Fractions and Decimals
5.17 Multiplication and division of decimals
(i) Multiplication of decimals by 10 or powers of 10
In the given decimal block, each cube
represents 0.1. Therefore, there are ten 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
times 0.1.
10 × 0.1 = 1 (block)
Again, let's take 10 times 0.2.
Here, 10 × 0.2 = 2 (blocks)
Thus, while multiplying a decimal number by 10 or powers of 10, the
decimal point is shifted as many digit to the right as there are zero in 10
or powers of 10.
Example 1: Multiply 0.6789 by 10, 100 and and 1000.
Solution:
0.6789 × 10 = 6.789 10 has one zero. So, the decimal point is
shifted one digit to the right. Of course,
multiplying by 10 means, the decimal
point is shifted to the right of tenths
place.
0.6789 × 100 = 67.89 100 has two zeros. So, the decimal
point is shifted two digits to the right.
Of course, the decimal point is shifted
to the right of hundredths place !!
0.6789 × 1000 = 678.9 1000 has three zeros. So, the decimal
point is shifted three digits to the right !
Actually, the decimal point is shifted to
the right of thousandths place !!
(ii) Multiplication of decimal numbers by whole numbers
Let's study the following illustrations and learn about the multiplication
of decimal numbers by whole numbers.
2 times 0.2 = 2 × 0.2 = 0.4
2 times 0.3 = 2 × 0.3 =- 0.6
5 times 0.2 = 5 × 0.2 = 1.0 (or 1)
(1 block)
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4 times 0.4 = 4 × 0.4 = 1.6
Similarly, 1 0.6
1.5 one digit after 1.5 × 9 means 9 times 1 whole block (or 1) and
×9 decimal point 9 times 5 small decimal cubes (or 0.5)
13.5
one digit after = 9 × 1 whole block + 9 × 5 decimal cubes
decimal point
in the product = 9 whole blocks + 45 decimal cubes
= (9 + 4) whole blocks + 5 decimal cubes = 13.5
(iii) Multiplication of decimal numbers by decimal numbers
Let's study the following illustrations and discover the rule of
multiplication of decimal numbers by decimal numbers.
a) Let's multiply 0.3 by 0.2. }0.06 } }0.3
0.2 × 0.3 means 0.2 of 0.3
}
From the shaded diagram given alongside,
0.2 of 0.3 is 6-hundredths = 0.06 }}
0.2
}0 . 3 There are two digits after decimal 0.2 = 2 and 0.3 = 3
point altogether in 0.3 and 0.2 10 10
×0.2 2 3
0 . 06 So, the product 0.06 has two digits 0.2 × 0.3 = 10 × 10 = 6 = 0.06
after decimal point. 100
b) Let's multiply 0.5 by 0.3. }0.15 }0.3
0.3 × 0.5 means 0.3 of 0.5
From the shaded diagram given alongside,
0.3 of 0.5 is 15-hundredths = 0.15
0.5
}0 . 5 There are two digits after decimal 0.3 = 3 and 0.5 = 5
point altogether in multiplicand 10 10
×0.3 and multiplier 3 5
0 . 15 0.3 × 0.5 = 10 × 10 = 15 = 0.15
So, the product 0.15 has two digits 100
after decimal point.
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Fractions and Decimals
Fractions and Decimals
Thus, when a decimal number is multiplied by decimal number (or
whole number), the product has as many decimal places as there are
total number of decimal places in multiplicand and multiplier.
Example 2 : Multiply : 7.25 by 1.3
Solution :
7.25 I understood ! 7.25 × 1.3 = 725 × 13
× 1.3 In 7.25 × 1.3, there are three 100 10
decimal places altogether in
2175 multiplicand and multiplier. So, = 9425 = 9.425
+7250 the product 9.425 also has three 1000
decimal places.
9.425
(iv) Division of decimals by 10 or powers of 10
In the given figure, each square room 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
represents 0.1. Let's study the illustrations
and discover the rule of division of decimal 1 (or whole)
numbers by 10 or power of 10. 0.1
a) Let's divide 0.1 by 10.
It means divide 0.1 into 10 equal parts.
From the diagram, 0.1 ÷ 10 = 0.01 0.01
b) Let's divide 0.3 by 10. 0.3
It means divide 0.3 into 10 equal parts.
From the diagram, 0.3 ÷ 10 = 0.03
0.03
Thus, when a decimal number is divided by 10 or powers of 10, we should
shift the decimal point as many digits to the left as there are zeros in the
powers of 10. Let’s learn more about it from the examples.
Example 3: Divide 529.4 by 10, 100 and 1000.
Solution:
529.4 ÷ 10 = 52.94 10 has one zero. So, the decimal point is
shifted one digit to the left !
529.4 ÷ 100 = 5.294 100 has two zeros. So the decimal
point is shifted two digits to the left !
529.4 ÷ 1000 = 0.5294 1000 has three zeros. So the decimal
point is shifted three digits to the left !
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Fractions and Decimals
Fractions and Decimals
Example 4: Divide 0.69 by 10,100 and 1000. I got it !
Solution : 0.69 is same as 00.69
or 000.69, etc.
0.69 ÷ 10 = 0.069
0.69 ÷ 100 = 00.69 ÷ 100 = 0.0069
0.69 ÷ 1000 = 000.69 ÷ 1000 = 0.00069
(v) Division of decimals by whole numbers 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
In the given diagram, each square room 1 (or whole)
represents 0.1. Let's study the illustrations
and investigate the rule of division of decimal
numbers by whole numbers.
a) Let's divide 0.1 by 2. 0.05
It means divide 0.1 into 2 equal parts.
From the diagram, 0.1 ÷ 2 = 0.05 0.1
0.04
2 0.1 0. o 2 0.1 0.0 o 2 0.10 0.05
– 10
0
b) Let's divide 0.2 by 5.
It means divide 0.2 into 5 equal parts.
From the diagram, 0.2 ÷ 5 = 0.04
5 0.2 0. o 5 0.2 0.0 o 5 0.20 0.04 0.2
– 20
0
Similarly, let's study the following examples and learn more about the
division of decimal numbers by whole numbers.
The division of a decimal number by a natural number is performed
similarly as usual division process. For example,
Example 5: Divide a) 24.15 ÷ 7 b) 6.66 ÷ 9
Solution:
7 24.15 3.45 9 6.66 0.74
–21 –0
31 66
28 63
35 36
35 36
0 0
? 24.15 ÷ 7 = 3.45 ? 6.66 ÷ 9 = 0.94
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Fractions and Decimals
(v) Division of whole number or decimal number by decimal number
In the given diagram, each cube represents
0.1. Let's study the following illustrations 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
and discover the rule to divide whole
numbers or decimal numbers by decimal 1 (or whole block)
numbers.
a) Let's divide 1 by 0.2. one two three four five
0.2 0.2 0.2 0.2 0.2
It means how many 0.2s there are in 1.
From the diagram, there are five 0.2s 1 (or whole block)
in 1.
1 ÷ 0.2 = 1 ÷ 2 = 1 × 10 = 5
10 2
b) Let's divide 1.2 by 0.3. one two three four
It means how many 0.3s there are 0.3 0.3 0.3 0.3
in 1.2. From the diagram, there are
four 0.3s in 1.2.
1.2 ÷ 0.3 = 12 ÷ 3 = 12 × 10 = 4 1 0.2
10 10 10 3
Thus, when a whole number or decimal number is divided by another deci-
mal number at first, remove the decimal point from the divisor and then
divide as usual process.
Example 6: Divide 8.664 ÷ 2.4
Solution: 24 86.64 3.61
–72
8.664 ÷ 2.4 = 8.664 × 10 To remove the decimal point from 146
2.4 × 10 2.4, it is multiplied by 10. In the 144
same time, the dividend 8.664 24
= 86.64 should also be multiplied by 10. 24
24 0
? 8.664 ÷ 2.4 = 3.61
EXERCISE 5.7
General Section – Classwork
1. Each square room represents 0.1. Let's write the multiplier, multiplicand,
and find the product by using the diagrams.
a) 0.1 0.1 0.1 0.1 0.1 0.1 b)
............ × ............ = ............
3 0.2 0.6............ × ............ = ............
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Fractions and Decimals d)
c)
............ × ............ = ............ ............ × ............ = ............
2. Each square room represents 0.01. Let's write the multiplier, multiplicand,
and find the product by using the shaded diagrams.
}
}
}
}a) 0.2 b)
}} } c)
}
0.3
........ × ........ = .......... ........ × ........ = ........
........ × ........ = ........
3. Each cube represents 0.1. Let's say and write the quotients by using the
diagrams.
0.5 0.5
a) How many 0.5s are there in 1?
1 (or whole block) 1 ÷ 0.5 = .............
0.2 0.2 0.2 0.2 0.2 How many 0.2s are there in 1?
1 ÷ 0.2 = .............
b)
1 (or whole block)
0.4 0.4 0.4 0.4
0.6
c) How many 0.4s are there in 1.6?
1.6 ÷ 0.4 = .............
1 (or whole block)
4. Let’s tell and write the products as quickly as possible.
a) 0.374 × 10 = ......., 0.374 × 100 = ......., 0.374 × 1000 = .........
b) 0.0729 × 10 = ........, 0.0729 × 100 = ......., 0.0729 × 1000 = .........
c) 0.0008 × 10 = ....... 0.0008 × 100 = ......., 0.0008 × 1000 = .........
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Fractions and Decimals
5. Let’s tell and write the quotients as quickly as possible.
a) 264.5 ÷ 10 = ......, 264.5 ÷ 100 = ......., 264.5 ÷ 1000 = .............
b) 57.3 ÷ 10 = ........, 57.3 ÷ 100 = .........., 57.3 ÷ 1000 = .............
c) 0.7 ÷ 10 = .........., 0.7 ÷ 100 = ........, 0.7 ÷ 1000 = .............
6. Let’s tell and write the products as quickly as possible.
a) 0.3× 2 = ............, 0.03 × 2 = ......... 0.003 × 2 = ..............
b) 0.7 × 5 = ............, 0.4 × 8 = ........., 0.6 × 9 = ..............
c) 0.3 × 0.2 = .........., 0.2 × 0.4 = ........., 0.3 × 0.5 = ..............
7. Let’s tell and write the quotients as quickly as possible.
a) 0.6 ÷ 2 = ............, 6 ÷ 0.2 = ............, 0.6 ÷ 0.2 = ..............
b) 0.9 ÷ 3 = ............, 0.09 ÷ 3 = ............, 0.09 ÷ 0.3 = ..............
c) 1.2 ÷ 4 = ............, 0.12 ÷ 4 = ............, 0.12 ÷ 0.4 = ..............
Creative Section
8. Find the products or quotients as shown.
1.5 × 50 = 1.5 × 10 × 5 = 15 × 5 = 75
4.8 ÷ 20 = (4.8 ÷ 2) ÷ 10 = 2.4 ÷ 10 = 0.24 e) 1.3 × 200
j) 1.5 ÷ 30
a) 1.2 × 20 b) 1.5 × 30 c) 1.7 × 40 d) 2.1 × 50
f) 1.8 × 300 g) 2.4 × 400 h) 3.2 × 600 i) 1.4 ÷ 20
9. Let’s learn the process and express the following fractions in decimals.
1200 = 12 = 5 12 = 2.4 = 0.24
5000 50 × 10 10
a) 6 , 60 , 60 b) 16 , 160 , 160 c) 540 , 540 , 540
20 200 2000 40 400 4000 900 9000 90000
10. 4.5% means 4.5 = 0.045. Similarly, express the following percents in
100
decimals.
a) 0.6% b) 1.9% c) 2.5% d) 36.4% e) 73.6% f) 98.7%
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Fractions and Decimals
11. 1 cm = 10 mm. a) How many millimetres are there in
(i) 2.5 cm (ii) 5.6 cm ?
b) How many centimetres are there in
(i) 37 mm (ii) 84 mm ?
12. 1 m = 100 cm. a) How many centimetres are there in
(i) 3.15 m (ii) 6.8 m ?
b) How many metres are there in
(i) 225 cm (ii) 75.5 cm ?
13. Re 1 = 100 p a) How many paisa are there in
(i) Rs 4.50 (ii) Rs 8.25
b) How many rupees are there in
(i) 250 p (ii) 745 p ?
14. 1 km = 1000 m a) How many metres are there in
(i) 1.4 km (ii) 2.75 km ?
b) How many kilometres are there in
(i) 1800 m (ii) 4250 m ?
15. 1 Kg = 1000 g. a) How many grams are there in
(i) 1.25 kg (ii) 3.674 kg?
b) How many kilograms are there in
(i) 1500 g (ii) 2450 g ?
16. 1 l = 1000 ml. a) How many millilitres are there in
(i) 4.50 l (ii) 8.25 l ?
b) How many litres are there in
(i) 350 ml (ii) 1265 ml?
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Fractions and Decimals
EXERCISE 5.8
Creative Section - A
1. Let's find the products.
a) 2.6 × 7 b) 3.48 × 9 c) 6.03 × 14 d) 7.34 × 16
g) 1.2 × 0.3 h) 3.6 × 2.4
e) 0.432 × 18 f) 0.068 × 27 k) 0.73 × 0.6 l) 0.057 × 0.004
i) 5.68 × 3.7 j) 3.69 × 4.52 d) 5.04 ÷ 24
h) 4.8 ÷ 0.8
2. Let's find the quotients. l) 0.224 ÷ 1.4
a) 3.6 ÷ 3 b) 5.64 ÷ 4 c) 4.248 ÷ 18
e) 0.624 ÷ 8 f) 0.054 ÷ 15 g) 2.5 ÷ 0.5
i) 12.42 ÷ 0.9 j) 6.25 ÷ 2.5 k) 0.35 ÷ 0.7
3. Puzzle time
Let’s fill in the blanks by missing numbers.
Start ÷2 × 10 + 0.5 – 0.8 ×2
0.6 – 2.5 + 0.8 – 2.8 + 0.5
+ 4.4 ÷ 0.9
+ 4.2 6
– 1.5
×2 ÷ 3 – 0.6
4. Let's simplify. b) 3 × (0.9 – 0.4) c) (0.2 + 0.6) ÷ 2
a) 2 × (0.4 + 0.3) e) 2.36 – 1.28 ÷ 1.6 f) 1.3 × 0.4 + 2.4 × 0.3 – 1.8 × 0.2
d) (3 × 0.8) ÷ (2 × 0.2)
5. a) If the cost of a pen is Rs. 15.25, find the cost of 9 pens.
b) 1.35 m of cloth is required for 1 shirt. What will be the total length of
cloth required for 6 shirts of the same size ?
6. a) If the cost of one dozen of exercise books is Rs. 225.72, find the cost of 1
exercise book.
b) The length of a rope is 11.55 m. If it is cut into 7 equal pieces, find the
length of each piece of rope.
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Fractions and Decimals
7. a) The area of a rectangle is length × breadth (A = l × b). If the length of a
rectangle is 7.8 cm and its breadth is 5.4 cm. find its area in sq. cm.
b) The length of a rectangle is Area A . If the area of a rectangle is
Breadth b
28.35 sq. cm and its breadth is 4.5 cm, find its length.
Creative Section - B
8. a) Mother recorded the fuel consumption of her car as:
City: 21.4 km per litre Highway: 23.5 km per litre
The fuel tank of car holds 40.2 litre of petrol.
(i) How long could she drive on a full tank of petrol in the city.
(ii) How long could she drive on a full tank of petrol in the highway?
b) According to money exchange rate announced by Nepal Rastra Bank on
2nd November 2020, the buying and selling rates of 1 Australian dollar are:
Buying rate: Rs 77.96 Selling rate: Rs 78.30
(i) How much Nepalese rupees do you need to buy 25.5 Australian dollar?
(ii) How much Nepalese rupees do you get while selling 80 Australian
dollars?
9. a) In a supermarket a 1.4 kg bag of apples costs Rs 196.70.
(i) What is the actual cost of 1 kg of apples?
(ii) If you spend Rs 491.75 for apples, what would be the weight of the
apples?
b) Last week, Mr. Lama worked 38.5 hours and earned Rs 5563.25.
(i) How much money did he earn per hour?
(ii) How many hours would he work to earn Rs 6574.75?
10. a) Anamol, Shadeep and Bishal got different answer for their problem:
12 × (4.8 ÷ 0.3) – 3.5 × 3.64. Anamol’s answer was 39.12, Shadeep’s
answer was 179.26 and Bishal’s answer was 659.26.
(i) Simplify and then identify which student had correct answer.
(ii) Find the mistake done by other two students and explain.
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Fractions and Decimals
b) Evaluate by showing all steps:
{2.5 × (4.3 – 3.4) ÷ (1.5)2} + 0.1
It’s your time - Project Work!
11. a) Let's search and write any two pairs of decimal numbers (multiplier
and multiplicand) to get each of the following products.
Multiplier × Multiplicand Product
(i) 0.2 × 0.2 0.04
(ii) 0.1 × 0.4
(i) ............. × ............. 0.06
(ii) ............. × .............
(i) ............. × ............. 0.18
(ii) ............. × .............
(i) ............. × ............. 0.18
(ii) ............. × .............
b) Let's cut a few number of rectangular paper strips each of 10 cm long.
Divide each strip into 10 equal parts. Shade the parts with different
colours to show the following multiplications of decimal numbers.
Then, find the products.
(i) 3 × 0.2 (ii) 2 × 0.3 (iii) 4 × 0.2 (iv) 3 × 0.4 (v) 4 × 0.6
c) Let's visit to the available website and find today's exchange rates of the
following currencies with Nepali currency. Then calculate how much
Nepali currency is required to exchange the given foreign currencies?
(i) 1 U.S.$ = Rs ............ 10 U.S.$ = Rs ........... 100 U.S.$ = Rs ...........
(ii) 1 AU$ = Rs ............ 10 AU$ = Rs ........... 100 AU$ = Rs ...........
(iii) 1 QAR = Rs ............ 10 QAR = Rs ........... 100 QAR = Rs ...........
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Fractions and Decimals
5.18 Rounding off decimal numbers (Approximation)
The process of rounding off a number to the nearest convenient figure is called
rounding off number. It is also called approximation.
We can round off a whole number to the nearest tens, hundreds, thousands,
etc. Similarly, a decimal number can also be rounded off to the nearest whole
number, tenths, hundredths, thousandths, etc.
Let's learn the following rules of rounding off decimal numbers.
Rule 1: If the digit which is to be rounded off is less than 5, its place is
considered zero and the digit at the higher place remains unchanged.
Example 1: Round off 23.2134 to 3, 2 and 1 decimal places.
Solution:
Rule 2: If the digit which is to be rounded off is 5 or greater than 5, it is
considered as zero and 1 is added to the digit at the higher place.
Example 2: Round off 35.4685 to 3, 2 and 1 decimal places.
Solution:
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Fractions and Decimals
EXERCISE 5.9
General Section – Classwork
1. a) Round off these numbers to the nearest tens. Tell and write the answers
as quickly as possible.
(i) 346 ............. (ii) 221 ............. (iii) 685 ............. (iv) 904 .............
b) Round off these numbers to the nearest hundreds. Tell and write the
answers as quickly as possible.
(i) 287 ........... (ii) 536 ........... (iii) 1495 ........... (iv) 7312 ...............
c) Round off these numbers to the nearest thousands. Tell and write the
answers as quickly as possible.
(i) 1250 ............. (ii) 1940 ............ (iii) 3678 ............ (iv) 7355 ..........
2. Round off as instructed. Tell and write the answers as quickly as
possible.
a) 2.43 (to 1 d.p.) .................. 5.69 (to 1 d.p.) ..................
b) 4.576 (to 2 d.p.) .................. 8.384 (to 2 d.p.) ..................
c) 7. 2483 (to 3 d .p.) .................. 6.4635 (to 3 d.p.) ..................
Creative Section
3. Round off to 1 decimal place (d.p.)
a) 3.43 b) 6.47 c) 17.142 d) 25.258 e) 0.386
4. Round off to 2 decimal places (d.p).
a) 3.416 b) 8.321 c) 21.9832 d) 0.1647 e) 0.5273
5. Round off to 3 decimal laces (d.p)
a) 5.1023 b) 12.6275 c) 15.47562 d) 0.27438 e) 0.08956
6. Round off to 3, 2 and 1 decimal places.
a) 4.2134 b) 10.6579 c) 8.10925 d) 0.39768 e) 0.095487
7. Find the product and write the answers to 3 decimal places.
a) 0.236 × 0.6 b) 0.7934 × 0.12 c) 1.053 × 2.4
8. Find the quotients to 4 decimal places and write the answers to 2 decimal
places.
a) 1 b) 2 c) 5 d) 22 e) 25 f) 28
3 3 6 7 12 15
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Unit Unitary Method
6
6.1 Unitary Method - Looking back
Classwork - Exercise
Let’s tell and write the answers as quickly as possible.
1. a) Cost of 1 pencil is Rs 7, cost of 2 pens is .....................
b) Cost of 1 sweet is Rs 15, cost of 3 sweets is .....................
c) Cost of 2 kg of rice is Rs 200, cost of 1 kg of rice is .....................
d) 3 jars can hold 60 l of water 1 jar holds .....................
2. a) 1 man does a piece of work in 6 days, 2 men do the work in .....................
b) 1 pipe fills a tank in 4 hours, 4 pipes fill the tank in .....................
1
3. a) 2 part of the distance between two places is 10 km, whole
distance is .....................
b) 1 part of a tank contains 400 l of water, the capacity of tank is ................
3
6.2 Unit number of quantity and unit value
Suppose, the cost of 1 pen is Rs 40.
Here, 1 pen is the unit number of item and Rs 40 is its unit value.
Now, the cost of 2 pens is Rs 80, 3 pens is Rs 120, and so on.
Here, the values of more number of items are obtained multiplying the unit
value by the number of items.
Similarly, suppose the cost of 2 kg of potato is Rs 90.
Then, the cost of 1 kg of potato is Rs 45.
Here, the unit value is obtained dividing the total value by the number of
quantity.
6.3 More number of quantity and more value
Suppose, the cost of 1 kg of sugar is Rs 90.
Then, the cost of 2 kg of sugar = 2 × Rs 90 = Rs 180
The cost of 3 kg of sugar = 3 × Rs 90 = Rs 270, and so on.
Thus, when we know the value of unit quantity, we can find the value of more
quantity by multiplication.
Of course, number of any item and its cost are in direct proportion. In the case of
direct proportion, unit value is obtained by division and value of more numbers
of quantity is obtained by multiplication.
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Unitary Method
Thus, the mathematical method that we use to find unit value or the value of
any number of quantities is known as unitary method.
In unitary method, Total cost
Number of items
(i) unit cost =
(ii) total cost = Number of items × Unit cost
Total cost
(iii) number of items = Unit cost
Worked-out examples
Example 1 : If the unit cost of a book is Rs 150, find the cost of 6 books.
Solution :
Here, the unit cost of a book = Rs 150
Number of books = 6
Total cost = Number of books × unit cost
= 6 × Rs 150 = Rs 900.
Hence, the required cost is Rs 900.
Example 2 : Calculate the unit cost a pen if the cost of 7 pens is Rs 385.
Solution :
Here, the number of pen = 7
The total cost = Rs 385
Total cost Rs 385
∴ Unit cost = Number of items = 7 = Rs 55
Hence, the required unit cost is Rs 55.
Example 3: Suntali spends Rs 5,400 to buy a few number of pendrives at
Rs 450 each. How many pendrives does she buy ?
Solution :
Here, total cost = Rs 5,400
Unit cost = Rs 450 Rs 5,400
Total cost Rs 450
∴ Number of pendrives = Unit cost = = 12
Hence, she buys 12 pendrives.
Example 4 : Which is the best buy, 12 sweets for Rs 48 or 15 sweets for Rs 75 ?
Solution : = Total cost = Rs 48 = Rs 4.
In the first case, unit cost Number of sweets 12
In the second case unit cost = Total cost = Rs 75 = Rs 5.
Number of sweets 15
Hence, the best buy is 12 sweets for Rs 48.
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Unitary Method
Example 5: If the cost of 1 dozen of bananas is Rs 78, find the cost of
15 bananas.
Solution : Alternative process:
Here, 1 dozen bananas= 12 bananas
Now, cost of 12 bananas = Rs 78 Unit cost = Total cost
No. o bananas
78
Cost of 1 banana = Rs 12 = Rs 6.50 Rs 78
12
Cost of 15 bananas = 15 × Rs 6.50 = = Rs 6.50
= Rs 97.50 Cost of 15 bananas = 15 × unit cost
Hence, the required cost is Rs 97.50. = 15 × Rs 6.50
= Rs 97.50
Example 6: The cost of 6 kg of sugar is Rs 444. How many kilograms of sugar
can be bought for Rs 666 ?
Solution: Alternative process:
Here, Rs 444 is the cost of 6 kg of sugar. Unit cost = Total cost
Quantity of sugar
6
Re 1 is the cost of 444 kg of sugar. = Rs 444 = Rs 74
6
Rs 666 is the cost of 6 × 666 kg of sugar Total cost
444 Quantity of sugar = Unit cost
= 9 kg of sugar. = Rs 666
Hence, the required quantity of sugar is 9 kg. Rs 74
= 9 kg
Example 7: The average speed of a bus is 48 km per hour. How much distance
does it cover in 40 minutes?
Solution :
1 hour = 60 minutes
In 60 minutes, the bus covers 48 km.
In 1 minute, the bus covers 48 km
60
In 40 minutes, the bus covers 48 × 40 km = 32 km
60
Therefore, the required distance covered by the bus is 32 km.
Example 8: A pipe can fill a water tank of capacity 10,000 l completely in 5
hours. How much water does it fill in 2 hours?
Solution :
Vedanta Excel in Mathematics - Book 6 116 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unitary Method
In 5 hours, the pipe can fill 10,000 l.
In 1 hour, the pipe can fill 10,000 l
5
10,000
In 2 hours the pipe can fill 5 × 2 l = 4,000 l
Hence, the required amount of water filled by the pipe is 4,000 l.
Example 9: If 2 parts of the distance between two places is 30 km, find 3
3 5
parts of the distance between these places.
Solution :
Here, 2 parts of the distance = 30 km
∴ 3 = 30
Whole (1) distance 2 km
= 3 3 km = 45 km
30 × 2
Again, 3 parts of the distance = 3 of 45 km = 3 × 45 km = 27 km
5 5 5
Hence, the required distance is 27 km.
EXERCISE 6.1
General Section – Classwork
1. Let’s tell the answers and write in the blanks as quickly as possible
Articles Unit cost Cost of 2 Cost of 3 Cost of 4
articles articles articles
a) Eraser Rs 4 ....................... ....................... .......................
b) Pencil ....................... RS 16 ....................... .......................
c) Pen ....................... ....................... RS 90 .......................
d) Copy ....................... ....................... ....................... RS 100
2. a) If 1 of a sum is Rs 10, then the sum is .....................
2
b) If 1 of a distance is 5 km, then the distance is .....................
3
c) If 1 of a weight is 3 kg, then the weight is .....................
4
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Unitary Method
Creative Section - A
3. Find the total cost in the following cases.
Number of items Unit cost
a) 4 Rs 20
b) 9 Rs 45
c) 12 Rs 60.25
d) 25 Rs 75.50
4. Find the unit cost in the following cases.
Number of items Total cost
a) 5 Rs 60
b) 8 Rs 192
c) 18 Rs 990
d) 20 Rs 1110
5. Find the number of items in the following cases.
Unit cost Total cost
a) Rs 12 Rs 84
b) Rs 15 Rs 210
c) Rs 25 Rs 425
d) Rs 48 Rs 1248
6. Which is the best buy in the following cases.
a) 4 pens for Rs 56 or 6 pens for Rs 90
b) 5 kg of rice for Rs 500 or 9 kg of rice for Rs 810
c) 1 dozen of eggs for Rs 120 or 30 eggs for Rs 360
Creative Section - B
7. a) If the cost of 1 dozen of pencils is Rs 96, find the cost of 18 pencils.
b) The wages of a labour for a week is Rs 3,500, find the wages of the labour
for 12 days.
c) If the cost of 1 quintal of onions is Rs 4,000, what is the cost of 15 kg of
onions?
d) If the cost of carpeting 1 sq. m of a floor is Rs 120, what is the cost of
carpeting 96 sq. m of floor ?
8. a) The speed of a bus is 40 km per hour. How much distance does it cover
in (i) 45 minutes (ii) 3 hours (iii) 2 hours 30 minutes.
b) A car can cover a distance of 240 km in 4 hours.
(i) Find the speed of the car in km per hour.
(ii) How many kilometer does it cover with the same speed in 7 hours?
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Unitary Method
9. a) The cost of 6 kg of oranges is Rs 540. How many kilograms of oranges
can be purchased for Rs 720?
b) The cost of 5 story books is Rs 800. How many books can be bought for
Rs 1,440?
10. a) If 3 of the distance between two places is 30 km, how much is 2 of the
4 5
distance of these places ?
b) The cost of carpeting 2 parts of a floor is Rs 6,000. What is the cost of
3 3 floor ?
carpeting 5 parts of the
11. a) A pipe can fill a water tank of capacity 7,500 l completely in 4 hours.
How much water does it fill in 3 hours?
b) A tap can fill 6,000 l of water in 3 hours. How long does it take to fill a
water tank of capacity 4,000 l completely?
12. a) A computer can download a 200 megabytes (MB) of a game file in
20 seconds. How many megabytes does it download in 1 second? What
is the download speed of the internet?
b) A computer can download a 2.5 GB of a movie file in 50 seconds. Find
the download speed of the internet in megabytes (MB) per second.
It's your time - Project work!
13. a) Let's make groups of friends and visit a local market in your locality. List
the rates of cost of the following groceries and find the cost of the given
quantities of each grocery.
Grocery Rate of cost Quantity Cost
Rice ............. per kg 5 kg of rice .................................
Sugar ............. per kg 3 kg of sugar .................................
Kitchen oil ............. per litre 2 l of oil .................................
Milk ............. per litre 2 l of milk .................................
b) Do you have any internet connection in your house? If so, what is the
speed of the internet? How long does it take to download a file of size
1.5 GB in this internet speed?
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Unit Percent
7
7.1 Percent - Looking back
Classwork - Exercise
1. Let’s count the coloured blocks and write it as a fraction.
a) b) c)
o 3 o ......... o .........
10
d) e) f)
o ......... o ......... o .........
2. Let’s tell and write the correct answers as quickly as possible.
a) Express these fractions in percents.
a) 7 = ......... b) 19 = ......... c) 63 = ......... d) 80 = .........
100 100 100 100
b) Express these percentages in fractions.
a) 9% = ........... b) 33% = ............. c) 71% = ............. d) 97% = .........
c) Express these decimals in percents.
a) 0.05 = ......... b) 0.23 = ........... c) 0.65 = ........... d) 0.99 = .........
d) Express these percents in decimals.
a) 7% = ......... b) 39 % = ........... c) 45% = ........... d) 98 % = ........
7.2 Percent and percentage
Percent means per hundred or out of 100.
10 percent, or 10% means 10 per hundred or 10 out of 100.
A student obtained 85 percent marks in mathematics means that he/she obtained
85 marks out of 100 full marks.
The symbol % is usually written instead of the words percent.
For example, 25 percent is written as 25%, 76 percent is written as 76% and so on.
The word percent and percentage are closely related. The word 'percent'
(or the symbol %) accompanies a specific number, whereas the more general
word 'percentage' is used without a number.
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For example:
88 percent (or 88%) students of class secured A+ grade in maths test.
A large percentage of students of a class secured A+ grade in maths test.
7.3 Conversion of fractions or decimals to percent
To convert a fraction or decimal to a percent, we multiply it by 100 and affix the
symbol % to the product.
For example:
1 = 1 × 100% = 50%, 3 = 3 × 100% = 75%
2 2 4 4
0.06 = 0.06 × 100% = 6%, 0.87 = 0.87 × 100% = 87%
7.4 Conversion of percent to fractions or decimals
To covert a percent to a fraction or decimal we divide it by 100 and remove the
symbol of percent (%).
For example,
6% = 6 = 3 , 20% = 20 = 1
100 50 100 5
6% = 6 = 0.06 20% = 20 = 0.2
100 100
7.5 To express a given quantity as the percent of whole quantity
In this case, we first express the given quantity as the fraction of the whole
quantity. Then, we multiply the fraction by 100%.
For example: 16
40
16 as the percent of 40 = × 100% = 40%
45 as the percent of 60 = 45 × 100% = 75%
60
Worked-out examples
Example 1: 3 of the number of students of a school are boys.
5
(a) How many percentage of the students are boys?
(b) How many percentage of the students are girls?
Solution: 3 3 To convert a fraction
5 5 into percent, we should
(a) Here, = × 100% = 60% multiply it by 100%.
Hence, 60% of the students are boys.
(b) Also, percent of girls = 100% – 60% Total number of quantity
= 40 % is always considered as
100%.
Hence, 40% of the students are girls.
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Percent
Example 2: Pinky obtained 18 marks out of 20 full marks in a maths test.
Express her marks in percent.
Solution:
Here, 18 as the percent of 20 = 18 × 100% = 90%
20
Hence, she obtained 90% marks.
Example 3: Mr. Tharu earns Rs 25,000 in a month. He spends Rs 2,400 on his
children's education and Rs 5,600 to run his family. How many
percentage of his income does he save in a month?
Solution:
Here, total income = Rs 25,000
total expense = Rs 2,400 + Rs 5,600 = Rs 8,000.
∴ His saving in a month = Rs 25,000 – Rs 8,000 = 17,000.
17,000
Now, Rs 17,000 as the percent of Rs 25,000 = 25,000 × 100% = 68%
Hence, he saves 68% of his income.
EXERCISE 7.1
General Section – Classwork
1. The following graph shows some of the water-rich fruits and vegetables.
Let’s observe the graph and answer the following questions.
100%
90%
80%
Water content 70%
60%
50%
40%
30%
20%
10%
0 Orange Cucumber Apple Grapes Potato
Fruits
a) Percentage of water in orange = .........
b) Percentage of water in potato = .........
c) Percentage of water in cucumber = .........
d) Percentage of water in apple = .........
e) Percentage of water in grapes = .........
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2. Let’s tell and write the correct answers as quickly as possible.
a) Convert these fractions into percents.
a) 1 = ......... b) 1 = ......... c) 1 = ......... d) 1 = .........
2 4 5 10
b) Convert these decimals into percents. d) 0.99 = ........
a) 0.05= ........ b) 0.27 = .......... c) 0.45 = ..........
c) Convert these percents into fractions.
a) 7% = ......... b) 36% = .......... c) 76% = ........... d) 89% = ..........
d) Convert these percents into decimals. d) 88% = .........
a) 4% = ......... b) 25% = .......... c) 61% = ............
Creative Section - A
3. Convert the following fractions and decimals into percents.
3 4 7 9 12 43
a) 4 b) 5 c) 10 d) 20 e) 25 f) 50
g) 0.06 h) 0.33 i) 0.45 j) 0.62 k) 0.75 l) 0.93
4. Convert the following percents into fractions of their lowest terms.
a) 20% b) 25% c) 45% d) 64% e) 75% f) 80%
5. Express the parts of whole as the percentage of the whole.
a) 5 as a percent of 10 b) 6 as a percent of 15
c) Rs 40 as a percent of Rs 250 d) 60 marks as a percent of 75 marks
e) 9 months as a percentage of a year f) 250 gram as a percentage of 1 kg
6. Express the following as percents.
a) 4 out of 20 b) 60 marks out of 80 full marks
c) 15 girls out of 75 pupils d) Rs 120 out of Rs 480
2
7. a) 5 of the number of students of a school are girls ?
(i) How many percentage of the students are girls ?
(ii) How many percentage are boys ?
13
b) Mrs. Pyakurel spends 20 parts of her monthly income every month.
(i) How many percentage of the income does she spend ?
(ii) How many percentage of the income does she save?
8. a) There are 360 students in a school. 198 of them are boys.
(i) Find the percentage of the boys.
(ii) Find the percentage of the girls.
b) In the S.E.E. Examination, Bishwant got a total marks of 640 out of 800
full marks. Express his marks in percent.
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Percent
c) There are 240 girls and 260 boys in a school.
(i) Find the percentage of girls in the school.
(ii) Find the percentage of boys in the school.
Hint : Percentage of girls = No. of girls × 100%
Total no. of students
d) Mrs. Rawal earns Rs 374 from business and Rs 306 from job everyday.
Express her earnings from business and job in percents.
Creative Section - B Animals Number
Cow 24
9. In an educational excursion, the students of Buffalo 16
class 6 of a school visited to an animal-farm and Goat 40
recorded the list of animals alongside. Express
each animal as percentage of the total animals.
10. There are 25 students in a class. The number of students who were present
in the last week are given below:
Days Sunday Monday Tuesday Wednesday Thursday Friday
No. of present 20 23 18 25 21 19
students
(i) Express the number of present students of everyday in percent.
(ii) Find the percentage of absent students of each day.
11. Look at the monthly progress report of Dipti Shah and solve the given
problems.
a) Express her marks of every Subjects Full Marks Marks Obtained
subject in percent.
English 40 34
b) In which subject did Nepali 20 15
she show the better Maths 50 45
performance? Science 25 21
c) Find the percent of her total marks out of total full marks.
It’s your time - Project work
12. a) Let’s ask to your 5 friends about the number of children, teenagers, adults
and grand parents of their families. Find the total number of children,
teenagers, adults and grand parents that you record and express them in
percentage separately.
b) Let's conduct a survey in your school and collect the data of number of
girls and boys in each class of 4 to 6. Then express each of the numbers of
girls and boys as the percentage of total number of students of the class.
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Percent
7.6 To find the value of the given percent of a quantity
To find the value of the given percent of a quantity, we multiply the quantity by
the given percent. Then, we convert the percent into the fraction and simplify.
For example: 25
100
25% of Rs 480 = 25% × Rs 480 = × Rs 480 = Rs 120
7.7 To find a quantity whose value of certain percent is given
In this case, we consider the whole quantity as 100 percent. Then, by using
unitary method, we can find the whole quantity. For example:
If 25% of a sum is Rs 55, find the sum.
Here, 25% of a sum is Rs 55 Direct process
55
1% of the sum is Rs 55 The sum = Rs 25%
25
55 5255 100 = Rs 220
100% of the sum is Rs 25 × 100 = Rs = Rs 55 × 25
= Rs 220 100
Alternatively, we can consider the whole quantity by a variable such as x. Then
we form an equation and by solving the equation we can find the value of x. For
example:
In the above example, let the required sum be Rs x.
Now, 25% of Rs x = Rs 55
or, 25 ×x = Rs 55
100
or, x = Rs 55
4
or, x = 4 × Rs 55 = Rs 220
Hence, the required number is Rs 220.
Worked-out examples
Example 1: There are 480 students in a school. If 60% of them are boys, find
(i) the number of boys. (ii) the number of girls.
Solution:
(i) Here, the number of boys = 60% of 480
= 60 u 480 = 6 u 48 = 288
100
Hence, there are 288 boys in the school.
(ii) Again, the number of girls = 480 – 288 = 192
Hence, there are 192 girls in the school.
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Example 2: In a monthly test, Binita obtained 80% marks in mathematics out
of 25 full marks. How many marks did she obtain?
Solution:
Here, the marks obtained by Binita = 80% of 25
= 80 u 25 = 20
100
Hence, she obtained 20 marks.
Example 3: Bikram spends Rs 5400 every month which is 60% of his monthly
income. What is his monthly income?
Solution:
Let his monthly income be Rs x. Direct process
Now, 60% of x = Rs 5400 Monthly income = Rs 5400
60%
or, 60 ux = Rs 5400 Rs 5400
100 = 60
3x
or, 5 = Rs 5400 100
or, 3x = 5 u Rs 5400 = Rs 5400 × 100
60
5 × Rs 5400 = Rs 9,000
or, x = 3
= 5 u Rs 1800
= Rs 9000.
Hence, his monthly income is Rs 9000.
Example 4: A shopkeeper bought a laptop for Rs 36,000 and sold it at 15%
profit. Find his profit amount. At what price did he sell the laptop?
Solution:
Here, profit percent = 15%
Now, profit amount = 15% of Rs 36,000
= Rs 5,400
Again, his selling price = Rs 36,000 + Rs 5,400 = Rs 41,400
Hence, his profit amount is Rs 5,400 and he sold it at Rs 41,400.
Example 5: In a school 45% of the students are girls. If there are 110 boys, find
(i) the total number of students.
(ii) The number of girls.
Solution:
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Percent
(i) Let the total number of students be x.
Here, the percentage of boys = 100% – 45% = 55%
Now, 55% of x = 110 Alternatively, number of girls = 45% of 200
or, 55 u x = 110 = 45 × 200
100 100
or, 11x = 110 = Rs 90
20
or, 11x = 110 u 20
or, x = 110 × 20 = 200
11
Hence, there are 200 students in the school.
(ii) Again, the number of girls = 200 – 110 = 90
Hence, there are 90 girls in the school.
EXERCISE 7.2
General Section – Classwork
1. t1h0e%va=lu1e10s00. = 1 . So, 10% of 50 = 1 of 50 = 5. Similarly, tell and write
10 10
a) 10% of 40 = ............, 10% of 60 = ............ 10% of 200 = .........
b) 20% of 50 = ............, 20% of 80 = ........... 20% of 150 = ...........
c) 25% of 60 = ............, 25% of 120 = .......... 25% of 400 = ............
d) 50% of 30 = ............, 50% of 70 = ............ 50% of 500 = ............
2. 20% of a number is 5. Then, the number = 5 × 100 = 25. Similarly, tell
20
and write the required number as quickly as possible.
a) 10% of a number is 4, the required number = ...............
b) 20% of a number is 10, the required number = ...............
c) 25% of a number is 5, the required number = ...............
d) 50% of a number is 20, the required number = ...............
Creative Section - A
3 Find the value of the following.
a) 5% of Rs 300 b) 12% of Rs 450 c) 33% of Rs 700
d) 56% of 450 girls e) 75% of 720 boys f) 99% of 1,500 people
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4. Find the whole quantity whose value of certain percent is given.
a) 6% of a sum Rs 24 b) 10% of a sum is Rs 64
c) 25% of people are 220 men d) 48% of students are 336 girls
e) 55% of income is Rs 2,420 f) 75% of saving is Rs, 2,700.
5. a) Bimala got 85% marks in S.E.E. Examination. If the total full marks
was 800, how many marks did she obtain?
b) There are 40 students in class six. 30% of them are girls.
(i) How many are girls ? (ii) How many are boys ?
c) Out of 40 students of grade VI, 80% of them are attending their virtual
classes in zoom regularly. Find the number of students who are not
attending the virtual classes.
d) Mrs. Bhatta earns Rs 5,400 and spends 65% every week.
(i) How much does she spend every week ?
(ii) How much does she save every week?
(iii) How much does she save in 1 year (52 weeks) ?
e) Father earns Rs 21,600 in a month. He spends 15% of his earning on
the education of his children and 25% on food.
(i) Find his total expense on education and food.
(ii) How much does he save in a month ?
(iii) How much does he save in 1 year ?
(iv) If he spends 80% of his annual saving to buy a piece of land, What
is the cost of the land ?
6. a) A shopkeeper purchases a watch for Rs 900 and sells it at 20% profit.
Find his profit amount. [Hint: profit = 20% of Rs 900]
b) Mr Chaudhari buys a mountain-bike for Rs 15,300. He sells it at a
profit of 10%.
(i) How much profit does he make?
(ii) At what price does he sell the bicycle?
c) Mrs Gurung bought a mobile phone for Rs 7,200. She sold it at a loss
of 25%. Find her loss amount.
d) A man purchased a piece of land for Rs 4,50,000. He sold it at a loss
of 5%.
(i) Find his loss amount.
(ii) At what price did he sell the land?
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Percent
7. a) The listed price (or marked price) of a pendrive is Rs 1,250 and the
shopkeeper gives 10% discount.
(i) How much is the discount amount ?
(ii) How much do you need to pay to buy the pendrive after discount?
b) In the occasion of ‘Dashai festival’, 20% discount was given on
each of the following items. Find the discount amount. Find the selling
price of each item after discount.
(i) (ii) (iii)
Rs 1,500 Rs 1,150 Rs 5,000
8. a) If 20% of an income of a man is Rs 4,800, find his income.
b) Mr. Subba walked 24 km which is 75% distance between Phikkal and
Ilam. Find the distance between these two places.
c) There are 120 girls in a school which is 30% of the total number of
students. How many students are there in the school ?
d) Saloni Rai got 30 marks in mathematics. If it is 75% of the full marks,
how much is the full marks ?
9. a) 55% of the students in a school are girls. If there are 135 boys, find
(i) the total number of students (ii) the number of girls.
(Hint : Percent of the no. of boys = 100% – 55% = 45%, then 45% of x = 135)
b) Mrs. Gurung spends 60% of her monthly income and saves Rs 8,240
every month.
(i) Find her saving percent.
(ii) Find her monthly income.
(iii) How much does she spend in a month?
Creative Section - B
10. Pragya got 21 marks out of 25 full marks in a spelling test. Pooja got 80%
marks in the same test. Who got the more mark? By what percent was the
marks more than the lower marks?
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11. Are the numbers 10 and 10% same? By what percent is the number smaller
than the greater one?
12. In a mock test of mathematics, the maximum marks is 80. Bodhraj gets 85%
marks and cannot get A+ grade by at least 4 marks. Find:
a) least marks for A+ grade
b) least percent for A+ grade
13. Out of 50 students in a class, 60% are boys. If 50% of boys and 40% of girls
wear spectacles, find:
a) number of boys and girls.
b) number of boys and girls who wear spectacles.
c) number of students who wear spectacles.
d) percentage of students who wear spectacles in all.
14. Shreyasha spends 40% of her income and saves Rs 18,000 in a month.
Sahayata spends 60% of her income and saves Rs 12,800 in a month.
a) Who has more monthly income? Find by calculation.
b) Who spends more money and by how much?
It’s your time- Project work!
15. a) Let’s list out the name of your family members. Then calculate the
percentage of
(i) Males (ii) Females (iii) Children
b) Let’s collect name list of the students of your class. Then find the
percentage of
(i) Boys (ii) Girls (iii) Students wearing spectacles
c) Let’s visit nearby vegetable store or vegetable market or see the newspaper
which provides the prices of vegetables. Collect the information about
the change in price of any 5 types of vegetables of 1 week. Then find the
percentage in change in the price of each type of vegetables.
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Unit Profit and Loss
8
8.1 Cost price (C. P.) and selling price (S. P.) - Looking back
Classwork - Exercise
Let’s tell and write the answers as quickly as possible.
1. Buying price of a notebook is Rs 100 and its selling price is Rs 120.
a) How much is the cost price (C.P.) of the notebook ? ........................
b) How much is the selling price (S.P.) of the notebook ? ........................
c) In this case, is there profit or loss ? How much ? ........................
d) Well done ! Now, please write the formula to find profit ........................
2. You buy a pen for Rs 20 and sell to your friend for Rs 15.
a) What is your cost price (C.P.) ? .......................
b) What is your selling price (S.P.) ? .......................
c) Do you make profit or loss ? How much ? .......................
d) Could you please write the formula to find loss ........................
3. Discuss and Investigate, under what conditions one can make profit or loss.
............................................................................................................................
............................................................................................................................
8.2 Profit and loss
We pay money for the cost of an article while buying it.
So, it is called the cost price (C.P.) of the article. Similarly, the money for which
the article is sold is called its selling price (S.P.)
A profit is made when the selling price of an item is higher than its cost price.
In this case, profit is calculated as the difference between the selling price and
the cost price. Profit = S.P. – C.P.
On the other hand, when the selling price of an item is less than its cost price,
a loss is made. In this case, loss is calculated as the difference between the cost
price and the selling price.
Loss = C.P. – S.P.
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Profit and Loss
8.3 Discount
Have you ever seen the price of a book printed on it?
The price of a Lehenga taged on it?
The price of television in its catalog?
These are all about marked price. Thus, the listed price of an item is called its
marked price (M. P.).
Sometimes the listed price of any item is reduced and sold to customers. The
reduced amount is called discount.
For example,
Suppose, the marked price (M.P) of an article is Rs 340.
If the shopkeeper allows a discount of Rs 50, then
S.P. of the article = Rs 340 – Rs 50 = Rs 290.
Thus, S.P. = M.P. – Discount
And, Discount = M.P. – S.P.
EXERCISE 8.1
General Section – Classwork
Let's tell and write the answers as quickly as possible.
1. a) C.P. = Rs 20, S.P. = Rs 25, profit = ................
b) C.P. = Rs 68, S.P. = Rs 60, loss = ................
c) C.P = Rs 30, profit = Rs 5, S.P. = ................
d) C.P. = Rs 80, loss = Rs 10, S.P. = ................
e) S.P = Rs 50, profit = Rs 6, C.P. = ................
f) S.P. = Rs 72, loss = Rs 8, C.P. = ................
Let's tell and write the answers as quickly as possible.
2. a) M.P. = Rs 100, discount = Rs 10, S.P. = ..................
b) M.P. = Rs 300, discount = Rs 50, S.P. = ..................
c) M.P. = Rs 450, S.P. = Rs 400, discount = ..................
d) M.P. = Rs 700, S.P. = Rs 660, discount = ..................
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Profit and Loss
Creative Section - A
3. Find profit or loss in the following cases.
a) C.P. = Rs 50, S.P. = Rs 65 b) C.P. = Rs 135, S.P. = Rs 175
c) C.P. = Rs 575, S.P. = Rs 540 d) C.P. = Rs 780, S.P. = Rs 750
4. Find C.P. or S.P. in the following cases.
a) C.P. = Rs 120, profit = Rs 25 b) C.P. = Rs 275, loss = Rs 50
c) S.P. = Rs 200, profit = Rs 20 d) S.P. = Rs 600, loss = Rs 45
5. Find S.P. in the following cases.
a) M.P. = Rs 480, discount = Rs 40 b) M.P. = Rs 870, discount = Rs 75
6. Find the discount in the following cases.
a) M.P. = Rs 380, S.P. = Rs 345 b) M.P. = Rs 1,525, S.P. = Rs 1,500
7. a) Mr. Yadav buys a calculator for Rs 375 and sells it for Rs 450. Find his
profit.
b) Mrs. Shakya bought a saree for Rs 1,650 and sold it for Rs 1,485. Find her
loss.
8. a) A shopkeeper sold a camera for Rs 2,550 and made a loss of Rs 480. At
what price did he/she purchase the camera ?
b) Mr. Sharma bought a bicycle for Rs 5,995 and sold at a profit of Rs 1,001.
At what price did he sell the bicycle ?
c) A stationer sold 1 dozen of pens at the rate of Rs 25 each and gained
Rs 120 altogether. How much was the cost price of the pens ?
d) A fruit seller purchased 10 kg of apples at the rate of Rs 75 per kg and sold
at the total loss of Rs 50. How much was the selling price of the apples?
Creative Section - B
9. A merchant bought a quintal of potatoes for Rs 4,000. He sold all potatoes at
Rs 390 per 10 kg. Find his profit or loss.
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Profit and Loss
10. Mr. Poudel buys 60 bananas at Rs 80 per score and sells all the bananas at
Rs 60 per dozen. What is his profit or loss? (1 score = 20)
11. A shopkeeper buys a table fan for Rs 1,200 and expenses Rs 40 on its repairs.
If he sells the fan for Rs 1,550, find his profit.
12. Mr. Thapa bought a gross of copies for Rs 2,880 and made a profit of 288,
find the selling price of each copy? (1 gross = 12 dozens)
13. A retailer bought 6 watches for Rs 3,330 and sold them at a gain of Rs 111
from each watch. At what price did he sell each watch?
It’s your time- Project work!
14. a) Let’s visit the nearby shops or stores and collect the information of the
prices of any five types of daily useful goods of same quality. Compare
the prices of the goods and find which shop or store is best for shopping
and why? Prepare a short note and present in your classroom.
b) Let’s suppose that you are a stationer. List out the cost prices of any five
items and assume their selling prices and find the profit or loss amount
on each item.
15. a) Let’s become a problem maker and problem solver.
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 = ..........
Find profit Find loss Find S.P.
C.P. = .......... S.P. = .......... S.P. = ..........
Loss = .......... Profit = .......... Loss = ..........
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 = Rs 180 Loss = Rs 270
C. P. = ............................ C. P. = ............................
S. P. = ............................ S. P. = ............................
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Unit Algebra
9
9.1 Constant and variables - Looking back
Classwork - Exercise
Let’s tell and tick the correct answers as quickly as possible.
1. a) x represents the number of provinces of Nepal. x is (constant/variable)
b) x represents the heights of the students in your class. x is (constant/
variable)
2. a) 2ab is a (monomial/binomial) expression.
b) 2a + b, is a (monomial/binomial) expression.
3. a) In x2 , coefficient is .........., base is .........., power is .............
b) In 7y3, coefficient is ............., base is .........., power is ..............
4. a) If x = 2, y = 3, then, x + y = ................. and x y = .................
b) If l = 8, b = 5 then l × b = ................. and 2(l + b) = .................
What does the number 5 represent? Does it represent six, four, or any other
number of things?
The numbers such as 1, 2, 3, 4, 5, ... etc. always represent the fixed number of
things. These numbers are called constants. We may also use letters like x, y, z, a,
b, c,... in the place of numbers. If a letter represents a fixed value (number), it is
considered as constant. However, if the letter represents many values (numbers),
it is called variable. For example,
(i) x represents the sum of 7 and 3. Here x = 10. Here, x represents a fixed
value. So, x is a constant.
(ii) x represents the prime numbers less than 10. Here, the values of x may be 2,
3, 5, or 7. Thus, x represents many numbers. Therefore, it is a variable.
9.2 Algebraic terms and expressions
Let’s study the following illustrations and investigate the idea of algebraic
expressions.
‘5 times x’ is an expression o 5x is an algebraic expression.
‘y is added to 3’ is an expression o y + 3 is an algebraic expression.
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Thus, an algebraic expression is a mathematical statement formed by the four
fundamental operations (+, –, ×, ÷) on constants and variables.
Let’s consider an algebraic expression 2a + 3b – 6.
Here, 2a, 3b and 6 are called the terms of the expression. In this expression,
there are 3 terms.
Depending on the number of terms, an expression may be monomial, binomial,
trinomial or polynomial.
An algebraic term itself is a monomial expression. So,
Monomial expression a monomial expression contains only a single term.
a
For example, 5x, 6ab, b , 8p÷q etc. are monomial
expressions.
An algebraic expression containing two unlike terms is
Binomial expression called a binomial expression. a + b, 3p – 4q, ab + 7xy,
etc. are binomial expressions.
An algebraic expression containing three unlike terms
Trinomial expression is called a trinomial expression. 2x + 3y – 5, a + 4b – 9c,
xy – yz + zx, etc. are trinomial expressions.
An algebraic expression containing two or more than two unlike terms is also
called a multinomial expression.
Binomial and trinomial are also the multinomial. For example:
x + y, xy – 5, abc + a, etc. are multinomial of two terms.
a + b + c, xy + x + y, mn – 5 m + 7, etc are multinomial of three terms.
2a – 3b + c – 8, 2xy – 4yz + 7zx + 9, etc are multinomial of four terms.
9.3 Coefficient, base and exponent of algebraic terms
Let’s consider an algebraic term 3x2.
Her, 3 is called the coefficient, x is the base and 2 is the exponent. The coefficient
3 in 3x2 tells that the term x2 is added 3 times.
i.e. x2 + x2 + x2 + = 3x2
Similarly, the exponent 2 of x in 3x2 tells that the base x is multiplied 2 times.
i.e. 3 u x u x = 3x2 .
A power is a product of repeated multiplication of the same base. The exponent
of a power is the number of times the base is multiplied.
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5a3 x×x×x
Exponent
Base x3 exponent
Coefficient base
Let’s take another algebraic term y. power
Here, y = 1.y1. So, in y, its coefficient is 1 and exponent is also 1.
If the coefficient of a base is a number, it is called a numerical coefficient. In 3x,
3 is the numerical coefficient of x. If the coefficient of a base is a letter, it is called
a literal coefficient. In ax, a is the literal coefficient of x.
EXERCISE 9.1
General Section – Classwork
1. Let’s tell and write whether the letters represent ‘constant’ or ‘variable’.
a) x represents the natural numbers between 3 and 5. x is a ........................
b) x represents the natural numbers between 6 and 9. x is a ........................
c) y represents the even prime numbers. y is a ........................
d) y represents the composite numbers less than 10. y is a ........................
2. Let’s tell and write whether the following expressions are monomial,
binomial or trinomial.
a) 7xyz is ......................................... expression.
b) 2a + 5b – c is a ......................................... expression.
c) xy – ab is a ......................................... expression.
pq ......................................... expression.
d) r is a ......................................... expression.
e) 5x ÷ y is a
3. a) In 5a2, coefficient is .............. base is .............. power is ..............
b) In 4x3, coefficient is .............. base is .............. power is ..............
c) In y, coefficient is .............. base is .............. power is ..............
d) In 6ax, numerical coefficient is ............... literal coefficient is ..............
4. Let’s tell and write the algebraic expressions as quickly as possible.
a) Sum of a and b = .........................................
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b) Difference of a and b = .........................................
c) product of a and b = .........................................
d) The quotient of a divided by b = .........................................
e) Two times the sum of x and y = .........................................
f) Five tomes the product of m and n = .........................................
Creative Section
5. Let’s take the terms x, y and z. Make monomial, binomial and trinomial
expressions of your own using these terms.
6. Rewrite the following statements in algebraic expressions.
a) Product of x and y is added to z.
b) Three times the sum of x and y is increased by 5.
c) Two times the difference of p and q is decreased by 4.
d) Product of p and q is subtracted from r.
e) Five times the product of a and b is increased by x
f) The sum of x and y is divided by 2 and decreased by 7.
7. a) The present age of Anamol is x years.
(i) How old was he 2 years before?
(ii) How old will he be after 2 years ?
(iii) If his father is four times older than him, how old is his father ?
b) The breadth of a room is b metre. If its length is 5 metres longer than its
breadth, represent the length of the room by an expression.
c) The marks obtained by A in maths is x. The marks obtained by B is double
than that of A and marks obtained by C is double than that of B. Represent
the marks obtained by B and C by expressions.
8. Rewrite the following formulae in algebraic expressions.
a) The perimeter of a triangle is sum of its three sides a, b and c of the
triangle. What is the formula of perimeter of the triangle ?
b) The area of a triangle is half of the product of base (b) and height (h).
What is the formula of area of the triangle ?
c) The perimeter of a square is four times of its length (l). What is the
perimeter of the square?
d) The area of a square is the square of its length (l). What is the area of
the square?
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e) The perimeter of a rectangle is two times the sum of its length (l) and
breadth (b). What is the perimeter of the rectangle ?
f) The area of a rectangle is the product of its length (l) and breadth (b).
What is the area of the rectangle ?
It's your time - Project work!
9. Let's write any two different monomial expressions. Write the coefficient,
base and exponent of each expression.
Expressions Coefficient Base Exponent
9.4 Evaluation of algebraic expressions
We obtain the value of a term or an expression by replacing the variable/s
of the term or expression with numbers. It is called evaluation of a term or
expression. For example:
If x = 2 and y = 3, then 4x = 4 × 2 = 8, 2(x + y) = 2(2 + 3) = 2 × 5 = 10
(xy)2 = (2 × 3)2 = 62 = 36, (x – y)2 = (2 – 3)2 = (– 1)2 = 1 and so on.
Worked-out examples
Example 1: If y = 2x + 1 and x is a variable of the set B = {1, 2, 3}, find the
possible values of y.
Solution :
When x = 1, then y = 2x + 1 = 2 × 1 + 1 = 2 + 1 = 3
When x = 2, then y = 2x + 1 = 2 × 2 + 1 = 4 + 1 = 5
When x = 3, then y = 2x + 1 = 2 × 3 + 1 = 6 + 1 = 7
So, the required values of y are 3, 5 and 7.
Example 2: If a = 3b, express 4a + 5b in terms of b and evaluate the
expression when b = 2.
Solution :
Here, a = 3b
∴ 4a + 5b = 4 × 3b + 5b = 12b + 5b = 17b
Now, when b = 2, then 17b = 17 × 2 = 34
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Example 3: If x = 2y = 4z, express x + 3y + z in terms of z and evaluate the
expression when z = 3.
Solution :
Here, x = 2y = 4z
∴ x = 4z and 2y = 4z
4z
y = 2 = 2z
Now, x + 3y + z = 4z + 3 × 2z + z = 11z
Again, when z = 3, then 11z = 11 × 3 = 33
EXERCISE 9.2
General Section – Classwork
1. Let's input x = 1, 2, 3, ... Then tell and write outputs.
Input (x) Outputs
1
X+1 X+5 2X 3X
................. .................
................. .................
2 ................. ................. ................. .................
3 ................. ................. ................. .................
4 ................. ................. ................. .................
5 ................. ................. ................. .................
2. Let's tell and write the values as quickly as possible.
a) If x = 2 and y = 3, then 4x = ..........., 5y = ........... , 2xy = ...............
b) If x = – 1 and y = 2, then x + y = ................, x – y = ...............
c) If a = 3 and b = 2, then (ab)2 = .............., (a + b)2 = ...............
d) If l = 5 and b = 4, then l × b = ................, 2 (l + b) = ...............
3. Let’s tell and write the answers as quickly as possible.
a) If x = 2y, then x + y in terms of y = .......................
b) If x = 3y, then 2xy in terms of y = ......................
c) If a = 2b and b = 3, then a + b = ................
d) If p = 3q and q = 2, then p – q = ....................
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Creative Section - A
4. If x = 2 and y = 3 and z = 4, evaluate the following expressions.
a) x + y + z b) 2x + 3y – z c) 3(x – y + z) d) 5xy
e) x2 + y2 f) y2 + z2 – x2 g) x+ 2y h) (x–y+z)2
5. If = 3, z y
= 4 and = 5, find the values of following expressions.
a) 2 – b) 3 + 2 – c) ( + ) d) ( + ) ÷
6. a) The area of a square = l2. Find the area of squares in sq. cm.
(i) l = 4cm (ii) l = 7cm (iii) l = 3.5 cm (iv) l = 4.2 cm
b) The perimeter of a square = 4l. Find the erimeter of squares in cm.
(i) l = 6cm (ii) l = 9cm (iii) l = 8.5 cm (iv) l = 4.3 cm
7. a) Area of rectangle = l × b. Find the area of rectangles in sq. cm.
(i) l = 7cm, b = 4cm (ii) l = 8cm, b = 6cm (iii) l = 7.5 cm, b = 4cm
b) perimeter of rectangle = 2 (l + b). Find the perimeter of rectangles in cm.
(i) l = 6cm, b = 4cm (ii) l = 9cm, b = 5cm (iii) l = 6.4cm, b = 3.7 cm
8. The area of four walls of a room = 2h(l + b). Find the area of four walls in sq. m.
(i) l = 10 m, b = 8 m, h = 5 m (ii) l = 12 m, b = 7.5 m, h = 4 m
9. a) The area of a circle = Sr2, where S = 22 . Find the area of circles in sq. cm.
7
(i) r = 7 cm (ii) r = 14 cm (iii) r = 21 cm
b) The perimeter of a circle = 2Sr, where S = 22 . Find the perimeter of circle
in cm. 7
(i) r = 7 cm (ii) r = 14 cm (iii) r = 21 cm
10. a) The volume of a cube = l3. Find the volume of cubes in cubic cm.
(i) l = 2 cm (ii) l = 4 cm (iii) l = 5 cm
b) The area of a cube = 6l2. Find the area of cubes in sq. cm.
(i) l = 3 cm (ii) l = 5 cm (iii) l = 6 cm
11. a) The volume of a cuboid = l u b u h. Find the volume of cuboids in cubic cm.
(i) l = 5 cm, b = 4 cm, h = 3 cm
(ii) l = 10 cm, b = 7 cm, h = 5.5 cm
b) The area of a cuboid = 2(lb + bh + lh). Find the area of cuboids in sq. cm.
(i) l = 8 cm, b = 5 cm, h = 2 cm
(ii) l = 10 cm, b = 6 cm, h = 4 cm
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Creative Section - B
12. If x = 5 cm, find the length of the following line segments.
13. Write the algebraic expressions to represent the perimeters of the following
figures. If x = 2, y = 3 and z = 5, find the perimeters of the figures.
14. a) x is a variable on the set A = {1, 2, 3}, that is, x can be replaced by 1, 2
and 3. Evaluate the expressions (i) x + 5 (ii) 2x – 1
b) If y = 2x + 1 and x is a variable on the set B = {2, 4, 6}, find the possible
values of y.
15. a) If x = 2y, express 2x + 5y in terms of y and evaluate the expression when
y = 3.
b) If a = b + 3, express a + b in terms of b and evaluate the expression when
b = 3.
c) If x = 2a + 1, show that 3x – 6a + 7 = 10
It's your time - Project work!
16. a) Let's write a monomial expression. Replace the variable of the expression
by the natural numbers less than 4 and evaluate the expression.
b) Let's write a binomial expression of the form ax + b, where a is a numeral
coefficient and b is an integer. Replace the variable by the whole numbers
less than 4 and evaluate your expression.
9.5 Like and unlike terms
1 apple and 2 apples are like (same) type of things. Similarly, x and 2x are like terms.
3 pens and 4 pens are like (same) type of things. Similarly, 3y and 4y are like terms.
The algebraic terms having the same base and equal power are called like terms.
For examples:
3x, 5x, 8x, etc. are like terms because they have same base x and equal power 1.
5a2, 9a2, 12a2, etc. are like terms because they have the same base a and equal
power 2.
2 apples and 2 pens are unlike (different) types of things. Similarly, 2x and 2y are
unlike terms. 2x and 2y have unlike bases.
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Again, 2x, 3x2, 5x2 etc. have the same base x but each base does not have equal
power. So, they are also unlike terms. Similarly, 5x2, 3y2, 4z2, etc. are unlike
terms because these terms do not have the same base, even though each base
has the equal power.
Thus, the algebraic terms which have different bases or different powers are
called unlike terms.
9.6 Addition and subtraction of algebraic terms
Let's study the following illustrations and learn to add coefficients while adding
like terms.
2 apples + 3 apples = 5 apples 2x + 3x = 5x
4 oranges + 5 oranges = 9 oranges 4y + 5y = 9y
6 pens – 2 pens = 4 pens 6a – 2a = 4a
8 books – 3 books = 5 books 8p – 3p = 5p
Thus, to add or subtract algebraic terms, we should add or subtract the coeffi-
cients of like terms. For example:
2x + 3x = 5x, 4x2 + 6x2 = 10x2, 3ab + 2ab + ab = 6ab
8y – 5y = 3y, 9y3 – 4y3 = 5y3, 7pqr – 6pqr = pqr
But, we cannot add or subtract the coefficients of unlike terms.
For example:
2x + 3y + x + 4y = 3x + 7y, 4a + 5a + 8 = 9a + 8 and so on.
EXERCISE 9.3
General Section A – Classwork
1. Let’s tell and write whether these terms are like or unlike.
a) 4x, 7x, – 3x ................. terms b) 3x, 2x2, xy ................. terms
c) 3a2, 4a3, 6a ................. terms d) p2, 3p2, 8p2 ................. terms
e) 2xy, 5xy, 3xy ................. terms f) 2x2y, 5xy2, 3 2y2 ................. terms
2. Let’s tell and write the answers as quickly as possible.
a) a + 2a = ........... b) x2 + x2 = ............ c) 3xy + 2xy = ...........
d) 5x – 3x = ........... e) 4a2 – a2 = ............ f) 9pq2 – 4pq2 = ..........
3. a) What should be added to 2x to get 5x? ..................
b) What should be subtracted from 8x to get 3x ? ..................
Creative Section
4. Add or subtract.
a) 4x + 7x b) 5a2 + 9a2 c) 5xy + 3xy d) 3abc + 6abc
e) 10p3 – 3p3 f) 12mn – 7mn g) 9x2y – 4x2y h) 8xy2 – 7xy2
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5. Simplify b) 2p2 + 3p2 + p2 + 4p2
a) a + 2a + 3a
c) 2x + 3y + x + y d) 6m + 2n + n – 4m
e) 3xy + yz – xy – 2yz – 5 f) 5p2 – 2q2 + 2p2 – 3q2
6. a) What should be added to 9x to get 15x ?
b) What should be added to p to get p + q ?
c) What should be subtracted from 12a to get 5a ?
d) What should be subtracted from 3x to get 3x – 2y ?
Game Time!
7. Complete this obstacle course.
START x Add 4x Subtract 6x Add 5x
Subtract 3x Same answer in Subtract x
Add 7x both ways? If
not, start again.
Add 6x
Subtract 2x Add 9x Subtract 3x FINISH
It's your time - Project work!
8. Let's write any two like terms with the variable x. Find the sum and difference
of your like terms.
Like terms Sum Difference
..............., ............... ............................ ............................
..............., ............... ............................ ............................
9.7 Addition and subtraction of algebraic expressions
While adding or subtracting two or more binomials, trinomials or multinomial,
the like terms should be arranged in the same column, then, their coefficients
should be added or subtracted. Alternatively, we can arrange the expressions
horizontally and the addition or subtraction of like terms can be performed.
Worked-out examples
Example 1: Add 7x2 + 4xy + 5y2 and 2x2 – 3xy – 8y2.
Solution:
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Addition by horizontal arrangement Addition by vertical arrangement
7x2 + 4xy + 5y2 + 2x2 – 3xy – 8y2 7x2 + 4xy + 5y2
= 7x2 + 2x2 + 4xy – 3xy + 5y2 – 8y2 2x2 – 3xy – 8y2
= 9x2 + xy – 3y2 9x2 + xy – 3y2
Example 2: Subtract 2ab – 4bc + 7 from 5ab – 4bc – 2.
Solution:
Subtraction by horizontal arrangement Subtraction by vertical arrangement
5ab – 4bc – 2 – (2ab – 4bc + 7) 5ab – 4bc – 2
= 5ab – 4bc – 2 – 2ab + 4bc – 7 ±2ab 4bc ± 7
= 5ab – 2ab – 2 – 7 = 3ab – 9 3ab – 9
Example 3: What should be added to 3x – 4 to get 8x + 5?
Solution: Let’s think, what should be
added to 7 to get 10? It’s 3 and
The required express to be added is it is 10 – 7.
8x + 5 – (3x – 4) It’s my good idea to work out
such problems!
= 8x + 5 – 3x + 4
= 8x – 3x + 5 + 4 = 5x + 9
Example 4: What should be subtracted from 8x – 3y + 2 to get 5x + 2y – 1?
Solution:
The required expression to be subtracted is
8x – 3y + 2 – (5x + 2y – 1) Let’s think, what should be
= 8x – 3y + 2 – 5x – 2y + 1 subtracted from 9 to get 7? It’s 2
= 8x – 5x – 3y – 2y + 2 + 1 and it is 9 – 7.
= 3x – 5y + 3
I can also work out such questions.
Example 5 : If x = m + 2 and y = n – m, show that x + y = n + 2.
Solution :
Here, x = m + 2 I got it !
y=n–m In x + y, x is replaced by m + 2
Now, x + y = m + 2 + n – m and y is replaced by n – m.
= n + 2 proved.
EXERCISE 9.4
General Section – Classwork
Let’s tell and write the correct answers as quickly as possible.
1. a) Sum of x + 1 and x + 2 = ............................................
b) Sum of 3x + 5 and x – 4 = ............................................
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c) Sum of a2 + b2 and a2 – b2 = ............................................
d) Difference of 2x + 3 and x + 1 = ............................................
e) Difference of 4y + 7 and 3y + 2 = ............................................
f) Difference of 8a + 5 and 4a – 3 = ............................................
2. Complete the following table.
+ a+2 a–1 a+b
a+5 2a – b + 2 2a – 4 2a + b + 5
a–3
a–b
Creative Section - A
3. Add.
a) x + 3 and x + 2 b) 2x + 5 and 3x + 1 c) 3a + 8 and 4a – 5
d) 4 + 3a and 5 – 8a e) 9m – 2 and m – 7 f) 11 – 2n and 4 – 5n
g) 3a + 4b and 7a + 2b h) 8a – 5b and a – 3b i) p2 + q2 and 2p2 + 3q2
j) 3a2 – 5b2 and 2a2 + 7b2 k) x2 – xy + y2 and 2x2 + 6xy – y2
l) x2 + x – 2 and 2x2 – 5x + 7 m) 2ab – 3bc + 4 and ab + 4bc – 5
n) 8abc – 5ab + 3a and 2a + 7ab – 4abc
4. Subtract.
a) x + 1 from 5x + 4 b) 2x + 3 from 5x + 1
c) 6a – 7 from 9a + 2 d) 2a + 5 from 6a – 1
e) 3p – 2 from 5p – 7 f) 8 – 3p from 1 – p
g) 5m + 3n from 7m – 2n h) 9mn – 5p from 6mn – p
i) 4x2 – 3y2 from 9x2 + 4y2 j) 3a3 + 5b3 from 8a3 – b3
k) 2x3 – 5x2 + x – 6 from 4x3 + 2x2 – 3x + 7
l) 4ax – 6by + 7 from 6ax + 2by – 3
m) abc + 2ab – 3bc – ca from 3ab – ab + bc + 2ca
n) xyz – 4xy + 5 from 5xyz + xy + 3yz – 4
5. a) What should be added to 2x – 3 to get 5x + 7?
b) What should be added to 3a + 4b to get 7a – 5b?
c) What should be subtracted from 5m – 3n to get 2m + 5n?
Vedanta Excel in Mathematics - Book 6 146 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Algebra
d) What should be subtracted from 5x – 3y to get 2x – 5y?
e) To what expression 2a – 3b + 1 must be added to get 4a + 7b – 3?
f) From what expression x2 + 5y2 – 3xy must be subtracted to get 2x2 – y2 + 4xy?
Creative Section - B
6. a) If x = a + 7 and y = b – a, show that x + y = b + 7.
b) If x = 2m – n and y = m + n, show that x – y = m – 2n.
c) If x2 = 2a2 – b2, and y2 = 2b2 – c2 and z2 = 2c2 – a2,
show that: x2 + y2 + z2 = a2 + b2 + c2.
7. Simplify:
a) 4x + 3y – (3x + y) b) 5a – 3b – (a + 6b)
c) 8p + q – (5p – 3q) d) 6a – 5x – (2x – 3a)
e) – 7m – 2n – 5 – (–8m – n +1) f) –a – 2t – (t – 4a) – (–3a – 5t)
g) –(b – z) – (–2b + z) – (–3b – z) h) –(a – b) – (–4b + 3c) – (c – 2a)
8. a) The sides of a triangle are x + 4, 2x – 3 and 3x + 1, find its perimeter.
b) The sides of a triangle are 2x + 3y, x + 2y and 7x – 2y, find its perimeter.
It's your time - Project work!
9. a) Let's write two binomial expressions of the form ax + b, where a is a
natural number and b is an integer. Find the sum and difference of your
expressions.
b) Let's write two trinomial expressions of the form ax + by + c, where a, b,
and c are integers. Find the sum and difference of your expressions.
9.8 Multiplication of algebraic terms
While multiplying the algebraic terms, the coefficients of the terms are multiplied
and the exponents of the same bases are added. For example:
Example 1: Multiply 3x by 2x. 3 × 2 = 6 (Coefficient are
multiplied.)
Solution: x × x = x1+1 = x2 (Exponents of
the same bases are added.)
Here, 3x u 2x = 6x2
x2 x2 x2 x
The multiplication of 3x × 2x can also be 2x
shown diagrammatically.
From diagram, x2 x2 x2 x
3x × 2x = x2 + x2 + x2 + x2 + x2 + x2 = 6x2 x xx
3x
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 147 Vedanta Excel in Mathematics - Book 6
Algebra
Example 2: Multiply 4a2b by – 5ab2 I got it!
Solution:
Here, 4a2b u (– 5ab2) = – 20a3b3 4 × (–5) = –20
a2 × a = a2 + 1 = a3
b × b2 = b1 + 2 = b3
9.9 Multiplication of polynomials by monomials
Of course, monomial, binomial and trinomial with positive exponents of variables
are also called polynomials. To multiply a polynomial by a monomial, each term
of the polynomial is separately multiplied by the monomial. For example:
Example 3: Multiply: (a + b) × c. ac + bc c
Solution:
ac bc
(a + b) × c = a × c + b × c ab
= ac + bc a+b
Example 4: Multiply 4a2 + 3b2 by 2ab
Solution: It’s easier!
Here, 2ab × (4a2 + 3b2) = 2ab × 4a2 + 2ab × 3b2 Each term of 4a2 + 3b2 is
separately multiplied by 2ab
= 8a3b + 6ab3.
Example 5: Multiply 7x2y2 – 4xy + 3 by – 2xy
Solution:
Here, –2xy(7x2y2 – 4xy + 3) = –2xy × 7x2y2 – 2xy(–4xy) – 2xy × 3
= –14x3y3 + 8x2y2 – 6xy
Example 6: If x = p + 3 and y = 2p, show that xy = 2p2 + 6p.
Solution: It’s easier!
Here, x = p + 3 and y = 2p In xy, x is replaced by p + 3 and
∴ xy = (p + 3) × 2p y is replaced by 2p.
= 2p × p + 2p × 3 = 2p2 + 6p proved.
Example 7: If x = 2a, y = 3a and a = 4, find the value of 6xy .
Solution:
Here, x = 2a and y = 3a
∴ 6xy = 6 × 2a × 3a = 36a2 = 6a
When a = 4, then 6a = 6 × 4 = 24 148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Vedanta Excel in Mathematics - Book 6
Algebra
EXERCISE 9.5
General Section – Classwork
1. Area of rectangle = length (l) × breadth (b). Let’s find the area of the
following rectangles.
a) b)
3y Area (A) = ............ 5q Area (A) = ............
4x
c) d) 2p (2x+1) 2a Area (A) = ............
2x Area (A) = ............ 5a Area (A) = ............
3x
3x
e) f)
x Area (A) = ............
(2x + 1)
2. Volume of cuboid (V) = length (l) × breadth (b) × height (h). Let’s find
the volume of following cuboids.
a) b) 3z
c
b 2y
a 4x
V = ..................... V = .....................
c) d)
x 2p
2x 3p
2x 5p
V = ..................... V = .....................
3. Let's tell and write the products as quickly as possible.
a) x × x = ........... b) 2x × x = .......... c) 3a × 2a = ...........
d) p2 × p = .......... e) 3b2 × 4b2 = ........... f) 2x3 × 5x2 = ...........
g) x (x + 1) = ................... h) 2x(x – 3) = ...................
i) 3x(x2 + 2) = ................... j) 2a(a2 – 5) = ...................
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149 Vedanta Excel in Mathematics - Book 6
Algebra
4. Look at the targets and investigate the operation. Tell and write the
answers as quickly as possible. a
ay
x xa ay y
2x 4a aa
x2 a4
a
x 2p y 2x
2y 3y
2x
xx p 3x
xx 3x x
Creative Section - A
5. Multiply.
a) x × x b) 2x × 3x c) 3x × 4x
d) x × 2x2 e) 2y × 3y2 f) 3xy × 2xy
g) 4a2b × 3ab2 h) –5xyz × 4xyz i) 3xy × (–5x2y2)
6. Simplify.
a) x u x u x2 b) 2a2 u 3a u 5a c) 3x u 2y u 2x u y
d) p2 u 5q u 2p u 3q2 e) 3b3 u c2 u 2b2 u 4c3 f) xyz u 2xy u yz u 5zx
g) (– 3ab) u (– 2bc) u (– abc) h) (–5qr) × (– 2pqr) u pq
i) 6x2y u (– xy2) u xyz u (– 3yz2)
7. Simplify.
a) 2x × 2x + 2x × 5 b) 3a × 2a – 3a × 4 c) 5x2 × x2 + 5x2 × 8
d) ab × 2a – ab × 3b e) ab × 5a2 – ab × 3b2 f) 10x2y2 × 3x – 10x2y2 × 2y
g) –2x × 4x – (–2x) × 3 h) –pq × 2p – (–pq) × 3q i) –3abc × 2a2 + (–3abc) × 5bc
8. Multiply.
a) 3x (2x + 5) b) 4a (5a – 6) c) 6p (8p2 – 1)
d) 2y2 (y2 + 7) e) 6m2 (2 – 3m2) f) 7b3 (8 + 3b2)
g) xy (x – y) h) 2xy (2x + 3y) i) 2ab (3a2 – 4b2)
j) – 3pq (2p + 5q) k) – 2bc (7b – 3c) l) – 5xy (5x2 – 4y2)
m) – 4x (x2 – 2x + 1) n) – 5a (3a2 + 3a – 5) o) – 3xy (4x2 – 5xy – 3y2)
Creative Section - B
9. a) If x = a + 5 and y = 2a, show that xy = 2a2 + 10a.
b) If p = 5x and q = 3x – 1, show that 2pq = 30x2 – 10x.
c) If a = x – 3, b = 3x and x = 2, find the value of 4ab.
Vedanta Excel in Mathematics - Book 6 150 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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