 30 30 – 20 = 10 Rs 7 10×Rs 7 = Rs 70
31 – 50 45 – 30 = 15 Rs 8.50 15×Rs 8.50 = Rs 127.50
25 Vedanta Excel in Mathematics Teachers' Manual - 9
Total charge of electricity with service charge = energy charge + service charge
= Rs 60 + Rs70 + Rs 127.50 + Rs 75 = Rs 332.50
Since, the payment was made on 10th day of meter reading. So, there was neither rebate nor
fine.
Hence, the required payment was Rs332.50
4. Mrs.Bajracharya’s house has a 15A meter. She uses 5 LED bulbs of 10 watt each for
4 hours, 2 televisions of 60 watt each for 5 hours and a refrigerator of250 watt for 2
hours every day. Find the cost of payment of the bill of the month at the rate of Rs4
per unit up to 20 units, Rs 7 per unit from 21 to 30 units and Rs 8.50 from 31-50 units
with Rs100 service charge, if the payment is made on 35th day of the meter reading.
Solution:
Here, consumption of electricity in 1 day = (5×10×4 + 2×60×5 + 1×250×2) watts
= 1300 watts
Consumption of electricity in 1 months = 30×1300 watts = 39000 watts = 39 kW
? Number of consumed units = 39 units because the time duration of use of electric
appliances were measured in hours.
Now,
Consumption No. of units Rate of charge Electricity charge
block
0 – 20 20 – 0 = 20 Rs4 20×Rs4 = Rs 80
21 – 30 30 – 20 = 10 Rs 7 10×Rs 7 = Rs 70
31 – 50 39– 30 = 9 Rs 8.50 9×Rs 8.50 = Rs76.50
Total charge of electricity with service charge = energy charge + service charge
= Rs 80 + Rs 70 + Rs76.50 + Rs100 = Rs 326.50
Since, the payment was made on 35th day of meter reading. So, 10% fine was added
Hence, the required payment was Rs326.50 + 10% of Rs 326.50 = Rs 359.15
5. The meter box of a family house is 15 A. If the family made the payment of Rs
1336.50 with service charge of Rs125 on 36th day of meter reading, how many units
of electricity was consumed in the month? Calculate it under the following rates.
Units 0 – 20 21 – 30 31 – 50 51 – 150
Rate of charge per unit Rs 4 Rs 7 Rs 8.50 Rs 11
Payment up to 40th day from the meter reading – 10% fine
Solution:
Let the number of consumed units in the month be x units
Now,
Consumption block No. of units Rate of charge Electricity charge
0 – 20 20 – 0 = 20 Rs 4 20×Rs 4 = Rs 80
21 – 30 30 – 20 = 10 Rs 7 10×Rs 7 = Rs 70
31 – 50 50 – 30 = 20 Rs 8.50 20×Rs 8.50 = Rs170
51 – 150 Rs 11 (x – 50)×Rs 11 = Rs (11x – 550)
x – 50
Total charge of electricity with service charge = energy charge + service charge
Vedanta Excel in Mathematics Teachers' Manual - 9 26
= Rs 80 + Rs 70 + Rs170 + + 11x – Rs 550 + Rs 125 = Rs(11x – 105)
Also, extra fine = 10% of Rs (11x – 105) = Rs (1.1x – 10.5)
According to question,
Payment of bill with fine = Rs 1336.50
or,Rs (11x – 105) + Rs (1.1x – 10.5) = Rs 1336.50 ? x = 120
Hence, the required payment number of consumed units is 120.
Extra Questions
1. The rate of electricity charge up to 20 units is Rs 3 per unit and Rs 7 per unit from 21
to 30 units. Find the charge of consumption of 24 units with Rs 50 service charge.
[Ans:Rs138]
2. The rate of electricity charge up to 20 units is Rs 3 per unit and Rs 7 per unit from 21 to
30 units. If a family paid the bill of Rs 166 with Rs 50 service charge, how many units
of electricity was consumed in the month? [Ans: 28]
3. The meter readings of Krishna’s house in 1st Asar was 02967 and 1stSharwan was 03015.
Find the electricity charge for the month of Asar according to the given information if
the bill was made on 5th day from meter reading.
KWh (units) Service charge Energy charge per unit
0 – 20 Rs 30 Rs3
21 – 30 Rs 50 Rs 7
31 – 50 Rs 75 Rs 8.50
Payment within 7 days of meter reading – 3% rebate [Ans: Rs347.26]
4. The meter box of a family house is 15 A. If the family made the payment of Rs1331 with
service charge of Rs125 on 36th day of meter reading, how many units of electricity was
consumed in the month? Calculate it under the following rates.
Units 0 – 20 21 – 30 31 – 50 51 – 150
Rate of charge per unit Rs 4 Rs 7 Rs 8.50 Rs 11
Payment up to 40th day from the meter reading – 10% fine [Ans: 110 units]
B. Telephone bills
Teaching Activities
1. With a recharge card, discuss about the mobile network service
2. Provide some telephone bills to the students, and ask the following questions:
(i) What is concerned authority for the implementation of this bill?
(ii) To whom is the telephone bill issued?
(iii) What is the address of the telephone line? (iv) What type of telephone is it?
(v) What status of telephone is mentioned? (vi) What type of telephone is it?
(vii) What is the previous reading of telephone given in the bill?
(viii) What is the current reading of telephone given in the bill?
(ix) How many telephone calls is made in the month? (x) What is the rental amount?
(xi) How many extra calls are made in the month?
27 Vedanta Excel in Mathematics Teachers' Manual - 9
(xii) What is the sub-total amount given in the bill?
(xiii) What amount is to be paid for TSC?
(xiv) What amount of VAT is to be paid?
(xv) What is the grand total given in the bill?
3. Explain the telephone billing system implemented by Nepal Doorsanchar Company Ltd.
4. List the following formulae after discussion
(i) Tariff = Sub-total = Minimum charge + Extra charge
(ii) Telecom service charge (TSC) = 10% of Sub-total
(iii) Total = Sub-total + TSC (iv) VAT amount = VAT% of Total = 13% of Sub-total
(v) Grand total = Total + VAT amount
Solution of selected problems from Vedanta Excel in Mathematics
1. The minimum charge up to 175 calls is Rs 200. If the charge for each additional call
is Re 1, how much will be the charge for 475 calls with 10% TSC and 13% VAT?
Solution:
Here, minimum charge up to 175 calls = Rs 200
The additional number of calls = 475 – 175 = 300
The charge for additional calls = 300 × Re 1 = Rs 300
Now, sub-total = Minimum charge + additional charge = Rs 200 + Rs 300 = Rs 500
TSC = 10% of Rs 500 = Rs 50
Also, total = sub-total + TSC = Rs 500 + Rs 50 = Rs 550
Again, grand-total = Total + VAT% of Total = Rs 550 + 13% of Rs 550 = Rs 621.50
Hence, the total charge for 375 calls is Rs 621.50
2. The minimum charge of telephone calls up to 175 calls is Rs 200. The charge for each
extra call of 2 minutes duration is Re 1. If the household paid Rs 633.93 with 10%
TSC and 13% VAT to clear the bill of the month, find the total number of calls made
in the month?
Solution:
Let the charge of telephone calls without VAT be Rs x.
Then, grand total = Total + VAT% of total
or, Rs 633.93 = x + 13% of x ? x = Rs 561
Also, let the charge of telephone calls without TSC be Rs y.
Then, total = sub-total + TSC or, Rs 561 = y + 10% of y ? y = Rs 510
Again, the charge for extra calls = Rs 510 – Rs 200 = Rs 310
Rs 310
Now, the number of extra calls = Re 1 = 310
Hence, the total number of calls of the month is 175 + 310 = 485 calls
Extra Questions
1. The minimum charge up to 175 calls is Rs 200. If the charge for each additional call is
Re 1, how much will be the charge for 375 calls with 10% TSC and 13% VAT?
[Ans: Rs 497.20]
2. The reading of the Baishakh-1 of local calls in Supriya’s house is 5270 and that of the
Jestha-1 is 5605.The minimum charge up to 175 calls is Rs 200 and charge for each
additional call is Re 1.
Vedanta Excel in Mathematics Teachers' Manual - 9 28
(i) How many calls are made in the month of Baishakh?
(ii) What is the bill amount of telephone callswith 10% TSC and 13% VAT?
[Ans: (i) 635 calls, (ii) Rs 820.38]
3. The minimum charge of telephone calls up to 175 calls is Rs 200. The charge for each
extra call of 2 minutes duration is Re 1. If the household paid Rs 559.35 with 10% TSC
and 13% VAT to clear the bill of the month, find the total number of calls made in the
month? [Ans: 425 calls]
C. Water bills
Teaching Activities
1. Provide some water bills to the students, and ask the following questions:
(i) What is concerned authority for the implementation of this bill?
(ii) How many units of water is consumed?
(iii) What is the size of pipe?
(iv) What is the previous reading of water consumption?
(v) What is the present reading of water consumption?
(vi) What is the minimum charge?
(vii) What is the additional charge?
(viii) What is the total charge?
2. Discuss about 1 unit of water consumption
3. Show the water tariff rules in the chart paper implemented by Nepal Water Supply
Corporation.
S.N. Size of pipe Tap with meter Taps without meter
Main Tap Branch Tap
Minimum Minimum Additional
consumption Charge consumption per Charge(Rs) Charge (Rs)
1. 1" (litre) 110 1000 litre (Rs) 560 200
2 10,000 1490 25 3360 1600
3420 9200 2700
2. 3" 27, 000 40
4
56,000 40
3. 1"
The compulsory provision of sewerage service charge = 50% of the water consumption
charge.
Payment schedule of the bill
S.N. Payment is made after the bill issued Rebate/ Fine
1. Within the 1st and the 2nd month 3% rebate
2. Within the 3rd month No rebate and no fine
3. Within the 4th month 10% fine
4. Within the 5th month 20% fine
5. After 5th month 50% fine
4. Present the water tariff rules implemented by Kathmandu Upatyaka Khanepani
Limited (KUKL)in the chart paper.
29 Vedanta Excel in Mathematics Teachers' Manual - 9
S.N. Size of Tap with meter Taps without
pipe meter
Minimum Minimum Additional Branch Tap
Charge (Rs)
consumption Charge consumption per
785
(litre) 1000 litre (Rs)
4595
1. 1" 10,000 100 32
2 9540
2. 3" 27, 000 1910 71
4
3. 1" 56,000 3960 71
Solution of selected problems from Vedanta Excel in Mathematics
1. 127 units of water is consumed by using 34"pipe in a hotel. If the payment of the bill
is made within the fifth month after the bill issued, how money is required to clear
the bill with 50% sewerage service charge?
Solution:
Here, 3"
4
According to the water tariff provisions of NWS for the use of pipe of the size
The minimum charge up to 27000 litres i.e., 27 units = Rs1490
The charge of additional units = Rs 40 per unit
The additional number of units = 127 – 27 = 100 units
The charge for additional units = 100 × Re 40 = Rs4000
Now, total charge = Rs 1490 + Rs 4000 = Rs 5,490
Again, the charge including 50% sewerage service = Rs 5490 + 50% of Rs 5490 = Rs 8235
1"
2. A household uses 2 of water pipe. The meter reading of the household on 1st of Asar
was 1260 units and on 1st of Shrawan was 1330 units. Calculate the charge to be
paid including 50% sewerage service charge if the payment of the bill is made in the
following schedule.
(i) Within the first month after the bill issued
(ii) Within the third month after the bill issued
(iii) Within the fifth month after the bill issued
(iv) Within the seventh month after the bill issued
Solution:
Here,
The meter reading of 1st Asar = 1260 and that of 1st Shrawan = 1330
? Consumed units of water = 1330 – 1260 = 70 1"
2
According to the water tariff provisions of NWS for the use of pipe of the size
The minimum charge up to 10000 litres i.e., 10 units = Rs110
The charge of additional units = Rs 25 per unit
The additional number of units = 70 – 10 = 60 units
The charge for additional units = 60 × Re 25 = Rs1500
Now,
Vedanta Excel in Mathematics Teachers' Manual - 9 30
Total charge = Rs110 + Rs 1500 = Rs 1610
Again, the charge including 50% sewerage service = Rs 1610 + 50% of Rs 1610 = Rs 2415
(i) When the payment of the bill is made within the first month after the bill issued, 3%
rebate is allowed
? Required payment = Rs 2415 – 3% of Rs 2415 = Rs 2342.55
(ii) When the payment of the bill is made within the third month after the bill issued, there
is no rebate no fine.
? Required payment = Rs 2415
(iii) When the payment of the bill is made within the fifth month after the bill issued 20%
fine is charged
? Required payment = Rs 2415 + 20% of Rs 2415 = Rs 2898
(iv) When the payment of the bill is made within the seventh month after the bill issued,
50% fine is charged
? Required payment = Rs 2415 + 50% of Rs 2415 = 3622.50
Extra Questions
1. 18 units of water is consumed by using 12"pipe in Rameshwor’s house. If the payment of
the bill is made within the second month after the bill issued, how money is required to
clear the bill with 50% sewerage service charge? [Ans:Rs 451.05]
3
2. 147 units of water is consumed by using 4 inch pipe in Everest Hotel. If the payment
of the bill is made within the fourth month after the bill issued, how much amount is
required to pay as bill? [Ans: Rs 10,378.50]
1"
3. A household uses 2 of water pipe. The meter reading of the household on 1st of Kartik
was 1420 units and on 1st of Mansir was 1480 units. Calculate the charge to be paid
including 50% sewerage service charge if the payment of the bill is made in the following
schedule.
(i) Within the second month after the bill issued
(ii) Within the third month after the bill issued
(iii) Within the fifth month after the bill issued
(iv) Within the sixth month after the bill issued
[Ans: (i) Rs 1076.70, (ii) Rs 1110, (iii) 1332 (iv) 1665]
D. Taxifare
Teaching Activities
1. Discuss upon taxi fare paid by parents or students themselves
2. Present the rules and regulations implemented and monitored by Nepal Bureau of
Standards and Metrology (NBSM) regarding the taxi fare as shown in the following table.
Time Minimum fare Fare of per Waiting charge per
200 meters 2 minutes
6:00 am to 9:00 pm Rs 14 Rs 7.20 Rs 7.20
9:00 pm to 6:00 am Rs 21 Rs 10.80 Rs 10.80
31 Vedanta Excel in Mathematics Teachers' Manual - 9
Solution of selected problems from Vedanta Excel in Mathematics
1. Mr. Kattel travelled 15 km by a hired taxi at 5:00 am. The minimum fare of Rs 21
appeared immediately after the meter was flagged down. Then, the fare went on at
the rate of Rs 10.80 per 200 metres. An additional waiting charge of Rs 10.80 per 2
minutes was charged for waiting of 10 minutes. Calculate the total of the taxi fare
paid by him.
Solution:
Here, the minimum fare = Rs 21, distance travelled = 15 km = 15000 m
Now,
The fare of 200 metres = Rs 10.80 or, the fare of 1 m = Rs 0.054
? The fare of 15000 m = 15000×Rs 0.054 = Rs 810
Also,
Waiting charge of 2 minutes = Rs 10.80 ? Waiting charge of 10 minutes = Rs 54
Hence, the total fare = Rs 21 + 810 + Rs 54 = Rs 885
2. Rita hired taxi and travelled a certain distance at 8:00 a.m. She paid the total fare
of Rs 194. If the minimum fare is Rs 14 and the fare per 200 metres is Rs 7.20, find the
distance travelled by her.
Solution:
Here, minimum charge = Rs 14
? The taxi fare excluding minimum fare = Rs 194 – 14 = Rs 180
200
Now, Rs 7.20 is the fare of 200 metres or, Re 1 is the fare of 7.20 metres
? Rs 180 is the 200 km
fare of 7.20 × 180 = 5000 meters = 5
Hence, she travelled 5 km.
Extra Questions
1. Mr. Koirala travelled 7 km by a hired taxi at 1:30 p.m. The minimum fare of Rs 14
appeared immediately after the meter was flagged down. Then, the fare went on at the
rate of Rs 7:20 per 200 metres. Calculate the total of the taxi fare paid by him.
[Ans: Rs 266]
2. Mrs. Maharjan travelled 10 km by a hired taxi at 4:45 am. The minimum fare of Rs 21
appeared immediately after the meter was flagged down. Then, the fare went on at the
rate of Rs 10.80 per 200 metres. An additional waiting charge of Rs 10.80 per 2 minutes
was charged for waiting of 10 minutes. Calculate the total of the taxi fare paid by him.
[Ans: Rs 615]
3. Smriti hired taxi and travelled a certain distance at 7:15 a.m. She paid the total fare of
Rs 158. If the minimum fare is Rs 14 and the fare per 200 metres is Rs 7.20, find the
distance travelled by her. [Ans: 4 km]
4. Rupesh hired taxi and travelled a certain distance at 10:00 p.m. He paid the total fare
of Rs 1155 including the waiting charge of 10 minutes. Find the distance travelled by
him. [Ans: 4 km]
Vedanta Excel in Mathematics Teachers' Manual - 9 32
Unit Mensuration
5
Competency Allocated teaching periods 22
- To find the area of plane surface and surface area and volume of solids then solving
real life problems based on cost estimation.
Learning Outcomes
- To estimate the cost of carpeting, constructing and gravelling the paths, painting,
papering, plastering etc. related to real life situations
- To find the problems based on area of walls, floor and ceiling of the room
- To find the cross sectional area, LSA , TSA and volume of prisms and solve the related
problems
- To prepare reports and project works on problems of area related to daily life situations
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
1. Knowledge (K) - To define the area of plane surface
2. Understanding - To tell the formula of finding the area of cross, inner and
3. (U) outer paths
4. - To relate the area, rate of cost per sq. units and the total
Application (A)
cost of working (carpeting, plastering etc.)
High Ability (HA) - To tell the formula of finding the area of four walls of
room
- To define cross-sectional area of prism
- To recall the general formulae of finding the LSA, TSA
and volume of the prism
- To area of plane figures (rectangle, square, parallelogram,
triangle, quadrilateral, trapezium and circle)
- To calculate the area of paths (cross, inner and outer
paths)
- To find the area of floor, walls and ceiling of room
- To find the volume of prism
- To find the surface area of prisms
- To find the area of paths and cost of graveling, covering
with stones
- To find the area of floor and cost of carpeting
- To find the area of walls and estimate the cost of
painting, papering, carpeting
- To solve the problems related to the volume and surface
area of prisms
- To estimate the number of bricks and cost required to
construct the walls etc.
- To mathematize the daily life problems related to area
and volume then solve them.
- To prepare the projects works and reports and present in
the class
33 Vedanta Excel in Mathematics Teachers' Manual - 9
Required Teaching Materials/ Resources
Colourful chart-paper with definitions and formulae, scale, scissors, pencil, geo-board,
tangram, graph paper, models of various prisms, box, cartoon, measuring tape, ICT tools etc.
Pre-knowledge: Perimeter and area of rectangle, parallelogram, volume of cube, cuboid etc.
A. Area of plane surfaces and paths
Teaching Activities
1. Divide the students into 4/5 groups and give them the figure of rectangle, square,
parallelogram, rhombus, circle, quadrilateral, trapezium etc. on the sheet of paper
and tell them to recall the perimeter and area formula of related figures and call for
presentation.
2. Discuss upon the following formulae with related figures and models.
(i) Area of rectangle = length (l) × breadth (b)
(ii) Area of square = (side)2 or 1 (diagonal)2
2
(iii) Area of parallelogram = base (b) × height (h)
(iv) Area of triangle = 1 ×base (b) × height (h)
(v) Area of 2 triangle = 21× base (b) ×
right angled perpendicular (p)
(vi) Area of equilateral triangle = 3 a2 where ‘a’ is the side length
4
(vii) Area of triangle = s(s – a) (s – b) (s – c) where a, b and c are the lengths of sides
and s is the semi-perimeter.
aAAAnrrrdeeeaaahoooifffstrkrthihatoepeme=hzbei12uuigsmh=dt=112of12tdhd2e1hwtrh(aadep2r+eewzbdhiu)1emwraen.hdde1rdea2n‘aadr’eadnt2hdaer‘ebd’itaahgreeodntihaaelgsoloennfagtlhstheoskfoitthfepe. arrhaollmelbsuisd.es
(viii)
(ix)
(x)
(xi) oAnretaheofsaqmuaedbrialsaetewrahli=ch21thedd(ihag1 o+nahl2()d‘h) 1o’faqnuda‘dhr2i’laatreertahl e heights of two triangles
1
(xii) Area of circle = 2 Sd2 or Sr2
(xiii) Area of 1
semi-circle = 8 Sd2 or Sr2
3. With discussion, derive the following formula of finding the area of pathways
(i) Area of path running outside the rectangle = 2d (l + b + 2d)
(ii) Area of path running inside the rectangle = 2d (l + b - 2d)
(iii) Area of crossing paths = d (l + b – d)
(iv) Area of path running outside the circle = Sd (2r + d) or S (R2 – r2)
(v) Area of path running inside the circle = Sd (2r – d)
4. With examples, discussion upon the following formulae
(i) Total cost (T) = Length (l) × Rate (R) per unit
(ii) Total cost (T) = Area (A) × Rate (R) per square unit
(iii) Total cost (T) = Number of bricks/stone (N) × Rate (R) per brick/stone
Area of path (A)
(iv) Number of bricks/stone (N) = Base area of each brick/stone (a)
Vedanta Excel in Mathematics Teachers' Manual - 9 34
Solution of selected problems from Vedanta Excel in Mathematics
1. Calculate the area of shaded region.
14 cm
Solution: = 1 × b × p = 1 × 10 cm × 14 cm = 70 cm2
Area of triangular region (A1) 2 2
Area of semi-circular region (A2) = 81Sd2 = 1 × 2 × (14 cm)2 = 77 cm2
8 7
? Area of the shaded region (A) = A1 + A2 = 70cm2 + 77cm2 = 147 cm2
2. The adjoining figure is a park in the shape of a trapezium. 225m
Calculate the cost of paving the park with stones at 90m
Rs 80 per sq.m.
Solution: 1 1 75m
2 2
Area of the park (A) = × h(a + b) = × 90 m (225 + 75)m = 13,500 m2
Cost of paving the park with stones (T) = Area (A)× Rate (R) = 13,500Rs 80 = Rs 10,80,000
3. The frame of the given window is made up of a rectangle with a
semi-circular top. Find its perimeter and area.
Solution: 2.1 m
1.4 m
(i) Perimeter of lowermost rectangular shape (P1) = l + 2b = 1.4 m + 2 2.1 m = 5.6 m
1 1 22
Perimeter of uppermost semi-circular top (P2) = 4 × 2Sr2 = 2 × 7 × 0.7 m = 2.2 m
?The perimeter of the frame (P) = P1 + P2 = 5.6m + 2.2m = 7.8 m
Area of the frame (A) = l × b+ = = 3.71 m2
4. The shape alongside is a one-quarter of a circle with radius of 14 cm. find
(i) the length of the arc AB. (ii) the perimeter of the figure. A B
(iii) the area of the figure. (iv) the area of ∆AOB.
(v) the area of the shaded segment.
Solution: 1 O
4
(i) The length of the arc AB = × circumference of circle with radius OA
(ii) The perimeter of the figur=e (41P)×=2lSern=gth21 × 22 × (14 cm) = 22 cm 214cm = 50 cm
of a7rc AB + 2r = 22cm +
tt'hhAeeOsfiBhgau=dreed21=s×e14g1m×4ecSnmrt2=×=111454×4ccm2m72=2×–99(8184ccmmcm22 =)2
(iii) The area of = 154 cm2
56 cm2
(iv) The area of
(v) The area of
5. A rectangular hall is 12 m long and 10 m broad. Find the length of carpet 2 m wide
required for covering its floor. If the rate of carpet is Rs 110 per meter, find the cost
of carpeting the floor.
Solution:
35 Vedanta Excel in Mathematics Teachers' Manual - 9
Here, length of hall (l) = 12 m and breadth of hall (b) = 10 m
?Area of the floor of the hall (A) = l b = 12 m 10 m = 120 m2
Also, width of carpet (b1) = 2 m, length of carpet (l1) =?
We know, area of carpet required = area of floor of the hall
or, l1 b1 = 120 m2 or, l1 2 m = 120 m2 ? b1 = 60 m
Again, rate of carpeting the floor (R) = Rs 110 per meter
?Total cost of carpeting the floor of the hall (T) = l1 R = 60 Rs 110 = Rs 6600
6. A rectangular court is twice as long as its breadth and its perimeter is 540 m. Find
the number of bricks of size 20 cm × 12 cm to pave the court. If the rate of cost of
bricks is Rs 950 per 1000, find the cost of paving the court.
Solution:
Let, the breadth of the court (b) be x m then length (l) = 2x m.
Now, perimeter (P) = 2 (l + b) or, 540 m = 2 (2x + x) ?x = 90 m and 2x = 180 m i.e., length
(l) = 180 m and breadth (b) = 90 m
Also, area of the court (A) = l b = 180 m 90 m = 16200 m2
Base area of each brick (a) = 20 cm12 cm = 0.2 m 0.12 m = 0.024 m2
Area of path (A) 16200 m2
?No. of bricks required (N) = Base area of each brick/stone (a) = 0.024 m2 = 675000
Again, cost of 1000 bricks = Rs 950 or, cost of each brick = Rs 0.95
Hence, the cost of 675000 bricks = 675000 × Rs 0.95 = Rs 6,41,250.
7. A rectangular park is 250 m long and 140 m broad. A path 2 m wide is running
around inside the park.
(i) Calculate the cost of paving the path with stones at Rs 45 per sq. metre.
(ii)Calculate the cost of covering the empty space with turfs at Rs 25 per sq. metre.
Solution: 2m
Here, length of the park (l) = 250 m,
breadth of the park (b) = 140 m 2m 140 m
Width of the path running around inside the park (d) = 2 m
Now,
(i) Area of the inner path (A) = 2d(l + b – 2d) 250 m
= 22 m(250m + 140 m –2 2 m)= 1544 m2
Rate of cost of paving the path (R) = Rs 45 per sq. m
?Total cost of paving the park with stones (T) = Area (A) Rate (R) = 1544 Rs 45
= Rs 69,480
Hence, the total cost of paving the park with stones is Rs 69,480.
(ii) Area of entire park = 250 m 140 m = 35000 m2
? Area of empty space = area of the park – area of the path
= 35000 m2 – 1544 m2 = 33,456 m2
Again, rate of covering the empty space with turfs (R) = Rs 25 per sq. m
?Total cost of covering empty space with turfs (T) = Area (A) Rate (R)
= 33,456 Rs 25 = Rs 8,36,400
Hence, the total cost of covering empty space with turfs is Rs 8,36,400.
Vedanta Excel in Mathematics Teachers' Manual - 9 36
8. The length of the side of a square land is 70 feet. For the purpose of the real-estate
business, a path 10 feet wide inside the boundary of the land is made.
(i) Find the area of the path.
(ii) How many stones each of 2 feet long and 1.5 feet breadth are required to pave
the path?
(iii) If a stone costs Rs 105, calculate the cost of paving the path.
Solution: 10 ft.
Here,
Length of the square land park (l) = 70 feet 10 ft. 70 ft.
Width of the path running around inside the boundary (d) = 10 feet
Now,
(i) Area of the inner path (A) = 2d (l + b – 2d) 70 ft.
= 210 feet (70 feet + 70 feet –2 10 feet) = 2400 feet2
(ii) Area of each stone (a) = 2 feet 1.5 feet = 3 feet2
? No. of stones required = Area of path (A) (a) = 2400 feet2 = 800
Base area of each brick 3 feet2
(iii) Cost of each stone (R) = Rs 105
?Total cost of paving the path (T) = No. of stones (N) Rate (R)
= 800 Rs 105 = Rs 84,000
Hence, the total cost of paving the path with stones is RS 84,000.
9. The area of a square pond is 5625 m2 and a 2 m wide path is made around the pond.
(i) Find the area of the path.
(ii) Calculate the number of tiles each of 40 cm × 20 cm required to pave the path.
(iii) If the cost of a tile is Rs 35, find the cost of paving the path.
Solution:
Here,
(i) Area of the square pond (A) = 5625 m2
or, l 2 = 5625 m2
?l = 75 m
(ii) Width of path around the pond (d) = 2 m
? Area of the outer path (A) = 2d (l + b + 2d) = 22 m (75 m + 75 m –2 2 m) = 616 m2
(iii) Area of each tile (a) = 40 cm 20 cm = 0.4 m 0.2 m = 0.08 m2
Area of path (A) 616 m2
? Required no. of tiles = Base area of each brick (a) = 0.08 m2 = 7700
(iv) Cost of each tile (R) = Rs 35
? Total cost of paving the path (T) = No. of tiles (N)Rate (R) = 7700Rs 35 = Rs 2,69,500
Hence, the total cost of paving the path with tiles is Rs 2,69,500.
10. The cost of construction of s path 5 m broad inside the boundary of a square lawn
at Rs 36.25 per sq. metre is Rs 90,625. What is the cost of covering the empty space
with turfs at the rate of Rs 20 per sq. metre?
Solution:
37 Vedanta Excel in Mathematics Teachers' Manual - 9
Here, width of the path (d) = 5 m
Rate of cost of construction of the path (R) = Rs 36.25
Total cost of construction of the path (T) = RS 90,625
Now, area of the path (A) = T = Rs 90625 = 2500 m2
R Rs 36.25
or, 4d (l – d) = 2500 m2 or, 45m (l – 5) = 2500 m2 ? l = 130 m
Again, length of empty space = l – 2d = 130 m - 25 m = 120 m
Area of empty space (A) = (12m)2 = 144 m2
Thus, the cost of covering the empty space with turfs = Area (A) Rate (R)
= 14400 Rs 20 = Rs 2,88,000
11. The given window frame is made up of iron plate of 7 cm wide. It is in the shape of
a rectangle with a semi-circular top. Find the cost of painting the frame at 60 paisa
per sq. cm. 7cm
Solution:
Here,
In the semi-circular top of the frame; 7cm 2.1m
7cm
External diameter = 1.4 m = 140 cm
?External radius (R) = 70 cm 7cm
Internal diameter = 1.4 m – 2d = 140 cm – 14 cm = 126cm 1.4m
?Internal radius (r) = 63 m 1 1 22
2 2 7
Now, area of semi-circular part of frame = S(R2 – r2) = × (702 – 632) = 1463 cm2
In the rectangular part of the frame;
Length of external rectangle (L) = 1.4 m = 140 cm and breadth (B) = 2.1 m – R
= 210cm – 70 cm = 140 cm
Length of internal rectangle (l) = 1.4m – 2d = 126 cm and breadth (b) = B – d
= 140 cm – 7 cm = 133 cm
Area of rectangular frame = LB – l b = 140 cm140cm – 126 cm 133 cm = 2842 cm2
Area of the frame (A) = 1463 cm2+ 2842 cm2 = 4305 cm2
Again, rate of painting (R) = 60 paisa per sq. cm = Rs 0.60 per sq. cm
?Total cost of painting the frame (T) = Area Rate = 4305 Rs 0.60 = Rs 2583
12. A wire in the form of rectangle 25.6 cm long and 18.4 cm wide is bent and reshaped
into the form of circle. Calculate the change in the area in percent.
Solution:
Let the radius of the circle be r cm.
Then, perimeter of rectangular shape = circumference of circular shape
or, 2(l + b) = 2 d
or, 2(25.6 cm + 18.4 cm) = 2 × 272r
Now, area of rectangular shape = l b = 25.6 cm 18.4 cm = 471.04 cm2
Area of circular shape = 22 × (14 cm)2 = 616 cm2
7
Vedanta Excel in Mathematics Teachers' Manual - 9 38
Change in area = 616 cm 2 – 471.04 cm 2 = 144.96 cm 2
Area is increased by = 144.96 cm2 × 100% = 30.8%
471.04 cm2
13. The length of the adjoining rectangular park is two times its
breadth. The cost of constructing two crossing paths running
across the middle of the park at Rs 20 per sq. m is Rs 9000. If the
cost of plastering the shaded portion of the park at Rs 35 per sq.
m is Rs 315, find the cost of growing grasses in the empty spaces
at Rs 12 per sq. metre.
Solution:
Let the breadth of the park be x m then length = 2x m.
Also, let the width of the crossing paths be d m.
Rate of constructing crossing paths (R) = Rs 20
Total cost of constructing crossing paths (T) = Rs 9000
Now, area of the crossing paths (A) =
or, d (l + b – d) = or, d (2x + x– d) = 450 ? d (3x– d) = 450 … (i)
Also, rate of plastering the square portion with side‘d m’ (R) = Rs 35
Total cost of plastering (T) = Rs 315 T Rs 315
R Rs 35
We have, area of the squared portion (A) = =
or, d 2 = 9 ?d = 3m
Putting the value of‘d’ in equation (i), we get
3 (3x– 3) = 450 or, x = 51 ?breadth (b) = 51 m and length (l) = 2x = 102 m
Again, area of empty space = area of whole park – area of crossing paths
= lb – d (l + b – d) = 102 m 51 m – 3m (102m + 51m – 3m) =4752 m2
Thus, the total cost of growing grasses in the empty spaces = Area (A) Rate (R)
= 4752 × Rs 12 = Rs 57,024
Extra Questions
1. A rectangular room is 8 m long and 5 m broad. Find the length of carpet 2.5 m wide
required for carpeting the floor. If the rate of cost of carpet is Rs 375 per meter, find the
cost of carpeting the floor. [Ans: 16m, Rs 6000]
2. The length of a rectangular room is two times its breadth and its perimeter is 78 feet.
Find the cost of carpeting its floor at Rs 25 per sq. feet. [Ans:Rs 8450]
3. A rectangular garden 60 m long and 50 m broad is surrounded by 2.5 m wide path. Find
the cost of paving the path at Rs 140 per sq. m [Ans: Rs 80,500]
4. The cost of gravelling a path 3 m broad inside the boundary of a square psrk at Rs 35 per
sq.m. is Rs 50,400. Find the cost of covering the empty space with turfs at Rs 25 per sq. m.
[Ans:Rs 3,42,225]
5. A rectangular garden is 62 m long and 48 m broad. Two paths each of 2 m wide, running
across the middle of the garden, are at right angle. Calculate the cost of paving the paths
by the bricks of size 18 cm by 12 cm at Rs 15 per brick. [Ans:Rs 1,50,000]
39 Vedanta Excel in Mathematics Teachers' Manual - 9
B. Area of 4 walls, floor and ceiling
Teaching Activities
1. Recall formula of finding the area of rectangle.
2. Tell the students to measure the internal and external lengths and breadths of the frame
of white board and find the area of frame.
3. By taking classroom as an example, discuss about floor, ceiling and walls.
4. Make the group of students to measure the length, breadth and height of the room by
measuring tape and tell them to find the area of floor, walls and criling.
5. Guide the students to discover the formula of area of 4 walls and ceiling of the room.
6. List the following formulae after discussion of the parts of the rooms with length = l,
breadth = b and height = h:
(i) Area of floor = l b
(ii) Area of ceiling = l b
(iii) Area of four walls = 2lh + 2bh = 2h (l + b)
(iv) Area of four walls = 2 (l + b)h = Perimeter of floor (P) h
(v) Area of 4 walls and ceiling = 2h (l + b) + l b
(vi) Area of 4 walls, floor and ceiling = 2h (l + b) + l b + l b = 2h (l + b) + 2l b
Solution of selected problems from Vedanta Excel in Mathematics
1. A rectangular room is 10 m long, 8 m wide and 5 m high. Find the cost of colouring
its walls and ceiling at Rs 65 per sq. metre.
Solution:
Here, length of room (l) = 10 m, breadth (b) = 8 m and height (h) = 5 m
Now, area of walls and ceiling = 2h (l + b) + l b = 25m (10 m+ 8 m) + 10 m 8 m = 260 m2
Again, rate of colouring (R) = Rs 65 per sq. metre
?Total cost of colouring 4 walls and ceiling (T) = Area (A)Rate (R) = 260Rs 65 =Rs 16,900
2. The cost of plastering the walls and ceiling of a room at Rs 20 per sq. feet is
Rs 14,400. Find the cost of colouring the walls and the ceiling at Rs 16 per sq. feet.
Solution:
Here, rate of plastering the 4 walls and ceiling (R) = Rs 20 sq. feet
Total cost of plastering the 4 walls and ceiling (T) = Rs 14,400
?Area of 4 walls and ceiling (A) = T = Rs 14,400 = 720 m2
Again, R Rs 20
Rate of colouring the 4 walls and ceiling (R) = Rs 16 sq. feet
Total cost of plastering the 4 walls and ceiling (T) = Area (A) Rate (R) =720Rs 16
= Rs 11,520
3. A rectangular room is 8 m long, 6 m broad and 4 m high. It contains 2 windows of size
2 m 1.5 m each and a door of size 1 m 4 m, find the cost of painting its walls and
ceiling at Rs 54 per sq. metre.
Solution:
Vedanta Excel in Mathematics Teachers' Manual - 9 40
Here, the length of room (l) = 8 m, the breadth (b) = 6 m and height (h) = 4m
The rate of painting the walls and ceiling (R) = Rs 54 per sq. metre
Now, the area of four walls = 2h (l +b) = 24m (8m + 6m) = 112 m2
The area of ceiling = lb = 8 m 6 m = 48 m2
The area of its 2 windows = 2(2 m 1.5 m) = 6 m2
The area of its 1 door = 1 m 4 m = 4 m2
?The area of 4 walls and ceiling excluding 2 windows and a door = (112 + 48 – 6 – 4) m2
= 150 m2
Again, rate of painting its walls and ceiling (T) = Area (A) Rate (R) =150Rs 54 = Rs 8,100
4. A rectangular room is 10 m long and 8 m wide. It has 2 windows each of size 2 m 2
m and a door of size 1.5 m × 4 m. If the cost of plastering its walls at Rs 55 per sq.
metre is Rs 9,130, find the height of the room.
Solution:
Here, the length of room (l) = 10 m and the breadth (b) = 8 m
The rate of plastering the walls (R) = Rs 55 per sq. metre
Total cost of plastering the walls (T) = Rs 9,130
The area of its 2 windows = 2(2 m 2 m) = 8 m2
The area of its 1 door = 1.5 m 4 m = 6 m2 T Rs 9,130
R Rs 55
Now, the area of four walls excluding windows and door = =
or, 2h (l +b) – 8 m2 – 6m2 = 166 m2
or, 2h (10 +8) – 14 = 166 ?h = 5 m
Hence, the height of the room is 5 m.
5. The cost of carpeting a square room at Rs 110 per sq. metre is Rs 5,390. If the cost of
plastering its walls at Rs 56 per sq. metre is Rs 7,840, find the height of the room.
Solution: T 5,390
R 110
The area of floor of the square room = = m2
or, l2 = 49m2 ?l = 7 m and b = 7 m
Again, area of 4 walls = T = Rs 7,840
R Rs 56
or, 2h (7 + 7) = 140 m ?h = 5 m
Hence, the height of the room is 5 m.
6. A rectangular room is twice as long as it is broad and its height is 4.5 m. If the cost
of papering its walls at Rs 40 per sq. metre is Rs 6,480, find the cost of paving on its
floor at Rs 150 per sq. metre.
Solution:
Let the breadth of the room (b) = x m then length (l) = 2x, height (h) = 4.5 m
Now, area of the walls (A) = T = 6,480 m2
R 40
or, 2h (l +b) = 162 or, 24.5 (2x + x) = 162 or, x = 6
41 Vedanta Excel in Mathematics Teachers' Manual - 9
? Length of room (l) = 2x = 12 m and breadth (b) = 6 m
Again, area of floor (A) = lb = 12m 6 m = 72 m 2
Hence, the total cost of paving the floor (T) = Area (A) Rate (R) = 72Rs 150 = Rs 10,800
7. The length of a rectangular room is twice its breadth and thrice its height. If the cost
of carpeting the floor ar Rs 112 per sq. metre is Rs 12,600, find the cost of plastering
its walls and ceiling at Rs 60 per sq. metre.
Solution:
Let the length of the room (l) = x m. x x
2 3
Then the breadth of the room (b) = m and height (h) = m
Now, area of the floor (A) = T = 12,600 m2
R 112
x
or, lb = 112.5 m 2 or, x × 2 = 112.5 ? x = 15.
The length of the room (l) = 15 m,
(tbh)e=w1a2l5lsman=d h2ehig(lh+t (bh))+=lx3bm 15
The breadth 7.5 m and the = 3 m = 5 m
Again, area of the ceiling =
= 25m (15m + 7.5m)+15m7.5 m = 337.5 m2
Hence, the total cost of plastering the walls and ceiling (T) = Area (A) Rate (R)
= 337.5Rs 60 = Rs 20,250
Extra Questions
1. A rectangular room is 8 m long, 5 m broad and 4 m high. Find the cost of plastering its
four walls at the rate of Rs 140 per sq. metre. [Ans: Rs 14,560]
2. A rectangular room is 15 m long, 10 m broad and 5 m high. It contains two windows of
size 2 m 1.5 m each and a door of size 1 m 4 m, find the cost of painting its walls at
Rs 40 per sq. metre [Ans:Rs 9,600]
3. A rectangular room is 10 m long and 8 m wide. It has 2 windows each of size 2 m 2 m
and a door of size 1.5 m 4 m. If the cost of plastering its walls at Rs 55 per sq. metre is
Rs 9,130, find the height of the room. [Ans: 5 m]
4. A house has three rooms. The length, breadth and height of the first room are 10 m, 9
m and 4.5 m, that of second rooms are 8 m, 6 m and 4.5 m and third room are 12 m, 7.5
m and 4.5 m. find the cost of painting the walls and ceilings of the rooms at the rate of
Rs 50 sq. metre. [Ans:Rs 35,025 ]
5. The cost of carpeting a square room at the rate of Rs 75 per sq. metre is Rs 10,800. If
the cost of plastering the walls at Rs 25 per sq. metre is Rs 6000, find the height of the
room. [Ans: 5 m]
6. The length of a dining hall is twice its breadth and the breadth is twice its height. If the
cost of carpeting the room at Rs 80 per sq. metre is Rs 10,240, what will be the cost of
plastering its walls at Rs 40 per sq. metre? [Ans: Rs 8,640]
C. Area and volumes of solids
Teaching Activities
1. With solid shapes like cuboid and cube, ask the number of vertices, faces and edges.
2. Show the nets of solid object made by straws/ match sticks or through Geo-Gebra
Vedanta Excel in Mathematics Teachers' Manual - 9 42
3. Tell the students to draw the figure of cuboid and cube with corresponding formulae to
find the volume and surface area on the chart paper
4. Discuss upon the volume and surface area of cuboid and cube as
(i) Base area of cuboid = lb
(ii) Lateral surface area of cuboid = area of 4 walls = 2h (l + b)
(iii) Total surface area of cuboid = 2 (lb+ bh + lh)
(iv) Total surface area of lidless cuboid = 2h (l + b) + lb h
(v) Total surface area of hollow cuboid = 2h (l + b) b
(vi) Volume of cuboid = lb h l
(vii) Base area of cube = l2
(viii) Lateral surface area of cube = area of 4 walls = 4l2 l
(ix) Total surface area of cube= 6l2
(x) Total surface area of lidless cube = 5l2
(xi) Total surface area of hollow cube= 4l2 l
(xii) Volume of cuboid = l3 l
5. Give the solid prisms and their model/ nets to the students or show it through OHP using
Geo-Gebra tool ask the definition and properties of prism
6. Discuss about its base, cross-section and lateral surface etc.
7. Conclude the following general formulae for prisms
(i) Volume of prism = Area of cross-section height
(ii) Lateral surface (rectangular faces) area (L.S.A.) of prism = perimeter of cross
section height
(iii) Total surface area of prism = lateral surface area + 2area of cross section
Solution of selected problems from Vedanta Excel in Mathematics
1. A rectangular metallic block is 40 cm long, 24 cm broad and 10 cm high. How many
pieces of rectangular slices each of 8 mm thick can be cast lengthwise from the
block?
Solution:
Here, length of block (l) = 40 m, breadth (b) = 24 m and height (h) = 10 m
Now, volume of block (V) = l b h= 40 cm 24 cm 10 m = 9600 cm3
Again, volume of each slice (v) = 0.8 cm 24 cm 10 m = 192 cm3
?Number of slides (N) = Volume of block (V) = 9600 = 50
Volume of each slice (v) 192
Hence, the required number of rectangular slices is 50. E
2. Calculate the cross sectional area, the lateral surface 3m 3m
area, the total surface area and volume of the given prism. D
C
Solution: 5m 5m
Here A 4m B 6m
(i) The area of cross section = area of rectangle ABCD + area of 'EDC
The area of rectangle ABCD = l b = 4m 5m = 20 m2
a + b+ c 3 m + 3m+4 m
Semi-perimeter of 'EDC = 2 = 2 = 5m
43 Vedanta Excel in Mathematics Teachers' Manual - 9
Area of 'EDC = s(s – a) (s – b) (s – c) = 5(5 – 3) (5 – 3) (5 – 4) = 20
? The area of cross section (A) = area of rectangle ABCD + area of 'EDC
= 20 m2 + 4.47 m2 = 24.47 m2
(ii) Perimeter of the cross section (P) = 3m + 3m + 5 m + 4 m + 5 m = 20 m
Length of prism (h) = 6 m
? Lateral surface area (L.S.A.) = P h = 20 m 6 m = 120 m2
(iii) Total surface are of prism = L.S.A. + 2A = 120m2 + 224.47 m2 = 168.94m2
(iv) Volume of the prism (V) = Ah = 24.74 m2 6m = 146.82 m3
3. A cubical water tank is filled in 1296 seconds at the rate of 1 litre per 6 seconds.
(i) Calculate the internal volume and length of side of tank.
(ii) Calculate the total internal surface area of the tank.
Solution:
Here,
(i) Amount of water filled in 6 seconds = 1 litre
1
or, Amount of water filled in 1 second = 6 litre
1
? Amount of water filled in 1296 seconds = 6 × 1296 litres = 216 litres
We know that, 1 litre = 1000 cm3
?216 litres = 216 1000 cm3 = 216000 cm3
Also,
Internal volume of the tank = 216000 cm3
or, l3 = 216000 cm3 ? l = 60 cm
So, the internal length of the tank is 60 cm.
(ii) The total internal surface area of the tank = 6l2 = 6602 = 21,600 cm2
Hence, the total internal surface area of the tank is 21,600 cm2.
4. A lidless rectangular water tank made of zinc plate is 2 m long, 1.5 m broad and 1 m
high. (i) How many square metres of zinc plates are used in the tank?
(ii) How many litres of water does it hold when it is full?
(iii) Find the cost of zinc plates at Rs 1200 per sq.m.
Solution:
Here, the length of tank (l) = 2 m, breadth (b) = 1.5 m and height (h) = 1 m
(i) Surface area of lidless tank = 2h(l + b) + lb = 21m(2 m + 1.5 m) + 2 m1.5 m = 10 m2
Thus, 10 sq. metre of zinc are used in the tank.
(ii) Volume of the tank (V) = lbh = 2 m 1.5 m 1 m = 3 m3
? Capacity of the tank = 31000 litre = 3000 litres
(iii) Total cost of zinc (T) = Area (A) Rate (R) = 10 Rs 1200 = Rs 12,000
Hence, the cost of zinc plates is Rs 12,000.
5. A rectangular carton is 80 cm 60 cm 40 cm.
(i) How many packets of soaps each of 10 cm 5 cm 4 cm can be kept inside the
carton?
(ii) By how many centimetres should the height of the carton be increased to keep
1200 packets of soaps?
Solution:
Vedanta Excel in Mathematics Teachers' Manual - 9 44
(i) Volume of carton (V) = 80 cm 60 cm 40 cm = 1,92,000 cm 3
Volume of each soap (v) = 10 cm 5 cm 4 cm = 200 cm 3 1,92,000
V 200
? No. of packets of soap that can be kept inside the carton= v = = 960
(ii) Let the height of carton to be increased be x cm.
Then, height of the carton (h) = (40 + x) cm
Now, volume of carton = 1200volume of each soap
or, 80 cm 60 cm (40 + x) cm = 1200200 cm3
or, x = 10
Hence, the height of the carton should be increased by 10 cm to keep 1200 packets of soap.
6. A rectangular metallic block is 50 cm 20 cm 8 cm. If it is melted and reformed in
to a cubical block, find the length of edge of the cube.
Solution:
Let the edge of the cube be l cm.
Then, volume of cube = volume of the rectangular block
or, l3 = 50 cm 20 cm 8 cm = 8000 ? l = 20 cm
Hence, the edge of the cube is 20 cm.
7. In the given figure, a cubical vessel of length 20 cm is completely filled with water.
If the water is poured in to a rectangular 20 cm
vessel of length 32 cm, breadth 25 cm and 20 cm
15 cm
height 15 cm, find the height of water level
in the rectangular vessel. How much more 20 cm 32 cm 25 cm
water is required to fill the rectangular
vessel completely? (1 l = 1000 cm3)
Solution:
Volume of cubical vessel = (20 cm)3 = 8000 cm3
Volume of water in rectangular vessel = 32 cm 25 cm h
Now, 32 cm 25 cm h = 8000 cm3 ?h = 10 cm
Again, remaining height of rectangular vessel to be filled = 15 cm – 10 cm = 5 cm
Then, volume of empty space of vessel = 32 cm 25 cm 5 cm = 4000 cm3
We know, 1000 cm3 = 1 l 4000 cm3 = 4 l
Hence, the height of water level in the rectangular vessel is 10 cm and 4l more water is
required to fill the rectangular vessel completely.
Extra Questions 2cm 2cm
3cm 3cm 2cm
1. Calculate the total surface area and volume of the
given prism. [Ans: 136 cm2, 80 cm3]
4cm
2. A rectangular metallic block is 24 cm long, 20 cm broad and 15 cm high. How many
pieces of rectangular slices each of 6 mm thick can be cast lengthwise from the block?
[Ans:40]
3. Three metallic cubes of edges 3 cm, 4 cm and 5 cm respectively are melted and reformed
a bigger cube, what is the edge of bigger cube. [Ans: 6 cm]
45 Vedanta Excel in Mathematics Teachers' Manual - 9
4. Two cubes having volumes 125 cm3 each are joined to form a cuboidal, find the total
surface area of cuboidal. [Ans:250 cm2]
5. Vegetable gee is stored in a rectangular vessel of internal dimensions 16 cm 12 cm
9 cm. It is transferred in to the identical cubical vessels. If the internal length of each
cubical vessel is 6 cm, how many vessels are required to empty the rectangular vessel?
[Ans: 8]
D. Estimation of number of cubes and cost required for building wall
Teaching Activities
1. Discuss upon the following formulae
(i) Volume of wall = lbh
(ii) Volume of each bricVkol=uml1ebo1fwh1all (V)
No. of bricks (N) = Volume of each brick (v)
Total cost of bricks (T) = No. of bricks (N) × Rate (R) per brick
Space occupied by window = length × height of window × thickness of wall
Space occupied by door = length × height of door × thickness of wall
Solution of selected problems from Vedanta Excel in Mathematics
1. The dimensions of a part of compound wall are 30 m 20 cm 5 m. It contains 3
windows each of 2 m 1.5 m. How many bricks of size 20 cm 10 cm 5 cm are
required to build the wall leaving 10% of the space for the cement work? Also, find
the cost of bricks at the rate of Rs 18,000 per 1000 bricks.
Solution:
Here, the length of wall (l) = 30 m = 3000 cm, the breadth (b) = 20 cm and
the height (h) = 5 m = 500 cm
Now, volume of wall (V) = l b h= 3000 cm 20cm 500 cm = 3,00,00,000 cm3
Also, the length of window (l1) = 2 m = 200 cm,
the breadth (b1) = width of wall= 20 cm and the height (h1) = 1.5 m =150 cm
?Volume space occupied by of 3 windows = 3(200 cm 20 cm 150 cm) = 18,00,000 cm3
Volume of wall excluding 3 windows = 3,00,00,000 cm3 – 18,00,000 cm3 = 2,82,00,000 cm3
Space of wall for cement work = 10% of 2,82,00,000 cm3 = 28,20,000 cm3
?Volume of wall excluding the space occupied by windows and the space for cement work
= 2,82,00,000 cm3 – 28,20,000 cm3 = 2,53,80,000 cm3
Again, volume of each brick (v) = 20 cm 10cm 5 cm = 1000 cm3
?Number of bricks (N) = Volume of wall (V) = 2,53,80,000 = 25,380
Volume of each brick (v) 1000
Now, cost of 1000 bricks = Rs18,000 or, cost of 1 brick = Rs 18
?The total cost of 25,380 bricks = 25,380Rs 18 = Rs 4,56,840
Hence, the required number of bricks is 25,380 and cost of bricks is Rs 4,56,840
2. One of the four walls of a hall is 20 m long and 20 cm thick. If 4,700 bricks each
of size 25 cm 16 cm 10 cm are required to construct the wall containing two
windows of size 2m 1.5 m, find the height of the wall.
Vedanta Excel in Mathematics Teachers' Manual - 9 46
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