Fraction and decimal
9. 1 kg = 1000 g
a) How many grams are there in i) 1.8 kg ii) 7.35 kg iii) 0.05 kg iv) 0.48 kg?
b) How many kilograms are there in (i) 50 g (ii) 200 g (iii) 1260 g (iv) 2575 g?
10. 1 l = 1000 ml
a) How many millilitres are there in i) 2.3 l ii) 4.55 l iii) 0.075 l iv) 0.125 l?
b) How many litres are there in i) 5 ml ii) 20 ml iii) 350 ml iv) 1590 ml?
11. a) Find the cost of 10.5 kg of rice if the rate of cost of rice is Rs 110.80 per kg.
b) If the cost of 7.5 kg of sugar is Rs 566.25, find the rate of cost of sugar.
Creative Section - B
12. Area of rectangle = l × b, area of square = l2 = l × l, perimeter of rectangle
= 2(l + b) and perimeter of square = 4 × l. Find the area and perimeter of the
following rectangles and squares.
b=3.2cm
l=2.7cm
b=4.3cm
l=3.2cm
a) b) c) d)
l = 5.8 cm l=2.7cm l=2.5cm l=3.2cm
13. a) A rectangular park is 60.48 m long and 40.75 m wide.
(i) Find its perimeter (ii) Find its area.
b) A square garden is 50.5 m long. Find its perimeter and area.
c) A rectangular field is 36.75 m long and its area is 937.125 m2, find
(i) its breadth. (ii) Its perimeter
d) The perimeter of a square ground is 147.2 m. Find its length and area.
14. a) The circumference of the wheel of a bus is 2.8 m. How many revolutions
does it make to cover 3.5 km ?
b) The diameter of a circular coin is 1.14 cm. How many coins of the same
size are required to place in a straight row to cover 2.85 m length (without
leaving any gap between each coin) ?
c) The radius of a circular plate is 6.25 cm. How many plates of same size
are required to place in a straight row to cover 10.625 m length (without
leaving any gap between each plate.)
It's your time - Project work!
15. Let's search today's exchange rates of the following foreign currencies from
any daily news papers or by visiting reliable website.
a) US dollars ($) b) Sterling pound (£) c) Eure (¤)
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Fraction and decimal
d) Indian currency e) Chinese Yuan f) Australian dollar (AUD)
(i) Now, calculate, how much Nepali currency is required to exchange 100
units of each currency?
(ii) How many rupees do you need to exchange $ 1,500?
(iii) How many Indian rupees can you exchange with Nepali Rs 8,000?
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative. 58, 23, 7 and 10 ?
1. Which is the largest fraction amongst 9 13
(A) 85 (B) 79 (C) 23 (D) 10
13
2. An integer is 20 more than its one-third part. The integer is:
(A) 15 (B) 18 (C) 30 (D) 45
3. What should be added to 1 to get 7 ?
4 12
12 (B) 31 (C) 61 (D) 1
(A) 12
4. Mrs. Sharma bought 321 kg of apples and 243 kg of pomegranates from a fruit
shop. What was the total weight of the fruits bought by her?
(A) 516 kg 2 (B) 514 kg (C) 612kg (D) 641kg
3 3 (C) 250 kg (D) 150 g
5. What is the value of of 8 of 1 kg?
(A) 200 g (B) 250 g
6. I2n041holwitrems aonfywtaitmerestoagjeutgthheavjairngemcapptiaecdi?ty 3 litre can empty a jar containing
4
(A) 25 (B) 26 (C) 27 (D) 30
7. Which of the following sets of decimals are in ascending order?
(A) 0.02, 0.2, 0.002, 0.0002 (B) 0.0002, 0.002, 0.02, 0.2
(C) 0.2, 0.02, 0.002, 0.0002 (D) 0.002, 0.02, 0.0002, 0.2
8. The value of 0.582 – 0.79 + 0.6 is
(A) 0.392 (B) -0.808 (C) 1.972 (D) 0.772
9. The decimal number in the blank space of 0.9 + 0.09 + ... = 0.999 is
(A) 0.009 (B) 0.09 (C) 0.99 (D) 0.099
10. What should be subtracted from 0.9 to get 0.81?
(A) 0.09 (B) 0.009 (C) 0.8 (D) 0.08
Vedanta Excel in Mathematics - Book 7 100 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 6 Ratio and Proportion
6.1 Ratio - Review
Let’s discuss the answers of the following questions.
1. The price of a pencil is Rs 10 and the price of a pen is Rs 30.
a) By how much is the price of pencil less than the price of the pen?
b) By how many times is the price of pencil of the price of pen?
Here, the price of pencil is (Rs 30 – Rs 10) = Rs 20 less than the price of pen.
And, the price of pencil is Rs 10 = 1 time of the price of pen.
Rs 30 3
We can compare two quantities of the same kind and in the same unit by using a
ratio. A ratio shows how many times a quantity is greater or smaller than another
quantity of the same kind.
Suppose, Hari has Rs 40 as his pocket money and Laxmi has Rs 20 as her pocket
money.
Here, the ratio of Hari's money to Laxmi's money = 40 :20 = 40 = 2 : 1
20
The ratio 2 : 1 shows that Hari's pocket money is two times that of Laxmi's.
Also, the ratio of Laxmi's pocket money to Hari's money = 20 : 40 = 20 = 1 : 2
40
The ratio 1 : 2 shows that Laxmi's pocket money is half of Hari's pocket money.
Thus, if a and b are any two quantities of the same kind and in the same unit, then,
the ratio of a to b = a : b and read as 'a is to b'.
Also, the ratio of b to a = b : a and read as 'b is to a'.
In a : b, the first term a is called antecedent and the second term b is called
consequent.
Similarly, in b : a, the first term b is antecedent and the second term a is consequent.
(i) A ratio compares two or more quantities of the same kind and in same unit.
(ii) A ratio is made by dividing one quantity by another quantity of the same unit.
(iii) A ratio does not have any unit.
(iv) A ratio is usually expressed in the simplest form of the terms.
101Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Ratio and Proportion
Worked-out examples
Example 1: Each part of the given number line represents 1 cm.
ABC DEFGH I J K
0 1 2 3 4 5 6 7 8 9 10
Find the ratio of AD and AG.
Solution:
Here, AD = 3 cm and AG = 6 cm
Now, AD : AG = 3cm : 6 cm = 3 cm = 12= 1 : 2
6 cm
Example 2: A table is 1 m long and 80 cm broad. Find the ratio of the length
and breadth of the table.
Solution: To make the ratio, the quantities should have the
Here, length (l) = 1 m = 100 cm same unit. So, 1 m is converted into 100 cm.
Breadth (b) = 80 cm
Now, ratio of length to breadth = 100 : 80 = 100 = 5: 4
80
Hence, the required ratio of length to breadth is 5 : 4.
Example 3 : Express 2 : 3 and 3 : 4 in the lowest common denominator and
compare them.
Solution:
2 : 3 = 2 and 3: 4 = 43
3
L.C.M. of the denominators 3 and 4 is 12.
Now, 23= 2 × 4 = 8 and 43= 3 × 3 = 9
3 × 4 12 4 × 3 12
So, 8 < 9 , i.e. 2 : 3 < 3 : 4
12 12
Example 4: Rekha obtained 60 marks out of 75 full marks in Science and 75
Solution: marks out of 100 full marks in Mathematics. In which subject did
she obtain better marks?
Here, the ratio of marks obtained and full marks in Science = 60:75 = 60 = 4
75 5
75 3
The ratio of marks obtained and full marks in Mathematics = 75:100 = 100 = 4
Now, L.C.M. of the denominators 5 and 4 = 20
Vedanta Excel in Mathematics - Book 7 102 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio and Proportion
∴ 4 = 4×4 = 2160 and 3 = 3×5 = 15
5 5×4 4 4×5 20
Since 16 > 1205 , she obtained better marks in Science.
20
Example 5: The ratio of number of girls and boys in a class is 5 : 4. If there are
15 girls, find the number of boys.
Solution:
Let the number of boys be x.
Now, 15 = 5 5:4
x 4 2×5:2×4
or, 5x = 15 × 4
135 × 4
or, x = 51
or, x = 12 3×5:3×4
Hence, the required number of boys is 12.
Example 6: The ratio of the ages of a father and his son is 8 : 3. If the son is
15 years old, find the age of the father.
Solution:
Let the father’s age be x years.
Now, x = 8 5 5555555 555
15 3
or, 3x = 15 × 8 F FFFFFFF SSS
or, x = 155 × 8 = 40 Father’s age = 8 times 5 years Son’s age = 3 times 5 years
31
Hence, the father is 40 years old.
Example 7: Mrs. Rai has 15 sweets. If she divides these sweets between her
daughter and son in 3 : 2 ratio, how many sweets does each of
them get?
Solution:
Here, the ratio of sweet received by her daughter and son is 3 : 2.
Let the daughter gets 3x and the son gets 2x sweets.
Now, 3x + 2x = 15 3+2=5 6 + 4 = 10 9 + 6 = 15
or, 5x = 15
15
or, x = 5 = 3
∴ Number of sweets received by the daughter = 3x = 3 × 3 = 9
Number of sweets received by the son = 2x = 2 × 3 = 6
Example 8: The angles of a triangle are in the ratio 4 : 5 : 6. Find the size of
each angle of the triangle.
Solution:
103Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Ratio and Proportion
Here, the angles of the triangle are in the ratio 4 : 5 : 6. 5x°
Let the angles of the triangle are 4x°, 5x° and 6x°.
Now, 4x° + 5x° + 6x° = 180° 4x° 6x°
or, 15x° = 180°
or, x° = 180° = 12°
15
∴ The first angle of the triangle = 4x° = 4 × 12° = 48°
The second angle of the triangle = 5x° = 5 × 12° = 60°
The third angle of the triangle = 6x° = 6 × 12° = 72°
Example 9: A bag contains 1 rupee coins, 50 paisa coins and 25 paisa coins
in the ratio of 2 : 5 : 8. If the total amount of money in the bag is
Rs 65, find the number of each kind of coins.
Solution:
Let the number of coins of Re 1, 50 p and 25 p be 2x, 5x and 8x respectively.
5x
Then, 5x number of 50 p = Rs 2 = Rs 2.5 x 2 coins of 50 p = Re 1
8x number of 25 p = Rs 8x = Rs 2x 4 coins of 25 p = Re 1
4
According to the question,
Rs (2x + 2.5x + 2x) = Rs 65 Answer checking:
or, 6.5 x = 65 No. of Re 1 coins = 20 = Rs 20
or, x = 65 = 10 No. of 50 p coins = 50 = Rs 25
6.5 No. of 25 p coins = 80 = Rs 20
Now, the number of Re 1 coins = 2x = 2 × 10 = 20 ∴ Rs 20+Rs 25+Rs 20 = Rs 65
the number of 50 p coins = 5x = 5 × 10 = 50
the number of 25 p coins = 8x = 8 × 10 = 80 which is given in the question.
EXERCISE 6.1
General Section - Classwork
1. Let's say and write the ratios as quickly as possible.
a) 2 m t0 3 m ......................... b) Rs 50 to Rs 40 .........................
c) 10 kg to 20 kg ......................... d) Re 1 to 50 p .........................
e) 70 cm to 1m ......................... f) 300 g to 1 kg .........................
2. Let's say and write the answers as quickly as possible.
a) In p : q, antecedent is ......................... and consequent is .........................
b) In 3 : 8, antecedent is ......................... and consequent is .........................
c) If the antecedent is 9 and consequent is 5, the required ratio is .......................
3. a) If x : 2 = 3 :1, x = ..................... b) If 3 : x = 1 : 4, x = .....................
c) If 1 : 4 = 1 : y, y = ..................... d) If 5 : 1 = m : 1, m = .....................
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Ratio and Proportion
Creative Section - A
4. Answer the following questions.
a) What is a ratio? Write with examples.
b) Is it possible to make a ratio between 2 kg and 3 m? Write with reason.
c) Define with examples, the antecedent, and consequent of a ratio.
d) The ratio of the number of girls and boys in a class is 1 : 2. What does it
mean? Write with examples.
e) The number of girls in a school is three times the number of boys. Express
it in a ratio.
5. Each part of the given number line represents 1 cm.
ABC D EFG H I JK
0 1 2 3 45 6 78 9 10
a) AC : AD b) AE : AF c) AC : AG d) AE : AI
6. Find the ratios and reduce them in their lowest terms.
a) 15 cm and 12 cm b) 8 cm and 8 mm c) 1 m and 75 cm
d) 9 months and 2 years e) 2 kg and 750 g f) 900 ml and 3 l.
7. Express the following ratios in the lowest common denominator and compare
them.
a) 1 : 2 and 3 : 4 b) 2 : 5 and 1 : 3 c) 2 : 3 and 5 : 6
d) 4 : 7 and 3 : 8 e) 7 : 9 and 5 : 8 f) 7 : 12 and 4 : 9
8. a) There are 40 students in a class and 16 of them are girls.
(i) Find the ratio of the girls and the total number of students.
(ii) Find the ratio of the boys and the total number of students.
(iii) Find the ratio of the girls and boys.
b) An alloy contains 120 g of copper and 150 g of zinc.
(i) Find the ratio of copper and zinc in the alloy.
(ii) Find the ratio of copper and the total weight of alloy.
(iii) Find the ratio of zinc and the total weight of alloy.
9. a) Harka obtained 80 marks out of 100 full marks in Mathematics and 60 marks
out of 80 full marks in English. In which subject did he obtain better marks?
b) Out of 50 questions in History, Chhiribo answered 45 questions correctly
and out of 36 questions in Geography, she answered 33 questions correctly.
In which subject did she have better performance?
10. a) A ratio of two numbers is equal to 3 : 8. If the smaller number is 12, find the
greater number.
b) The ratio of two numbers is 5 : 2. If the smaller number is 14, find the greater
number.
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Ratio and Proportion
c) The ratio of two numbers is 7 : 2. If the greater number is 56, find the smaller
one.
d) Two numbers are in the ratio 5 : 4. If the smaller number is 32, find the
greater one.
11. a) The ratio of the number of boys and girls in a class is 4 : 3. If there are 18
girls, find the number of boys.
b) The ratio of the ages of a mother and her son is 9 : 4. If the mother is 54 years
old, find the age of the son.
c) The ratio of monthly income to the monthly saving of a family is 9 : 2. If the
saving is Rs 4,320, find the income and expenditure of the family.
12. a) The ratio of length and breadth of a room is 3 : 2. If the room is 18 feet long,
find the following:
(i) the breadth of the room (ii) the perimeter of the room
(iii) the area of the floor of the room
b) The ratio of length and breadth of a piece of land is 5 : 3 and it's breadth is
48 m. Find the perimeter and area of the land.
13. a) Divide Rs 65 in the ratio of 2 : 3.
b) Divide 192 kg in the ratio of 7 : 5.
c) There are 32 students in a class. If the ratio of the number of boys and girls
is 5 : 3, find the number of boys and girls.
d) Mr. Yadav divides a sum of Rs 25,000 between his son and daughter in the
ratio of 2 : 3. Find the sum obtained by each of them.
e) Pratik and Debasis invested a sum of Rs 84,000 in a business. If the ratio of
their shares is 7 : 5, how much money did each of them invest?
f) There are 28 teachers in a school. If the ratio of the male and female teachers
is 4 : 3, find their numbers.
g) The population of a rural municipality is 9,966. If the ratio of the adult and
children population is 5 : 6, Find their numbers.
h) Electrum is an alloy of gold and silver. If the ratio of gold and silver in 200 g
of electrum is 11 : 9, find the weight of each metal in the alloy.
14. a) If a pair of complementary angles are in the ratio 2 : 3, find them.
b) If a pair of supplementary angles are in the ratio 3 : 7, find them.
c) The angles of a triangle are in the ratio 2 : 3 : 4. Find the size of each angle.
Creative Section - B
15. a) The ratio of number of girls and boys in a class of 30 students is 7 : 8. If 5
new boy students admit in the class, what is the ratio of number of girls and
boys?
Vedanta Excel in Mathematics - Book 7 106 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio and Proportion
b) The ratio of milk and water in the mixture of 45 litres is 7 : 2. If 4 litres
more water is added in the mixture, what will be the new ratio of milk and
water?
16. a) A bag contains 1 rupee, 50 paisa and 25 paisa coins in the ratio 1 : 2 : 4. If the
total amount is Rs 90, find the number of each kind of coins.
b) Mrs. Sharma exchanged Rs 5,600 into the number of 10 rupee, 20 rupee and
50 rupee notes in the ratio of 5 : 4 : 3 at a bank for Dashain Tika, find the
numbers of each type of rupee notes.
It's your time - Project work!
17. a) How many teachers are there in your school? How many male teachers
and female teachers are there?
(i) Write the ratio of male teaches to the female teachers.
(ii) Write the ratio of male teachers to the total number of teachers.
(iii) Write the ratio of total number of teachers to the female teachers.
b) How many students are there in your class? How many boys and girls are
there in your class?
(i) Find the ratio of the number of girls to boys.
(ii) Find the ratio of the number of girls to the total number of students.
(iii) Find the ratio of the total number of students to the number of boys.
(iv) If 2 more new girls and 3 more new boys are admitted in your class,
what is the new ratio of the number of boys to the girls?
18. a) Let's draw as many blue and white circles so that the total number of
circles are 18 and the ratio of blue to white circles is 4 : 5.
b) Let's draw as many triangles and rectangles so that the total number of
these plane shapes are 20 and the ratio of the number of triangles to the
number of rectangles is 3 : 2.
6.2 Proportion
Let’s take two ratios 2 : 3 and 6 : 9.
Here, 6 : 9 = 6 2 = 2 : 3
9 3
Thus, 2 : 3 and 6 : 9 are two equal ratios. The equality of two ratios is called a
proportion. Here, the terms 2, 3, 6, and 9 of the equal ratios are called proportional.
Again, let a, b, c and d are in proportion.
It is written as a : b = c : d or a : b :: c : d.
Here, the terms a, b, c, and d are called the first, second, third, and the fourth
proportional respectively.
Furthermore, the first and the fourth proportional are called extremes. The second
and third proportional are called means.
107Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Ratio and Proportion
Extremes
Means
a:b c:d
1st 2nd 3rd 4th
Proportional Proportional
In a:b = c:d or a = c ; so, a×d=b×c
b d
∴ In a proportion, the product of extremes = the product of means
6.3 Types of proportions
There are two types of proportions- direct proportion and inverse proportion.
(i) Direct proportion
Suppose the cost of 1 pen = Rs 20
Then, the cost of 2 pens = 2 × Rs 20 = Rs 40.
Here, the ratio of the number of pens = 1 : 2
Also, the ratio of the cost of pens = 20 : 40 = 1 : 2
Thus, the ratio of the number of pens and the ratio of their cost are equal. So,
the ratios are in proportion. Furthermore, if the number of pens increases (or
decreases), their cost also increases (or decreases). Such type of proportion is
called direct proportion.
(ii) Inverse proportion
Suppose, 1 pipe can fill a water tank in 4 hours.
Then, 2 pipes can fill the tank in 2 hours.
Here, the ratio of the number pipes = 1 : 2
The ratio of time taken to fill the tank = 4 : 2 = 2 : 1
Thus, the ratio of the number of pipe and the ratio of the time taken to fill the
tank are exactly opposite. So, they are in proportion but inversely. Such type
of proportion is called inverse proportion.
Worked-out examples
Example 1: Test whether the ratios 4 : 5 and 12 : 15 are in proportion.
Solution:
Here, the product of extremes = 4 × 15 = 60
the product of means = 5 × 12 = 60
∴ Product of extremes = Product of means
∴ 4 : 5 = 12 : 15
Hence, 4 : 5 and 12 : 15 are in proportion.
Vedanta Excel in Mathematics - Book 7 108 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio and Proportion
Example 2: If the terms 3, 8, and 9 are in proportion, find the fourth proportional.
Solution:
Let, the fourth proportional be x.
Now, if 3, 8, 9, and x are in proportion,
3 : 8 = 9 : x If 3, 8, 9, and x are in proportion, the ratio of the first
two terms is equal to the ratio of the last two terms.
or, 83 = 9
x
or, 3x = 8 × 9
or, x = 8 × 93 = 24
31
Hence, the required fourth proportional is 24.
Example 3: If the terms 4, 6, and 18 are in proportion, find the third proportional.
Solution: We got it!
Let, the third proportional be x. To find third proportional
we should suppose the
Now, if 4, 6, x, and 18 are in proportion, third term as x!
4 : 6 = x : 18
64 = x
or, 18
or, 6x = 4 × 18
or, x = 4 × 183 = 12
61
Hence, the required third proportional is 12.
Example 4: If the cost of 8 kg of apples is Rs 1,000, find the cost of 5 kg of
apples by proportion method.
Solution:
Let, the required cost of 5 kg of apples be Rs x.
Then, quantity of apples cost When the quantity of
8 kg Rs 1,000 apples decreases, cost
5 kg Rs x also decreases.
Since, the quantities of apples and their cost are in direct proportion,
8 : 5 = 1000 : x
or, 85 = 1000
x
or, 8x = 5 × Rs 1000
or, x = 5 × Rs 1010205
81
= 5 × Rs 125
= Rs 625
Hence, the required cost is Rs 625.
109Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Ratio and Proportion
Example 5: If 20 workers can complete building a house in 24 days, in how
many days would 15 workers complete the same work?
Solution:
Let, the required number of working days be x.
Then, number of workers number of working days When the number of
20 24 workers decreases,
15 x working days increases.
Since the number of workers and their working days are in inverse proportion,
20 : 15 = x : 24
or, 20 4 x
15 24
3 =
or, 3x = 4 × 24
or, x = 4 × 248 = 32
31
Hence, the required number of working days is 32 days.
Example 6: In a hostel, 50 students have food enough for 54 days. How many
Solution: students should be added in the hostel so that the food is enough
for only 45 days?
Let, the required number of students to be added be x.
Then, total number students = (50 + x).
no. of students no. of days For fewer students, the food is enough
50 54 for longer days. For more students,
the food is enough for fewer days.
( 50 + x ) 45
The number of students and food enough for number of days are in inverse
proportion.
So, 50 : (50 + x) = 45 : 54
or, 50 x = 455
50 + 546
or, 5 (50 + x) = 50 × 6
or, 250 + 5x = 300
or, 5x = 300 – 250
or, 5x = 50
or, x = 5010
51
Hence, 10 students should be added.
Vedanta Excel in Mathematics - Book 7 110 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio and Proportion
EXERCISE 6.2
General Section – Classwork
\1. Let's say and write the ratios which are in proportion.
a) 1 : 2 and 3 : 6 b) 2 : 3 and 4 : 9 Ratios which are in proportion
c) 2 : 5 and 1 : 2 d) 3 : 4 and 6 : 8
2. Let's say and write whether these quantities are in direct or in inverse
proportion. Write 'Direct' or 'Inverse' in the blank spaces.
a) Number f books and their cost .....................................
b) Rate of cost of a pen and number of pens .....................................
c) Rate of cost of a pen and the total cost of pens .....................................
d) Internet speed and time taken to download .....................................
an Application file
3. Let's say and write the values of x in these proportions.
a) x : 2 = 3 : 1, x = ............... b) 2 : x = 1 : 4, x = ...............
c) 5 : 1 = x : 3, x = ............... d) 1 : 5 = 5 : x , x = ...............
Creative Section - A
4. a) Define proportion with an example.
b) Write the types of proportions with examples.
c) Define extremes and means of a proportion with an example.
d) How do we say the rate of cost of rice and the total cost of rice are in direct
proportion write with an example.
e) How do we say the rate of cost of rice and the quantity of rice that we
purchase are in inverse proportion? Write with an example.
5. Let's test whether the following ratios are in proportion.
a) 2 : 3 and 8 : 12 b) 5 : 4 and 15 : 12 c) 6 : 8 and 24 : 16
6. Let's find the value of x in each of the following proportions.
a) 4 : 3 = 12 : x b) 5 : 7 = 20 : x c) 6 : 9 = x : 36
d) 8 : 10 = x : 50 e) 9 : x :: 36 : 48 f) 4 : x :: 28 : 42
g) x : 9 :: 30 : 27 h) x : 8 :: 35 : 40 i) 14 : 15 :: 42 : x
111Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Ratio and Proportion
7. a) If the following terms are in proportion, find the fourth proportional.
(i) 2, 3, 8 (ii) 4, 7, 12 (iii) 9, 10, 36 (iv) 10, 15, 30
b) If the following terms are in proportion, find the third proportional.
(i) 4, 6, 18 (ii) 7, 3, 21 (iii) 5, 9, 45 (iv) 9, 10, 40
c) If the following terms are in proportion, find the second proportional.
(i) 6, 24, 36 (ii) 10, 30, 21 (iii) 12, 60, 40 (iv) 15, 45, 18
d) If the following terms are in proportion, find the first proportional.
(i) 5, 12, 30 (ii) 8, 21, 24 (iii) 12, 18, 24 (iv) 7, 24, 21
Let's solve these problems using proportion methods.
8. a) If the cost of 7 kg of rice is Rs 672, find the cost of 4 kg of rice.
b) If a bus covers 385 km in 7 hours, how many kilometres does it cover in
10 hours with the same speed?
c) In an internet service of 50 MBps speed, a mobile application file is
downloaded in 1 minute. Find the size of the file in GB.
d) If an internet service can download a 2 GB file in 50 seconds, find the
internet speed in MBps.
e) If 6 packets of tea cost Rs 1,260, how many packets of tea can be purchased for
Rs 1,890?
f) The cost of 5 kg of oranges is same as the cost of 3 kg of apples. How many
kilograms of oranges are required to exchange 9 kg of apples?
9. a) If 10 pipes can fill a tank in 16 minutes, in how many minutes would 8
pipes fill the same tank?
b) If a group of 30 workers can complete a piece of work, in 21 days, in how
many days would 45 workers complete the same work?
c) When the rate of cost of rice is Rs 85 per kg, 18 kg of rice can be purchased
for a certain sum of money. How many kilograms of rice can be purchased
for the same sum if the rate is increased to Rs 90 per kg?
d) Mother can buy 15 kg of sugar at the rate of Rs 80 per kg. If the rate is
reduced by Rs 5 per kg, how much sugar can she buy for the same amount
of money?
10. a) 10 workers can complete a piece of work in 18 days. How many workers
are required to complete the work in 12 days?
b) 12 workers can complete a piece of work in 20 days. How many workers
are needed to complete the work in 16 days?
c) A garrison of 200 men has provisions for 45 days. For how many men
would the provisions last only for 30 days?
d) 60 boys in a hostel have food for 30 days. How long would the food last for
72 boys?
Vedanta Excel in Mathematics - Book 7 112 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio and Proportion
Creative Section - B
11. a) 15 labourers were employed to build a wall in 28 days. How many more
labourers should be employed to finish the construction in 21 days?
b) If 18 men can complete a piece of work in 42 days, how many more men
should be added to complete the work in 36 days?
12. a) 15 women can complete a piece of work in 16 days. How many women
should leave so that the work would be finished in 20 days?
b) In a hostel, 30 students have food enough for 40 days. How many students
should leave the hostel so that the food would be enough for 100 days?
13. a) A piece of work can be completed in 30 days working 8 hours a day. In
how many days would the work be completed working 6 hours a day?
b) A certain number of workers are employed to construct a road in
24 days working 9 hours a day. How many hours a day should they work
to complete the construction 6 days earlier?
c) A barrack has enough provisions to last 200 soldiers for 30 days. How
many soldiers must be transferred elsewhere to last the provisions for 40
days ?
It's your time - Project work!
14. a) Let's search and write any three pairs of quantities in our real life situations,
which are in direct proportion.
b) Let's search and write any three pairs of quantities in our real life situations,
which are in inverse (or indirect) proportion.
15. a) Let's write any three sets of four numbers such that the numbers in each
set are in proportion. Find the product of extremes and the product of
means in each proportion. Then, show that:
product of extremes = product of means.
b) Let's write the rate of cost of few items such as rice, sugar, potatoes, milk,
etc. in your local market. Write short reports on the following cases.
(i) How does change in the rate of cost of these items affect the total cost
of each items that we purchase?
(ii) How does change in the rate of cost of these items affect the amount
of quantities of these items that we purchase?
113Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Unit 7 Unitary Method
7.1 Unitary method - Review
Let's have some discussions on the following questions.
a) What are the rates of cost of the following items in your local markets?
(i) Milk per litre (ii) Rice per kg (iii) Sugar per kg (iv) Potatoes per kg
b) Now, will you please calculate the cost of the following quantities of these items?
(i) 2 l of milk (ii) 5 kg of rice (iii) 3 kg of sugar (iv) 10 kg of potatoes
c) How did you find the cost of these items? Discuss in the class.
Let the rate of cost of milk is Rs 90 per litre.
Then, the cost of 2 l of milk = 2 × Rs 90 = Rs 180
Here, 1 litre is the unit quantity and Rs 90 is the unit cost (or unit value). 2 l is more
quantity and Rs 180 is more value.
Thus, in the case of direct proportion, more value is obtained multiplying the unit
value by the given quantity.
Again, if the cost of 3 kg of sugar is Rs 240,
then, the cost of 1 kg of sugar = Rs 240 ÷ 3 = Rs 80
Thus, in the case of direct proportion, unit value is obtained dividing more value by
the given quantity.
Furthermore,
Let 1 pipe can fill a water tank in 60 minutes.
Then, 2 pipes of the same size can fill the tank in 60 ÷ 2 = 30 minutes.
Also, if 3 pipe can fill the tank in 20 minutes
Then, 1 pipe can fill the tank in 3 × 20 minutes = 60 minutes.
Thus, in the case of inverse (or indirect) proportion, more value is obtained by
division and unit value is obtained by multiplication.
In this way, the mathematical method that we apply to find the unit value or more
value by multiplying or by dividing is known as unitary method.
Vedanta Excel in Mathematics - Book 7 114 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unitary Method
Worked-out examples
Example 1: The cost of 5 litres of petrol is Rs 640.
a) Find the cost of 12 l of petrol
b) How much petrol can be bought for Rs 1152?
Solution:
a) The cost of 5 l of petrol = Rs 640 Alternative process:
The cost of 1 l of petrol = Rs 640 Quantity Cost
5
5 l Rs 640
= Rs 128 12 l Rs x
y l Rs 1152
The cost of 12 l of petrol= 12×Rs 128
= Rs 1,536 a) 5 : 12 = 640 : x
Hence, the required cost of 12 l or, x = 640 5:×111252= Rs 1536
of petrol is Rs 1,536. b) 5 : y = 640
b) Rs 640 is the cost of 5 l of petrol. or, y = 5 × 1152 =9l
5 640
Re 1 is the cost of 640 l of petrol.
Rs 1152 is the cost of 5 × 1152 l of petrol = 9l
640
Hence, the required quantity of petrol is 9 l.
Example 2: The floor of a hall is 15 m long and 8.4 m broad. Find the cost of
carpeting the floor at Rs 85 per sq. m.
Solution:
Here, the length of the floor (l) = 15 m
the breadth of the floor (b) = 8.4 m
∴ Area of the floor = l × b = 15 m × 8.4 m = 126 m2
Now, the cost of carpeting 1 m2 is Rs 85.
the cost of carpeting 126 m2 is 126 × Rs 85 = Rs 10,710.
Hence, the required cost of carpeting the floor is Rs 10,710.
Example 3: A married person should pay Re 1 as the social security tax for the
annual income of Rs 100. How much tax does the person pay if his
annual income is Rs 3,60,900?
Solution:
Here, when the annual income is Rs 100, the tax = Re 1
When the annual income is Re 1, the tax = Rs 1
100
1
When the annual income is Rs 3,60,900, the tax = Rs 100 × 3,60,900 = Rs 3,609.
Hence, the required social security tax is Rs 3,609.
Example 4: If 2 parts of a land costs Rs 45000, find the cost of 4 parts of the
3 5
land.
115Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Unitary Method
Solution: 2
3
Here, the cost of parts of a land = Rs 45000
The cost of 1 (whole) land = Rs 45000 = 45000 × 3 = Rs 67,500.
2 2
3
4 4
The cost of 5 parts of the land = Rs 67,500 × 5 = Rs 54,000
Hence, the required cost of the land is Rs 54,000.
Example 5: If a computer can download 200 megabyte (MB) of a movie file in
20 seconds.
a) Find the download speed of the internet in per second.
b) How long does the computer take to download the whole movie
file of size 2.4 gigabyte (GB)?
Solution:
a) Here, the internet speed = rate of download of the file
= downloaded size of file = 200 MB = 10 MB per second
time taken 20 seconds
Hence, the download speed of the internet is 10 MB per second (or 10 mbps)
b) Again, 10 MB of file is downloaded in 1 second.
1 MB of file is downloaded in 1 second.
10
2.4 GB or 2400 MB of file is downloaded in 1 × 2400 seconds = 240 seconds
10
= 240 minutes
60
= 4 minutes
Hence, the computer takes 4 minutes to download the whole movie file.
Example 6: At an average internet speed of 20 megabyte per second (mbps),
a computer takes 6 minutes to download an application file. How
long does the computer take to download the file if its speed slows
down to 15 mbps?
Solution:
a) Here, 20 mbps speed takes 6 minutes to download the file.
1 mbps speed takes 20 × 6 minutes to download the file.
15 mbps sped takes 20 × 6 = 8 minutes to download the file.
15
Hence, the computer takes 8 minutes to download the file.
Example 7: 24 students in a hostel had provisions enough for 30 days. If 16
more students join the hostel, how long would the provisions last?
Solution:
Here, after joining 16 more students, total number of students in the hostel = 24 + 16 = 40.
24 students had provisions for 30 days.
1 students had provisions for 24 × 30 days.
40 students had provisions for 24 × 30 days = 18 days.
40
Hence, the provisions would last for 18 days.
Vedanta Excel in Mathematics - Book 7 116 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unitary Method
EXERCISE 7.1
General Section - Classwork
Let's say and write the answer as quickly as possible.
1. a) Cost of 1 kg of sugar is Rs 80, cost of 2 kg of sugar is .....................
b) Cost of 3 kg of sugar is Rs 240, rate of cost of sugar is .....................
c) If the rate of cost of rice is Rs 110 per kg, the cost of 5 kg of rice is ................
2. a) 1 pipe can fill a tank completely in 2 hours. 2 pipes can fill in the tank
in ......................
b) 3 pipes can fill a tank completely in 2 hours. 1 pipe can fill the tank
in ......................
c) 2 workers can build a wall in 7 days; then, 1 worker can build it
in ......................
3. a) If 1 part of a sum is Rs 50, the whole (1) sum is ......................
2
b) If 1 part of a land worth Rs 20,000, the value of the whole land
3
is ......................
c) If 1 part of a tank is filled in 10 minutes, the whole tank is filled
4
in ......................
Creative Section - A
4. a) Define unitary method with examples.
b) Define unit quantity and unit value with an example.
c) How do we find unit value if the quantities are in direct proportion? Write
with an example.
d) How do we find unit value if the quantities are in inverse proportion? Write
with an example.
5. a) The cost of 9 litres of milk is Rs 810.
(i) Find the cost of 10 litres of milk.
(ii) How much milk can be bought for Rs 360?
b) A taxi covers 330 km distance in 6 hours.
(i) Find the speed of the taxi in km per hour.
(ii) How many kilometres does it travel in 8 hours at the same speed ?
c) A motorcycle covers 315 km distance with 7 l of petrol.
(i) How many kilometres does it covers with 12 l of petrol in the same
mileage?
(ii) How much petrol is needed to cover 270 km in the same mileage?
117Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Unitary Method
d) The rent of a room for 4 months is Rs 20,000.
(i) What is the rent for 7 months?
(ii) How long would a tenant hire the room for Rs 55,000 at the same rate
of rent?
6. a) The annual income of an individual is Rs 3,85,200. How much tax should
she pay at the rate of Re 1 per Rs 100 as social security tax?
b) Mrs. Nepali is a school teacher. She should pay Re 1 as the social security
tax for the annual income of Rs 100. If she pays Rs 3,348 as the annual tax,
find her annual income.
c) Mr. Rawal is a service man. His monthly salary is Rs 32,500. How much
social security tax should he pay in a year at Re 1 per Rs 100 annual income?
7. a) The interest of Rs 100 in one year is Rs 8. Find the interest of Rs 5,400
(i) in 1 year (ii) in 3 years.
b) The interest of Rs 100 in one year in a bank is Rs 10. Mr. Tharu borrowed
some loan from the bank and paid an interest of Rs 2,000 after 1 year. How
much loan did he borrow?
8. a) The floor of a rectangular room is 12m long and 7.5m broad.
(i) Find the area of the floor.
(ii) Find the cost of carpeting the floor at Rs 99 per sq. m.
b) A rectangular floor is 15m long and 10m broad. Find the cost of plastering it at
Rs 75.50 per sq.m.
9. a) If the exchange rate of 1 dollar is Rs 115.80, how many dollars can be
exchanged for Rs 5,790 ?
b) The cost of a bicycle is ICRs 7,450. If the exchange rate of Indian Currency
is ICRe 1 = NC Rs 1.60, find the cost of cycle in Nepali Currency.
10. a) (35iip) aFrtins dofthaelacnodstcoofst23Rpsa3r6ts,0o0f0t.he (i) Find the cost of the whole land.
land.
b) If the cost of 4 parts of 1 ropani land is Rs 2,10,000, find the cost of 2 parts
7 3
of the land. 2
5
c) Mr. Pandey spends Rs 4,800 on his children’s education which is part of
his monthly income. (i) Find his monthly income.
3
(ii) If he spends 8 parts of the income on food, how much does it amount to?
Creative Section -B
11. a) A computer can download 250 megabyte (MB) of a movie file in 25 seconds.
(i) Find the download speed of the internet in per second.
(ii) How long does the computer take to download the whole file of size
1.8 GB?
Vedanta Excel in Mathematics - Book 7 118 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unitary Method
b) At an average internet speed of 30 megabyte per second (30 MBps), a
computer can download an application file in 3 minutes. Find the size of
the file in gigabyte (GB).
12. a) At an average internet speed of 15 MBps, a computer takes 4 minutes to
download an application file. If the speed slows down to 10 MBps, how
long does the computer take to download the file?
b) A computer can download 360 MB of a movie file from YouTube in
30 seconds. If the average internet speed increases by 8 MBps, how long
does the computer take to download the whole movie file of size 2.4 GB?
c) 5 machines of a noodle factory complete the required production of
noodles in 15 days. If 2 machines are shut down due to loadshadding, in
how many days would the remaining number of machines complete the
required production of noodles?
d) 2 printing machines of a printing house can complete the printing work of
the required number of Excel in Mathematics textbooks in 20 days. After
printing the books for 8 days, if 1 more machine was added, in how many
days would the remaining work be completed?
13. a) 50 students in a hostel had provisions for 60 days.
(i) How long would the provisions last for 1 student ?
(ii) How long would the provisions last for 40 students ?
b) A garrison of 60 soldiers had provisions for 45 days. If 15 more soldiers
joined the garrison, how long would the provisions last ?
14. a) With an average speed of 45 km per hour, a bus takes 10 hours to arrive
Kathmandu from Biratnagar. In how early does it arrive Kathmandu, if its
average speed is increased by 5 km per hour?
b) A contractor hired 18 workers to complete the construction of a road in 60
days. How many more workers should he hire to complete the construction
work 15 days earlier?
15. a) The transportation cost of 35 kg of potatoes is Rs 105.
(i) What is the transportation cost of 2.5 quintals of potatoes at the same rate
for the same distance?
(ii) How many quintals of potatoes can be transported for Rs 1,500 at the same
rate for the same distance?
b) Prabin types 765 words in 9 minutes. How many words would he type
in half an hour? How long does he take to type 12,750 words at the same
speed?
c) Dinesh is a mason and he takes Rs 3,250 for 5 days working everyday. If
he received only Rs 7,800 in two weeks, how many days was he absent in
his work?
119Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Unitary Method
It's your time - Project work!
16. a) Let's make groups of students and conduct a survey to find the rate of cost
of the following items in your local market.
(i) Milk per litre (ii) Rice per kg (iii) Sugar per kg (iv) Kitchen oil per litre
(v) Seasonal vegetables per kg (vi) Seasonal fruits per kg
Now, find the cost of three different quantities of each of the items and
discuss in the class.
b) Again, let's increase or decrease the rate of cost of the above surveyed items
and find the cost of different quantities of each item. Write a short report
about the impacts of rates of cost on the cost and quantities.
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. The ratio of 1 metre and 75 centimetre is
(A) 3:4 (B) 4:3 (C) 1:75 (D) 75:1
2. In three ratios 1 : 2, 2 : 3 and 3 : 4, which one of the following is true?
(A) 1 : 2 > 2 : 3 (B) 2 : 3 > 3 : 4 (C) 3 : 4 > 1 : 2 (D) 1 : 2 = 3 : 4
3. The number of boys and girls of a class are in the ratio 5: 4. If there are 24
girls, how many boys are there in the class?
(A) 36 (B) 40 (C) 30 (D) 28
4. Two numbers are in the ratio 3: 4. When 5 is added to each number, the new
ratio becomes 4: 5. the numbers are
(A) 12 and 16 (B) 15 and 20 (C) 21 and 28 (D) 30 and 40
5. Which of the following pair of ratios are in proportion?
(A) 1: 3 and 4: 9 (B) 2: 5 and 6: 15 (C) 7: 4 and 21: 20 (D) 24: 36 and 2: 5
6. In p : q = r: s, the means are :
(A) p and r (B) p and s (C) q and s (D) q and r
7. If w : x and y : z are in proportion, the extremes are:
(A) x and y (B) x and z (C) w and z (D) w and y
8. The terms 9, 10 and 36 are in a proportion, then the fourth term is
(A) 40 (B)45 (C) 60 (D) 72
9. A computer can download 1200 mb of an application file in 1 minute, the
internet speed in mbps is
(A) 25 mbps (B) 20 mbps (C) 10 mbps (D) 15 mbps
10. Which are of the following is the best buying?
(A) 3 sweets for Rs 30 (B) 4 sweets for Rs 36
(C) 5 sweets for Rs 55 (D) 8 sweets for Rs 64
Vedanta Excel in Mathematics - Book 7 120 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 8 Profit and Loss
8.1 Profit and Loss – Looking back
Classwork - Exercise
1. Let's answer these questions as quickly as possible.
A stationer buys an exercise book for Rs 50 and sells it for Rs 60.
a) What is the cost price (C.P.) of the exercise book? .............................
b) What is the selling price (S.P.) of the exercise book? .............................
c) Does the stationer make profit or loss? .............................
How much and how did you find it? .............................
d) Which one should be greater between C.P and S.P to make profit ? ...............
e) If the S.P. of the exercise book were Rs 48,
how much loss would he make? How did you find it? ..............................
f) Now, let's state the mathematical rule to find profit. Profit = ..........................
g) Also, let's state the mathematical rule to find loss. Loss = ..........................
2. Let's say and write the answers as quickly as possible.
a) If C.P = Rs 40, S.P = Rs 50, profit = ..............................
b) If C.P = Rs 85, S.P = Rs 70, loss = ..............................
c) If C.P = Rs 50, profit = Rs 5, S.P. = ..............................
d) If S.P = Rs 75, profit = Rs 10, C. P = ..............................
e) If C.P = Rs Rs 120, loss = Rs 20, S.P. = ..............................
f) If S.P = Rs 250, loss = Rs 25, C.P = ..............................
3. a) If C.P = Rs = 100, Profit = Rs 15, profit percent = ..............................
b) If C.P = Rs 100, loss = Rs 20, loss percent = ..............................
When we purchase something, we need to pay the cost of the thing. This cost is
called cost price (C.P.) Similarly, the price at which something is sold is called
selling price (S.P.)
When selling price of an article is higher than its cost price, a profit is made.
Thus, when S.P. > C.P., profit = S.P. – C.P.
Also, S.P. = C.P. + Profit and C.P. = S.P. – Profit
However, if selling price of an article is less than its cost price, there is loss to the
seller.
Thus, when S.P. < C.P., loss = C.P. – S.P.
Also, S.P. = C.P. – Loss and C.P. = S.P. + Loss
121Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Profit and Loss
8.2 Profit and loss per cent
Let the C.P. of an article is Rs 100 and it is sold at a profit of Rs 10.
Here, the profit of Rs 10 is out of the C.P. of Rs 100. Therefore, it is called 10% profit.
Also, let the C.P. of an article be Rs 100 and it is sold at a loss of Rs 5.
Here, the loss of Rs 5 is out of the C.P. of Rs 100. Therefore, it is called 5% loss.
Thus, profit or loss is always on C.P. If C.P. is taken as Rs 100 and profit or loss is
calculated on it, it is called profit percent or loss percent.
Suppose, C.P. of a pen is Rs 40 Suppose, C.P. of a pen is Rs 40
and it is sold at a profit of Rs 8. Then, and it is sold at a loss of Rs 4. Then,
on C.P. of Rs 40, profit is Rs 8 on C.P. of Rs 40, loss is Rs 4
on C.P. of Re 1, profit is Rs 8 on C.P. of Re 1, loss is Rs 4
40 40
8 4
on C.P. of Rs 100, profit is Rs 40 × 100 on C.P. of Rs 100, loss is Rs 40 × 100
= Rs 20 = Rs 10
Here, Rs 20 is the profit out of the C.P. Here, Rs 10 is the loss out of the C.P.
of Rs 100. So, it is 20% profit. of Rs 100. So, it is 10% loss.
Now, from the above illustrations, we can generalise the following formulas to
calculate profit or loss percent.
Profit percent = Profit × 100 % S.P. – C.P. × 100 %
C.P. C.P.
Loss percent = Loss × 100 % C.P. – S.P. × 100 %
C.P. C.P.
Worked-out examples
Example 1: A shopkeeper buys a T-shirt for Rs 600 and sells it for Rs 678. Find
his profit percent.
Solution:
Here, C.P. of the T-shirt = Rs 678 and S.P. of the T-shirt = Rs 600
∴ Profit = S.P. – C.P. = Rs 678 – Rs 600 = Rs 78
Now, profit percent = Profit × 100 % = Rs 78 × 100 % = 13 %
C.P. Rs 600
Hence, the required profit percent is 13 %.
Example 2: Mr. Chamling bought 150 eggs at the rate of Rs 8 each. 10 of them
were broken and he sold the remaining at the rate of Rs 9 each.
Find his profit or loss percent.
Vedanta Excel in Mathematics - Book 7 122 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
Solution:
Here, the remaining number of eggs = 150 – 10 = 140
C.P. of 150 eggs = 150 × Rs 8 = Rs 1,200
S.P. of 140 eggs = 140 × Rs 9 = Rs 1,260
∴ Profit = S.P. – C.P. = Rs 1,260 – Rs 1,200 = Rs 60
Now, profit percent = profit × 100% = Rs 60 × 100% = 5%
C.P. Rs 1200
Hence, his profit percent is 5%.
Example 3: A trader purchased a mobile for Rs 4,800 and sold it at a profit of
20%. How much profit did she make?
Solution:
Here, C.P. of the mobile = Rs 4,800
profit percent = 20% 20
100
∴ Amount of profit = 20% of C.P. = × Rs 4,800 = Rs 960
Hence, she made a profit of Rs 960.
8.3 Calculation of S.P. when C.P. and profit or loss per cent are given
In this case, C.P. is given and profit or loss percent is given. Then, we apply the
following processes to find S.P.:
Actual profit = profit percent × C.P.
Actual loss = loss percent × C.P.
Then, we calculate S.P. as:
S.P. = C.P. + Actual profit or S.P. = C.P. – Actual loss
Example 4: A shopkeeper bought a pair of shoes for Rs 1,620 and sold it at a
profit of 25%.
a) Find his profit amount. b) Find the selling price of the shoes.
Solution:
Here, C.P. of the shoes = Rs 1,620
Profit percent = 25 % 25
100
a) ∴ Actual profit = 25 % of C.P. = × Rs 1,620 = Rs 405
Hence, the required amount of profit is Rs 405.
b) Again, S.P. of the shoes = C.P. + profit = Rs 1,620 + Rs 405 = Rs 2,025
Hence, the required selling price of the shoes is Rs 2,025.
8.4 Calculation of C.P. when S.P. and profit or loss per cent are given
In this case, S.P. is given and profit or loss percent is given. Then, we apply the
following process to find C.P.
123Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Profit and Loss
Let's consider, C.P. = Rs x
Then, actual profit = P% of C.P. = P% of Rs x
Also, actual loss = L% of C.P. = L% of Rs x
Now, C.P. = S.P. – Profit OR C.P. = S.P. + Loss
or, x = S.P. – P% of Rs x x = S.P. + L% of x
Then, solving the equation, we find the value of x which is the required C.P.
Example 5: If S.P. = Rs 400 and loss percent is 20 %, find C.P.
Solution:
Let, the required C.P. be Rs x. Profit or loss percent is always out of C.P. So, actual
Here, actual loss = 20 % of C.P. profit or actual loss is always calculated from C.P.
= 20 × Rs x = Rs x
100 5
Now, C.P. = S.P. + loss x
or, 5
or, x = Rs 400 + Rs
x
or, x– 5 = Rs 400
or, 45x = Rs 400
5 × Rs 400
x = 4 = Rs 500
So, the required C.P. is Rs 500.
Example 6: Suntali sold a bag for Rs 832 at 4 % profit. At what price did she
purchase the bag?
Solution:
Here, S.P. of the bag = Rs 832
Profit percent = 4 %
Let the C.P. of the bag be Rs x. 4
100
Now, actual profit = 4 % of C.P. = × Rs x Alternative process:
= Rs x C.P. = 100 × S.P.
25 (100 + p)
Again, C.P. = S.P. – profit
= 100 × Rs 832
x (100 + 4)
or, x = Rs 832 – 25
= 100 × Rs 832
or, x+ x = Rs 832 104
25
26x = Rs 800
25
or, = Rs 832
or, x = 25 × Rs 832 = Rs 800
26
Hence, she purchased the bag for Rs 800.
Vedanta Excel in Mathematics - Book 7 124 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
EXERCISE 8.1
General Section - Classwork
Let’s say and write the answers as quickly as possible.
1. a) If C.P = Rs 300, S.P. = Rs 330, then profit = .........................
b) If C.P. = Rs 420, S.P. = Rs 400, then loss = .........................
c) If C.P. = Rs 450, Profit = Rs 100, then S.P = .........................
d) If C. P. = Rs 200, Loss = Rs 20, then S.P = .........................
e) If S.P. = Rs 800, profit = Rs 100, then C.P = .........................
f) If S.P. = Rs 700, loss = Rs, 70, then C. P = .........................
2. a) If C.P. = Rs 100, profit = Rs 5, then profit percent = .........................
b) If C.P. = Rs 100, loss = Rs 10, then loss percent = .........................
c) If C.P. = Rs 100, S.P. = Rs 120, then profit percent = .........................
d) If C.P. = Rs 100, S.P. = Rs 80, then loss percent = .........................
3. a) If C. P = Rs 100, profit percent = 12% , then S.P. = .........................
b) If C.P = Rs 100, loss percent = 8% , then S.P. = .........................
Creative Section - A
Let's find the unknown variables in the following cases.
4. a) If C.P. = Rs 150 and profit = Rs 12, find profit percent.
b) If C.P. = Rs 440 and loss = Rs 22, find loss percent.
c) If C.P. = Rs 360, S.P. = Rs 396, find profit and profit percent.
d) If C.P. = Rs 700, S.P. = Rs 665, find loss and loss percent.
e) If C.P. = Rs 590 and profit = Rs 75, find S.P.
f) If C.P. = Rs 630 and loss = Rs 48, find S.P.
5. a) If C.P. = Rs 280 and profit percent = 10%, find profit amount.
b) If C.P. = Rs 810 and loss percent = 20%, find loss amount.
c) If C.P. = Rs 900 and profit percent = 8%, find profit amount and S.P.
d) If C.P. = Rs 1,500 and loss percent = 7%, find loss amount and S.P.
125Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Profit and Loss
6. a) If S.P. = Rs 210, profit percent = 5%, C.P. = Rs x, find C.P.
b) If S.P. = Rs 315, loss percent = 10%, C.P. = Rs x, find C.P.
7. If S.P. = Rs 480, profit percent = 20%, then
C.P. = S.P. = Rs 480 = Rs 480 × 100 = Rs 400
120% 120 120
100
If S.P. = Rs 240, loss percent = 20%, then
C.P. = S.P. = Rs 240 = Rs 240 × 100 = Rs 300
80% 80 80
100
Let’s apply the similar process and find C.P.
a) S.P. = Rs 132, profit percent = 10%, find C.P.
b) S.P. = Rs 325, profit percent = 25%, find C.P.
c) S.P. = Rs 264, loss percent = 12%, find C.P.
d) S.P. = Rs 425, loss percent = 15%, find C.P.
8. a) A stationer buys 1 dozen of pens at Rs 20 each and sells them at Rs 25 each.
Find his profit and profit percent.
b) A grocer purchased 5 dozen of eggs at Rs 12 each. 10 eggs were broken and
he sold the remaining at Rs 14 each. Find:
(i) his total profit or loss. (ii) Profit or loss percent.
c) A fruit seller sold 50 kg of oranges at the rate of Rs 80 per kg and gained
Rs 800. Calculate (i) his purchasing rate. (ii) Profit percent.
d) A vegetable seller purchased 1 quintal of potatoes at Rs 35 per kg and sold
at a loss of Rs 350. Find his (i) rate of selling price (ii) Loss percent.
9. a) Mrs. Shrestha bought a fancy item for Rs 720 and sold it at 10 % profit.
(i) Find her profit amount. (ii) Find the selling price of the item.
b) A retailer purchased a mobile for Rs 3,250 and sold it at 4% loss.
(i) Find his loss amount. (ii) At what price did he sell the mobile?
c) Anamol buys an old bicycle for Rs 4,500 and he spends Rs 500 to repair it.
If he sells it at 25% profit, find the selling price of the bicycle.
d) Ashma purchased a second hand scooter for Rs 1,52,000. She spent Rs 8,000
to repair it and sold it at 3% loss. At what price did she sell the scooter?
10. a) Mr. Agrawal has an electrical shop. He sold a tube light for Rs 240 at 20%
profit. Find the cost price of tube light.
b) Mrs. Sunuwar sells digital items. She sold a digital watch for Rs 1,260 and
made 10% loss. At what price did she purchase the watch?
Vedanta Excel in Mathematics - Book 7 126 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
Creative Section - B
11. a) A grocer bought 200 eggs at the rate of Rs 10 each. 20 eggs were broken
and he sold the remaining eggs at 8% profit. Find the rate of selling price
of each egg.
b) Suntali bought 50 kg oranges at the rate of Rs 80 per kg. She paid Rs 5 per
kg as the fare to bring the oranges to her shop. If she sold all the oranges
for Rs 4,675. Find her profit or loss percent.
c) A stationer bought 100 books at Rs 80 each. He donated 10 books in a
school's library and sold rest of the books at Rs 88 each. Calculate his gain
or loss percent.
12. a) A shopkeeper purchased a bag for Rs 800 and an umbrella for Rs 400. If
she sold the bag at 15% profit and umbrella at 15% loss, find her profit or
loss percent in whole transaction.
b) A farmer bought a goat for Rs 15000 and a cow for Rs 35000. If he sold the
goat at 10% profit and cow at 20% loss, find his profit or loss percent in
whole transaction.
It’s your time - Project work!
Let’s become a problem maker and problem solver.
13. a) Write the values of the variables of your own. Then, solve each problem to
find unknown variable.
C.P. = ................... C.P. = ................... C.P. = ..........
S.P. = ................... S.P. = ................... Profit percent = ..........
Find profit percent Find loss percent Find S.P.
C.P. = .......... S.P. = .......... S.P. = ..........
Loss percent = .......... Profit percent = .......... Loss percent = ..........
Find S.P. Find C.P. Find C.P.
b) Let’s write appropriate amount of C.P. and S.P. to get the given profit or loss
percent.
Profit percent = 10% Loss percent = 5%
C. P. = ............................ C. P. = ............................
S.P.=............................ S. P. = ............................
127Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Profit and Loss
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. When the selling price of an article is more than its cost price, then article is said to
be sold at a
(A) loss (B) gain (C) both (D) none
2. The selling price of a mobile is Rs 9,900 and loss is Rs 900, the buying price of the
mobile is
(A) Rs 9,000 (B) Rs 10,800 (C) Rs 9,990 (D) Rs 9,800
3. The profit or loss is expressed as the percentage of
(A) the cost price (B) the selling price
(C) either (A) or (B) (D) neither (A) nor (B)
4. A shopkeeper bought an article for Rs 80 and sold it for Rs 100. Which one of the
following is true?
(A) 20% profit (B) 25% loss (C) 20% loss (D) 25% profit
5. By selling a laptop for Rs 66,000; a dealer earns the profit of Rs 6,000, then the profit
percent is
(A) 5% (B) 10% (C) 9.09% (D) 11.1%
6. A shopkeeper sold a pen for Rs 120 and made a profit of 20%. If he had sold it for
Rs 115, his profit percent would be:
(A) 10% (B) 15% (C) 20% (D) 25%
7. Mrs. Karki sold a packet of crayons for Rs 90 and made a loss of Rs 10. At what price
should she sell it to gain 10%?
(A) Rs 110 (B) Rs 100 (C) Rs 95 (D) Rs 105
8. If C.P. = Rs x, S.P. = Rs y and profit = Rs z, then profit percent is:
(A) x × 100% (B) z × 100% (C) y × 100% (D) z × 100%
z y z x
9. Mr. Jha buys some sweets at the rate of 10 sweets for Rs 100 and he sells all sweets at
the rate of 5 sweets for Rs 60. Which one of the following is true?
(A) 20% loss (B) 10% loss (C) 20% profit (D) 10% profit
10. A man bought some umbrellas for Rs 8000 and sold them for Rs 9000 by making
Rs 50 profit in each, how many umbrellas did he buy and sell?
(A) 10 (B) 15 (C) 20 (D) 25
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Vedanta Excel in Mathematics - Book 7 128 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 9 Laws of Indices
9.1 Laws of indices (or exponents)
Let’s consider an algebraic term 2x3.
Here, 2 is called the coefficient, x is the base and 3 is the exponent of the base. The
exponent is also called the index of the base. The plural form of index is indices.
An index of a base shows the number of times the base is multiplied. For example:
x × x → x is multiplied two times = x2 (x squared)
x × x × x → x is multiplied three times = x3 (x cubed)
x × x × x × x → x is multiplied four times = x4 (x raise to the power 4)
While performing the operations of multiplication and division of algebraic
expressions we need to work out indices of the same bases under the certain rules.
These rules are also called laws of indices.
1. Product law of indices
Study the following illustrations and investigate the idea of the product law of
indices.
1 2 = 4 unit squares 123
3 4 2 = 21 × 21 = 21 +1 = 9 unit squares
22 32 4 5 6 3 = 31 × 31 = 31 +1
2 789
3
Similarly,
x2 x = x1 × x1 y2 y = y1 × y1
= x1 +1 = y1
+1
x y
Again,
3 = 8 unit cubes
2 = 27 unit cubes
14 = 31 × 31 × 31
2 = 21 × 21 × 21
23 7 8 = 21 + 1 + 1 3
56 2 33 = 31 + 1 + 1
2
3 3
x3 x = x1 × x1 × x1 y3 y = y1 × y1 × y1
= x1 + 1 + 1 = y1 + 1 + 1
xx y
y
129Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Vedanta Excel in Mathematics - Book 7
Laws of Indices
Furthermore, 23 = 21+2 I understood!
21 × 22 = 2 × (2 × 2) 25 = 22+3 When the same bases are
27 = 23+4 multiplied, we should
22 × 23 = (2 × 2) × (2 × 2 × 2) a5 = a2+3 add their indices!!
23 × 24 = (2 × 2 × 2) × (2 × 2 × 2 × 2)
Similarly, a2 × a3 = (a × a) × (a × a × a)
Thus, if am and an are any two terms with the same base a and the indices m and n
respectively, then, am × an = am + n
2. Quotient law of indices
Let's study the following illustrations and try to investigate the idea of quotient law
of indices.
22 ÷ 2 = 22 = 2×2 = 2 = 22 – 1
2 2
23 ÷ 2 = 23 = 2 × 2 × 2 = 22 = 23 – 1
2 2
I also understood!
25 ÷ 22 = 25 = 2×2×2×2×2 = 23 = 25– 2 When a base is divided
22 2×2 by another same base,
indices should be
24 ÷ 26 = 24 = 2×2 × 2 × 2 = 1 = 1 subtracted.
26 2×2×2 × 2 × 2×2 22 26 – 4
Similarly, a5 ÷ a2 = a5 = a×a×a×a×a = a3 =a5-2
a2 a×a
Thus, if am and an are any two terms with the same base a and the exponents m and
1
n respectively, then, am ÷ an = am – n if m > n and am ÷ an = an – m if n > m
3. Power law of indices
(i) Let's study the following illustrations and try to investigate the idea of power
law of indices.
(22)3 = 22 × 22 × 22 = 22 + 2 + 2 = 26 = 22 × 3 When a base with some
(22)4 = 22 × 22 × 22 × 22 = 22 + 2 + 2 + 2 = 22 × 4 = 28 power has another power,
Similarly, the indices are multiplied.
(a3)2 = a3 × a3 = a3 + 3 = a2 × 3 = a6
Thus, if am is any term with the base a and the index m, then, (am) n = am × n
(ii) Let's study the following illustrations.
(2 × 3)2 = 22 × 32, (4 × 5)3 = 43 × 53
Similarly, (a × b)3 = a3 × b3
Thus, if a and b are any two terms, then, (a × b)m = am × bm Also, am bm = (ab)m
(iii) Let's study the following illustrations.
2 2 22 , 4 3 43 , Similarly, a 4 a4
3 32 5 53 b b4
= = =
Thus, if a and b are any two terms, then a m = bamm
b
Vedanta Excel in Mathematics - Book 7 130 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Laws of Indices
4. Law of zero index
Let's study the following illustrations and investigate the result when a base has
zero index. It’s interesting!
The value of a base with
20 = 21– 1 = 21 ÷ 21 = 21 =1 exponent 0, is always 1!
21
30 = 31 – 1 = 31 ÷ 31 = 31 =1
31
Thus, if a0 is any term with base a and exponent 0, then, a0 = 1 where a ≠ 0.
Worked-out examples
Example 1: Which one is greater 25 or 52 ?
Solution:
Here, 25 = 2 × 2 × 2 × 2 × 2 = 32 and 52 = 5 × 5 = 25
Since 32 > 25, 25 is greater than 52.
Example 2: Express the numbers as a product of prime factor in exponential
form. a) 432 b) 675
Solution:
a) 2 432 b) 3 675
2 216 3 225
2 108 3 75
2 54 5 25
3 27 5
39
3 ∴ 675 = 3 × 3 × 3 × 5 × 5
∴432 = 2 × 2 × 2 × 2 ×3 × 3 × 3
= 24 × 33 = 33 × 52
Example 3: Factorize the number and express in exponential form.
a) 216 b) 2744
Solution: b) 2 2744
a) 2 216
2 108 2 1372
2 54 2 686
3 27 7 343
3 9 7 49
37
∴ 216 = 2 × 2 × 2 × 3 × 3 × 3 ∴ 2744 = 2 × 2 × 2 × 7 × 7 × 7
= 23 × 33 = (2 × 3)3 = 63 = 23 × 73 = (2 × 7)3 = 143
131Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
Example 4: Express the numbers as the products of powers of 10.
a) 600 b) 900000 c) 11000000
Solution:
a) 600 = 6 × 100 = 6 × 10 × 10 = 6 × 102
b) 900000 = 9 × 100000 = 9 × 10 × 10 × 10 × 10 × 10 = 9 × 105
c) 11000000 = 11 × 1000000 = 11 × 10 × 10 × 10 × 10 × 10 × 10 = 11 × 106
Example 5: Find the product in their exponential forms.
a) 52 × 54 × 56 b) −3 3 ×−34 4 × −34 −5 c) 34 × 93 × 272
Solution: 4
a) 52 × 54 × 56 = 52+4+6 = 512
b) −3 3 × −34 −2 × −34 5 = −3 3 – 2 + 5 = −3 6 = 36
4 4 4 4
c) 34 × 93 × 272 = 34 × (32)3 × (33)2 = 34 × 36 × 36 = 3 4+6+6 = 316
Example 6: Find the quotients in their exponential forms.
a) 35 ÷ 32 b) (2x)7 ÷ (2x) –2 c) 85 ÷ 44 d) 34 ÷ 93
Solution:
a) 35 ÷ 32 = 35–2 = 33
b) (2x)7 ÷ (2x) –2 = (2x)7–(–2) = (2x) 7+2 = (2x)9
c) 85 ÷ 44 = (23)5 ÷ (22)4 = 215 ÷ 28 = 215–8 = 27
1
d) 34 ÷ 93 = 34 ÷ (32)3 = 34 ÷ 36 = 1 4 = 32
36 –
Example 7: Find the value of a) 4 1 b) 8 2
2 3
Solution: 9 27
a) 4 1 22 1 = 2 2×12 = 21 = 2
9 32 2 3 3 3
2=
b) 8 2 = 23 2 = 2 3× 2 = 2 2 22 = 4
3 3 3 3 3 32 9
=
27 33
Example 8: Simplify a) 25 3 b) 33 × 35 × 37 c) 253 × 52
Solution: 23 32 × 95 54 × 1252
a) 25 3 = (25 – 3)3 = (22)3 = 22 × 3 = 26 = 64
23
b) 33 × 35 × 37 = 33 + 5 + 7 = 315 = 315 = 315 – 12 = 33 = 27
32 × 95 32 × (32)5 32 × 310 312
c) 253 × 52 = (52)3 × 52 = 56 × 52 = 56 + 2 = 58 = 1 = 1 = 1
54 × 1252 54 × (53)2 54 × 56 54 + 6 510 510 – 8 52 25
Example 9: Simplify a) xb – c × xc – a × xa – b b) xa + b × xb + c × xc + a
Solution: x2a × x2b × x2c
Vedanta Excel in Mathematics - Book 7 132 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Laws of Indices
a) xb – c × xc – a × xa – b = xb – c + c – a + a – b = x° = 1.
b) xa + b × xb + c × xc + a = xa + b + b + c + c + a
x2a × x2b × x2c x2a + 2b + 2c
= x2a + 2b + 2c = x2a + 2b + 2c – 2a –2b – 2c = x° = 1
x2a + 2b + 2c
EXERCISE 9.1
General Section - Classwork
1. Let's say and write the answers as quickly as possible.
a) a × a2 = ......... b) a2 × a3 = ......... c) a5 × a4 = .........
d) am × an = ......... e) x2 ÷x = ......... f) x5 ÷ x2 = .........
g) x10 ÷ x7 = ......... h) xm ÷ xn = ......... i) (y2)3 = .........
j) (y3)4 = ......... k) (y6)3 = ......... l) (ya)b = .........
m) x° = ......... n) (2x)° = ......... o) (p) xm–m = .........
2. Let's say and write the products in their exponential forms.
a) 7 × 7 × 7 = ......... b) (–2)×(–2)×(–2)×(–2) = .........
c) (5p) × (5p) × (5p) × (5p) × (5p) = ......... d) −23 × −32 × −23 = .........
Creative Section - A
3. a) Define coefficient, base and index of an algebraic term with an example.
b) Define product law of indices taking two terms xa and xb.
c) Define quotient law of indices taking two terms px and py.
d) Define power law of indices taking (xa)b.
e) Write the law of zero index with an example.
4. Let's evaluate and identify which one is greater?
a) 23 and 32 b) 43 and 34 c) 25 and 52
d) 210 and 102 e) 54 and 45 f) 53 and 35
5. Let's factorize the following numbers and express in exponential form.
a) 16 b) 27 c) 128 d) 343 e) 625 f) 400 g) 864 h) 1125
6. Let's find the prime factors and express in exponential forms.
a) 36 b) 100 c) 216 d) 225 e) 196 f) 441 g) 1000 h) 1296
7. Let's express the following numbers as the products of power of 10.
a) 200, 2000, 20000, 200000 b) 5000, 50000, 500000, 5000000
8. Let's find the products in their exponential forms.
a) 2 ×22 × 23 b) 22 × 42 × 82 c) 273 × 92 × 3
d) 54 × 252 × 125 e) (3a)4 ×(3a)3 × (3a)–2 2 −3 45
f) 4 × 4 × 7
7 7
133Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
9. Let's find the quotients in their exponential forms.
a) 27 ÷ 23 b) 39 ÷ 34 c) (2x)8 ÷ (2x)2 d) (3a)4 ÷ (3a)–3
e) 95 ÷ 36 f) 163 ÷ 82 g) 4 8 ÷ 4 3 h) 4 5 26
5 5 9 3
÷
10. Let's find the quotients in their exponential forms.
a) 34 ÷ 39 b) 53 ÷ 510 c) (5a) 4 ÷ (5a)7 d) (4p) –3 ÷ (4p)2
e) 22 ÷ 43 f) 35 ÷ 94 g) 54 ÷ 253 h) 7 ÷ 492
11. Let's evaluate. 1 2 3 3
5
a) 4 1 25 1 4 2 d) 8 3 e) 16 4 f) 32
9 27 81 243
2 b) 2 c)
12. Let's simplify.
a) a5 × a7 b) p6 × p4 c) x7 ×x −2 d) y10 × y −4 e) 4x5 × 3x
a9 p3 x3 y4 6x2
13. Let's simplify.
a) 22 3 b) 34 3 c) 57 2 92 3 e) 43 4 82 5 272 2
54 32 23 29 310
d) f) g)
Creative Section - B
14. Let's simplify.
a) 23 × 4453××18642 b) 32 × 93 × 821724 c) 22 × 34 × 123
22 × 33 × 94 × 25 × 63
d) 64 × 92 × 125563 e) 43 2 34 2 92 2 f) 53 2 × 82 3 54
32 × 42 × 25 32 272 252 42 2
× × ×
15. Let's simplify.
a) xa − b × xb – a b) xa – b × xb – c × xc – a
c) (xa)b – c × (xb)c – a × (xc)a – b d) (xp − q)r × (xq − r)p × (xr − p)q
e) xa + b × xb + c × xc + a f) (x2)a + b × (x2)b + c × (x2)c + a
x2a × x2b × x2c (xa × xb × xc)4
16. a) If a = 1 and b = 2, find the value of (i) ab (ii) ba (iii) (a + b)a + b
b) If x = 5 and y = 3, find the value of (i) xy (ii) yx (iii) (x – y)x – y
c) If p = 10 and q = 1, find the value of p2 + 2pq + q2
p+q
d) If m = 15 and n = 5, find the value of = m2 − n2 .
m−n
17. a) If x = 5a and y = 5b, show that xy = 5a + b
b) If x = 7a – b and y = 7b − a, show that xy = 1.
c) If x + y = x, show that xy = 1.
d) If a = 2 and b = 1, show that (a + b)2 = a2 + 2ab + b2
e) If a = 5 and b = 3, show that (a – b)2 = a2 – 2ab + b2
Vedanta Excel in Mathematics - Book 7 134 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Laws of Indices
It's your time - Project work
18. a) Let's fill in the blanks with correct numbers.
1° = ......... 11 = ......... 12 = ......... 13 = ......... 14 = .........
15 = ......... 16 = ......... 17 = ......... 18 = ......... 19 = .........
i) What idea did you investigate ?
ii) What would be the value of: 150 , 11000 and 1x ?
b) Lets find the value of :
(–1)1 = ......, (–1)2 = ......, (–1)3 = ......, (–1)4 = ......, (–1)5 = ......,
(–1)6 = ......, (–1)7 = ......, (–1)8 = ......, (–1)9 = ......, (–1)10 = ......,
i) What idea did you investigate?
ii) What would be the value of (–1)20 , (–1)99 and (–1)500?
c) Write a short report about your investigation on a) and b), then present in
the class.
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. What is the product of a×a×a×a?
(A) a (B) 4a (C) 4a4 (D) a4
2. For any two positive integers m and n, am × an is equal to
(A) a2mn (B) am + n (C) am – n (D) amn
3. Which of the following laws is used in xa÷ xb = xa –b ?
(A) Product law (B) Quotient law (C) Power law (D) Zero index law
4. (23)4 equals to
(A) 27 (B) 64 (C) 212 (D) 646
5. Which of the following statements is true?
(A) 23 is greater than 32 (B) 24 is less than 42
(C) 52 is equal to 25 (D) 34 is greater than 43
6. What is the value of (2x) o, x≠0?
(A) 1 (B) 2 (C) 2x (D) 0
(D) 1/9
7. The value of 35 ÷ 92 is
(A) 1 (B) 3 (C) 9
8. The product of (10a)4 ×(10a)3 × (10a)–5 is
(A) 10a2 (B) 100a2 (C) 10a (D) 100a
9. The value of 8 1
27
3 is
(A) 32 (B) 23 (C) 4 (D) 9
9 4
10. If a + b + c = 0, then xa+b× xb+c× xc+a is
(A) 0 (B) 1 (C) x (D) x2
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Unit 10 Algebraic Expressions
10.1 Algebraic terms and expressions – Looking back
Classwork - Exercise
1. Let's say and write the answers as quickly as possible.
a) How many terms are there in each of the following expressions? Also write
the terms.
(i) In 2xy, number of terms: .................., terms are .......................................
(ii) In 2x + y, number of terms: .................., terms are ...................................
(iii) In 2+x+y, number of terms: .................., terms are ..................................
b) In 5x3 , coefficient is ............... base is .............. and power is ......................
c) If l = 3 and b = 2 then (i) 2 (l + b) = .................... (ii) l × b = ....................
2. a) The mathematical expression for the sum of 2x and 3y is .............................
b) The mathematical expression for the difference of 7ab and 3bc is ................
c) The mathematical expression for the product of 5p and 2q is .......................
3. Let's say and write the sum, difference, or product.
a) 7x + 2x = ......... 7x − 2x = ......... 7x × 2x = .........
b) 5p2 + 2p2 = ......... 5p2 − 2p2 = ......... 5p2 × 2p2 = .........
x, 2x, 3ab, p2q, etc. are algebraic terms. An algebraic expressions is a collection of
one or more terms. Which are separated to each other by either addition (+) or
subtraction (–) sign.
For example: 3xyz, 7x – 2y, x + y – z, etc are algebraic expressions. We can represent
the terms and factors of the terms of an expression by a tree diagram.
For example: Expression (5x + yz) (2xy − 6)
Terms 5x yz 2xy −6
Factors 5x yz 2 x y −2 3
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Algebraic Expressions
10.2 Types of algebraic expressions
Algebraic expressions are categorized according to the number of terms contained
by the expressions. The table given below shows the types of expressions.
Monomial An algebraic expression with only one term is called a
Binomial monomial . For example : 5xy, – 8m, 9x2y, 11 etc are monomials.
Trinomial
An algebraic expression with two unlike terms is called a
binomial. For example : x + y, x – 4, 2xy + 3x, a2 – b2, pq – r,
x y
a + b etc are binomials.
An algebraic expression with three unlike terms is called a trinomial.
For example : 2x +y – 1, a2 + ab + b2, xy + x + y etc. are trinomial.
10.3 Polynomial
An algebraic expression with one or more terms and powers of the variables being
whole numbers in each term is called polynomial.
For example : x2y3z, x2 – 4, x + y + 7 etc. are polynomials.
However, x2 + 1 , x2/3 – y2/3 are not polynomials. Because the powers of the variables
x2
in these expressions are not whole numbers.
10.4 Degree of polynomials
Let's study the illustrations given in the table and learn about the degree of
polynomials.
Polynomials Degree of polynomials
2x Power of the variable x is 1. So, its degree is 1.
3y2 Power of the variable y is 2. So, its degree is 2.
x2yz The sum of the powers of the variables x, y, and z
= 2 + 1 + 1 = 4. So, it's degree is 4.
2p3 – 3p2 + 5 The highest power of the variable p is 3. So, its degree is 3.
a2b2 + 2a2b – 4ab2 The highest sum of the powers of ab = 2 + 2 = 4. So, its
degree is 4.
10.5 Evaluation of algebraic expressions
Let's take an algebraic expression 2x – 3y and evaluate it when x = 2 and y = 1.
Here, if x = 2 and y = 1, then 2x – 3y = 2 × 2 – 3 × 1 = 4 – 3 = 1
In this way, the process of finding the value of an algebraic expression by replacing
the variables with numbers is called evaluation.
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Worked-out examples
Example 1 : Which of the following expressions are polynomials? Write with
reason.
x 3
a) 2 + 5 b) 3p2 – p c) √ 5 y2 + 3 d) 2√ x – 1
Solution:
x
a) 2 + 5 is a polynomial because the power of is 1, which is a whole number.
b) 3p2 – 3 is not a polynomial because the power of the term 3 is –1, which is not
p p
a whole number.
c) √ 5 y2 + 3 is a polynomial because the power of y is 2, which is a whole number.
d) 2√ x – 1 is not a polynomial because the power of x is 1 , which is not a whole
number. 2
Example 2 : Find the degree of a) 4x2 b) 3x2y c) 7x5y2 − 9x2y3 + 4xy5
d) (xy)2 + x2 – y2
Solution:
a) The degree of 4x2 is 2
b) The degree of 3x2y is 2 + 1 = 3
c) In 7x5y2 , the sum of powers of variables = 5 + 2 = 7
In − 9x2y3, the sum of powers of variables = 2 + 3 = 5
In 4xy5, the sum of powers of variables = 1 + 5 = 6
Since the highest sum of powers of variables is 7,
the degree of 7x5y2 –9x2y3 + 4xy5 is 7.
d) Here, (xy)2 + x2 – y2 = x2y2 + x2 + y2
The highest sum of the powers of variables is 2 +2 = 4. So, its degree is 4.
Example 3 : If l = 5 and b = 3, evaluate 2(l + b).
Solution:
Here, when l = 5 and b = 3, then 2(l + b) = 2(5 + 3) = 2 × 8 = 16
EXERCISE 10.1
General Section - Classwork
1. Let's tick (√ ) the correct answer.
a) The terms of expression 5x2 – 3xy are
(i) 5x2 and 3xy (ii) x2 and xy (iii) 5x2 and – xy (iv) 5x2 and –3xy
b) The number of terms in the expression x2 + y2 + z2 is
7
(i) 2 (ii) 3 (iii) 4 (iv) 7
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c) An algebraic expression with three unlike terms is called a
(i) Monomial ii) binomial (iii) trinomial (iv) all of these
d) Which of following expressions is a binomial?
(i) 3m2n – mn2 + m2n (ii) pq + qr + pr
(iii) √ 3 xy + z – 7 (iv) x2y –3x – x2y
e) Which one of the following expressions is not a polynomial? x3 4
x3 y3 2 x2
(i) √ 5 x2y + z (ii) 2x3 – xy + y2 (iii) 2 + 3 (iv) +
f) The degree of polynomial 2x4yz3 is
(i) 2 (ii) 8 (iii) 4 (iv) 3
g) The degree of polynomial 5x2y – xy + y2 is
(i) 2 (ii) 3 (iii) 5 (iv) 7
2. Let's say and write the value of the expressions quickly.
a) If x = 3, y = 2, then (i) x + y = ........... (ii) x – y = ........... (iii) xy = ...........
b) If a = 2, b = 3, then (i) 2(a + b) = ........ (ii) a2 = ............ (iii) b2 = ............
Creative Section
3. a) Define algebraic expressions with examples.
b) What are monomial, binomial, and trinomial expressions? Write with
examples.
c) Is x – y + 2x a trinomial expression? Why?
d) What is a polynomial? Give an example of a polynomial.
1
e) Why is x2 a polynomial, but x2 is not a polynomial?
4. Let's identify and then classify the given expressions as monomial, binomial
or trinomial.
a) x2y + xy2 b) 9 – x2 c) xyz d) pq + p + q
2
e) x2 + y2 f) a2 + a + 1 g) 3x2 + 7xy + 6y2 h) 3x + xy – 8y2
(i) x2 + x (j) – 6x2 k) 1 + x + xy l) 64
y2
5. Let's state with reason whether the given expressions are polynomials.
a) x3 + x2 b) 3 c) ab −a + b d) x2 + x –2
x2 3
e) √ 3 x2 – xy f) x1/2 + y1/2 g) √ x + 2x + 1 h) 5x3 – 4x2 + 6xy – 7
6. Let's find the degree of the following polynomials.
a) 3x2 b) –2xy c) 4x2yz
d) x2 + 5x + 6 e) 3y3 – 2y2 + 5y – 6 f) x2yz + xyz – 6
g) 2x2y2 + x2y – xy2 – 3xy + 4 h) x – x2y3 + (xy)3 i) (xy)2 + (xy)3 + (xy)4
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7. a) If l = 6, b = 4 and h = 2, evaluate the following expressions.
(i) l × b (ii) l × b × h (iii) 2(l + b)
(iv) 2h(l + b) (v) l2 (vi) 6l2
8. If x = 2 and y = 3, show that: a) (x + y)2 = x2 + 2xy + y2
b) (x – y)2 = x2 – 2xy + y2 c) x2 – y2 = (x + y) (x – y)
10.6 Addition and subtraction of algebraic expressions
x, 2x, 5x, etc. are the like terms.
The sum of x and 2x = x + 2x = 3x (1 + 2)x = 3x
The sum of 2x and 5x = 2x + 5x = 7x (2 + 5)x = 7x
The sum of x, 2x and 5x = x + 2x + 5x = 8x (1 + 2 + 5)x = 8x
The difference of 5x and 2x = 5x – 2x = 3x (5 – 2)x = 3x
The difference of 2x and x = 2x – x = x (2 – 1)x = x
Thus, when we add or subtract like terms, we should add or subtract the coefficients
of the like terms.
On the other hand, x, x2, y, 2y2 are unlike terms.
Sum of x and x2 = x + x2, difference of x and x2 = x – x2
Sum of x and y = x + y, difference of x and y = x – y
Thus, we do not add or subtract the coefficient of unlike terms.
Worked-out examples
Example 1: Add (i) 2a, 3a2, 4a and a2 (ii) 3x2y, 2x2y, 3xy2 and – 2xy2.
Solution:
(i) 2a + 3a2 + 4a + a2 = 2a + 4a + 3a2 + a2
= 6a + 4a2
(ii) 3x2y + 2x2y + 3xy2 + (– 2xy2) = 5x2y + 3xy2 – 2xy2
= 5x2y – xy2
Example 2: Add 4ab + 7bc – 5, 3bc – 8ab + 6 and 9ab – bc – 2.
Solution:
Addition by vertical arrangement Addition by horizontal arrangement
4ab + 7bc – 5 (4ab + 7bc – 5) + (3bc – 8ab + 6) + (9ab – bc – 2)
– 8ab + 3bc + 6 = 4ab + 7bc – 5 + 3bc – 8ab + 6 + 9ab – bc – 2
9ab – bc – 2 = 4ab + 9ab − 8ab + 7bc + 3bc – bc – 5 + 6 – 2
5ab + 9bc − 1 = 5ab + 9bc − 1
= 5ab + 9bc − 1
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Example 3: Subtract 2a2 + 5ab – b2 from 5a2 – ab + 3b2.
Solution:
Subtraction by vertical arrangement Subtraction by horizontal arrangement
5a2 − ab + 3b2
± 2a2 ± 5ab b2 (5a2 – ab + 3b2) − (2a2 + 5ab – b2)
= 5a2 – ab + 3b2 − 2a2 − 5ab + b2)
= 5a2 − 2a2 – ab − 5ab + 3b2 + b2
= 3a2 − 6ab + 4b2
3a2 – 6ab + 4b2
Example 4: What should be added to 3a + 4x to get 7a – 2x ?
Solution:
Here, the required expression to be added is Let’s think, what should be
(7a − 2x) − (3a + 4x) added to 3 to get 7.
= 7a – 2x – 3a – 4x It’s 4 and it is 7 – 3. It’s my
= 7a – 3a – 2x – 4x investigation to work out
= 4a − 6x such problems.
Example 5: What should be subtracted from 8a – 5b + 2 to get 2a + 3b – 9?
Solution:
Here, the required expression to be subtracted is
(8a – 5b + 2) – (2a + 3b – 9) Let’s think, what should be
= 8a – 5b + 2 – 2a – 3b + 9 subtracted from to 8 to get 5.
= 8a – 2a − 5b − 3b + 2 + 9 It’s 3 and it is 8 – 5. It’s my rule
= 6a − 8b + 11 to work out such problems.
EXERCISE 10.2
General Section - Classwork
1. Let's say and write the sums or differences as quickly as possible.
a) 4x + 3x = ............... (b) 3xy + 6xy = ............... (c) 8a2 + 5a2 = .............
(d) 8p – 3p = ............... (e) 9ab2 – 6ab2 = ............... (f) 10x3 – 3x3 = .............
2. a) What is the sum of a2 and a ? ...............................
b) What is the difference of 2x and 3y? ...............................
c) What should be added to 3x2 to get 7x2? ...............................
d) What should be subtracted from 11a3 to get 6a3? ...............................
Creative Section - A
3. a) What are like and unlike algebraic terms? Write with examples.
b) How do we add or subtract like terms? Write with examples.
c) Can we add or subtract the coefficients of unlike terms? Write with examples.
d) In what way the sum of x + x and the product of x × x different with each
other?
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4. Let's add.
a) 3x, 5y, 4x and 6y b) 6ab, 8bc, (–2ab) and (–3bc)
c) 7at2, (–2at), (–4at2) and at d) (–5p3q2), (–9p2q3), 6p3q2 and 10p2q3
e) 3t – 2tz + 4 and 5tz + 2t – 10 f) 2x + 3y – 6 and 5x – 4y + 1
g) 7a + 4b – 8 and 3b – 5a + 9 h) 5x2 + 3x + 4 and x2 + 2x – 7
i) 8a2 + 3ab – 2b2, 3a2 – ab + 5b2 and 5ab – 7a2 – b2
j) a2 + 3ab – bc, b2 + 3bc – ca and c2 + 3ca – ab
5. Let's subtract.
a) 5pq from 8pq b) 2x2y from 9x2y
c) –3ab from 5ab d) 2a2b3 from – 3a2b3
e) 4m + 5n from 6m + 9n f) 7p2 – 6q2 from 5p2 + 2q2
g) 4x – y – (2x – 3y) h) 2a + 3b – (a + 5b)
i) 4x2 + 5x – 3 – (2x2 – 3x + 6) from 4x2 + 5x – 3
j) y3 – 5y2 + y – 11 from 4y3 – 3y2 – y – 6
k) a3b3 – 2a2b2 + 3ab – 4 from –5a3b3 – a2b2 – 4ab – 7
l) 2.6x4 – 3.8x3 – 1.2x2 + 4.6x – 5.4 from 6.2x4 + 8.3x3 –2.1x2 + 6.4x – 4.5
6. Let's simplify.
a) 3x + 2y + 5x – 9y b) 9a – 2b – 4a + 7b
c) 4x2 – 2y2 + 2x2 – 5y2 d) 9a2 − 5a – 6a2 + 3a – 2
e) 7p2 + p + p2 – 6p + 3 f) 5x2 – (2x2 – y2) – 4y2
g) 10a + 4b – (3a + 2b) h) 7p – 5q – (2p – 8q)
i) 12x – (5x + 4y) – (2y + 3x) j) 13a2 – (3b2 – 4c2) + a2 – (8a2 – 5b2 + 7c2)
7. a) What should be added to 5xy to get 9xy ?
b) What should be added to −ab + bc + ca to get ab − bc − ca ?
c) What should be subtracted from 9x2y to get 4x2y?
d) What should be subtracted from 3p2 + 2p – 1 to get p2 – 3p − 4 ?
8. a) To what expression must 5a2 – 4a + 3 be added to make the sum zero?
b) From what expression must x2 + 5x – 7 be subtracted to make the difference
unity?
9. a) If a = x + y and b = x – y, show that (i) ( a + b)2 = 4x2 (ii) (a –b)2 = 4y2
b) If x = p + 2 and y = p – 3, show that (i) x + y + 1 = 2p (ii) x − y – 5 = 0
It's your time – Project work
10. a) Let's write any three different pairs of like terms. Then, find the sum and
difference of each pair.
b) Write any three binomial algebraic expressions and denote them by A, B and C
respectively.
Then, find (i) A +B – C (ii) A – (B – C) (iii) (A + B) – (A – C)
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10.7 Multiplication of algebraic expressions
While multiplying algebraic expressions, the coefficients of the terms are multiplied
and the power of the same bases are added. For example:
Example 1: Multiply: 4x2 by 3x
Solution: 4 × 3 = 12 (Coefficients are multiplied.)
Here, 4x2 × 3x = 12x3 x2 × x = x2 + 1 = x3 (Power of the same bases are added.)
(i) Multiplication of polynomials by monomials
While multiplying a polynomial by a monomial, we multiply each term of a
polynomial separately by the monomial. For example:
Example 2: Multiply: (x + y) by a xy a
Solution: ax ay
Here, a × (x + y) = ax + ay
x+y
Example 3: Multiply (3m2 – 2n2) by 5mn
Solution: Each term of 3m2 – 2n2 is separately
Here, 5mn × (3m2 – 2n2) = 5mn × 3m2 – 5mn × 2n2 multiplied by 5mn.
= 15m3n – 10mn3
(ii) Multiplication of polynomials
While multiplying two polynomials, each term of a polynomial is separately
multiplied by each term of another polynomial. Then, the product is simplified.
For example: ab
Example 4: Multiply (a + b) by (x + y) x ax bx
Solution: y ay x+y
by
Here, (x + y) × (a + b) = x (a + b) + y (a + b) a+b
= ax + bx + ay + by
Example 5: Multiply (3x2 + 2x – 4) by (2x – 3)
Solution:
By horizontal arrangement By vertical arrangement
(2x – 3) (3x2 + 2x – 4)
= 2x (3x2 + 2x – 4) – 3 (3x2 + 2x – 4) 3x2 + 2x – 4
= 6x3 + 4x2 – 8x – 9x2 – 6x + 12
= 6x3 – 5x2 – 14x + 12 × 2x – 3
6x3 + 4x2 – 8x
– 9x2 – 6x + 12
6x3 – 5x2 – 14x +12
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EXERCISE 10.3
General Section - Classwork
Let's say and write the products as quickly as possible.
1. a) 2x × 3x × 4x = ............. b) 4y × (– 5y3) = .............
c) (− 3ab × (− 4a2b) = ............. d) (– 2xy) × y × (– 3x2yz) = .............
e) p (p + 5) = ............. f) 2x (3x − 7) = .............
g) −5x2 (3x − 2y) = ............. h) − ab (a2 + b2) = .............
i) x(xy + x + y) = ............. j) ab (ab − a + 1) = .............
2. a) If a = x and b = 2x2, then 2ab = ..................
b) If x = 2p2 and y = 3p3, then 3xy = ..................
3. Let's investigate the tricky process of multiplication shown below.
1+ 2
x×x 1× 2
(x + 1)(x + 2)= x2... (x + 1)(x + 2) = x2 + 3x... (x + 1)(x + 2) = x2 + 3x + 2
a× a 2–3 2× –3
(a + 2 )(a–3) = a2 ... (a + 2)(a – 3) = a2 – 1.a ... (a + 2)(a – 3) = a2 – a – 6
Now, apply the tricky process shown above. Then, say and write the products as
quickly as possible.
(a) (x + 2) (x + 1) = ......................... (b) (x + 2) (x + 3) = .........................
(c) (a + 3) (a + 4) = ......................... (d) (a + 3) (a – 2) = .........................
(e) (x – 3) (x + 5) = ......................... (f) (x + 2) (x – 5) = .........................
(g) (x – 3) (x − 2) = ......................... (h) (a − 6) (a – 3) = .........................
Creative Section - A b) (–3x2) × 4x × (–x)
4. Let's simplify: d) (–3pq) × (–5qr) × pqr
a) x × 2x × 3x2 f) (– 4x) × (–x2yz) × (–3y) × (–2z)
c) (–2ab) × (–5a) × (–b)
e) mn × (–m2) × (–3n2)
5. Let's find the products.
a) x (2x + 4) b) 2x (3x – 5) c) 3p (2p2 – 1)
d) – 3p (p2 – 2p + 3) e) – 3t2 (2t2 + t – 2) f) xyz (x2 – y2 – z2)
6. Let's simplify.
a) x (x + 2) + 3 (x + 2) b) 2x (x – 3) + 5 (x – 3)
c) 3x (x + y) + y (x + y) d) 2a (2a + b) – b (2a + b)
e) x (x2 – xy + y2) + y (x2 – xy + y2) f) a (a2 + ab + b2) – b (a2 + ab + b2)
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Algebraic Expressions
7. Let's multiply.
a) (a + b) (a + ) b) (a – b) (a – ) c) (a + b) (a – )
f) (2a + 5) (3a − 4)
d) (x + 2) (x +3) e) (y + 4) (y –5) i) (x + y) (x – y)
l) ( x + 2) (x2 + 2x + 3)
g) (4p – 3) (p + 2) h) (3c − 2d) (2c – 3d ) o) (x –y) (x2 + xy + y2)
j) (2x + 3y) (2x − 3y) k) (7m − 3n) (7m + 3n) c) x
y
m) (x − 3) (2x2 – 4x + 5) n) (a + b) (a2 − ab + b2) p
Creative Section - B
8. Let's find the areas of the following rectangles.
a) b) x
n
m ab
d) e) a f) 3x
2a b 8x y
2y
3a b 2a b
9. a) The length of a rectangular garden is (2x + 3y)m and its breadth is
(4x – y)m . Find the area of the garden.
b) A rectangular piece of carpet has length (3x + y)m and breadth (2x – y)m.
Find its area.
c) The floor of a bedroom is (5a –3)m long and (3b +2)m wide.
(i) Find the area of the floor.
(ii) Find its actual area if a = 2 and b = 1
d) The length of a play ground is (7p + 2q)m and width is (5p – 3q)m.
(i) Find the area of playground
(ii) If p = 5 and q = 2, find its actual area.
10. a) If x = (a + 7) and y = (a – 7), show that a2 = xy + 49
b) If p = (3x + 1) and q = (3x – 1), show that: pq + 1 = x2
c) If a = (2x + 1) and b = (4x2 – 2x + 1) show 9
: ab – 1= x3
that 8
It's your time - Project work!
11. a) Let's write any three pairs of monomial expressions and find the product
of each pair.
b) Let's write any three pairs of binomial expressions and find the product of
each pair.
c) Let's write any three pairs of monomial expressions such that the product
of each pair is 12x4y4.
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Algebraic Expressions
10.8 Some special products and formulae
Da bC
ab a + b
(i) The product of (a + b) and (a + b) a a2
(square of binomials)
Let’s multiply (a + b) by (a + b) b ab b2 b
(a + b) × (a + b) = a (a + b) + b (a + b) A a+b b B
(a + b)2 = a2 + ab + ab + b2
Area of ABCD = a2 + ab + ab + b2
= a2 + 2ab + b2
(a + b)×(a + b)=(a + b)2 = a2 + 2ab + b2
Thus, (a + b)2 = a2 + 2ab + b2
Here, if (a + b)2 = a2 + 2ab + b2, then, Vedanta ICT Corner
Please! Scan this QR code or
a2 + b2 = (a + b)2 – 2ab browse the link given below:
https://www.geogebra.org/m/sasrnnqc
(ii) The product of (a – b) and (a – b) D a–b b C
Let’s multiply (a – b) by (a – b)
(a – b) × (a – b) = a (a – b) – b (a – b) (a – b)2 b(a – b) a–b
(a – b)2 = a2 – ab – ab + b2 a
= a2 – 2ab + b2 b(a – b) b2 b
Thus, (a – b)2 = a2 – 2ab + b2
A a–b b B
Here, if (a – b)2 = a2 – 2ab + b2, then,
a2 + b2 = (a – b)2 + 2ab In the figure, the length and breadth of
the square ABCD are decreased by b.
Vedanta ICT Corner Area of square
Please! Scan this QR code or
browse the link given below: = (a – b)2 + b (a – b) + b (a – b) + b2
a2 = (a – b)2 + ab – b2 + ab – b2 + b2
https://www.geogebra.org/m/rfh9n84s a2 = (a – b)2 + 2ab – b2
∴ (a – b)2 = a2 – 2ab + b2
Now, let's recall the following important formulas,
1. (a + b)2 = a2 + 2ab + b2 = a2 – 2ab + b2 + 4ab = (a – b)2 + 4ab
2. (a – b)2 = a2 – 2ab + b2 = a2 + 2ab + b2 – 4ab = (a + b)2 – 4ab
Worked-out examples
Example 1: Find the squares of (x + 2) a) by geometrically
b) without using formula c) using formula
Solution: D x 1 1C
a) By geometrical process: x2
The square of (x + 2) x 1.x
(x + 2)2 = x2 + x + x + x + x + 1 + 1 + 1 + 1 1.x
x+2
= x2 + 4x + 4
1 1.x 1 1
b) Without using formula: 1 1.x 1 1
A B
The square of (x + 2) = (x + 2)2 x+2
= (x + 2) (x + 2) = x (x + 2) + 2 (x + 2)
= x2 + 2x + 2x + 4 = x2 + 4x + 4
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Algebraic Expressions
c) By using formula: Consider, x = a and 2 = b
The square of (x + 2) = (x + 2)2 Now, (a + b)2 = a2 + 2ab + b2
∴ (x + 2)2 = (x)2 + 2.x.2 + (2)2
= (x)2 + 2. x . 2 + (2)2
= x2 + 4x + 4
Example 2: Expand: a) (2p2 + 3q2)2 b) 3x 12
Solution: 3x
Here,
a) (2p2 + 3q2)2 Let, 2p2 = a and 3q2 = b, then,
(a + b)2 = a2 + 2ab + b2
= (2p2)2 + 2. 2p2 . 3q2 + (3q2)2 ∴ (2p2 + 3q2)2 = (2p2)2 + 2. 2p2 . 3q2 + (3q2)2
= 4p4 + 12p2q2 + 9q4
b) 3x 1 2 = (3x)2 – 2 . 3x . 1 + 12 It is in the form of ( )2. So, using
3x 3x 3x the formula, (a – b)2 = a2 – 2ab + b2,
= 9x2 – 2 + 1 3x 1 2 = (3x)2 – 2 . 3x. 1 + 12
9x2 3x 3x 3x
Example 3: Simplify (2x + 3y)2 – (2x − 3y)2.
Solution:
(2x + 3y)2 – (2x – 3y)2 = (2x)2 + 2. 2x.3y + (3y)2 – [(2x)2 − 2. 2x.3y + (3y)2]
= 4x2 + 12xy + 9y2 – (4x2 − 12xy + 9y2)
= 4x2 + 12xy + 9y2 – 4x2 + 12xy – 9y2 = 24xy
Example 4: Express 9p2 + 24pq + 16q2 as a perfect square.
Solution: 9p2 + 24pq + 16q2
9p2 + 24pq + 16q2 (3p)2 + ………. + (4q)2
= (3p)2 + 2. 3p. 4q + (4q)2
= (3p + 4q)2 (3p)2 + 2. 3p. 4q. + (4q)2
It is in the form a2 + 2ab + b2 which is equal to (a + b)2.
Example 5: Find the squares of a) 99 b) 101
Solution:
a) Square of 99 = 992 = (100 – 1)2 = (100)2 – 2.100.1 + 12 = 10000 – 200+1 = 9801
b) Square of 101 = (101)2 = (100 + 1)2 = (100)2 + 2. 100.1 + 12
=10000 + 200 + 1 = 10201
Example 6: If x + y = 5 and xy = 6, find the value of x2 + y2.
Solution:
Here, x + y = 5
On squaring both sides, we get
(x + y)2 = (5)2
or, x2 + 2xy + y2 = 25 Using (a + b)2 = a2 + 2ab + b2
or, x2 + 2 × 6 + y2 = 25 Putting xy = 6 [ xy = 6]
or, x2 + 12 + y2 = 25
or, x2 + y2 = 25 – 12
∴ x2 + y2 = 13
147Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Algebraic Expressions
Example 7: If p – 1 = 3, find the value of p2 + 1 .
Solution: p p2
Here, p – 1 = 3
p
On squaring both sides, we get,
p 1 2 = (3)2
p
or, p2 – 2 × p × 1 + 1 = 9 Using (a – b)2 = a2 – 2ab + b2
p p2
or, p2 + 1 = 9 + 2
p2
∴ p2 +p12 = 11
Example 9: If m – 1 = 7, prove that : a) m2 + 1 = 51 b) m + 1 2 = 53
Solution: m m2 m
Here, a)
m – 1 = 7
m
or, m − 1 2 = 72 Squaring on the sides
m
or, m2 – 2 × m × 1 + 1 = 49
m m2
∴ m2 + 1 = 49 + 2 = 51 Proved.
m2
b) m + 1 2 = m − 1 2 + 4m × m1 (a + b)2 = (a – b)2 + 4ab
m m
= 72 + 4
= 49 + 4
∴ m + 1 2 = 53 Proved
m
EXERCISE 10.4
General Section – Classwork
1. Let's say and write the answers in the expanded forms.
a) (m + n)2 = .................................... b) (m – n)2 = ....................................
c) (p + q)2 = .................................... d) (p – q)2 = ....................................
e) (a + 1)2 = .................................... f) (a – 1)2 = ....................................
g) (x + 2)2 = .................................... h) (x – 2)2 = ....................................
Vedanta Excel in Mathematics - Book 7 148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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Vedanta Excel in Mathematics Book 7 Final (2079)
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