Excel in Mathematics - BOOK 8 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 51 Vedanta Excel in Mathematics - Book 8 19. a) Let's take a point A(1, 1) in the coordinate axes of the graph paper and join with the origin O(0, 0) as shown in the given graph. Then, find the length of OA by using the distance formula. Now, draw a circle with radius OA. (i) Is the number obtained as the length of OA an irrational number? (ii) Can the irrational number be shown in the number line? b) Let's take a few more points such as P(1, 2), Q(2, 2), R(3, 1), etc. in the coordinate axes of separate graph papers and repeat the same activities mentioned above. Then, show the irrational numbers 5 , 8 and 10 in number lines. Demonstrate and discuss about your activities in the class. 3.12 Scientific notation of numbers It is somehow tedious to read, write and compute with very large or a very small numbers. In such cases, we express a very large or a very small number as the product of digit (or digits) and some power of 10. It is known as scientific notation of numbers. For example, 4500 = 4.5 × 103 , 299000 = 2.99 × 105 , 693000000 km = 6.93 × 108 km, 0.0072 = 7.2 × 10–3, 0.00000000954 kg = 9.54 × 10–9 kg and so on Now, let's study the following rules and learn to express numbers in scientific notation. Rule 1: The whole number part in scientific notation of a number should be of one digit number. For example, 59400000000 = 5.94 × 1010 (But 59.4 × 109 is incorrect scientific notation) 0.0000000826 = 8.26 × 10–8 (But 82.6 × 10–9 is incorrect scientific notation) Rule 2: In the case of a large number, after separating whole number part and decimal part by using decimal point, count the number of digits after the decimal point and express as the product of the same number of power of 10. For example: Insert decimal point here. 3275000000 = 3.275 × 109 1 2 3 4 5 6 7 8 9 9 digits There are 9 digits after the decimal point. So the power of 10 will be 109 . Rule 3: In the case of a very small number, after separating whole number part and decimal part by using decimal point, count the number of digits before the decimal point and express as the product of the same number of negative power of 10. For example: A (1, 1) O(0, 0) 148000000 km Earth Sun Real numbers

Vedanta Excel in Mathematics - Book 8 52 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 0.0000000653 = 6.53 × 10–8 1 2 3 4 5 6 7 8 8 digits Worked-out Examples Example 1: Write these numbers in scientific notation. a) 1700000 b) 2745.8 c) 0.000054 Solution: a) 1700000 = 1.7 × 106 b) 2745.8 = 2.7458 × 103 = 2.75 × 103 c) 0.000054 = 5.4 × 10–5 Example 2: Write the decimal values of the following scientific notation of numbers. a) 2.52 × 106 b) 5.723 × 102 c) 9.81 × 10–7 Solution: a) 2.52 × 106 = 2.52 × 1000000 = 2520000 b) 5.723 × 102 = 5.723 × 100 = 572.3 c) 9.81 × 10–7 = 9.81 10000000 = 0.000000981 Example 3: The average weight of an adult elephant found in Nepal is 34 quintal. a) Write the average weight of elephant in kilograms (kg) b) Express the weight in kilograms in scientific notation of number. c) Convert the weight in gram and express it in scientific notation. Solution: a) The average weight of elephant = 34 × 100 kg = 3400 kg b) 3400 kg = 3.4 × 103 kg c) 3400 kg = 1000 × 3400 g = 3400000 g = 3.4 × 106 g Decimal point is shifted after eight number of digits to make a one digit whole number. So, the power of 10 will be 10–8. The whole number part in scientific notation should have one digit. So, after the decimal, there are six digits, so 106 is written. 2745.8 = 274.58 × 101 274.58 = 27.458 × 102 27.458 = 2.7458 × 103 = 2.75 × 103 0.000054 = 0.00054 × 10–1 0.00054 = 0.0054 × 10–2 0.0054 = 0.054 × 10–3 0.054 = 0.54 × 10–4 0.54 = 5.4 × 10–5 Real numbers

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 53 Vedanta Excel in Mathematics - Book 8 Example 4: Simplify a) 3.6 × 104 + 4.5 × 105 b) 2.8 × 10–2 – 1.2 × 10–3 Solution: a) 3.6 × 104 + 4.5 × 105 = 0.36 × 105 + 4.5 × 105 = 4.86 × 105 b) 2.8 × 10–2 – 1.2 × 10–3 = 2.8 × 10–2 – 0.12 × 10–2 = 2.68 × 10–2 Example 5: Simplify a) (2.4 × 103 ) × (6.2 × 104 ) b) 4.5 × 106 + 2.55 × 107 1.4 × 103 + 1.1 × 103 Solution: a) (2.4 × 103 ) × (6.2 × 104 ) = (2.4 × 6.2) × 103 + 4 = 14.88 × 107 = 1.488 × 108 b) 4.5 × 106 + 2.55 × 107 1.4 × 103 + 1.1 × 103 = 0.45 × 107 + 2.65 × 107 2.5 × 103 = 3 × 107 2.5 × 103 = 1.2 × 107 – 3 = 1.2 × 104 EXERCISE 3.3 Section A – Classwork 1. Let's say the scientific notation of these numbers and write as quickly as possible. a) 16000 = .................... b) 3250000 = .................... c) 517000000 = .................... d) 0.294 = .................... e) 0.0058 = .................... f) 0.0000063 = .................... g) 84.6 = .................... h) 145.3 = .................... i) 9137.6 = .................... 2. Let's say and write the values of these numbers in decimal system. a) 3.2 × 10 = ................. b) 5.19 × 102 = ................. c) 7.4629 × 104 = ................. Creative Section - A 3. Let's express the following numbers in scientific notations. a) 324000000 b) 612000000000 c) 74835000000 d) 0.000003 e) 0.000000054 f) 0.000000000936 4. a) 1 square kilometres = 10000000000 square centimetres. Express it in scientific notation. b) If the volume of 1 litre of water is 1000 c.c., find the volume of 2000 litres of water and rewrite it in scientific notation. c) 1 square centimetres = 1 10000 = 0.0001 square metres. Express it in scientific notation. d) 1 square centimetres = 1 10000000000 square kilometres. Express it in scientific notation. e) 1 cubic centimetres = 1 1000000 cubic metres. Rewrite it in scientific notation. f) How many seconds are there in a month of 30 days? Express it in scientific notation. Real numbers

Vedanta Excel in Mathematics - Book 8 54 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5. The average weight of an adult one-horned male rhinoceros found in Nepal is about 21 quintal. a) Write the average weight of rhinoceros in kilograms (kg) b) Express the weight in kilograms in scientific notation of number. c) Convert the weight in gram and express it in scientific notation. 6. The blue whale is the largest animal on the earth. The weight of an adult blue whale is about 199 metric tons (1 metric ton = 1000 kg) a) Write the weight of blue whale in grams. b) Express the weight in grams in scientific notation of number. 7. Lactobacillus bacteria are useful for converting milk into curd. The average length of this type of bacteria is about 5 × 10–6 metre. a) Express the length of bacteria in decimal of metres (m). b) Rewrite the length in centrimeters (cm) and millimetres (mm). 8. The radius of the earth is 6.371 × 103 km. a) Express this scientific notation in decimal system of numeration. b) Convert this scientific notation in metre and express in decimal system of numeration. 9. The distance between the sun and the earth is bout 148000000 km. a) Express this distance in scientific notation. b) Rewrite the scientific notation of the distance in metre (m). 10. The distance travelled by light in 1 year is called a light year. The value of 1 light year is about 9460000000000 km. a) Express this distance in scientific notation. b) Convert the scientific notation of the distance in metre (m). 11. Let's express the value of the following scientific notation of numbers in general number system. a) 2.7 × 103 b) 4.5 × 104 c) 7.56 × 105 d) 8.45 × 109 e) 2.5 × 10–2 f) 5.6 × 10–3 g) 4.95 × 10–5 h) 7.83 × 10–8 12. Let's simplify. a) 2.6 × 103 + 1.8 × 103 b) 3.5 × 104 + 5.7 × 104 c) 4.3 × 105 + 6.7 × 105 d) 6.4 × 109 + 9.7 × 109 e) 3.4 × 102 + 2.6 × 103 f) 2 × 103 + 4 × 102 g) 6 × 105 + 3 × 104 h) 1.5 × 107 – 9.5 × 105 Real numbers

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 55 Vedanta Excel in Mathematics - Book 8 i) 1.3 × 105 – 4.9 × 104 j) 3.3 × 10– 4 – 5.2 × 10– 5 k) 4.8 × 10–5 – 3.5 × 10–6 l) 5.4 × 10–3 – 8.6 × 10–5 13. Let's simplify. a) (3.4 × 102 ) × (1.6 × 103 ) b) (2.7 × 103 ) × (3.9 × 104 ) c) (4.5 × 10–3) × (6.7 × 105 ) d) (3.6 × 10–8) × (5.4 × 10–5) Creative Section - B 14. Let's simplify. a) 1.5 × 104 × 4.5 × 103 2.5 × 102 b) 6.4 × 10 – 5 × 3.6 × 109 1.6 × 1010 × 1.8 × 10–3 c) 6.7 × 10–11 × 2.5 × 1020 × 3.6 × 1015 (3.35 × 106 ) 2 d) 7.8 × 107 + 6.5 × 107 1.3 × 103 e) 3.2 × 103 + 4.8 × 104 1.6 × 102 f) 9.6 × 106 – 7.2 × 105 2.4 × 103 15. a) A petrol tank has 3.6 × 103 litres of petrol. If 1.5 × 104 litres of petrol is added into it how much petrol does it have now? b) An underground water tank contains 1.8 × 104 litres of water. When 1.8 × 103 litres of water is used up how much water is left in the tank? c) Sound travels 3.43 × 102 metres in 1 second. Calculate the distance travelled by sound in 3.6 × 103 seconds. d) How many petrol tanks of capacity 1.2 × 104 litres each are required to empty 4.8 × 105 litres of petrol? e) The speed of light in Vacuum is 2.99 × 105 km in 1 second. Calculate the distance travelled by light in 3.6 × 103 seconds. It's your time - Project work and Activity section 16. a) Let's visit to the available website and search the distance of 7 planets from the earth. Express the distances in scientific notation. b) What is your body weight? Write it in grams and milligrams, then express in scientific notation. 17. a) Write any two 7-digit and 9-digit numerals and express them in scientific notation. b) Write any two decimal numbers with 6 and 7 decimal places. Express them in scientific notation. 18. How is the scientific notation of numbers useful in numeration or in measurements? Write a short report and present in your class. Real numbers

Vedanta Excel in Mathematics - Book 8 56 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Assessment - I 1. Given that a set A = {1, 2, 3, 4, 5}. List the elements of the following sets. B = {y: y = x + 5, x ∈ A} C = {z: z = 2x, x ∈ A} Answer the following questions. (a) State whether the sets A and B are disjoint or overlapping. Give reason. (b) Are the sets B and C equal or equivalent? Give reason. (c) Show the sets A and C in a Venn-diagram. (d) Write the universal sets representing the sets A, B and C. 2. Let A = {c, o, w} is a set of letters of the word ‘cow’. (a) Write the formula to find the number of subsets of the set A. (b) Find how many subsets of the set A are possible? (c) Write all possible subsets of the set A. (d) Which one is the improper subset? Give reason 3. P and Q are the subsets of a universal set U. If U = {x: x is a whole number less than 10}, P = {y: y is an odd number, y < 10} and Q = {z: z is a prime number, 2< z < 10}. (a) List element of these sets. (b) Show the relation of the sets P and Q in a Venn-diagram. (c) Is the set Q a proper subset of the set P? Give reason. 4. Spandan is a student of class VIII. She writes a binary number in the expanded form as: 1 × 24 +0 × 23 + 0 × 22 + 1 ×21 + 1 × 20 . (a) Write the short form of this binary number. (b) Convert this binary number into decimal number. (c) Convert this decimal number into quinary number. 5. The binary and quinary numbers are given by P = (10A010)2 and Q = (1B3)5 . (a) Find the value of A if P = 42(10). (b) Find the value of B if P + Q = 80 (10). (c) Convert the value of P – Q into binary number system. 6. Simplify: 9.6 × 106 – 7.2 × 105 2.4 × 103 7. A drinking water supply company has two tankers A and B. The tanker-A can carry 6.5 × 103 liters of water and tanker-B can carry 7.2 × 103 liters of water. (a) A person has a large empty water tank in his/her house. He/she calls both the tankers full of drinking water to fill the empty tank completely. How many liters of water does the tank hold? Write in scientific notation. (b) If the company supplies 10 full tankers using tanker-A and 15 full tankers using tanker-B every day, how many litres of drinking water does it supply in a day?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 57 Vedanta Excel in Mathematics - Book 8 4.1 Ratio – Looking back Classwork - Exercise 1. There are 18 girls and 12 boys in class VIII. a) Write the ratio of the number of girls to the number of boys. ................ b) Write the ratio of the number of boys to the number of girls. ................ 2. A desk is 2 m long and 30 cm broad. a) Write the ratio of the length to the breadth. ................ b) Write the ratio of the breadth to the length. ................ 3. Puja's height is 4 feet and that of Amar's is 5 feet. a) Write the ratio of the height of Amar to the height of Puja. ................ b) The antecedent of ratio is ................, the consequent is ................ 4.2 Ratio - A comparision of quantities of the same kind Let Ram has Rs 50 and Sita has Rs 100. Here, the ratio of Ram’s to Sita’s money = 50 100 = 1 : 2 It shows that Ram’s money is half of Sita’s money. Similarly, the ratio of Sita’s to Ram’s money = 100 50 = 2 : 1 It shows that Sita’s money is two times of Ram’s money. Thus, a ratio compares two or more quantities of the same kind. It shows how many times a quantity is less or more than another quantity of the same kind. Let’s take a ratio a:b. We read it as ‘a is to b’. Here, the first term a is called the antecedent and the second term b is called the consequent. For example, In 2 : 3 (read as ‘2 is to 3’), 2 is antecedent and 3 is consequent. In 3 : 2 (read as ‘3 is to 2’), 3 is antecedent and 2 is consequent. Remember that while making a ratio, the quantities must be of the same kind and they should have the same unit. The ratio itself does not have any unit. Unit 4 Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 58 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 4.3 Compounded ratio Let’s take any two ratios a : b and c : d. The compounded ratio of a : b and c : d = a : b × c : d = a b × c d = ac : bd Thus, a ratio which is obtained by multiplying two or more ratios is called a compounded ratio. For example, If 1 : 2 and 3 : 4 are any two ratios, the compounded ratio 3 8 1 2 3 4 = × = = 3 : 8 Worked-out Examples Example 1: A table in a class is 75 cm wide and 1 m long. a) Write the ratio of the width to the length of the table. b) Reduce the ratio in its lowest terms. Solution: a) Here, 1 m = 100 cm Now, ratio of the width to the length = 75 : 100 b) Again, 75 : 100 = 75 100 = 3 4 = 3 : 4 Example 2: There are 20 boys and 10 girls in a class. a) Write the ratio of the number of boys to the number of girls. b) Reduce the ratio in its lowest terms. c) How many times are the number of boys than the number of girls? Solution: a) The ratio of the number of boys to the number of girls = 20 : 10 b) Again, 20 : 10 = 20 10 = 2 1 = 2 : 1 c) The number of boys are 2 times the number of girls. Example 3: Find the compounded ratio of a) 3 : 2 and 1 : 4 b) 1 : 2, 4 : 5 and 10 : 3. Solution: The compounded ratio of 3 : 2 and 1 : 4 3 8 3 2 1 4 = × = =3 : 8 The compounded ratio of 1 : 2, 4 : 5 and 10 : 3 1 2 4 5 10 3 = × × = 4 : 3 Example 4: If x : y = 2 : 3 and y : z = 6 : 5, find a) x : z b) x : y : z Expressing the length into the same unit of width www.geogebra.org/classroom/drtjdb2t Classroom code: DRTJ DB2T Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 59 Vedanta Excel in Mathematics - Book 8 Solution: Here,x : y = 2 : 3 and y : z = 6 : 5 Now, x y × y z = 2 3 × 6 5 = 4 5 or, x : z = 4 : 5 Again, in the first ratio, y = 3 and in the second ratio, y = 6. The L.C.M. of 3 and 6 = 6 Now, 2 3 x y = = 2 × 2 3 × 2 = 4 6 And, y z = 6 5 Therefore, x : y : z = 4 : 6 : 5 Example 5: The ratio of the number of boys and girls in a school is 2 : 3. a) Who are smaller in number, boys or girls? Give reason. b)If there are 360 girls, find the number of boys. Solution: a) Here, the ratio of the number of boys and girls = 2 : 3 = 2 3 As the number of boys are only 2 3 (two-third) of the number of girls, boys are smaller in number. b) Let the number of boys be x. According to the question, x 360 = 2 3 or, 3x = 360 × 2 or, x = 360 × 2 3 = 240 Hence, the required number of boys is 240. Example 6: An alloy contains copper and zinc in the ratio 3 : 2. Find the mass of the metals in 450 g of alloy. Solution: Let the mass of copper be 3x g and that of zinc is 2x g. According to the question, 3x + 2x = 450 or, 5x = 450 or, x = 450 5 = 90 g ∴ Mass of copper= 3x = 3 × 90 g = 270g Mass of zinc = 2x = 2 × 90 g = 180 g It’s easier! boys : girls = 2 : 3 So, x : 360 = 2 : 3 Here, if copper is 3g, zinc is 2g If copper is 2 × 3g, zinc is 2 × 2g If copper is 3 × 3g, zinc is 3 × 2g If copper is x × 3g, zinc is x × 2g So, we consider copper is 3x g and zinc is 2x g. Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 60 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 7: A, B, and C invested a sum of Rs 2,70,000 on a business in the ratio of 2 : 3 : 4. (i) Find the sum invested by each of them. (ii) At the end of the year, if they gained Rs 90,000, divide the profit among them in the ratio of their shares. Solution: (i) Let the sums invested by A, B, and C are Rs 2x, Rs 3x and Rs 4x respectively. According to the question, 2x + 3x + 4x = Rs 2,70,000 or, 9x = Rs 2,70,000 or, 270000 9 x = Rs = Rs 30,000 ∴ The sum invested by A = 2x = 2 × Rs 30,000 = Rs 60,000 The sum invested by B = 3x = 3 × Rs 30,000 = Rs 90,000 The sum invested by C = 4x = 4 × Rs 30,000 = Rs 1,20,000 (ii) Again, the profit = Rs 90,000 and the ratio of sharing profit is 2 : 3 : 4 Now, 2x + 3x + 4x = Rs 90,000 or, 9x = Rs 90,000 or, x = Rs 90000 9 = Rs 10,000 ∴ The profit received by A = 2x = 2 × Rs 10,000 = Rs 20,000 The profit received by B = 3x = 3 × Rs 10,000 = Rs 30,000 The profit received by C = 4x = 4 × Rs 10,000 = Rs 40,000 Example 8: The monthly income of X is double than that of Y and the monthly income of Y is treble than that of Z. If the total monthly income of these three persons is Rs 1,00,000, find the monthly income of each person. Solution: Here, the ratio of the income of Y and Z = Y : Z = 3 : 1 = 3 1 the ratio of the income of X and Y = X : Y = 2 : 1 = 2 1 = 2 × 3 1 × 3 = 6 3 ∴ The ratio of X, Y and Z = X : Y : Z = 6 : 3 : 1 Let, the monthly incomes of X, Y and Z be Rs 6x, Rs 3x, and Rs x respectively. According to the question, 6x + 3x + x = Rs 1,00,000 or, 10x = Rs 1,00,000 or, x = Rs 100000 10 = Rs 10,000 ∴ The income of X = 6x = 6 × Rs 10,000 = Rs 60,000 The income of Y = 3x = 3 × Rs 10,000 = Rs 30,000 The income of Z = x = Rs 10,000. Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 61 Vedanta Excel in Mathematics - Book 8 Example 9: Last year, 50 students in a school appeared SEE. Among them, 16 students secured grade C, 4 secured grade D, and the rest of them secured grades A and B in the ratio 2 : 3. (i) Find the number of students who secured grades A and B. (ii) Find the ratio of the number of students who secured the grades A, B, C, and D. Solution: i) Here, the number of students who secured grades A and B = 50 – (16 + 4) = 30 Let the number of students who secured grade A is 2x and grade B is 3x. Now, according to the question, 2x + 3x = 30 or, 5x = 30 or, x = 6 Now, the number of students who secured grade A = 2x = 2 × 6 = 12 The number of students who secured grade B = 3x = 3 × 6 = 18 (ii) Again, the ratio of the number of students who secured grades A, B, C, and D = 12 : 18 : 16 : 4 = 6 : 9 : 8 : 2 Example 10: Two numbers are in the ratio 3 : 4. When 9 is subtracted from each of the numbers, the ratio becomes 2 : 3. Find the numbers. Solution: Let the required numbers be 3x and 4x. According to the question, 3x – 9 4x – 9 = 2 3 or, 3 (3x – 9) = 2 (4x – 9) or, 9x – 27 = 8x – 18 or, x = 9 Now, the required first number = 3x = 3 × 9 = 27 The second number = 4x = 4 × 9 = 36 Example 11: The ratio of the present age of a father and his son is 4 : 1. Four years hence, the ratio of their ages will be 3 : 1. Find their present ages. Solution: Let the present age of the father be 4x years and that of the son is x years. 4 years hence, age of the father = (4x + 4) years. 4 years hence, age of the son = (x + 4) years. According to the question, 4x + 4 x + 4 = 3 1 or, 4x + 4 = 3x + 12 or, x = 8 Now, the present age of the father = 4x = 4 × 8 years = 32 years The present age of the son = x = 8 years I got it! 9 is subtracted from 3x = 3x – 9 9 is subtracted from 4x = 4x – 9 And the new ratio = 3x – 9 4x – 9 = 2 3 Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 62 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 12: If the ratios of the angles of a quadrilateral are 1 : 2 : 3 : 4, find the size of each angle. Solution: Let the sizes of the angles of the quadrilateral be x°, 2x°, 3x°, and 4x° respectively. Now, x + 2x + 3x + 4x = 360° or, 10x = 360° or, 360° 10 x = = 36° Now, the first angle = x° = 36°, the second angle = 2x° = 2 × 36° = 72° the third angle = 3x° = 3 × 36° = 108° and fourth angle = 4x° = 4 × 36° = 144° Hence, the required angles of the quadrilateral are 36°, 72°, 108°, and 144°. EXERCISE 4.1 General Section - Classwork Let’s say and write the answers as quickly as possible. 1. a) Ratio of 3 to 4 is …………, antecedent is ………, consequent is ……… b) Ratio of 1 kg to 500 g is ………, antecedent is ………, consequent is ……… c) The compounded ratio of p : q and r : s is …………….. d) The compounded ratio of 1 : 4 and 3 : 1 is …………….. e) If a b = 1 2 , and b c = 2 3 , then a c = …………….. f) If x : y = 1 : 2 and y : z = 4 : 3, then x : z = …………….. 2. a) Two numbers are in the ratio 1 : 2. If the smaller one is 5, the greater one is …...... b) Two sums are in the ratio 3 : 1 and the first sum is Rs 30, the second sum is …...... c) Two weights are in the ratio 1 : 4 and the second weight is 20 kg, the first weight is …...... d) x : y = 2 : 1. If x = 10, then y = ………. e) a : b = 1 : 5. If a = 5, then b = ……… The sum of the angles of a quadrilateral is 360°. x° 2x° 4x° 3x° Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 63 Vedanta Excel in Mathematics - Book 8 Creative Section - A 3. Answer the following questions. a) The ratio of monthly income of Sita and Geeta is 1 : 2. (i) Define the meaning of ratio with the help of the above given ratio. (ii) What are antecedent and consequent in the given ratio? (iii) If the monthly income of Geeta is Rs 36,000, what is the monthly income of Sita? b) A rectangular meeting table is 4.5 m long and 1.5 m broad. (i) Find the ratio of the length to the breadth of the table. (ii) Reduce the ratio in its lowest terms. (iii) How many times is the length longer than the breadth? c) The weight of a lemon is two-fifth times the weight of an orange. (i) Express the weight of lemon and orange in ratio. (ii) If the weight of the orange is 175 g, find the weight of the lemon. 4. Let’s find the ratios in the lowest terms. a) 20 girls and 15 boys b) 1.5 m and 75 cm c) 800 g and 1 kg d) 750 m and 1.5 km 5. a) There are 15 teachers in a school and 10 of them are male teachers. (i) Find the ratio of the number of male and female teachers. (ii) Find the ratio of the number of female teachers and the total number of teachers. b) There are 30 students in a class and 18 of them are girls. (i) Find the ratio of girls and boys in the lowest terms. (ii) What are the consequent and antecedent terms in this ratio? (iii) If 2 girls were absent on a day, find the new ratio of the number of boys and the girls on that day. 6. a) One of the angles of a pair of complementary angles is 50°. Answer the following questions. (i) What is the sum of a pair of complementary angles? (ii) If the size of the complement of 50° is x°, find the value of x°. (iii) Find the ratio of these pair of complementary angles in the lowest terms. b) p° and 105° are a pair of supplementary angles. Answer the following questions. (i) What is the sum of p° and 105°? (ii) Find the size of p°. (iii) Find the ratio of these pair of supplementary angles in the lowest terms. 7. Find the compounded ratios. a) 1 : 2 and 4 : 3 b) 6 : 5 and 1 : 4 c) 7 : 9 and 6 : 5 d) 8 : 3 and 15 : 16 Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 64 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 8. a) If p : q = 1 : 2 and q : r = 4 : 3, find (i) p : r (ii) p : q : r b) If a : b = 2 : 3 and b : c = 6 : 5, find (i) a : c (ii) a : b : c 9. Let’s find the value of x in the following proportions. a) x : 3 = 4 : 6 b) 2 : x = 6 : 9 c) 5 : 4 = x : 12 d) 4 : 7 = 8 : x 10. a) Two numbers are in the ratio of 2 : 5 and the bigger number is 30. Find the smaller number. b) The ratio of length and breadth of a rectangular ground is 4 : 3 and the breadth is 75 m. Find the length of the ground. c) There are 18 girls in a class and the ratio of the number of girls and boys in the class is 3 : 2. Find the number of boys. d) The ratio of the monthly income and expenditure of a family is 7 : 4. If the income of the family in a month is Rs 21,000, find: (i) the monthly expenditure of the family. (ii) How much income does the family save in a year? 11. a) When a sum of Rs 100 is divided in the ratio of 2 : 3, find the share of each part. b) Mr. Karki divides a sum of Rs 1,80,000 between his son and daughter in the ratio of 4 : 5. Find the sum obtained by each of them. c) An alloy contains copper and zinc in the ratio 5 : 3. Find the mass of the metals in 960 g of alloy. 12. a) What is the actual distance between two places which is represented by 4.5 cm on a map if the map is drawn to the scale 1 : 100000 ? b) The distance between two places is 15 km. If the map scale is 1 : 500000, find the distance between the places in centimetres drawn in the map. 13. a) If the angles of a triangle are in the ratios 1 : 2 : 3, find the size of each angle. b) If the ratios of the angles of a quadrilateral are 2 : 3 : 4 : 6, find the size of each angle. c) If a pair of complementary angles are in the ratio 3 : 2, find the angles. d) If a pair of supplementary angles are in the ratio 1 : 2, find the angles. e) The acute angles of a right angled triangle are in the ratio 4 : 5, find the angles. Creative Section - B 14. a) The cost of a book is two times more than the cost of a box and the cost of the box is two times more than the cost of a pen. (i) If the cost of the pen is Rs x, what is the cost of the box? (ii) If the cost of the box is 2x, what is the cost of the book? (iii) Find the ratio of the cost of the book, box and the pen. (iv) If the total cost of these three items is Rs 350, find the cost of each item. b) The monthly income of A is double than that of B and the monthly income of B is treble than that of C. If the total income of three persons is Rs 80,000, find the monthly income of each of person. Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 65 Vedanta Excel in Mathematics - Book 8 c) Last year 60 students of a school appeared SEE. Among them 8 students secured grade C, 4 students secured grade D and the rest of them secured grades A and B in 3 : 5 ratio. (i) Find the number of students who secured grades A and B. (ii) Find the ratio of the number of students who secured the grades A, B, C and D. 15. a) Ram, Hari and Shyam invested a sum of Rs 1,80,000 on a business in the ratio of 2 : 3 : 4. Find the sum invested by each of them. b) Four businessmen invested a sum of Rs 4,50,000 in the ratio of 1 : 2 : 3 : 4 to start a new business. (i) Find the sum invested by each of them. (ii) At the end of the year, if they gained Rs 1,80,000, divide the profit among them in the ratio of their shares. 16. a) 6 and 9 are two given numbers. Answer the following questions. (i) What should be added to these numbers to get a ratio of 5 : 6? (ii) What should be subtracted from these numbers to get a ratio of 1 : 2? b) Two numbers are in the ratio 4 : 5. Answer the following questions. (i) If the smaller number is 4x, what is the bigger number? (ii) If 6 is added to each number, write the new numbers so formed. (iii) If the ratio of the new numbers is 5 : 6, find the numbers. c) Two numbers are in the ratio 7 : 5. When 10 is subtracted from each term, their ratio becomes 3 : 2. Find the numbers. 17. a) The ratio of the present age of a father and his son is 3 : 1. Answer the following questions. (i) If the present age of the son is x years, what is the present age of the father? (ii) How old will be the son and father after 5 years? (iii) If the ratio of their ages after 5 years is 5 : 2, express it in the form of an equation. (iv) Solve the equation and find the present age of the father and the son. b) The ratio of the present ages of A and B is 4 : 7. After three years, the ratio of their age will be 5 : 8. Find the present age of A and B. c) The present age of Bishwant and Sunayana are in the ratio 4 : 3. Three years ago, the ratio of their age was 7 : 5. Find their present age. d) The ratio of the present age of a mother and her daughter is 15 : 4. Four years ago, the ratio of their age was 13 : 2. Find the present age of the mother and the daughter. 18. a) The ratio of the length and breadth of a rectangle is 2 : 1. If the rectangle is 14 cm long, find its breadth. b) A rectangular field is 16 m broad and the ratio of the length and breadth is 3 : 2. (i) Find the length of the field. (ii) Find the perimeter of the field. (iii) Find the area of the field. Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 66 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) The ratio of the length and breadth of a rectangular ground is 4 : 3 and its perimeter is 126 m. (i) Write the formula of the perimeter of a rectangle. (ii) Find the length and breadth of the ground. (iii) Find the area of the ground. d) The ratio of the length and breadth of a rectangle is 5:4 and its area is 180 cm2 . (i) Write the formula of the area of a rectangle. (ii) Find its length and breadth (iii) Find its perimeter It’s your time - Project work and Activity section 19. a) How many students are there in your class? How many girls and boys are there? Let’s make the following ratios of these numbers. (i) Ratio of the total number of students to the number of girls. (ii) Ratio of the total number of students to the number of boys. (iii) Ratio of the number of girls to the number of boys. b) Let’s measure the length and breadth of the following objects and write the ratio of length and breadth. (i) Maths book (ii) desk or table (iii) floor of the classroom 4.4 Proportion Let's study the ratio of green and pink colored parts. Here, two ratios are 2 : 3 and 4 : 6. Also, 2 : 3 = 2 3 = and 4 : 6 = 4 6 = 2 3 = 2 : 3 Thus, the ratios 6 : 8 and 9 : 12 are equal ratios. The equality of ratios is called a proportion. 6 : 8 and 9 : 12 are said to be in proportion. Here, the terms 6, 8, 9, and 12 are called proportional. Suppose , the terms a, b, c and d are in proportion. We can write it as a : b = c : d or a : b :: c : d and we read it as ‘a is to b as c is to d.’ Here, the terms a, b, c, and d are the first second, third and fourth proportional respectively. In a proportion, the first and the fourth proportional are called extremes. The second and the third proportional are called means. Extremes Means 1st 2nd 3rd 4th Proportional Proportional a : b c : d We got it! If 6, 8, 15 and 20 are in proportion, then, 6 is the first, 8 is the second, 15 is the third and 20 is the fourth proportional! Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 67 Vedanta Excel in Mathematics - Book 8 In a : b= c : d or a b = c d , So, a × d = b × c We can use this fact to test whether the given ratios are in proportion or not. 4.5 Types of proportions There are two types of proportions: direct proportion and inverse proportion. (i) Direct proportion Suppose, the cost of 1 pen = Rs 30 Then, the cost of 2 pens = 2 × Rs 30 = Rs 60 Here, the ratio of the number of pens = 1 : 2 Also, the ratio of the cost of pens = 30 : 60 = 1 : 2 Thus, the ratio of the number of pens and the ratio of the cost of pens are equal. Therefore, these two ratios are in proportion. Here, if we increase or decrease the number of pens in a ratio, the cost also increases or decreases in the same ratio. Therefore, such proportion is called a direct proportion. Thus, in a proportion, if one ratio is increased (or decreased), the corresponding ratio also increases or decreases, such proportion is said to be a direct proportion. (ii) Inverse proportion Suppose 1 pipe can fill a tank in 12 hours. Then, 2 pipes of the same size can fill the tank in 6 hours. Here, the ratio of the number of pipes = 1 : 2 The ratio of time taken to fill the tank = 12 : 6 = 2 : 1 Thus, the ratio of the number of pipes and the ratio of time taken to fill the tank are oppositely equal. Therefore, these two quantities are in proportion with oppositely (inversely) equal ratios. In this case, the proportion is called an inverse proportion. In this way, in a proportion, if one ratio is increased (or decreased), the corresponding ratio decreases (or increases) inversely, such proportion is said to be an inverse proportion. Worked-out Examples Example 1: Test whether the terms 5, 6, 10, and 12 are in proportion. Solution: Here, the ratio of the first two terms is 5 : 6 and the last two terms is 10 : 12 Now, the product of extremes = 5 × 12 = 60 The product of means = 6 × 10 = 60 ∴ Product of extremes = Product of means Hence, 5, 6, 10 and 12 are in proportion. Interesting! Quantity and cost are always in direct proportion! I got it! Number of workers and working days are in inverse proportion! www.geogebra.org/classroom/tvrqvdws Classroom code: TVRQ VDWS Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 68 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: If 6, 9, and 42 are the terms of a proportion, find the fourth proportional. Solution: Let, the fourth proportional be x. Now, 6, 9, 42 and x are in proportion. ∴ 6 : 9 = 42 : x or, 6 9 = 42 x or, 2x = 126 or, x = 126 2 = 63 Hence, the required fourth proportional is 63. Example 3: If 4, 7, and 35 are the terms of a proportion, find the third proportional. Solution: Let, the third proportional be x. Now, 4, 7, x and 35 are in proportion. ∴ 4 : 7 = x : 35 or, 4 7 = x 35 or, 7x = 140 or, x = 140 7 = 20 Hence, the required third proportional is 20. Example 4: If the ratios 5 : x and 15 : 24 are in direct proportion, find the value of x. Solution: Here, 5 : x and 15 : 24 are in direct proportion. ∴ 5 : x = 15 : 24 or, 5 x = 15 24 or, 15x = 5 × 24 or, x = 5 × 24 15 = 8 Example 5: If the ratios 4 : 3 and x : 12 are in inverse proportion, find the value of x. Solution: Here, 4 : 3 and x : 12 are in inverse proportion. ∴ 4 : 3 = 12 : x or. 4 3 = 12 x or, 4x = 3 × 12 or, x = 3 × 12 4 = 9 Answer checking Here, 6 : 9 = 6 9 = 2 : 3 42 : 63 = 42 63 = 2 : 3 ∴ 6 : 9 = 42:63 6 : 9 = 42 : x Alternative process In, 4 : 7 = x : 35, product of means = product of extremes 7x = 4 × 35 or, x = 4 × 35 7 = 20 I got it! When 5 : x and 15 : 24 are in direct proportion, these ratios are directly equal. It’s easier! When 4 : 3 and x : 12 are in inverse proportion, the ratios are equal but oppositely. So, 4 : 3 = 12 : x Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 69 Vedanta Excel in Mathematics - Book 8 Example 6: If the cost of 9 kg of onions is Rs 360, how much onions can be purchased for Rs 480? Solution: Let x kg of onions can be purchased for Rs 480. Here, the ratio of the quantity of onions = 9 : x The ratio of the cost of onions = 360 : 480 The quantity of onions and the cost are in direct proportion. ∴ 9 : x = 360 : 480 or, 9 x = 360 480 or, 9 x = 3 4 or, 3x = 9 × 4 or, x = 9 × 4 3 = 12 Hence, the required quantity of onions is 12 kg. Example 7: If 9 workers can complete a piece of work in 20 days, how many workers should be added to complete the work in 15 days? Solution: Let, x number of workers can complete the work in 15 days. Here, the ratio of the number of workers = 9 : x The ratio of the number of working days = 20 : 15 The number of workers and their working days are in inverse proportion. ∴ 9 : x = 15 : 20 or, 9 x = 15 20 = 3 4 or, 3x = 9 × 4 or, x = 9 × 4 3 = 12 Hence, the required additional number of workers = 12 – 9 = 3. Example 8 : 40 students of a hostel have food enough for 30 days. If 10 more students join the hostel after 5 days, how long days the remaining food last ? Solution : Let, the remaining food will last for x days. Here, the total number of students after 5 days = 40 + 10 = 50 The remaining number of days = 30 – 5 = 25 days Here, the number of students and the number of days are in inverse proportion. Answer checking Cost of 9 kg of onions = Rs 360 Cost of 1 kg of onions = Rs 360 9 = Rs 40 Cost of 12 kg of onions= 12 × Rs 40 = Rs 480 Which is given in the question. Answer checking 9 workers complete the work in 20 days 1 worker complete the work in 9 × 20 days 12 workers complete the work in 9 × 20 12 = 15 days Which is given in the question. Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 70 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 40 : 50 = x : 25 or, 4 5 = x 25 or, x = 4 × 25 5 = 20 Hence, the remaining food will last for 20 days. Example 9 : A contractor employed 20 labourers to complete a construction work in 15 days working 8 hours a day. a) In how many days would 16 workers complete the construction working 6 hours a day? b) How many workers are required to complete the construction in 20 days working 5 hours a day? Solution : a) Let the required number of days be x Here, No. of days Working hours No. of labourers 15 8 20 x 6 16 Now, x : 15 = 8 : 6 × 20 : 16 or, x 15 = 8 6 × 20 16 or, x = 25 days Hence, the required number of days is 25 days. b) Let the required number of labourers be y Here, No. of days Working hours No. of labourers 15 8 20 20 5 y Now, y : 20 = 15 : 20 × 8 : 5 or, y 20 = 15 20 × 8 5 or, y = 24 Hence, the required number of labourers is 24. EXERCISE 4.2 General Section - Classwork 1. Let’s say and write whether these quantities are in direct or inverse proportions. a) Rate of cost and total cost of item ..................……………. b) Rate of cost and number of purchased items ..................……………. No. of workers, working hours and no. of days are in inverse proportion. No. of workers, working hours and no. of days are in inverse proportion. Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 71 Vedanta Excel in Mathematics - Book 8 c) Interval of time and distance covered ..................……………. d) Speed of a bus and time taken to cover certain distance ..................……………. e) Number of workers and number of working days ..................……………. 2. Let’s say and write the terms which are in proportion. a) 1, 2, 5, 10 b) 3, 4, 9, 15 c) 2, 3, 7, 9 d) 4, 5, 8, 10 3. Let’s say and write the values of x as quickly as possible. a) x : 2 = 2 : 1, x = …………… b) 3 : x = 1 : 3, x = …………………. c) 4: 1 = x : 2, x = …………….. d) 1 : 5 = 2 : x, x = …………………. Creative Section - A 4. Let’s find the products of extremes and means. Which of the following terms are in proportion? a) 2, 5, 4, 10 b) 3, 7, 6, 15 c) 4 : 5 and 12 : 15 d) 6 : 7 and 21 : 12 5. Find the values of x in the following proportions. a) 3:2 = 9 : x b) 4 : 5 = x : 15 c) 5 : x :: 10 : 12 d) x : 7 : : 18 : 14 6. a) If the following terms are in proportion, find the fourth proportional. (i) 3, 4, 6 (ii) 2, 3, 8 (iii) 4, 6, 12 (iv) 5, 7, 15 b) If the following terms are in proportion, find the third proportional. (i) 2, 5, 10 (ii) 5, 6, 18 (iii) 8, 3, 6 (iv) 9, 7, 28 c) If the following terms are in proportion, find the second proportional. (i) 3, 12, 28 (ii) 6, 24, 32 (iii) 7, 21, 15 (iv) 10, 30, 27 d) If the following terms are in proportion, find the first proportional. (i) 4, 15, 12 (ii) 16, 5, 4 (iii) 7, 30, 35 (iv) 6, 24, 18 7. a) If the following ratios are in direct proportion, find the value of x. (i) 3 : 5 and x : 15 (ii) 9 : x and 18 : 16 (iii) x : 10 and 28 : 40 b) If the following ratios are in inverse proportion, find the value of x. (i) 5 : 6 and x : 20 (ii) 12 : 9 and 36 : x (iii) x:20 and 60 : 45 Terms which are in proportion Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 72 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative Section - B Let’s use the rules of direct proportion and inverse proportion to workout the following problems. 8. a) The cost of a bag of 5 kg of rice is Rs 600 and Mrs. Bhattarai buys 3 bags of rice. (i) How many kilograms of rice does she buy? (ii) What type of proportion are the quantity and cost of rice? Give reason. (iii) How much money does Mrs. Bhattarai pay to buy 3 bags of rice? b) The rate of cost of per dozen of copies is Rs 360 and Prakash buys a few number of copies for Rs 540. (i) In what type of proportion are the number of copies and the cost? Give reason. (ii) How many copies does Prakash buy? c) A bus covered the distance of 240 km in every 4 hours of time interval and complete its destination of 420 km. (i) In what type of proportion are the time interval and distance covered? Give reason. (ii) How long did the bus take to cover the distance of its destination? 9. a) Last week, the bus fare of 8 passengers was Rs 640 and this week, the fare is decreased by Rs 5 per passenger. Write the types of proportion in the following cases. (i) Number of passengers and the cost of fare. (ii) The cost of fare and the rate of cost of fare. (iii) Rate of cost of fare and the number of passenger. (iv) How many passenger can travel for Rs 900? b) The bus fare of 12 passengers is Rs 540. If the fare is increased by Rs 5 per passenger, how many passengers can travel for Rs 750? 10. a) If 8 workers can complete a piece of work in 18 days, how many workers are required to complete the work in 12 days? b) 20 labourers complete the construction of a building in 24 days. In how many days would 16 labourers complete the construction? Ratio and Proportion

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 73 Vedanta Excel in Mathematics - Book 8 c) A bus completes its journey in 14 hours with an average speed of 40 km per hour. How long does the bus take to complete the journey if its average speed is increased to 56 km per hour? d) 15 workers were employed to complete a piece of work in 36 days. How many more workers should be added to complete the work in 30 days? e) 20 students in a hostel have provisions for 60 days. If 10 more students are admitted to the hostel, for how many days would the provisions be enough? f) A garrison of 60 men had provisions for 90 days. If 15 men left the garrison, how long would the provisions last? g) 60 students in a hostel have food enough for 30 days. If 20 more students join the hostel after 10 days, how long will the remaining food last? 11. a) A contractor employed 18 workers to complete a construction work in 30 days working 10 hours a day. (i) In what proportion are the number of workers, working days and working hours? (ii) In how many days would 20 workers complete the construction working 6 hours a day? (iii) How many workers are needed to complete the construction in 25 days working 8 hours a day? (iv) How many hours per day should 10 workers work to complete the construction in 45 days? b) 5 trucks can transport 500 quintals of rice to a remote region of Nepal in 3 days. In how many days would 4 trucks transport 800 quintals of rice to the same place? It’s your time - Project work and Activity section 12. a) Let’s write any three sets of four numbers such that each set of four numbers are in proportion. Find the product of extremes and means and verify that each set of numbers are in proportion. b) Let’s find the rate of cost of the following items in your local market. (i) rice (ii) sugar (iii) tomatoes (iv) onions (v) apples Let’s find the cost of different quantities of each of these items and verify that the cost and quantities are in direct proportion. c) Again, increase or decrease the rate of cost of items yourself. Then show that the rate of cost and number of quantities are in inverse proportion. Ratio and Proportion

Vedanta Excel in Mathematics - Book 8 74 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5.1 Unit quantity and unit value - Looking back Classwork - Exercise Let's say and write the answer of the following questions. 1. Mother buys 1 litre of milk for Rs 110. a) What is the unit quantity in this statement? ............................... b) What is the value of the unit quantity? ............................... c) If she buys 2 litres of milk, how much cost does she pay? ............................... d) If she buys 1 2 litre of milk, how much cost does she pay? ............................... 2. A fruit-seller purchased 50 kg of mangoes at the rate of Rs 90 per kg. a) What is the unit quantity in this statement? .......................... b) What is the value of the unit quantity? .......................... c) How much cost did the fruit-seller pay to buy the mangoes? .......................... 3. You have an internet service in your house. The download speed of your internet is 50 megabytes per second (Mbps). a) How many MB of a file can you download in 1 second? .......................... b) How many MB of a file can you download in 10 seconds? .......................... c) If the download speed is reduced to 25 MBps, how many seconds do you need to download 50 MB of a file? .......................... 5.2 Unitary method Suppose the cost of 1 kg of vegetables is Rs 40. Here, 1 kg of vegetable is unit quantity and Rs 40 is the unit cost (or unit value). Therefore, the value of unit number of quantity is called the unit value. Unit 5 Unitary Method

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 75 Vedanta Excel in Mathematics - Book 8 Unitary method in direct proportion In the case of direct proportion (or direct variation), unit value is obtained by division and the value of more number of quantity is obtained by multiplication. For example, The cost of 2 kg of vegetables is Rs 80. The cost of 1 kg of vegetables is Rs 80 2 . The cost of 5 kg of vegetables is Rs 80 2 × 5. Unitary method in inverse proportion However, in the case of inverse proportion (or inverse variation), unit value is obtained by multiplication and the value of more number of quantity is obtained by division. For example, 2 pipes can fill a cistern in 6 hours. 1 pipe can fill the cistern in 2 × 6 hours 3 pipe can fill the cistern in 2 × 6 3 hours. In this way, the method of finding the unit value or the value of more number of quantity by the process of division or multiplication is known as unitary method. Worked-out Examples Example 1: Mr. Thapa bought 4 packets of DDC milk for Rs 220 from a grocery store. a) At what rate of cost did he buy the milk? b) Find the cost of 10 packets of milk at the same rate. c) How many packets of milk can he bought for Rs 385? Solution: Here, cost of 4 packets of milk = Rs 220 a) Cost of 1 packet of milk = Rs 220 4 = Rs 55 Hence, the required rate of cost is Rs 55 per packet. b) Also, cost of 1 packet of milk = Rs 55 Cost of 10 packets of milk = 10 × Rs 55 = Rs 550 Hence, the required cost is Rs 550. Unit value is obtained by division. More values is obtained by multiplication. Unit values is obtained by multiplication. More values is obtained by division. www.geogebra.org/classroom/rxwtyhjc Classroom code: RXWT YHJC Vedanta ICT Corner Please! Scan this QR code or browse the link given below: By Proportion method: No. of packets Cost of milk 4 Rs 220 1 Rs x Here, 4 : 1 = 220 : x or, x = 220 4 = Rs 55 No. of packets Cost of milk 1 Rs 55 10 Rs y Here, 1 : 10= 55 : y or, y = 10 × 55 = Rs 550 Unitary Method

Vedanta Excel in Mathematics - Book 8 76 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) Again, Rs 55 is the cost of 1 packet of milk. Re 1 is the cost of 1 55 packet of milk. Rs 385 is the cost of 1 55 × 385 packets of milk. = 7 packets of milk Hence, the required number of packets of milk is 7 packets. Example 2: The download speed of an internet service is 60 megabytes per second (MBps) a) How many MB of a file can be downloaded in 1 second? b) How many gigabytes (GB) of file can be downloaded in 1 minute? (1 GB = 1000 MB) c) If the download speed is increased to 80 MBps, how long does it take to download 1 GB of file? Solution: a) 60 MB of file can be downloaded in 1 second. b) In 1 second 60 MB is downloaded. In 60 seconds 60 ×60 MB is downloaded = 3600 MB is downloaded = 3.6 GB is downloaded Hence, 3.6 GB of file can be downloaded in 1 minute (i. e. 60 seconds). c) 80 MB is downloaded in 1 second 1 MB is downloaded in 1 80 second 1000 MB is downloaded in 1 80 × 1000 seconds = 12.5 seconds Hence, the required time is 12.5 seconds. Example 3: When the average speed of a bus is 50 km per hour, it covers 300 km in a certain interval of time. If the speed of the bus is increased to 60 km per hour, find the distance covered by the bus in the same interval of time. Solution: No. of packets Cost of milk 1 Rs 55 z Rs 385 Here, 1 : z = 55 : 385 or, z = 7 By Proportion method: Download Size of time file 1s 60 MB 60 s x MB Here, 1 : 60 = 60 : x or, 1 60 = 60 x or, x = 3600 MB = 3.6 GB Download Size of time file 1s 80 MB y s 1000 MB Here, 1 : y = 80 : 1000 or, 1 y = 80 1000 or, y = 1000 80 = 12.5 seconds Unitary Method

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 77 Vedanta Excel in Mathematics - Book 8 When the speed is 50 km/hr, the bus covers 300 km When the speed is 1 km/hr, the bus covers 300 50 km = 6 km When the speed is 60 km/hr, the bus covers 60 × 6 km = 360 km Hence, the bus covers a distance of 360 km with the average speed of 60 km/hr. Example 4: If a pipe can fill 1000 l of water in 50 minutes, how much water does it fill in 1 hour? Solution: In 50 minutes a pipe can fill 1000 l of water. In 1 minute, the pipe can fill 1000 50 = 20 l of water In 60 minutes (1 h) the pipe can fill 60 × 20 l = 1200 l of water. Hence, the pipe can fill 1200 l of water in 1 hour. Example 5: If the cost of 2 3 parts of a land is Rs 1, 80,000, find the cost of 4 5 parts of the land. Solution: The cost of 2 3 parts of a land = Rs 1,80,000 The cost of 1 land (whole land) = Rs 1,80,000 2 3 = 1,80,000 × 3 2 = Rs 2,70,000 The cost of 4 5 parts of the land = 4 5 × Rs 2,70,000 = Rs 2,16,000. Hence, the required cost of 4 5 parts of the land is Rs 2,16,000. Example 6: The cost of 5 exercise books and 2 pens is Rs 250. If the cost of 1 exercise book is Rs 40, find the cost of 6 pens. By Proportion method: speed distance covered 50 km/h 300 km 60 km/h x km Here, 50 : 60 = 300 : x or, 50 60 = 300 x or, x = 360 km By Proportion method: amount time taken of water 1000 l 50 minutes x l 60 minutes Here, 1000 : x = 50 : 60 or, 1000 x = 50 60 or, x = 1200 l By Proportion method: Part of area cost land 2 3 Rs 1,80,000 4 5 Rs x Here, 2 3 : 4 5 = 1,80,000 : x or, 2 3 × 5 4 = 1,80,000 x or, x = Rs 2,16,000 Unitary Method

Vedanta Excel in Mathematics - Book 8 78 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Here, the cost of 1 exercise book = Rs 40 ∴ The cost of 5 exercise books = 5 × Rs 40 = Rs 200 The cost of (5 exercise books + 2 pens) = Rs 250 or, Rs 200 + cost of 2 pens = Rs 250 or, cost of 2 pens = Rs 250 – Rs 200 = Rs 50 ∴ cost of 1 pen = Rs 50 2 = Rs 25 And cost of 6 pens = 6 × Rs 25 = Rs 150 Hence, the required cost of 6 pens is Rs 150. Example 7: If 16 workers can construct a road in 45 days, how many additional numbers of workers should be employed to finish the construction in 30 days? Solution: In 45 days, 16 workers can construct the road. On 1 day, 45 × 16 workers can construct the road. In 30 days, 45 × 16 30 workers can construct the road. = 24 workers can construct the road. Hence, the additional number of workers = 24 – 16 = 8. Example 8: 32 students of a hostel had provisions for 30 days. If 8 more students were admitted after 5 days, how long would the remaining provisions last? Solution: Here, the remaining number of days = 30 – 5 = 25 days. Total number of students with 8 more students = 32 + 8 = 40 32 students had provisions for 25 days. 1 student had provisions for 32 × 25 days. 40 students had provisions for 32 × 25 40 days = 20 days. Hence, the remaining provisions would last for 20 days. By Proportion method: No. of working workers days 16 45 x 30 Here, 16 : x = 30 : 45 or, 16 x = 30 45 or, x = 24 workers Additional number of workers = 24 –16 = 8 By Proportion method: No. of Duration students of provisions 32 (30–5)=25 days (32+8)=40 x days Here, 32 : 40 = x : 25 or, 32 40 = x 25 or, x = 20 days Unitary Method

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 79 Vedanta Excel in Mathematics - Book 8 Problems with three variables Examples 9: 6 workers can do a piece of work in 15 days working 7 hours a day. a) In how many days would 1 worker complete the same work with the same working hours? b) In how many days would 5 workers complete the work with the same working hours? c) In how many days would 5 workers complete the work working 6 hours a day? Solution: a) 6 workers working 7 hours a day, complete the work in 15 days. 1 worker working 7 hours a day, complete the work in 6 × 15 days = 90 days Hence, 1 worker complete the work in 90 days. b) 1 worker working 7 hours a day complete the work in 90 days 5 workers working 7 hours a day complete the work in 90 5 days = 18 days Hence, 5 workers complete the work in 18 days. c) 5 workers working 7 hours a day complete the work in 18 days. 5 workers working 1 hour a day complete the work in 7 × 18 days 5 workers working 6 hours a day complete the work in 7 × 18 6 days = 21 days Hence, 5 workers complete the work in 21 days. Proportion method: No. of workers Working hours Working days 6 7 15 a) 1 7 x b) 5 7 y c) 5 6 z Solution: Here, number of workers, working hours and working days are in indirect proportion. a) ∴ 15 : x = 1 : 6 × 7 : 7 or, 15 x = 1 6 × 7 7 or, x = 90 days c) ∴ 18 : z = 5 : 5 × 6 : 7 or, 18 z = 5 5 × 6 7 or, z = 21 days Number of workers, working hours and working days are three variables! b) ∴ 90 : y = 5 : 1 × 7 : 7 or, 90 y = 5 1 × 7 7 or, y = 18 days Unitary Method

Vedanta Excel in Mathematics - Book 8 80 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Examples 10: An offset printing machine can print 15,000 story books in 6 days if it is operated 7 hours a day. a) In what proportion are the number of books and the printing days? b) How long does the machine take to print 50,000 books in the same operating hours? c) In what proportion are the per day printing hours and the printing days? d) How long does the machine take to print 50,000 books operating 10 hours a day? Solution: a) Number of books and printing days are in direct proportion. b) Operating 7 hours a day, 15,000 books are printing in 6 days Operating 7 hours a day, 1 book is printed in 6 15,000 days Operating 7 hours a day 50,000 books are printed in 6 × 50,000 15,000 days = 20 days Hence, the required number of days is 20 days. c) Per day printing hours and printing days are in inverse proportion. d) Operating 7 hours a day, 50,000 books are printed in 20 days Operating 1 hour a day, 50,000 books are printed in 7 × 20 days Operating 10 hours a day, 50,000 books are printed in 7 × 20 10 days = 14 days Hence, the required number of days is 14. EXERCISE 5.1 General Section - Classwork 1. Let’s say and write the answers in the blank spaces. a) Cost of 1 kg of onions is Rs 50, cost of 3 kg of onions is …………… b) Cost of 5 exercise books is Rs 200, cost of 1 exercise book is ……………. c) Cost of 3 pens is Rs 75, cost of 1 pen is ……….., cost of 4 pens is …………….. d) A bus covers 120 km in 3 hrs, it covers ……. km in 1 hr and ……… km in 6 hrs. 2. a) 1 worker can do a work in 6 days, 2 workers can do it in …………. b) 3 pipes can fill a cistern in 2 hours, 1 pipe can fill it in …………….. c) 2 labourers can do a work in 6 days, 6 labourers can do it in ……………. 3. a) 1 2 part of a sum is Rs 10, (i) the sum is ……… (ii) 1 4 part of the sum is ……… b) 1 3 part of a distance is 4 km, (i) the distance is ……… (ii) 1 4 part of the distance is ……… Unitary Method

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 81 Vedanta Excel in Mathematics - Book 8 Creative Section - A 4. a) The cost of 5 kg of rice is Rs 400. (i) Find the cost of 1 kg of rice and 12 kg of rice. (ii) How many kilograms of rice can be bought for Rs 240? b) A bus completed its journey of 320 km with an average speed of 40 km per hour. (i) How long did it take to complete the journey ? (ii) What would be the speed of the car if it covered only 240 km in the same interval of time? c) 5 litres of petrol is needed for a motorbike to cover 200 km. (i) What is the mileage of the bike? (ii) How much petrol is needed to cover 320 km? 5. a) A worker can do a piece of work in 14 days. (i) How much work does he do in 1 day ? (ii) How much work does he do in 7 days ? (iii) If he works for 2 days and leaves, how much work is left to finish? b) A factory completes the production of items in 12 days. (i) How much production does it do in 9 days. (ii) In how many days would it finish 1 3 part of the whole production? c) A computer can download 1 3 part of an application file in 40 second. (i) In how many minutes would it complete to download the whole file? (ii) How much file does it download in 1 minute? (iii) How much file is left to download if it downloaded only for 90 seconds? d) A pipe can fill a water tank in 1 hour. (i) What part of the tank is filled by the pipe in 20 minutes ? (ii) If the capacity of the tank is 1500 , how many litres of water does the pipe fill in 20 minutes ? 6. a) 3 4 parts of the distance between two places is 36 km. (i) Find the distance between the places. (ii) Find 2 3 parts of the distance between the places. b) A man spends Rs 7,500 every month to run his family, which is 1 3 part of his monthly income. (i) Find his monthly income. (ii) If he spends 3 25 parts of his income on his children’s education, how much money does he spend on education? Unitary Method

Vedanta Excel in Mathematics - Book 8 82 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative Section - B 7. a) The cost of 2 kg of sugar and 5 kg of rice is Rs 710 and the cost of 1 kg of sugar is Rs 80. (i) Find the cost of 2 kg of sugar (ii) Find the cost of 5 kg of rice (iii) Find the cost of 8 kg of rice b) The cost of 3 kg of apples and 2 kg of oranges is Rs 540. If the cost of 1 kg of oranges is Rs 90, find the cost of 6 kg of apples. 8. a) 3 pipes of the same size can fill a cistern in 90 minutes. (i) How long would 1 pipe take to fill the cistern? (ii) How long would 5 pipes take to fill the cistern? b) 4 machines of a factory can finish the required amount of production in 6 days. If one machine had machinery defect and stopped functioning, in how many days would the remaining machines finish the same quantity of production at the same rate of production? c) 6 workers complete a piece of work in 15 days. In how many days would 9 workers complete the same work? d) 12 workers were employed to complete the construction of a building in 60 days. How many additional numbers of workers should be employed to complete the construction in 45 days? e) 60 students of a hostel had provisions for 40 days. If 20 more students were admitted, how long would the provisions last? f) 120 students of a hostel have food enough for 60 days. If 30 more students join the hostel after 15 days, how long does the remaining food last? g) A garrison of 110 people had provisions for 30 days. If 22 people leave the garrison after 10 days, how long does the remaining provisions last? 9. a) A computer can finish to download an application file in 4 minutes at the rate of 600 MB per minute. (i) Find the size of the application file. (ii) How long does it take to download the file when the download rate increases to 800 MB per minute? b) The size of a movie file in YouTube is 1.8 gigabyte (GB) and the rate of download of the file is 900 megabyte (MB) per minute. (i) How long does it take to download the file? (ii) Find the rate of download of the file per second. 10. a) 12 workers finish to build compound walls in 20 days working 9 hours a day. (i) In how many days would 1 worker finish to build the walls with the same working hours? (ii) In how many days would 15 workers finish to build the walls with the same working hours? (iii) In how many days would 15 workers finish to build the walls working 6 hours a day? Unitary Method

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 83 Vedanta Excel in Mathematics - Book 8 b) 18 workers finish to renovate a building in 28 days working 8 hours a day. In how many days would 16 workers finish to renovate the building working 7 hours a day? c) 21 workers finish to construct a football ground in 32 days working 10 hours a day. How many workers are required to finish the construction in 35 days working 8 hours a day? 11. a) A printing press prints 10,000 textbooks in 3 days if it is operated 10 hours everyday. (i) In what proportion are the number of textbooks and the printing days? (ii) How long does the press take to print 40,000 books in the same operating hours? (iii) In what proportion are the everyday printing hours and the printing days? (iv) How long does the press take to print 40,000 books if it is operated 8 hours a day? b) An offset printing machine can print 12,000 textbooks in 5 days if it is operated 6 hours a day. (i) How many books can be printed in 15 days operating the press 9 hours a day? (ii) How many hours per day should the machine be operated to print 32000 textbooks in 8 days? 12. A tap can fill 6000 l of water in 3 days if it is opened for 2 hours everyday. a) In how many days does the tap fill 15,000 l of water if it is opened for 3 hours a day? b) How much water does it fill in 7 days if it is opened 4 hours a day? c) For how many hours is the tap opened everyday to fill 35000 l of water in 7 days? It's your time - Project Work and Activity Section 13. Let’s make a survey and find the rate of cost of the following items in your local market. (i) rice (ii) potatoes (iii) cooking oil (iv) milk a) Estimate the cost of these items consumed by your family per day. b) How much budget do your family allocate on these items for a month and for one year? 14. Let’s search the sizes of some mobile application files. Calculate and estimate the time taken to download these files in your available internet speed. 15. Let’s visit to the available website and search today’s exchange rate of foreign currencies with our Nepali currency. Then, find how much Nepali currency is needed to exchange the following foreign currencies? a) £ 1,500 b) $ 1,500 c) 50,000 d) AUD 1,500 Unitary Method

Vedanta Excel in Mathematics - Book 8 84 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6.1 Principal and Interest - Looking back Classwork - Exercise Let's discuss the answers of the following questions. 1. Mrs. Tharu borrowed a sum of Rs 10,000 from a bank. At the end of 1 year, she paid Rs 1,500 as an extra charge for the use of the borrowed sum. a) How much is the principal in the above statements? b) How much is the interest? c) How much money did she need to clear the loan after 1 year? 2. Mr. Poudel deposited a sum of Rs 20,000 in a bank. At the end of 1 year, his sum was increased to Rs 22,000. a) How much is the principal in the above statements? b) How much extra charge did the bank pay to him at the end of 1 year? From the above discussion, do you make some idea about the principal and interest? In the first case, the principal is Rs 10,000 and the interest is Rs 1,500. In the second case, the principal is Rs 20,000 and the extra charge (which is an interest) is Rs 2,000. 6.2 Simple Interest What do we do when we are living in a rented house? Of course, we should pay the rent to the house-owner for the use of the house. Similarly, when we use other’s money we should pay some charge for the use of money. This charge (an additional amount of money) is called interest. In this way, when we borrow money from a bank, we should pay interest to the bank. When we deposite money in a bank, the bank pays interest to us. The interest which is calculated from the original borrowed (or deposited) sum is called simple interest. Let’s review the following terms which are needed to calculate simple interest. Unit 6 Simple Interest

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 85 Vedanta Excel in Mathematics - Book 8 Principal (P) – It is the deposited or borrowed sum of money. Rate of interest (R) – It is the interest of Rs 100 for 1 year. It is expressed as the percent per year (or per annum, p.a.) Time (T) – It is the duration for which principal is deposited or borrowed. Interest (I) – It is the simple interest on Rs P at R% p.a. in T years. Amount (A) – It is the total sum of principal and interest. So, A = P + I 6.3 Rate of interest The interest on the principal of Rs 100 for 1 year is called rate of interest (R). Rate of interest (R) is usually expressed in percent per year or per annum. For example, If the interest of Rs 100 in 1 year is Rs 5, then R = 5% p.a. If the interest of Rs 100 in 1 year is Rs 12, then R = 12% p.a. and so on. 6.4 Formula of simple interest According to the definition of the rate of interest, On Rs 100, for 1 year, interest (I) = R On Re 1, for 1 year, interest (I) = R 100 On Rs P, for 1 year, interest (I) = P × R 100 On Rs P, for T years, interest (I) = P × T × R 100 Thus, the formula to calculate simple interest is (I) = PTR 100 Again, if I = PTR 100 , then, PTR = I × 100 ∴ P = I × 100 T × R → is the formula to calculate principal (P) T = I × 100 P × R → is the formula to calculate time (T) R = I × 100 P × T → is the formula to calculate rate (R) Also, from the definition of amount (A), A = P + I or, A = P + PTR 100 Simple Interest

Vedanta Excel in Mathematics - Book 8 86 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur or, A = P 1 + TR 100 or, A = P 100 + TR 100 or, P = A × 100 100 + TR Worked-out Examples Example 1: Mr. Jha borrowed a sum from a development bank at 12% per year interest. a) If the borrowed sum is Rs 100, how much interest should he pay in 1 year? b) If the borrowed sum is Rs 27,000, how much interest should he pay in 3 years? c) Find the amount required to clear the loan at the end of 3 years. Solution: a) He should pay the interest of Rs 12 in 1 year. b) Here, Principal (P) = 27,000 Time (T) = 3 years Rate (R) = 12% per year Now, Interest (I) = P × T × R 100 = Rs 27,000 ×3 ×12 100 = Rs 9,720 Hence, the required interest is Rs 9,720 c) Again, amount (A) = P + I = Rs 27,000+ Rs 9,720 = Rs 36,720 Hence, the required amount is Rs 36,720 Example 2: A farmer borrowed a sum of Rs 50,000 from an Agricultural Development Bank at the rate of 7.5% per year to promote his poultry farming. Calculate the amount required to clear the loan at the end of 5 years. Solution: Here, Principal (P) = Rs 50,000 Rate (R) = 7.5% per year Time (T) = 5 years Now, interest (I) = PTR 100 = Rs 50,000 × 5 × 7.5 100 = Rs 18,750 Again, amount (A) = P + I = Rs 50,000 + Rs 18,750 = Rs 68,750 Hence, the required amount is Rs 68,750. which is used to calculate principal when time, rate and amount are given. www.geogebra.org/classroom/pg2sea9w Classroom code: PG2S EA9W Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Simple Interest

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 87 Vedanta Excel in Mathematics - Book 8 Example 3: At what rate percent per annum does a sum of Rs 9,450 amount to Rs 12,474 in 4 years? Solution: Here, principal (P) = Rs 9,450 Amount (A) = Rs 12,474 ∴ Interest (I) = A – P = Rs 12,474 – Rs 9,450 = Rs 3,024 Time (T) = 4 years Now, rate (R) = I × 100 P × T = 3,024 × 100 9,450 × 4 = 8% Hence, the required rate of interest is 8% per year. Example 4: Find the sum that amounts to Rs 9,144 in 3 years at 9% per year simple interest. Solution: Here, amount (A) = Rs 9,144 Time (T) = 3 years Rate (R) = 9% per year Now, principal (P) = A × 100 100 + TR = 9,144 × 100 100 + 3 × 9 = 9,14,400 127 = Rs 7,200 Hence, the required sum is Rs 7,200. EXERCISE 6.1 General Section - Classwork 1. The table given below shows the known variables. Say and write the formulae to find the unknown variables. No. Known variables Unknown variables a) Principal (P), Time (T), Rate (R%) Interest (I) = b) Interest (I), Time (T), Rate (R%) Principal (P) = c) Principal (P), Interest (I), Time (T) Rate (R) = d) Principal (P), Interest (I), Rate (R%) Time (T) = e) Amount (A), Time (T), Rate (R%) Principal (P) = f) Principal (P), Interest (I) Amount (A) = Answer checking P = Rs 9,450, R = 8%, T = 4 years I = PTR 100 = Rs 9,450 × 4 × 8 100 = Rs 3,024 A = Rs 9,450 + Rs 3,024 = Rs 12,474 which is given in the question. Simple Interest

Vedanta Excel in Mathematics - Book 8 88 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 2. Let’s say and write the answers as quickly as possible. a) When P = Rs 100, T = 1 year, I = Rs 5, then R = ………… b) When P = Rs 100, T = 1 year, I = Rs 9, then R = …………. c) R = 10% p.a. means, P = …………, T = ………….. I = ………….. d) R = 12% p.a means, P = …………, T = ………… I = ………….. 3. a) If P = Rs 1,000, I = Rs 100, A = ……………… b) If A = Rs 2,000, P = Rs 1,800, I = ……………… c) If A = Rs 3,000, I = Rs 300, P = ……………… Creative Section - A Let’s calculate the variables which are not given. 4. a) P = Rs 1,000, T = 2 years, R = 4% per year , find I. b) I = Rs 525, T = 3 years, R = 7% per annum, find P. c) I = Rs 1,200, P = Rs 3,000, R = 10% p.a., find T. d) I = Rs 2,700, P = Rs 4,500, T = 5 years, find R. e) A = Rs 1,800, T = 2 years, R = 10%, find P and I. f) A = Rs 5,250, T = 5 years, R = 15% , find P and I. 5. Mrs. Bhandari lends a sum of money to Mr. Magar on the condition that he should pay Rs 18 as the interest of Rs 100 every year. a) What is the rate of interest in percent? b) How much interest should Mr. Magar pay her in 2 years if the loan is Rs 25,000? c) Find the amount paid by Mr. Magar to clear the loan at the end of 2 years. 6. Sher Bahadur deposit a sum of money in a commercial bank at 9% per year interest. a) If the deposited sum is Rs 100, how much interest does he get in 1 year? b) If the deposited sum is Rs 18,000, how much interest does he get in 5 years?? 7. a) Mrs. Chapagain borrowed Rs 9,900 from a bank at the rate of 8% per annum. Answer the following questions. (i) What do you mean by the rate of 8% p.a.? (ii) Write the formula to calculate the simple interest. (iii) How much interest does she pay at the end of 2 years? (iv) How much amount should she pay at the end of 2 years if she wants to clear the loan. Simple Interest

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 89 Vedanta Excel in Mathematics - Book 8 b) Mr. Shah deposited a sum of Rs 15,000 in a bank at the rate of 10% per year. Find the amount received by him at the end of 3 years 6 months. c) Mrs. Rai borrowed a sum of Rs 10,800 from Mr. Yadav at the rate of 13% per year. Calculate the amount received by Mr. Yadav in 9 months. 8. a) Mr. Gurung borrowed a sum of Rs 2,700 from his friend Mr. Gupta. He paid an interest of Rs 594 to Mr. Gupta at the end of 2 years. (i) Write the formula to find the rate of interest. (ii) Find the rate of interest at which Mr. Gurung borrowed the sum. (iii) At the same rate of interest calculate the interest in 3 years. (iv) If Mr. Gurung had not paid any interest till the end of 3 years, how much amount would he need to clear the loan? b) If a sum of Rs 5,600 amounts to Rs 8,400 in 5 years, find the rate of interest. c) At what rate percent per year will a sum of Rs 12,000 amount to Rs 12,720 in 6 months? 9. a) If the interest of a sum at 7.5% p.a. in 4 years is Rs 1,050, find the sum. b) Find the principal that yields an interest of Rs 2,240 at the rate of 10% p.a. in 3 years 6 months. c) In how many years will the interest of a sum of Rs 3,600 at the rate of 5.5% per year be Rs 594? Creative Section - B 10. a) Mrs. Joshi borrowed a sum of money from a bank at 12% p.a. simple interest. If she paid an interest of Rs 1,920 in 2 years, find the sum borrowed by her. b) A farmer borrowed a sum of Rs 10,000 from a Rural Development Bank at the rate of 7.5% p.a. If he/she paid an amount of Rs 15,250 to clear the debt, how long did he/she use the sum? c) A businessman deposited Rs 60,000 in a Commercial Bank at 4.5% p.a. How long should he deposit the sum to receive an amount of Rs 61,350? 11. a) Mr. Gharti borrowed a loan from Mrs. Gupta at 10% p.a. At the end of 2 years, he paid an amount of Rs 3,600 and cleared the debt. Find the sum borrowed by Mr. Gharti. b) What sum of money amounts to Rs 5,454 at 8% p.a. in 30 months? c) Pradeep lent a sum to Santosh at the rate of 12 1 2 % p.a. After 4 years, Santosh paid him Rs 7500 and clear the debt, what sum did Pradeep lend to Santosh. 12. a) Mr. Gurung borrowed a loan of Rs 25,000 from Mr. Pandey at 12% p.a. At the end of 3 years, if he agreed to pay Rs 26,300 with a goat to clear his debt. Find the cost of goat. b) Shashwat lent a sum of Rs 44,000 to his friend Rahul at 10% p.a. After 21 2 year, his friend paid him Rs 40,000 with a cow. What was the cost of the cow? Simple Interest

Vedanta Excel in Mathematics - Book 8 90 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 13. a) Sajina deposited Rs 20,000 at the rate of 8% p.a. in her saving account. After 2 years, she withdrew Rs 5,000 and the total interest of 2 years. How long should she keep the remaining amount to get total interest of Rs 6,800 from the beginning? b) Ramesh took a loan of Rs 50,000 from Urmila at the rate of 10% p.a. If he paid a half of the principal and all the interest at the end of 3 years, in how many years should he pay the remaining amount with total interest of Rs 20,000 from the beginning? 14. a) A man took a loan of Rs 12,000 with simple interest for as many years as the rate of interest per year. If he paid Rs 3,000 as interest at the end of loan period, what was the rate of interest? b) A teacher deposited Rs 66,000 in his/her saving account for as many years as the rate of interest per annum. If he/she received Rs 10,560 as interest at the end of saving period, find the time duration and rate of interest. It's your time - Project Work and Activity Section 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. P = ................ T = ................ R = ................ Find I and A I = ................ T = ................ R = ................ Find P and A I = ................ P = ................ R = ................ Find T P = ................ I = ................ T = ................ Find R A = ................ T = ................ R = ................ Find P and I b) Let’s make the groups of your friends and visit nearby banks or finance company. Collect the various information about the rate of simple interest in different types of accounts such as current account, saving account and fixed deposit account. In which account are they providing more interest? c) Ask your parents if they have any bank account. What type of account is it? Current, saving, or fixed deposit? What is the rate of interest given by the bank? Simple Interest

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 91 Vedanta Excel in Mathematics - Book 8 7.1 Profit and loss - Looking back 1. Mrs. Sharma is a retailer of stationery items. She bought a few number of diaries at Rs 200 each and sold at Rs 250 each. Let's discuss and answer the following questions. a) What is the cost price of a diary? …………….. b) What is the selling price of a diary? …………….. c) Did she make profit or loss? …………….. d) What is the formula to find the profit? ………………. e) How much profit did she make? ………………. f) If she had sold each diary giving Rs 25 discount on Rs 250, how much profit would she get? ………………. g) If she had sold each diary for Rs 190, how much loss would she make in one diary? ………………. In the above given case: Rs 200 is the price paid by Mrs. Sharma to buy a diary. It is called cost price (C. P.). Rs 250 is the price at which she sold a diary. It is called selling price (S. P.). In this case, she makes a profit of Rs 50 which is Rs 250 – Rs 200. Thus, when S. P. is higher than C. P., there is profit. However, if S. P. is lower than C. P., there will be a loss. Hence, when S. P. > C. P. , Profit = S. P. – C. P. Also, when S. P. < C. P., Loss = C. P. – S. P. Furthermore, Profit = S. P. – C. P., then C. P. = S. P. – Profit and S. P. = C. P + Profit Loss = C. P. – S. P., then C. P. = S. P. + Loss and S. P. = C. P. – Loss 7.2 Profit percent and loss percent Let C. P. of an article be Rs 100 and profit is Rs 20, then profit percent is 20%. Also, C. P. of the article is Rs 100 and loss is Rs 15, then loss percent is 15%. Thus, when profit or loss is expressed out of C. P. of Rs 100, it is called profit percent or loss percent. Again, let’s consider the above given example. C. P. of a diary = Rs 200 and S. P. of the diary = Rs 250 Now, profit = S. P. – C. P. = Rs 250 – Rs 200 = Rs 50 Classwork - Exercise Unit 7 Profit and Loss

Vedanta Excel in Mathematics - Book 8 92 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur When C. P. is Rs 200, profit is Rs 50. When C. P. is Re 1, profit is Rs 50 200 When C. P. is Rs 100, profit is Rs 50 200 × 100 = Rs 25 Here, Rs 25 is the profit of C. P. of Rs 100. Therefore, profit percent is 25%. Thus, profit percent = profit C. P. × 100% ⇒ S. P. – C. P. C. P. × 100% And, loss percent = loss C. P. × 100% ⇒ C. P. – S. P. C. P. × 100% 7.3 To find S.P. when C.P. and profit or loss percents are given We know that, S.P. = C.P. + Profit or C.P. – Loss. So, to calculate S.P., at first we should calculate profit amount or loss amount from C.P. For example, If C.P. = Rs 400 and profit percent = 20%, find S.P. Here, profit amount = 20% of C.P. = 20 100 × Rs 400 = Rs 80 Now, S.P. = C.P. + Profit = Rs 400 + Rs 80 = Rs 480 Again, if C. P. = Rs 400 and loss percent = 12%, find S. P. Here, loss amount = 12% of C. P. = 12 100 × Rs 400 = Rs 48 Now, S. P. = C. P. – Loss = Rs 400 – Rs 48 = Rs 352 7.4 To find C.P. when S.P. and profit or loss percent are given Profit or loss percent is always calculated on C.P. But, here, C.P. is unknown. So, we should consider C.P. as Rs x. Then, the profit or loss amount is calculated from Rs x. Study the following examples: If S.P. = Rs 550 and prfit percent = 10%, find C.P. Let, the required C.P. be Rs x Here, profit amount = 10% of Rs x = 10 100 × Rs x = Rs x 10 Now, C.P. = S.P. – profit or, x = 550 – x 10 or, x + x 10 = 550 or, 11x 10 = 550 or, x = 550 × 10 11 = Rs 500 Hence, the required C.P. is Rs 500. Direct process S. P. = (100 + 20)% of C. P. = 120% of Rs 400 = 120 100 × Rs 400 = Rs 480 Direct process S. P. = (100 – 12)% of C. P. = 88% of Rs 400 = 88 100 × Rs 400 = Rs 352 Direct process (100 + 10)% of C. P. = S.P. or, 110% of C. P. = Rs 550 or, 110 100 × C. P. = Rs 550 or, C. P. = Rs 550 × 100 110 or, C. P. = Rs 500 Profit and Loss

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 93 Vedanta Excel in Mathematics - Book 8 Facts to Remember Calculation of Given Process Profit or loss C. P. and S. P. Profit = S. P. – C. P. and Loss = C. P. – S. P. Profit or loss percent C. P. or C. P. and S. P. Profit% = profit C. P. × 100% and Loss% = Loss C. P. × 100% Profit or loss amount Profit % or Loss % and C. P. Profit = Profit% of C. P. and Loss = Loss% of C. P. S. P. Profit % or Loss % and C. P. S. P. = C. P. + Profit% of C. P. and S. P. = C. P. – Loss% of C. P. C. P. Profit % or Loss % and S. P. C. P. + profit % of C. P. = S. P. and C. P. – Loss % of C. P. = S. P. or, x+Profit% of x = S.P. or, x–Loss% of x=S.P. Direct process: C. P. = S. P. (100+Profit)% and C. P. = S. P. (100–Profit)% Worked-out Examples Example 1: A shopkeeper bought a watch for Rs 1,250 and sold it for Rs 1,500. a) Did the shopkeeper make profit or loss? Give reason. b) How much profit or loss did he make? c) Express profit or loss in percent. Solution: a) The shopkeeper made profit. Because his selling price is more than the cost price. b) Here, C. P. = Rs 1,250 S.P. = Rs 1,500 ∴ Profit = S. P. – C. P. = Rs 1,500 – Rs 1,250 = Rs 250 Hence, the required profit is Rs 250. c) Again, profit percent = Profit C.P. × 100% = 250 1250 × 100% = 20% Hence, the required profit percent is 20%. Example 2: A grocer bought 300 eggs at Rs 8 each. 25 eggs were broken and he sold the remaining eggs at Rs 9.60 each. Find his/her profit or loss percent. www.geogebra.org/classroom/bjayswbz Classroom code: BJAY SWBZ Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Profit = S. P. – C. P. Profit % = S. P. – C. P. C. P. × 100% = 1500 – 1250 1250 × 100% = 250 1250 × 100% = 20% Direct process Profit and Loss

Vedanta Excel in Mathematics - Book 8 94 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Here, the remaining number of eggs = 300 – 25 = 275 C.P. of 300 eggs = 300 × Rs 8 = Rs 2,400 S.P. of 275 eggs = 275 × Rs 9.60 = Rs 2,640 ∴ Profit = S.P. – C.P. = Rs 2,640 – Rs 2,400 = Rs 240 Now, profit percent = Profit C.P. × 100% = 240 2400 × 100% = 10% Hence, his profit percent is 10%. Example 3: A stationer bought 12 gel pens for Rs 192 and sold 8 gel pens for Rs 176. If he/she sold the remaining pens at the same rate of cost price, find her profit or loss percent. Solution: Here, C.P. of 12 gel pens = Rs 192 C.P. of 1 pen = Rs 192 12 = Rs 16 Also, S.P. of 4 pens = 4 × Rs 16 = Rs 64 And, S.P. of 8 pens = Rs 176 Now, total S.P.= Rs 64 + Rs 176 = Rs 240 Then, Profit= S.P. – C.P. = Rs 240 – Rs 192 = Rs 48 Again, profit percent = Profit C.P. × 100% = 48 192 × 100% = 25% Hence, his/her profit percent is 25%. Example 4: Mr. Shakya purchased a gold item for Rs 32,400. If the gold is slightly devaluated and he sold it at 5% loss, find the selling price of the item. Solution: Here, C.P. of the gold item = Rs 32,400 Loss percent = 5% Now, Loss amount = 5% of C.P. = 5 100 × Rs 32,400 = Rs 1,620 ∴ S.P. of the gold item = C.P. – Loss = Rs 32,400 – Rs 1,620 = Rs 30,780 Hence, the selling price of the gold item is Rs 30,780. Example 5: Mrs. Tamang sold a cosmetic item for Rs 522 at 16% profit. At what price had she purchased the item? Direct process S. P. = (100 – 5)% of C. P. = 95% of Rs 32,400 = 95 100 × Rs 32,400 = Rs 30,780 Profit and Loss

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 95 Vedanta Excel in Mathematics - Book 8 Solution: Let she had purchased the cosmetic item for Rs x. So, C. P. = Rs x Now, C.P. = S.P. – profit or, x = Rs 522 – 16% of x or, x = Rs 522 – 16 100 × x or, x = Rs 522 – 4x 25 or, x + 4x 25 = Rs 522 or, 29x 25 = Rs 522 or, x = Rs 522 × 25 29 = Rs 450 Hence, the cost price of the cosmetic item is Rs 450. Example 6: Suntali bought a mobile phone and sold to Dhurmus at 10% loss. Dhurmus again sold it for Rs 6,750 at 25% profit. (i) Find the cost price of mobile for Dhurmus. (ii) Find the cost price of mobile for Suntali. Solution: (i) Let the cost price of the mobile for Dhurmus be Rs x. Now, C.P. = S.P. – profit or, x = Rs 6,750 – 25% of x or, x = Rs 6,750 – 25 100 × x or, x = Rs 6,750 – x 4 or, x + x 4 = Rs 6,750 or, 5x 4 = Rs 6,750 or, x = Rs 6,750 × 4 5 = Rs 5,400 ∴C.P. of Dhurmus = Rs 5,400 (ii) Again, C.P. of Dhurmus = S.P. of Suntali = Rs 5,400 Let C.P. of Suntali be Rs y. Now, C.P. = S.P. + Loss or, y = Rs 5,400 + 10% of y Direct process (100 + 16)% of C. P. = S.P. or, 116% of C. P. = Rs 522 or, C. P. = Rs 522 116% = Rs 522 116/100 = Rs 522 × 100 116 = Rs 450 Direct process (100 + 25)% of C. P. = S.P. or, 125% of C. P. = Rs 6,750 or, C. P. = Rs 6,750 125% = Rs 6,750 125/100 = Rs 6,750 × 100 125 = Rs 5,400 Direct process (100 – 10)% of C. P. = S.P. or, 90% of C. P. = Rs 5,400 or, C. P. = Rs 5,400 90% = Rs 5,400 90/100 = Rs 5,400 × 100 90 = Rs 6,000 Profit and Loss

Vedanta Excel in Mathematics - Book 8 96 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur or, y = Rs 5,400 + 10 100 × y or, y = Rs 5,400 + y 10 or, y – y 10 = Rs 5,400 or, 9y 10 = Rs 5,400 or, y = Rs 5,400 × 10 9 = Rs 6,000 ∴ C.P. of Suntali is Rs 6,000. EXERCISE 7.1 General Section - Classwork Let’s say and write the answers as quickly as possible. 1. a) C.P. = Rs 200, S.P. = Rs 224, profit = …………. b) C.P. = Rs 400, S.P. = Rs 380, loss = …………. c) C.P. = Rs 500, profit = Rs 60, S.P. = …………. d) C.P. = Rs 700, loss = Rs 150, S.P. = …………. e) S.P. = Rs 600, profit = Rs 100, C.P. = ………….. f) S.P. = Rs 800, loss =Rs 50, C.P. = ………….. 2. a) C.P. = Rs 100, S.P. = Rs 120, profit percent = ………….. b) C.P. = Rs 100, S.P. = Rs 85, loss percent = ………….. c) C.P. = Rs 100, profit percent = 18%, S.P. = ………….. d) C.P. = Rs 100, loss percent = 5% , S.P. = ………….. e) S.P. = Rs 110, profit percent = 10% C.P. = ………….. f) S.P. = Rs 88, loss percent = 12% C.P. = ………….. Creative Section - A 3. a) If C.P. = Rs 480, S.P. = Rs 528, find profit and profit percent. b) If C.P. = Rs 640, S.P. = Rs 608, find loss and loss percent. c) If C.P. = Rs 750, loss percent = 8% find S.P. d) If C.P = Rs 1,560, profit percent = 20%, find S.P. e) If S.P. = Rs 2,065, profit percent = 18%, find C.P. f) If S.P. = Rs 1,869, loss percent = 11% find C.P. Profit and Loss

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 97 Vedanta Excel in Mathematics - Book 8 4. a) A grocer purchased 80 kg of rice at Rs 90 per kg and sold at Rs 99 per kg. (i) Find his profit. (ii) Find his profit percent. b) A stationer purchased 5 dozen of pens at Rs 25 per piece and sold at Rs 24 per piece. (i) Find her loss. (ii) Find her loss percent. c) Sunayana bought second-hand bicycle for Rs 4,350 and spent Rs 450 to repair it. If she sold it for Rs 5,280, find her profit or loss percent. d) A trader bought 1000 glass tumblers at Rs 8 each. 100 glass tumblers were broken and she sold the rest at Rs 10 each. Find her gain or loss percent. e) A grocer purchased 200 eggs at Rs 9 each. 20 of them were broken and he/she sold the rest at Rs 9.50 each. Find her profit or loss percent. f) A book-seller bought 7 story books for Rs 525 and sold 4 books for Rs 360. If he sold the remaining books at the same rate of cost price find his/her profit or loss percent. g) A shopkeeper bought 10 kg of tomatoes for Rs 240 and sold 7 kg of it for Rs 210. If he sold the remaining quantity of tomatoes at the rate of cost price, find his gain or loss percent. 5. a) A trader bought a pen drive for Rs 460 and sold it at 5% profit. (i) Find the profit amount. (ii) Find the selling price of the pen drive. b) A Shopkeeper bought an electric fan for Rs 1,520 and sold it at 10% loss. (i) Find the loss amount. (ii) Find the selling price of the item. c) Parbati buys a mobile for Rs 6,300 and sells it to Laxmi at 15% profit. How much does Laxmi pay for it? d) Shiva bought a second-hand motorcycle for Rs 1,53,200. He spent Rs 1,200 to repair it and sold it to Bishnu at 8% loss. How much did Bishnu pay for it? e) A stationer bought 80 gel pens at Rs 30 each. 20 gel pens were stolen and he/she sold the remaining pens at 5% loss. Find the selling price of each gel pen. f) A retailer purchased 1000 LED bulbs at Rs 80 each. 100 bulbs were damaged and he sold the remaining bulbs at 21.5% profit. Find the selling price of each bulb. 6. a) A trader sold a bag for Rs 440 at 10% profit. Find the cost price of the bag. b) A shopkeeper sold a T-shirt for Rs 665 at a loss of 5%. At what price did she buy the T-shirt? c) A glassware seller bought 1000 glass tumblers. 100 of them were broken and she sold the remaining tumblers at Rs 80 each. If she made a loss of 4%, at what price did she purchase each tumbler? Creative Section - B 7. a) Anita bought a bag for Rs 4,000 and sold to Shova at 20% profit. Again, Shova sold it to Laxmi at 10% profit. (i) Find the S. P. of Anita (or C. P. of Shova). (ii) Find the S. P. of Shova (or C. P. of Laxmi) Profit and Loss

Vedanta Excel in Mathematics - Book 8 98 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) A dealer bought a laptop for Rs 45,000 and sold it to a retailer at 12% profit. The retailer again sold it to a customer at 25% profit. How much did the customer pay for the laptop? 8. a) Sunayana bought a mobile phone and sold to Pratik at 8% loss. Pratik sold it to Debashis for Rs 11,040 and made 20% profit. (i) Find the cost price of mobile to Pratik (ii) Find the cost price of mobile to Sunayana b) Sayad bought a watch and sold to Dakshes at 10% profit. Dakshes again sold it to Sahayata for Rs 6,050 at 10% profit. (i) Find the cost price of Dakshes. (ii) Find the cost price of Sayad. c) A wholesaler sold an electric heater to a retailer at 20% profit. The retailer sold it for Rs 2,052 to a customer at 5% loss. (i) How much did the retailer pay for it? (ii) How much did the wholesaler pay for it? It's your time - Project Work and Activity Section 9. 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. = ................... S.P. = ................... Find profit percent C.P. = ................... S.P. = ................... Find loss percent C.P. = ................... Profit percent = ................... Find S.P. C.P. = ................... Loss percent = ................... Find S.P. S.P. = ................... Profit percent = ................... Find C.P. S.P. = ................... Loss percent = ................... 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% C. P. = ............................ S. P. = ............................ Loss percent = 5% C. P. = ............................ S. P. = ............................ Profit and Loss

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 99 Vedanta Excel in Mathematics - Book 8 7.5 Discount and discount percent The price on the label of an article or product is called the marked price or list price. When a shopkeeper reduces the marked price (M.P.) of any articles and sells them to the customer, the amount of reduction in the price is called discount. A shopkeeper allows discount from the marked price (M.P.) of any article. When M.P. is considered as Rs 100 and discount is calculated from it, it is called discount percent. Discount percent = Discount amount = Discount % of M.P. Discount M.P. Selling price (S.P.) = M.P. – Discount × 100% Discount = M.P. – S.P. 7.6 Value Added Tax (VAT) Value Added Tax (VAT) is a tax which is charged at the time of consumption of goods and services. VAT is levied on the selling price of goods. So, VAT is changed as a certain percentage of S. P. VAT amount = VAT percent of S.P. S.P. with VAT = S.P. + VAT% of S.P. Facts to remember Calculation of Given Process S. P. M. P. and discount S. P. = M. P. – discount Discount amount M. P. and discount % Discount amount = Discount % of M. P. S. P. M. P. and discount% Discount amount = discount % of M. P., then, S. P. = M. P. – discount amount Direct process S. P. = (100 – discount)% of M. P. Discount % M. P. and discount Discount % = discount M.P. × 100% VAT amount S. P. and VAT % VAT amount = VAT % of S. P. VAT % S. P. and VAT VAT % = VAT S.P. × 100% S. P. with VAT S. P. and VAT % S. P. with VAT = S. P. + VAT % of S. P. Direct process S. P. with VAT = (100 + VAT)% of S. P. Profit and Loss

Vedanta Excel in Mathematics - Book 8 100 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: The marked price of a fan is Rs 1,620 and the shopkeeper allows some discount to sell it for Rs 1,539. a) Write the formula to find the discount amount. b) Calculate the discount amount. c) Express the discount in percent. Solution: a) The required formula to find discount amount = M.P. – S.P. where, M.P. is marked price and S.P. is selling price. b) Here, M.P. = Rs 1,620 and S.P. = Rs 1,539 ∴ Discount amount = Rs 1,620 – Rs 1,539 = Rs 81 Hence, the required discount amount is Rs 81. c) Again, discount percent = discount amount M. P. × 100% = 81 1620 × 100% = 5% Hence, the discount percent is 5%. Example 2: The marked price of a mobile is Rs 5,650 and the shopkeeper allows 20% discount to the customer. (i) Calculate the discount amount. (ii) How much should a customer pay for it? Solution: Here, M.P. of the mobile = Rs 5,650 Discount percent = 20% (i) Now, discount amount = 20% of M.P. = 20 100 × Rs 5,650 = Rs 1,130 (ii) Again, S.P. of the mobile = M.P. – Discount amount = Rs 5,650 – Rs 1,130 = Rs 4,520 Hence, the discount amount is Rs 1,130 and the customer should pay Rs 4,520. Example 3: The marked price of an article is Rs 3,200 and 10% discount is allowed on it. a) Find the discount amount b) Find the selling price of the article after discount c) If 13% VAT is charged on S.P., find the VAT amount. d) How much should a customer pay for it with VAT? www.geogebra.org/classroom/sk7dcats Classroom code: SK7D CATS Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Profit and Loss