The Leading Maths - 8

ARITHMETIC 45 EXERCISE 2.3 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is the rational number? Define it. (b) Why is rational number different from fraction? (c) Are 3 and 1 0 rational numbers? Why? 2. Which of the following are rational numbers? (a) 8 9 (b) –31 2 (c) 5 (d) 0 (e) 0.32 (f) – .0337 (g) 9 (h) – 1 3 3. State whether the following statements are true or false. (a) Every natural number is a rational number. (b) Each number of a set of integers is not rational number. (c) Whole numbers are rational numbers. (d) Even numbers are rational numbers. (e) Fractional numbers are not rational numbers. (f) Terminating decimal numbers are rational numbers. (g) Repeating decimal numbers are not rational numbers. 4. Circle ( ) the correct answers. (a) Which is a rational number? i. p q ; p, q ∈ N ii. p q ; p, q ∈ W iii. p q ; p, q ∈ Z iv. p q ; p, q ∈ Z, q ≠ 0 (b) Which is not a rational number? i. – 2 ii. 0 iii. 4 5 iv. 1 0 (c) What is the reciprocal of – 1 2 ? i. 2 ii. – 2 iii. – 1 iv. 1 (d) What is the additive inverse of 3 4 ? i. – 4 3 ii. 4 3 iii. – 3 4 iv. 3 4

46 The Leading Maths - 8 (e) Which is greater than 11 3 ? i. 7 2 ii. 10 3 iii. 15 4 iv. 15 7 5. Locate the following rational numbers in the number line. (a) 1 2 (b) 3 1 2 (c) 1 1 2 (d) 3 3 4 (e) 1 8 (f) 2 3 8 (g) 5 5 8 (h) 2 1 4 6. Locate the following rational numbers by the method of construction. (a) 2 5 (b) 3 3 5 (c) 2 4 5 (d) 4 1 7 (e) 2 3 7 (f) 5 4 9 (g) 8 7 10 (h) 5 9 10 7. Insert any three rational numbers between the following pairs of numbers. (a) 2 and 3 (b) 2 1 2 and 3 (c) 3 1 2 and 3 3 4 (d) 5 and 6 (e) 4 1 8 and 4 7 8 (f) 5 4 7 and 5 6 7 8. Find 5 rational numbers in between the following pairs of numbers. (a) –2 and –1 (b) 1 2 and 1 (c) 3 1 4 and 3 1 3 (d) 3 1 5 and 3 7 4 (e) 5 9 and 7 12 (f) 5 8 and 5 12 9. How many rational numbers are there between? (a) 1 and 2 (b) 1 2 and 1 4 (c) 3 6 and 7 10 (d) 6 and 6 1 2 (e) Between any two numbers a and b. 10. The weight of bags of two students Hari of John are 4 2 5 kg and 51 2 kg respectively. (a) Show the weight of their bags in a number line. (b) Find two rational numbers between these two weights. ANSWERS Consult with your teacher. Project Work 2.3 Show the rational numbers 3 4 , 5 8 and 12 7 in the number line. Prepare a report and present it in your classroom.

ARITHMETIC 47 2.4 Irrational Numbers At the end of this topic, the student will be able to: ¾ introduce the irrational numbers. Learning Objectives WARM-UP ” Tell the statement of the Pythagoras theorem. ” What is rational number? Tell some examples. ” What is the length of the hypotenuse of a right triangle with perpendicular (P) and base (b)? ” What is the length of hypotenuse of a right triangle having perpendicular 4 cm and base 3 cm? ” What is the length of AC in the given figure ? Is it rational number? A B 1 cm C 1 cm I. Introduction An irrational number is a number that cannot be expressed as a ratio of integers, i.e. as a fraction. Therefore, irrational numbers, when written as decimal numbers, do not terminate, nor do they repeat. For example, the number 15 has a decimal number 3.872983346........., but no finite number of digits can represent it exactly and it does not end in a segment that repeats itself infinitely often. The same can be said for any irrational number. The set of irrational number is denoted by Q and Q'. The discovery of irrational number is usually attributed to a Greek mathematician Pythagoras (570 - 495 BC), more specially to the Pythagore Mippasus of Metapontum (530 - 450 BC), who produced a proof of the irrationality of the square root of 2. II. Representation of Irrational Number in Number Line In the adjoining figure, take a point P on the number line XOX' such that OP = 1 unit and draw AP perpendicular to OX such that OP = PA = 1 unit. Join OA. Here, ∆OPA is a right-angled triangle.

48 The Leading Maths - 8 X' X A 1 1 1 0 2 O P Q 2 2 Now, according to Pythagoras’ theorem, OA2 = OP2 + PA2 [ h2 = b2 + p2 ] or, OA = OP2 + PA2 [∴ a2 = b ⇒ a = b] or, OA = 12 + 12 or, OA = 1 + 1 or, OA = 2 unit We assume that O as the center and OA as radius, and draw an arc which cuts the number line XOX’ at Q. Since OA and OQ are radii of the same arc, so OA = OQ = 2 unit. Similarly, we can compute 3, 5, 6, etc. from the number line. 0 1 A P B Q C R D S E UT x x 2 3 2 3 5 6 O But 1, 4, 9, 4, 9 16, etc. are not irrational numbers. They are rational numbers. Why? III. Irrational number as non-terminating and non-repeating decimal number Consider an irrational number as 15. Compute the square root 6 by division method. We find the decimal number as 3.872983346............ as shown in previous lesson. It has

ARITHMETIC 49 no finite number and no repeat of digits. So, the irrational number is non-terminating and non-repeating decimal number. IV. Properties of Irrational Numbers There are several properties of irrational numbers. Some of them are written below: 1. Sum of the two irrational numbers is irrational. Example: 2 + 2 = 2 2, which is the irrational number. 2. Subtraction of the irrational number may be rational or irrational. Example: 1) 2 – 2 = 0 is a rational number. 2) 2 – 18 = 2 – 3 2 = – 2 2 is on irrational number. 3. The product of two irrational numbers may be rational or irrational. Example: 1) 2 is irrational. Then 2 . 2 = 2, which is a rational number. 2) 2 and 3 are two irrational numbers. Then, 2. 3 = 2 × 3 = 6 , which is an irrational number. 4. An irrational number divided by an irrational number equals rational or irrational number. Example: 1) 2 2 = 1, which is the rational number. 2) 2 3 is an irrational number. 5. Irrational numbers are also infinite. i.e., irrational numbers are uncountable. Example: All numbers without perfect square... like 2, 3 , 20..... V. Real Numbers Areal numberis a value that represents a quantity along a line. The real numbersinclude all the rational numbers such as the integer − 5 and the fraction 3 4 and all the irrational numbers, such as 2 (1.41421356…, the square root of 2, an irrational algebraic number. It is the union of rational number (Q) and irrational number (Q). A set of real numbers is represented by R. A German mathematician Richard Dedeking (1831-1916) became the first person to define real numbers in 1872 AD.

50 The Leading Maths - 8 The Real Number Line The Real Number Line is like a geometric line. A point is chosen on the line to be the "origin". Points to the right are positive, and points to the left are negative. – 1 3 – � 2 20 9 Origin – 2.5 – 4 – 3 – 2 – 1 0 1 2 3 4 e � 1 2 A distance is chosen to be "1", then whole numbers are marked off: {1, 2, 3,...}, and also in the negative direction: {...,− 3, − 2, − 1} EXERCISE 2.4 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Define irrational number with an example. (b) Why is the irrational number different from the rational number? (c) What type of decimal number is called irrational number? (d) What is real number? Define it. 2. Circle ( ) the correct answers. (a) The irrational number is in the form of ................ i. {x : x ≠ p q ; p, q, ∈ N} ii. {x : x = p q ; p, q, ∈ Z} iii. {x : x = p q ; p, q, ∈ Z, q ≠ 0} iv. {x : x ≠ p q ; p, q, ∈ Z, q ≠ 0} (b) Which is an irrational number? i. 4 ii. 7 iii. 83 iv. 25 (c) Which is not an irrational number? i. 2 3 ii. 4 iii. 1 2 iv. 8 (d) Which set is a real number? i. Z ∪ Q ii. Q ∪ Q iii. Q ∪ N iv. W ∪ N

ARITHMETIC 51 (e) What is the relation between the sets of rational number and irrational number? i. equal ii. subset iii. disjoint iv. overlapping 3. Which of the following are irrational numbers? (a) 5 (b) – 3 (c) – 4 (d) 25 (e) 5 7 (f) 0 (g) 4.6 (h) 0.45 (i) 9 (j) 12 17 4. Locate the following points in a number line. (a) 5 (b) – 3 (c) 4 5 (d) – 7 (e) 2 3 5. (a) 5 and 6 are irrational numbers. Then verify that their product is also an irrational number. (b) 3 and 12 are irrational numbers. Then prove that their product is also a rational number. (c) 5 + 3 and 5 – 3 are irrational numbers. Prove that their sum and product are rational numbers. (d) Prove that the addition and multiplication of the numbers 3 + 5 and 3 – 5 are irrational and rational numbers respectively. (e) 2 3 and 12 are irrational numbers. Then verify that their quotient is a rational number. 6. The shorter two sides of right triangular paper are 5 cm and 8 cm. (a) Calculate the longer sides of the paper. (b) Is the length of its longer side rational or irrational number? Why? (c) Show all the length of the sides of the paper on a number line. ANSWERS Consult with your teacher. Project Work 2.4 1. Show the irrational numbers 7, 8, 10 and 11 in the number line in an A4 paper. 2. Construct the chart of the classification of real number system and show it in a Venn diagram in a chart paper.

52 The Leading Maths - 8 2.5 Scientific Notation At the end of this topic, the student will be able to: ¾ write the number into scientific notation and scientific notation into the number. Learning Objectives I. Introduction Notations of numbers have occurred and recurred in many scientific studies. The notions of approximation, decimal, significant figure and powers are found to be very helpful in dealing with them. Look at the following: 1 hour = 60 minutes 1 minute = 60 seconds 1 second = 10 mili-seconds 1 mili-second = 10 micro-seconds 1 micro-second = 10 nano-seconds 1 nano-second = 10 pico-seconds 1 pico-second = 10 femto-seconds 1 femto-second = 10 atto-seconds and so on. From this, we have 1 hour = 60 minutes = 60 × 60 seconds = 3600 seconds That is, 1 hour = 36 × 100 = 36 × 102 seconds = 36 × 102 × 10 mili-seconds = 36 × 103 mili-seconds = 36 × 103 × 10 micro-seconds = 36 × 104 micro-seconds = 36 × 104 × 10 nano-seconds = 36 × 105 nano-seconds = 36 × 105 × 10 pico-seconds = 36 × 106 pico-seconds = 36 × 106 × 10 femto-seconds = 36 × 107 femto-seconds = 36 ×107 × 10 atto-seconds = 36 × 108 atto-seconds Written in full, it looks like 3600000000 atto-seconds. Two other large numbers of common interest are: Distance of the Sun from the Earth = 14,95,00,000 km.

ARITHMETIC 53 Mass of an electron = 0.0000000000000000000000000000009109 kg To write numbers in such forms is a tedious job and is difficult to manipulate. We therefore write them using multiples of powers of 10 as shown below: 149500000 km. = 1495 × 105 km and 0.0000000000000000000000000000009109 = 9109 10000000000000000000000000000000000 = 9109 1034 = 9109 103 × 1031 = 9109 1000 × 1031 = 9.109 1031 = 9.109 × 10–31 kg. A more commonly used standard form or scientific form or scientific notation or standers index form is to write a given number in the form N × 10n , where N is a number lying between 1 and 10 (including 1 but not 10, i.e., 1≤ N < 10) and n is an integer, positive or negative. We then write, 3600000000 as 3.6 × 109 149500000 as 1.495 × 109 and 0.0000000000000000000000000000009109 = 9.109 × 10–31. The exponent of 10 is the number of places. The decimal point must be shifted to give the number in long form. A positive exponent shows that the decimal point is shifted that number of places to the right. A negative exponent shows that the decimal point is shifted that number of places to the left. Hence, scientific notation is the way of expressing the very large numbers or very small numbers in the form of decimal number in which one-digit before decimal point i.e. integral part of the decimal number is smaller than 10. On scientific calculators it is usually known as “SCI” or S ⇔ D display mode. A Greek mathematician Archimedes of Syracuse (287-212 BC) developed the concept behind scientific notation, and a French mathematician Rene Descartes (1596-1650) developed the modern system of scientific notation. Rene Descartes

54 The Leading Maths - 8 During multiplication we simply have to count number of zeros and during division we have to subtract the number of zeros. For example, (3.6 × 109 ) × (2 × 103 ) = (3.6 × 2) × (109 × 103 ) = (7.2) ×(109 × 103 ) = 7.2 × 109 + 3 = 7.2 × 1012 and 3.6 × 109 2 × 103 = 3.6 2 × 109 – 3 = 1.8 × 106 . CLASSWORK EXAMPLES Example : 1 Express the number 0.00000000 145 in scientific notation. Solution: Here, 0.00000000145 = 1.45 1000000000 = 1.45 109 = 1.45 × 10–9 Example : 2 Express the number 4.05 × 107 in general form. Solution: Here, 4.05 × 107 = 4.05 × 10000000 = 40500000.00 = 40500000 Example : 3 Simplify: (a) 2.61 × 105 + 1.56 × 104 (b) 3.72 × 104 – 1.3 × 10–3 (c) 8.1 × 108 × 3.2 × 10–3 (d) 6.8 × 109 – 2.1 × 105 250000 × 2.5 × 102

ARITHMETIC 55 Solution: Here, (a) 2.61 × 105 + 1.56 × 104 = 2.61 × 100000 + 1.56 × 10000 = 261000 + 15600 = 276600 = 2.766 × 100000 = 2.766 × 105 “Alternatively” (2.61 × 10 + 1.56 ) × 104 = (26.1 + 1.56) × 104 = 27.66 × 104 = 2.766 × 10 × 104 = 2.766 × 105 (b) 3.72 × 104 – 1.3 × 10–3 = 3.72 × 104 – 1.3 103 = 3.72 × 107 – 1.3 103 = 37200000 – 1.3 103 = 37199998.7 103 = 3.71999987 × 107 103 = 3.71999987 × 107 – 3 = 3.71999987 × 104 (c) 8.1 × 108 × 3.2 × 10–3 = 8.1 × 3.2 × 108 – 3 = 25.92 × 105 = 2.592 × 101 × 105 = 2.592 × 106

56 The Leading Maths - 8 (d) 6.78 × 109 – 2.1 × 105 250000 × 2.5 × 102 = (6.8 × 104 – 2.1) × 105 2.5 × 105 × 2.5 × 102 = 67997.9 × 105 6.25 × 105 × 102 = 67997.9 6.25 × 102 = 6.79979 × 104 6.25 × 102 = 6.79979 6.25 × 104 102 = 1.0879664 × 102 EXERCISE 2.5 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What do you mean by scientific notation of number? (b) What is scientific notation of 100? (c) What is scientific notation of 200500? (d) What is scientific notation of 0.000250? 2. Circle ( ) the correct answers. (a) Which is the scientific notation of 28000? i. 28 × 104 ii. 2.8 × 104 iii. 2.8 × 105 iv. 0.28 × 10–6 (b) What is the scientific notation of 1 day in second? i. 24 ii. 3.6 × 102 iii. 864 × 104 iv. 8.64 × 104 (c) Which is the correct scientific notation of 0.0000029? i. 2.9 × 106 ii. 2.9 × 10–6 iii. 29 × 10–5 iv. 0.29 × 10–7 (d) 2.3 × 102 + 3.2 × 102 = ................. i. 5.5 ×102 ii. 5.5 × 104 iii. 5.5 × 100 iv. 5.5 × 10–2

ARITHMETIC 57 3. Express the following numbers in scientific notation. (a) 580000000 (b) 3782492000 (c) 39999200000 (d) 0.2385 (e) 0.0000001 (f) 0.0000000000124992 4. Convert the following scientific notation of numbers in general form. (a) 2.0 × 105 (b) 2.177 × 108 (c) 1.99999 × 103 (d) 4.7 × 10– 4 (e) 5.0 × 10– 5 (f) 1.00002 × 10– 5 5. Add. (a) 2.3 × 105 + 1.9 × 105 (b) 4.5 × 108 + 6.8 × 107 (c) 4.2 × 106 + 1.3 × 104 (d) 5.8 × 107 + 3.4 × 103 (e) 8.9 × 104 + 1.1 × 10 – 2 (f) 7.8 × 105 + 1.0 × 10 – 6 6. Subtract. (a) 4.7 × 106 – 3.2 × 106 (b) 2.8 × 108 – 1.9 × 108 (c) 8.2 × 108 – 5.7 × 106 (d) 7.2 × 105 – 8.5 × 105 (e) 2.8 × 10– 9 – 3.2 × 10– 7 (f) 3.2 × 10– 5 – 4.8 × 10– 7 7. Multiply. (a) (2.5 × 105 ) × (3.2) (b) (6.2 × 102 ) × (3.5 × 103 ) (c) (8.5 × 104 ) × (4.3 × 10–2) (d) (9.2 × 107 ) × (3.5 × 10– 4) (e) (9.7 × 10– 7) × (8.2 × 10– 4) (f) (8.2 × 10– 4) × (1.0 × 10– 6) 8. Simplify. (a) 2.4 × 105 + 1.5 × 105 3.0 × 102 (a) 3.9 × 108 – 1.7 × 108 1.1 × 105 (c) 8.2 × 103 + 2.4 × 105 5.3 × 105 (d) 7.8 × 106 – 4.2 × 108 1.2 × 109 (e) (3.2 × 107 ) (2.8 × 102 2.2 × 105 – 1.7 × 105 (f) 2.1 × 102 (8.2 × 103 – 1.6 × 102 ) 2.2 × 104 + 1.1 × 103 9. Answer the following questions for the given situations: (a) If the distance between the Sun and Earth is 150000000 km, write it in scientific notation as meter. (b) If 1 km2 = 1000000000000 cm2 , write 1 cm2 in scientific notation. (c) The weight of a molecule is 1.7 × 10–8 mg. Convert it in general form.

58 The Leading Maths - 8 10. (a) The weight of a boy is 25 kg. i. Convert his weight into milligram (mg). ii. Express in scientific notation. (b) The road distance between Kathmandu and Pokhara is 204.8 km. i. Convert this distance into millimeter (mm). ii. Express it in scientific notation. (c) The length of amoeba is 2.3 micrometer. i. Express this length in kilometer. ii. Convert into scientific notation. (1 km = 1000000000 micrometer) ANSWERS 3. (a) 5.8 × 108 (b) 3.782492 × 109 (c) 3.99992 × 1010 (d) 2.985 × 10– 1 (e) 1 × 10– 7 (f) 1.24992 × 10– 11 4. (a) 200000 (b) 217700000 (c) 1999.99 (d) 0.00047 (e) 0.00005 (f) 0.0000100002 5. (a) 4.2 × 105 (b) 5.18 × 108 (c) 4.213 × 106 (d) 5.80034 × 107 (e) 8.9000011 × 104 (f) 7.8 × 105 6. (a) 1.5 × 106 (b) 9 × 107 (c) 8.143 × 108 (d) – 1.3 × 104 (e) 7.954 × 10– 10 (f) 1.74 × 10– 9 7. (a) 8 × 105 (b) 2.17 × 106 (c) 3.655 × 103 (d) 3.22 × 104 (e) 7.954 × 10– 10 (f) 1.74 × 10– 9 8. (a) 1.3 × 103 (b) 2.0 × 103 (c) 7.3 × 10– 1 (d) – 3.435 × 10– 1 (e) 1.792 × 105 (f) 7.31 × 10 9. (a) 1.5 × 1011 (b) 1.0 × 10–12 (c) 0.000000017 10. (a) i. 25000000 mg ii. 2.5 × 107 mg (b) i. 204800000 mm ii. 2.048 × 108 mm (c) i. 0.0000000023 km ii. 2.3 × 10–9 km Project Work 2.5 1. Search the distance of the moon and Mars from the Earth in the google. Convert the distances into scientific notation form and then find the farthest object among moon or Mars by how much percent. Also, find their total distance. 2. Search the weight electron, newton and proton of an atom. Convert their weights in scientific notation and then sum of the atom. Which is more heavier and which is lighter? Calculate and present them in your classroom.

ARITHMETIC 59 CHAPTER 3 RATIO AND PROPORTION Lesson Topics Pages 3.1 Ratio 61 3.2 Proportion 66 ” How to read a:b? What is its meaning? What are the other forms of a:b? ” In a bucket of milk, a man mixes 6 l of pure milk and 2 l of water. What is the ratio of water and pure milk in the mixture of milk in the bucket? ” Sohana looks TV for 1 hour 15 minutes and Suchan looks TV for 45 minutes only. Compare the time of looking TV by Sohana and Suchan. ” What is the ratio of weight of 1 kg of cotton and potatoes? Which is heavier and which is lighter? Discuss. WARM-UP S O H A N A S U C H A N

60 The Leading Maths - 8 3.1 Ratio At the end of this topic, the student will be able to: ¾ solve the problems related to the ratio of any two quantities. Learning Objectives I. Review on Ratio The simplest way to understand the meaning of a ratio is to define it as follows: A ratio of two numbers a and b is the quotient obtained by dividing the first quantity or number by the second quantity or number of the same units i.e. a b, b ≠ 0, where the ratio a b or a : b read as a upon b or a over b. It means it shows how many times one number contains another. For instances, each of the following statements (i) there exist one part oxygen and two parts hydrogen in water. (ii) three out of four parts of the Earth is filled with water and (iii) there exist 2 atoms of hydrogen in 1 atom of oxygen in one molecule of water, tells how one number is related to the other. We agree to denote the ideas contained in the above examples by (i) 1 2 and is read “ 1 upon 2 ” or “ 1 over 2 ” (ii) 3 4 and is read “ 3 upon 4 ” or “ 3 over 4 ” (iii) 2 1 and is read “ 2 upon 1 ” or “ 2 over 1 ”. Quite often we write it as a b , read “ a oblique b ”; and also as a:b, read “ a is to b ”. Of the two numbers in a ratio, the first one above the horizontal bar (or before the slash or colon) is called the numerator or antecedent and the later below the horizontal bar (or behind the slash or colon) is called the denominator or consequent. A ratio is called “ greater in equality ” , if the numerator or antecedent is larger and

ARITHMETIC 61 “ lesser in equality ”, if the numerator is smaller less than the denominator or consequent. We thus have an in-equality between two numbers when the numerator and denominators are not equal. Sometimes we form a ratio from two or more given ratios just by taking the ratio of the product the numerators to the product of the denominators. Such a ratio is called a compound ratio of the given ratios. Thus, (i) Given the ratios 1 3 and 3 5, then their compound ratio is 1 × 3 3 × 5 or 3 15 . (ii) Given the ratios a c and b d , then their compound ratio is ab cd or ab:cd. CLASSWORK EXAMPLES Example : 1 Find the compounded ratio of 3:5 and 2:9 and then reduce it to the lowest term: Solution: Here, the given ratios are 3 5 and 2 3. So, the compound ratio = 3 5 × 2 3 = 3 5 × 2 3 = 3 × 2 5 × 3 = 2 5 Example : 2 Find the ratio in the simplest form (a) Rs. 25 to Rs. 80 (b) Paisa 75 to Re. 1 (c) 250 gm to 1 kg Solution: (a) The ratio of Rs. 25 to Rs. 80 = Rs. 25 Rs. 80 = 5 16 (b) The ratio of paisa 75 to Re. 1 = 75 paisa Rs. 1 = 75 paisa 100 paisa = 3 4 (c) The ratio of 250 gm to 1 kg = 250 gm 1 km = 250 gm 1000 gm = 1 4 Do you know ! Which ratio is greater between 1 3 and 3 5 ? 5 < 9 ∴ 3 5 > 1 3 I know the ratio has not units.

62 The Leading Maths - 8 Example : 3 The ratio of two quantities is 3 : 5. If the first quantity is 18, find the other. Solution: Let the second quantity be x. Then, 3 5 = 18 x or, 3x = 90 or, x = 90 3 = 30 Hence, the required quantity is 30. Example : 4 Two quantities are in the ratio 5 : 6. If 15 is added to both numbers, the new ratio becomes 8 : 9, find the two numbers. Solution: Let the two numbers be 5x and 6x. Then, by the question, 5x + 15 6x + 15 = 8 9 or, 48x + 120 = 45x + 135 or, 48x – 45x = 135 – 120 or, 3x = 15 or, x = 15 3 ∴ x = 5 Hence, the two numbers are 5x = 5 × 5 = 25 and 6x = 6 × 5 = 30. EXERCISE 3.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. Circle ( ) the correct answers. (a) A ratio a:b is read as i. b over a ii. a times b iii. a over b iv. a plus b (b) Which is smaller ratio 1 3 or 1 2 or 3 4 or 1 4 ? i. 1 3 ii. 1 2 iii. 3 4 iv. 1 4

ARITHMETIC 63 (c) In which condition does a b define? i. a, b ∈ R, b ≠ 0 ii. a, b ∈ W, b ≠ 0 iii. a, b ∈ Z, b ≠ 0 iv. a, b ∈ N (d) The ratio of 30 kg and 50 kg is ..................... i. 8:5 ii. 5:3 iii. 3:5 iv. 3:8 (e) Which quantities have the ratio 3:4? i. 10 kg: 12kg ii. 20 kg: 15 kg iii. 24 gm:20 gm iv. 9 l:12 l (f) Which is greater ratio among 1 2, 3 4, 5 6 and 7 8? i. 1 2 ii. 3 4 iii. 5 6 iv. 7 8 (g) Which is lesser ratio among 12 5 , 13 7 , 15 7 and 21 9 ? i. 12 5 ii. 13 7 iii. 15 7 iv. 21 9 2. Write down the antecedent and consequent of the following ratios: (a) 1/4 (b) 1 3 (c) 4:1 3. A shopkeeper mixes 1 kg 500 gm of pebbles in 45 kg of rice. (a) In which ratio does he/she mix the pebbles in rice? Find it. (b) If he/she mixes 2 kg 200 gm of pebbles in 66 kg of rice, in which mixture does he/she mix more pebbles? Compare in ratio. 4. Sabanum mixes 8 l of water in 10 l of pure milk, but Sabina mix 8 l of water in 12 l of pure milk. (a) Find the ratio of mixture of water and pure milk on both mixtures. (b) Whose ratio of the mixture of water and pure milk is more? (c) Find the compound ratio of these two mixtures of water and pure milk. (d) If both of mixtures are mixed, what will be the ratio of new mixture of water and pure milk? Find it. 5. (a) Find the compounded ratio of 13 45 and 15 52 and then reduce it to the lowest term. (b) Find the compound ratio of 8 9 and 27 20 and then reduce it to its lowest term.

64 The Leading Maths - 8 6. Find the ratio of the first quantity to the second quantity in their lowest term. (a) Rs. 25 and Rs. 60 (b) 25 cm and 75 cm (c) 15 minutes and 45 minutes (d) 25 kg to 15 kg (e) 250 ml to 800 ml (f) Re.1 and paisa 75 (g) 45 minutes and 1 hour (h) 725 ml and 1 litre (i) 500 m and 3 km (j) 35 cm and 1 m (k) 120 cm to 1 m (l) Rs. 5 to 75 paisa. 7. (a) Two quantities are in the ratio 3 : 4. If the first quantity is 360, find the other. (b) The ratio between two quantities is 5 : 8. If the first quantity is 350, find the second quantity. 8. (a) The ratio of boys to the girls in a school is 5 : 4. If the number of girls is 320, find the number of boys in that school. (b) The ratio of length of two rectangular fields is 12 : 13. If the length of the first rectangle is 156m, find the length of the second rectangle. 9. (a) A map is drawn to the scale of 1: 1250. If the scaled length is 3 m, find the actual length in the field. (b) A map is drawn to the scale of 1 : 25000. If the scaled length is 4.5 m, find the actual length in the field. (c) A map is drawn to the scale 1 : 1250. If the real length in the field is 25 km, find the actual length in the map. 10. (a) Two quantities are in the ratio 5 : 7. If 20 is added to both of them, the new ratio becomes 7 : 9. Find the two numbers. (b) Two quantities are in the ratio 2 : 5. If 10 is subtracted from each of them, the new ratio becomes 1 : 4. Find the two numbers. 11. (a) The age of father to the son is 5 : 2. After 10 years the ratio will be 2 : 1. Find their ages. (b) The age of father to the son is 9 : 4. After five years it will be 2 : 1. Find their ages. (c) 10 years ago the age of father to the son was 11 : 6, now it is 13 : 8. Find their ages.

ARITHMETIC 65 12. Given that a : b = 3 : 4 and b : c = 5 : 7. (a) Express b in term of a. (b) Express b in terms of c. (c) Find the ratio a : c. 13. Repeat question 13 if a : b = 5 : 7 and b : c = 9 : 11. 14. (a) Three persons started a business with Rs. 5,60,000. If their share is in the ratio 5 : 7 : 8, find the money invested by each partner. (b) Three persons started a business with Rs.15,60,000. If their share is in the ratio 2 : 3 : 5, find the amount invested by each. 15. (a) The ratio of distance from Kathmandu to Damauli and Dumre to Pokhara is 7 : 3. If the distance from Kathmandu to Pokhara is 200 km, calculate the distance between Kathmandu to Damauli and Damauli to Pokhara. (b) Some milk of 42 l contains the pure milk and water in the ratio 5 : 2. Find the amount of pure milk and water in the milk. ANSWERS 2. (a) antecedent – 1, Consequent → 4 (b) antecedent →1, consequent → 3 (c) antecedent → 4, consequent → 1 3. (a) 1 : 30 (b) mixture of (b) 4. (a) 4 5, 2 3 . (b) Sabanam (c) 8 15 (d) 8 11 5. (a) 1 : 12 (b) 6 : 5 6. (a) 5 12 (b) 1 3 (c) 1 3 (d) 5 3 (e) 5 16 (f) 4 3 (g) 3 4 (h) 3 4 (i) 1 6 (j) 7 20 (k) 6 5 (l) 20 3 7. (a) 480 (b) 560 8. (a) 400 (b) 169 9. (a) 3750 m (b) 11250 m (c) 20 cm 10. (a) 50, 70 (b) 20, 50 11. (a) 50 yrs, 20 yrs (b) 45 yrs, 20 yrs (c) 65 yrs and 40 yrs 12. (a) 4a 3 (b) 5c 7 (c) 15:28 13. (a) 7a 5 (b) 9c 11 (c) 45 : 77 14. (a) 14,0000, Rs. 1,96,000, Rs. 2,24,000 (b) Rs. 3,12,000, Rs. 4,68,000, Rs. 7,80,000 15. (a) 140 km, 60 km (b) 30 l, 12 l

66 The Leading Maths - 8 3.2 Proportion At the end of this topic, the student will be able to: ¾ solve the problems related to proportion. Learning Objectives I. Introduction Four numbers a, b, c and d are said to be in proportion or proportional if the ratio of the first to the second is equal to that of the third to the fourth, i.e., if a b = c d. We write this also as a : b :: c: d and is read “a is to b is as c is to d ”. In particular, (i) 1, 3, 2, 6 are in proportion since 1 3 = 2 6, (ii) 5, 3, 10, 6 are proportional since 5 3 = 10 6 , The idea of proportion is exhibited diagrammatically by the given figure: If, instead of four quantities, we have three quantities in succession such that the ratio of the first to the second is the same as that of the second to the third, the three numbers are said to be in continued proportion. In particular, three quantities a, b and c are said to be in continued proportion if and only if a b = b c Similarly, a, b, c and d are in continued proportion if and only if a b = b c = c d. Proportions are of two types: (i) Direct proportion and (ii) Indirect proportion. II. Direct proportion Proportions are said to be direct if the ratio of one kind (i.e., first pair of numbers) is equal to the ratio of another kind (i.e., second pair of numbers). 5 3 1 3 2 6 10 6

ARITHMETIC 67 In particular, suppose the quantities of sugar and the prices are shown alongside. Here, the ratio of the quantities of sugar = 10 kg 15 kg = 2 5, and ratio of the prices = Rs. 350 Rs. 525 = 2 5. This is therefore the case in which prices of sugar and quantities of sugar are directly proportional or quantities of sugar and prices are directly proportional. An alternative way of saying this is “ Prices of sugar varies directly as the quantities of sugar ” or, “ Quantities of sugar varies directly as the prices of sugar ” Now, we quote some properties of direct proportions or variations. Suppose A is directly proportional to B and their corresponding values are as shown alongside. Then, we have (a) a1 a2 = b1 b2 (b) a2 a1 = b2 b1 (c) a1 b1 = a2 b2 (d) b1 a1 = b2 a2 (e) a1 b2 = a2 b1 (f) a2 b1 = a1 b2 (g) a1 = a2 × b1 b2 (h) b1 = b2 × a1 a2 (i) a2 = a1 × b2 b1 (j) b2 = b1 × a2 a1 CLASSWORK EXAMPLES Example : 1 If the four numbers 3, x, 8 and 16 are proportional, find x. Solution: Since 3, x , 8 and 16 are proportional so, 3 x = 8 16, or, 8x = 3 × 16 [ ∴ cross-multiplication] or, x = 3 × 16 8 6 Sugar Price 10 kg Rs. 350 15 kg Rs. 525 A B a1 b1 a2 b2

68 The Leading Maths - 8 Example : 2 Show that the three numbers 3, 15 and 75 are in continued proportion. Solution: Given numbers are 3, 15 and 75. Here, 3 15 = 1 5 and 15 75 = 1 5 Thus, 3 15 = 1 5 = 15 75 Hence, 3, 15 and 75 are in continued proportion. Example : 3 If the cost of x kg of sugar is Rs. 350 and the cost of 15 kg of sugar is Rs. 525, find x. Solution: Here, Sugar Price x kg (Required) Rs. 350 (Given) 15 kg (Given) Rs. 525 (Given) Sincerely, the changes are directly proportional, we have x 15 = 350 525 or, x = 15 × 350 525 = 10 or, x = 350 525 × 15 or, x = 350 35 = 10 That is, the required quantity is 10 kg. Example : 4 Suppose 1 man earns Rs. 850 per day. Find how much will 3 men earn per day. Solution: The ratio of man and his income is 1:850. Suppose the income of 3 men is Rs. x. Then, 1 850 = 3 x ∴ x = 850 × 3 = Rs. 2550. Hence, 3 men will earn Rs. 2250. So, the amount earned by 3 men per day = Rs.100 × 3 = Rs. 300.

ARITHMETIC 69 Example : 5 The cost of 1 dozen pencils is Rs. 60. Find the cost of 7 pencils. Solution: Here, the cost of 1 dozen (12 pencils) = Rs. 60. So, the cost of 1 pencil = Rs 60 ÷ 12 = Rs. 5 ∴ The cost of 7 pencils = Rs. 5 × 7 = Rs. 35. Example : 6 A man travels 10 km in 2 hours. How far will he go in 5 hours? Solution: Here, in 2 hours, a man travels 10 km. So, in 1 hour, the man will travel 10 km ÷ 2 = 5 km. So, in 5 hours, he will travel 5 km × 5 = 25 km. III. Indirect proportion Proportions are said to be indirect or inverse if the ratio of one kind (i.e., first pair of numbers) is equal to the inverse of the ratio of another kind (i.e., second pair of numbers). In particular, suppose a man has to cover 60 km. He takes 20 minutes when his speed is 3 km per min. He covers the same distance in 15 minutes if he increases his speed to 4 km per min. Here, Time taken Speed 20 mins. 3 km/hr 15 mins. 4 km/hr. Clearly, ratio of times taken = 20 mins 15 mins = 4 3 and ratio of speeds = 3 km/min 4 km/min = 3 4 In this case, The ratio of time taken = 3 4 = 1 4/3 = inverse of the ratio of the speeds. This is therefore the case of inverse or indirect proportion. We can put this in the following way also “ Time taken varies inversely or indirectly as the speed ” or, “ Speed varies inversely or indirectly as the time taken ”

70 The Leading Maths - 8 Now, we quote some properties of indirect proportion or variations. Suppose A is indirectly proportional to B and their corresponding values are as shown alongside: Then, we have (a) a1 a2 = b2 b1 (b) a2 a1 = b1 b2 (c) a1 b2 = a2 b1 (d) b2 a1 = b1 a2 (e) a1 b1 = a2 b2 (f) a2 b2 = a1 b1 (g) a1 = a2 × b2 b1 (h) b2 = b1 × a1 a2 (i) a2 = a1 × b1 b2 (j) b1 = b2 × a2 a1 CLASSWORK EXAMPLES Example : 7 If the four numbers 3, x, 16 and 8 are indirectly proportional, find x. Solution: Since 3, x , 16 and 8 are indirectly proportional, 3 x = 1 16 8 or, 3 x = 8 16 or, 8x = 3 × 16 or, x = 3 × 16 8 or, x = 6 Example : 8 A store has food for 100 persons for 30 days. If the number of persons and the number of days vary inversely, find how long the food will last for 75 persons. Solution: Here, Number of persons Food 100 (Given) 30 (Given) 75 (Given) x (Required) A B a1 b2 a2 b1

ARITHMETIC 71 Since, the changes are inversely proportional, we have 100 75 = x 30 or, x = 100 75 × 30 or, x = 40 That is, the required number of days is 40. Example : 9 1 man can do a work in 8 days. In how many days will 4 men do the same work? Solution: Here, 1 man can do the work in 8 days. We know that more men can do the work in less time. So, 4 men can do the work in 8 days ÷ 4 = 2 days. Example : 10 10 men can do 1 work in 4 days. In how many days, can 1 man do the same work? Solution: Here, 10 men can do 1 work in 4 days. We know that lesser number of men take more time. So, 1 man can do the work in 4 × 10 = 40 days. Example : 11 A store has food-stuff for 5 men for 15 days. How long the same foodstuff will last for 3 men? Solution: Here, a store has food-stuff for 5 men for 15 days. We know that for lesser number of men the food will last for more days. So, for 1 man the food stuff will last for 15 days × 5 = 75 days. Again, for more men, the same food-stuff will last for lesser number of days. So, the food-stuff will last for 75 days ÷ 3 = 25 days. Example : 12 A man runs at the rate of 5 km per hour. He covers a distance of 10 km in 2 hours. He changes his speed to 2 km per hour. Find how long he will take to cover the same distance. Solution: Here, Traveling at 5km/hr, a man covers a distance of 10 km in 2 hours. We know that slow speed will require more time.

72 The Leading Maths - 8 So, traveling at 1 km/hr, He covers the 10 km distance in 2 hours × 5 = 10 hours. We know that greater speed will require less time. So, traveling at 2 km/hr, He will cover the 10 km distance in 10 hours ÷ 2 = 5 hours. EXERCISE 3.2 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Define proportion with an example. (b) What type of ratio is called proportion? (c) Write the different parts of the proportion a, b, c, and d. (d) What is continued proportion? Illustrate with an example. 2. Circle ( ) the correct answer. (a) If a, b, c and d are proportional, b and c are called i. Extremes ii. Means ii. Antecedents iv. Consequents (b) If x 3 = 4 6, what is the value of x? i. 2 ii. 3 iii. 2 9 iv. 2 3 (c) If a, b, c and d are in proportion, which is the name of a and d? i. means ii. limits iii. extremes iv. ends (d) Which is proportion? i. 4, 6, 8, 12 ii. 1, 2, 4, 8 iii. 1, 2, 3, 4 iv. 2, 3, 4, 8 (e) Which is continued proportion? i. 4, 6, 12 ii. 2, 4, 6 iii. 6, 9, 18 iv. 4, 6, 9 3. Find x if each of the following four numbers are proportional. (a) 1, 3, x and 9 (b) 1, x, 2 and 10 (c) x, 3, 6 and 9 (d) 2, – 15, 6 and x

ARITHMETIC 73 4. Show that the following triples of numbers are in continued proportion. (a) 4, 16 and 64 (b) – 7, 35 and –175 5. Find the mean proportional (i.e., x) of the following triples of numbers. (a) 3, x, 48 (b) – 7, x, –175 6. (a) 10 gm of hydrogen combines with 160 gm of oxygen to form water. Find how much hydrogen combines with 32 gm of oxygen to form water. (b) The ratio of the length to the breadth of a room is 4 : 3. If the breadth is 6 m, find the length of the room. 7. (a) A gold ornament is made of 18 carat gold (i.e. ratio of gold to copper is 3 : 1). If the ornament is made of pure gold, find what carat gold it will be. (b) If father’s age is four times that of his son, find the age of the father when his son is 20 years old. 8. (a) Suppose 1 man earns Rs. 400 per day. Find how much will 3 men earn per day. (b) Suppose a man earns Rs. 200 per day. Find how much will he earn in 5 days. 9. (a) The cost of 6 tables is Rs. 850. Find the cost of 1 table. (b) The cost of 5 chairs is Rs. 3000. Find the cost of 1 chair. 10. (a) The cost of 1 quintal of steel rod is Rs 53,000. Find the cost of 1 kg of steel rod. (b) The cost of 50 bags of cement is Rs 42, 500. Find the cost of 1 bag of cement. 11. (a) A man travels 15 km in 3 hours. How far will he go in 5 hours? (b) A tap can fill a 200 litres tank in 10 minutes. How long will it take to fill a 1000 litres tank? 12. (a) 1 man does a work in 15 days. In how many days will 5 men do the same work? (b) A room requires 12 m of 2 m broad carpet. Find the length of the carpet required if the breadth of a carpet is 3 m. (c) A food store has food for 200 men for 15 days. How long will the same food stuff have last if there were 150 men? (d) A journey takes 4 hours if a man drives at a rate of 40 km/hr. What should be his speed if he wants to complete in 6 hours? (e) If the rate of interest is 5 percent per annum, Suman has to pay a certain amount of interest in 6 years. If the rate of interest is 6 per cent per annum, in how many years will he have to pay the same amount of interest?

74 The Leading Maths - 8 13. (a) A man can do a work in 6 days. In how many days will 3 men do the same work? (b) One man can reap a field in 6 hours. In how many hours can 3 men reap the same field? 14. (a) 8 men can do 1 work in 3 days. In how many days will 1 man do the same work? (b) 6 men can do 1 work in 2 days. In how many days will 1 man do the same work? 15. (a) A store has food-stuff for 3 men for 30 days. How long will the same food-stuff last for 5 men? (b) A garrison has provision for 10 men for 10 days. How long will it have last for 15 men? 16. (a) A man runs at the rate of 8 km per hour. He covers a distance of 24 km in 3 hours. He changes his speed to 6 km per hour. Find how long will he take to cover the same distance. (b) The speed of a car is 60 km per hour. A driver travels a certain distance in 5 hours. How long will it take to drive the same distance if the speed of his car is 40 km per hour. ANSWERS 3. (a) 3 (b) 5 (c) 2 (d) – 45 4. (a) 12 (b) 35 5. (a) 2 gm (b) 8 m 7. (a) 24 carat (b) 80 year 8. (a) Rs. 1200 (b) Rs. 4250 9. (a) Rs. 1000 (b) Rs. 600 10. (a) Rs. 530 (b) Rs. 650 11. (a) 25 km (b) 50 mins 12. (a) 3 days (b) 8 m (c) 20 days (d) 26 2 3 km/hr 13. (a) 2 days (b) 2 hrs 14. (a) 24 days (b) 12 days 15. (a) 18 days (b) 6 2 3 days 16. (a) 4 hrs (b) 7 hrs 30 mins. Project Work 3.2 List out the number of boys and girls from class 1 to 10. Find out the ratios of boys and girls, as well as that of girls and boys. Identify which classes have the same ratios. What are they called? Prepare a report and present in your classroom.

ARITHMETIC 75 CHAPTER 4 PROFIT AND LOSS Lesson Topics Pages 4.1 Profit and Loss 76 ” What are cost price and selling price of an article? Introduce. ” How much does a shopkeeper take profit or loss for the given jacket? ” Do you know what are the cost price and selling price of an article? ” In which condition is there profit? ” In which condition is there loss? ” What to do to take profit? WARM-UP Cost Price : Rs. 1430 Selling Price : Rs. 1716 Cost Price : Rs. 925 Selling Price : Rs. 1032

76 The Leading Maths - 8 4.1 Profit and Loss At the end of this topic, the student will be able to: ¾ solve the simple behaviour problems related to profit and loss. Learning Objectives I. Introduction In business arithmetic, the money that a shopkeeper pays for an article is called the cost price (CP) and the price of the article sold by the shopkeeper, or the price paid by a customer for buying the article is called the selling price (SP). There is certain gain or profit is the deal is the shopkeeper gets more than he pays for, and there is a loss if he gets less than, the amount he pays for. We use the following formula to work out the profit or less in the business. (i) When SP > CP, Profit or gain = selling price – cost price i.e., P = SP – CP or, SP = CP + P and CP = SP – P (ii) When SP > CP, Loss = cost price – selling price i.e., L = CP – SP or, SP = CP – L and CP = SP + L The related profit on CP of an article is called profit percentage and that if has of loss is called loss percentage. They are giving by profit percentage = profit CP × 100% = SP – CP CP × 100% and, Loss percent = loss CP × 100% = CP – SP CP × 100% Profit = P % of CP and, Loss = L% of CP.

ARITHMETIC 77 Furthermore, SP = CP + P = CP + P% of CP = CP(1 + P%) = CP 1 + P 100 = CP 100 + P 100 ∴ SP = CP 100 + P 100 and CP = SP × 100 100 + P Similarly, SP = CP – L = CP – L% of CP = CP (1 – L%) = CP 1 – L 100 = CP (100 – L) 100 ∴ SP = CP (100 – L) 100 and CP = SP × 100 100 – L Problems including the unit cost and the selling price If N articles are bought at the price of Rs. x per article and sold for Rs. y per article then the total cost price and selling price may be obtained as; Cost price (CP) = N × unit cost = N × x Selling price (SP) = N × unit selling price = N × y

78 The Leading Maths - 8 CLASSWORK EXAMPLES Example : 1 A fruit seller buys some apples for Rs. 45 per kg and sells them for Rs. 52 per kg. What is his/her profit in each kg ? Solution : Here, Cost price (CP) = Rs. 45 Selling price (SP) = Rs. 52 Now, we have, SP > CP, there is a profit. So, Profit (P) = SP – CP = Rs. 52 – Rs. 45 = Rs. 7 Example : 2 A grocers buys one crate of eggs at the rate of Rs. 5 each and sells them at the rate of Rs. 5.25 each. What is his profit in the selling of 1 crate of eggs ? Solution: 1 carrot = 30 pcs. ∴ Number of eggs (N) = 30 Unit cost price (x) = Rs. 5 ∴ Total cost price (CP) = N × x = 30 × Rs. 5 = Rs. 150 Unit selling price (y) = Rs. 5.25 ∴ Total selling price (SP) = N × y = 30 × Rs. 5.25 = Rs. 157.50 Here, SP > CP hence a profit is made. Therefore, Profit (P) = SP – CP = Rs. 157.50 – Rs. 150 = Rs. 7.50

ARITHMETIC 79 Example : 3 A fruit seller bought 400 oranges for Rs. 1200. 50 of them were rotten. He sold the remaining oranges for Rs. 3.25 each. Find his percentage of profit or loss. Solution: Here, Cost price (CP) = Rs. 1200 No. of good oranges (N) = 400 – 50 = 350 Unit selling price (y) = Rs. 3.25 ∴ Total selling price (SP) = N × y = 350 × Rs. 3.25 = Rs. 1137.50 Here, SP < CP. There is a loss in this deal. Therefore, Loss (L) = CP – SP = Rs. (1200 – 1137.50) = Rs. 62.50 Now, Loss% = Actual loss Cost price × 100% = Rs. 62.50 Rs. 1200 × 100% = 5.20% Example : 4 A shopkeeper pays Rs. 1200 for a watch. He puts the marked price 20% above the cost price and sells it with a discount of 10%. What is his profit in this transaction ? Solution: Here, Cost price (CP) = Rs. 1200 Marked price is 20% above cost price Hence, 20% of CP = 20 100 × Rs. 1200 \ Marked price = Cost price + 20% of CP = Rs. 1200 + Rs. 240 = Rs. 1440 Discount = 10% of marked price = 10 100 × Rs. 1440 = Rs. 144

80 The Leading Maths - 8 Hence, the selling price = Marked price – Discount = Rs. 1440 – Rs. 144 = Rs. 1296 ∴ Profit (P) = SP – CP = Rs. 1296 – Rs. 1200 = Rs. 96 EXERCISE 4.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is profit? (b) In which condition is the loss when an article is sold? (c) If a calculator is bought for Rs. a and is sold for Rs. b, how much is the profit in it? (d) If a pair of shoes bought for Rs. p and is sold for Rs. q, how much percent is loss in it? 2. Circle ( ) the correct answer. (a) When the cost price and selling price of an article are equal, what happens on profit or loss? i. profit ii. loss iii. neither profit nor loss iv. none of the above (b) When an article of the cost Rs. 200 is sold for Rs. 190, what is the profit or loss on it? i. loss ii. profit iii. neither profit nor loss iv. none of the above (c) A salesman sold a packet of biscuits of the cost Rs. 360 for Rs. 480. Which is profit amount? i. Rs. 20 ii. Rs. 840 iii. Rs. 120 iv. Rs. 220 (d) A shopkeeper sold a watch of the cost Rs. 840 for Rs. 790. Which is profit or loss amount? i. Profit Rs. 50 ii. Loss Rs. 50 iii. Profit Rs. 1630 iv. Loss Rs. 100

ARITHMETIC 81 (e) A girl sold a calculator of the cost Rs. 1000 for Rs. 1200. What percent profit does she get? i. 25% ii. 30% iii. 15% iv. 20% 3. Find the percentage of profit or loss in each of the following cases. Cost Price (CP) Selling Price (SP) (a) Rs. 500 Rs. 520.00 (b) Rs. 650 Rs. 725.00 (c) Rs. 300 Rs. 240 (d) Rs. 575 Rs. 500 (e) Rs. 1422 Rs. 1400 (f) Rs. 4000 Rs. 3333.33 4. (a) A stationery buys 1 dozen pencil for Rs. 30, and sells them for Rs. 3 each. i. Find the selling price of all pencils. ii. What is his loss or profit percentage on the transaction of these pencils? (b) A grocers bought 1 crate eggs for Rs. 150 and sells them at the rate of Rs. 5.50 each. i. Find the selling price of all eggs. ii. What is his loss or profit percentage on the transaction of selling eggs? 5. (a) An electric store buys 1000 bulbs for Rs. 40 each. 100 of them are broken. He sells the remaining for Rs. 45 each. i. How many bulbs are good in condition? ii. Find the selling price of good bulbs. iii. What is his profit or loss percentage? (b) A grocers buys 1000 eggs for Rs. 5000. 200 of the eggs were broken. He sells the remaining eggs for Rs. 5.50 each. i. How many eggs are not broken? ii. What is the selling price of the good eggs? Find it. iii. What is his loss or profit percentage ? 6. By selling a watch for Rs. 1260, a shopkeeper gains 20%. (a) Find the cost price of the watch. (b) Find the selling price if he wants to make a profit of 30%.

82 The Leading Maths - 8 7. By selling a transistor for Rs. 5000 a shopkeeper makes a profit of 25%. (a) What is his cost price? (b) If Rs. 200 was discounted on this selling price what would be his profit percentage? 8. (a) A stationer buys an article for Rs. 2000 and marks the price above 20%. i. Find the marked price of the article. ii. If he sells it at a discount of 10%, find its selling price. iii. Find his profit percentage. (b) A retailer makes the price of radio 25% above the cost price of Rs. 4000. i. Find the marked price of the ratio. ii. If he sell it at a discount of 15%, find its selling price. iii. What is his profit percentage? (c) A publisher marks the price of the books 60% above the cost price of Rs. 120000. i. Find its selling price. ii. If he provides 30% discount to the bookseller, find the selling price of the books. iii. What is his profit percentage? ANSWERS 3. (a) Profit 4% (b) Profit 11.54% (c) Loss 20% (d) Loss 13.04% (e) Loss 1.58% (f) Loss 16.67% 4. (a) i. Rs. 36 ii. Profit 20% (b) i. Rs. 165 ii. Profit 10% 5. (a) i. 900 ii. Rs. 40500 iii Profit 1.25% (b) i. 800 ii. Rs. 4400 iii. Profit 12% 6. (a) Rs. 1050 (b) Rs. 1365 7. (a) Rs. 4000 (b) 20% 8. (a) i. Rs. 2400 ii. Rs. 2160 iii 8% (b) i. Rs. 5000 ii. Rs. 4250 iii. 6.25% (c) i. Rs. 192000 ii. Rs. 134400 iii. 12% Project Work 4.1 Go to any shop and ask to the shopkeeper the cost price and selling price of any five articles. List them and find profit or loss and profit percent or loss percent for each articles. Also, find his total profit or loss percent on these five articles. Prepare a report for the above information and present it in your classroom.

ARITHMETIC 83 CHAPTER 5 UNITARY METHOD Lesson Topics Pages 5.1 Unitary Method 84 ” What is the cost of 10 pencils at Rs. 8 per piece? ” The cost of 50 note books is Rs. 5000. What is the cost of one note book? ” If the cost of 10 books is Rs. 1000, what is the cost of 6 books? ” What is the relation between the quantity of goods and their price? WARM-UP

84 The Leading Maths - 8 5.1 Unitary Method At the end of this topic, the student will be able to: ¾ solve the simple problems of unitary method based on direct and indirect variations. Learning Objectives I. Introduction Life is full of one or other kind of relation. It may be that between man and man, man and work, work and wage, time and work and so on. As concrete examples, we observe the following situations: i) If a man earns Rs. 500 per day, we may have to find how much the man can earn in 7 days. ii) If we need 4 teachers for every 80 students, we may have to find how many teachers are needed for 60 students. iii) If 4 men complete a work in 15 days, we should find how many men complete the same work in 20 days. In the first case, how one unit of one quantity is related to more than one unit of another; but in the second and third cases, more than one unit of one quantity is related to more than one units of another quantity. The unitary method is a technique in mathematics for solving a problem by finding the value of single unit, i.e., 1, (by dividing) and then finding the necessary value by multiplying the single unit value. As noted above, we meet many kinds of relations. But, we consider only those cases in which two quantities are related to each other such that i) One quantity increases when the other increases and vice versa. ii) One quantity increases when the other decreases and vice versa. II. Direct Variation or Proportion Look at the table of the income of a man in various days: No. of days 1 2 4 6 10 8 7 5 3 Income (Rs.) 500 1000 2000 3000 5000 4000 3500 2500 1500 In the above table, when the number of days is increased from the days 1 to 10, the income of the man is also increased from the amount Rs. 850 to Rs. 8500. But when

ARITHMETIC 85 the number of days is decreased from the days 10 to 3, the income of the man is also decreased from the amount Rs. 8500 to Rs. 2550. Show the above information in the graph as alongside; The above graph represents the straight line. If one quantity increases or decreases in a certain ratio, the other quantity also increases or decreases in the same ratio then such type of quantities are called in direct variation or proportion. Working Rule For two quantities related to each other such that one increases in the same ratio as the other: Given: 1 unit of a quantity is related to a unit of another quantity, Then, x units of the first is related to x × a units of the second. In the above table, the income of a man of 1 day is given. The number of days increased from 1 to 2, so did the price. That is, 1 day 2 days = Rs. 500 Rs. 1000 or, 1 2 = 500 1000 . We may therefore take the following as an alternative way of expressing the rule described above: Variables: A B First value: a (Given) b (Given) Second value: x (Required) d (Given) Then,a x = b d . By cross multiplication, we get x = a × d b x × b = a × d III. Indirect Variation or Proportion Look at the table of the time taken to work by certain men, No. of Men 1 2 4 6 10 8 7 5 3 Days 60 30 15 10 6 7 1 2 8 4 7 12 20 500 1 2 3 4 5 6 7 8 9 10 1000 1500 2000 2500 3000 3500 4000 5000 4500 Number of days Income in Rs

86 The Leading Maths - 8 In the above table, when the number of men is increased from the days 1 to 10, the working days is decreased from 60 to 6. But when the number of men is decreased from the days 10 to 3, the working days is increased from 6 to 20. Show the above information in the graph as alongside; The above graph represents the curve line. If one quantity increases or decreases in a certain ratio, the other quantity decreases or increases in the same ratio then such type of quantities are called in indirect variation or proportion. Working Rule For two quantities related to each other such that one increases in the same ratio as the other decreases: Given: 1 unit of a quantity is related to a unit of another quantity, Then, x units of the first is related to a x units of the second. In the above table, the working days by certain men is given. The number of men increased from 1 to 2, so the working days decreased from 60 days to 30 days. That is, 1 man 2 man = 30 days 60 days or, 1 2 = 30 60 It is reverse or reciprocal relation among number of men and the working days. We may therefore take the following as an alternative way of expressing the rule described above: Variables A B First value a (Given) d (Given) Second value x (Required) b (Given) Then, a x = d b . By cross multiplication, we get x = a × b d x × d = a × b 6 1 2 3 4 5 6 7 8 9 10 12 18 24 30 36 42 49 60 54 Number of Men Days

ARITHMETIC 87 CLASSWORK EXAMPLES Example : 1 If the cost of 15 kg of sugar is Rs. 525, find the cost of 25 kg of sugar. Solution: Here, The cost of 15 kg of sugar is Rs. 525. or, The cost of 1 kg of sugar is Rs. 525 15 . or, The cost of 25 kg of sugar is Rs. 525 × 25 15 = Rs. 875. Alternatively Let the required cost of 25 kg of sugar be Rs. x. Then Quantity of Sugar (kg) Cost of Sugar (Rs.) 15 525 25 x The quantity of Sugar (kg) and the cost of Sugar (Rs.) are in direct variation. So, 15 25 = 525 x or, x = 525 × 25 15 or, x = 875 ∴ The cost of 25 kg of sugar is Rs. 875. Example : 2 A store has food for 100 persons for 30 days. Find how long the food will last for 75 persons. Solution: Here, A store has food for 100 persons for 30 days. or, The store has food for 1 person for 30 × 100 days. ∴ The store has food for 75 persons for 30 × 100 75 = 40 days. Alternatively Number of persons Food 100 (Given) 30 (Given) 75 (Given) x (Required)

88 The Leading Maths - 8 Since, the changes are inversely proportional, we have 100 75 = x 30 or, x 30 = 100 75 or, x = 100 75 × 30 = 40 That is, the required number of days is 40. EXERCISE 5.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. Identify whether each of the following is a problem of direct or indirect variation. (a) Number of books and the cost of books (b) Length and breadth of the rectangle (c) Number of laborers and working days (d) Capacity of pipe to fill water and time (e) Apple trees and number of apples (f) Goats and their legs 2. Circle ( ) the correct answer. (a) The cost of 5 pens is Rs. 100. What is the cost of 1 pen? i. Rs. 50 ii. Rs. 25 iii. Rs. 20 iv. Rs. 105 (b) If the cost of 10 pencils is Rs. 100, the cost of 7 pencils is ................ i. Rs. 70 ii. Rs. 10 iii. Rs. 700 iv. Rs. 180 (c) If 1 man can do a work in 5 hours, in how many hours can 5 men do the same work? i. 25 hours ii. 5 hours iii. 1 hour iv. 2 hours (d) A girl can do 1 3 work in 2 days. In how many days can she do 1 work? i. 1 6 days ii. 3 days iii. 6 days iv. 3 2 days (e) If 2 boys can do 1 2 work in 3 days, how many boys can do 2 work in 6 days? i. 3 boys ii. 4 boys iii. 5 boys iv. 6 boys

ARITHMETIC 89 3. (a) If 1 man earns Rs. 100 per day, find how much will 3 men earn per day? (b) If the cost of a pen is Rs. 65, find the cost of 1 dozen pens. 4. (a) If 1 man can do a work in 8 days, in how many days will 4 men do the same work? (b) A pipe can fill a water tank in 40 minutes. How many pipes can fill the same tank in 10 minutes? 5. (a) The cost of 4 books is Rs. 200. Find the cost of 1 book. (b) The cost of 1 dozen copies is Rs. 600. Find the cost of 1 copy. 6. (a) The cost of 1 quintal of steel rod is Rs. 53,000. Find the cost of 1 kg of steel rod. (b) The cost of 1 gross of pencils is Rs. 864. Find the cost of 1 pencil. 7. (a) A man travels 10 km in 2 hours. How far will he go in 5 hours ? (b) A tap can fill a 200 liters tank in 10 minutes. How long will it take to fill a 1000 liter tank? 8. (a) A store has food-stuff for 5 men for 15 days. How long will the same food-stuff last for 3 men? (b) A food store has food for 200 men for 15 days. For how many men will the same food stuff be enough for 75 days ? 9. (a) The weight of 56 books is 7 kg. i. What is the weight of 90 such books? ii. How many such books do weigh 7.5 kg? (b) Shova bought a dozen pens for Rs. 144. i. Find the cost of 15 such pens. ii. How many pens can be bought for Rs. 84? 10. (a) The bus fare of 2 students for traveling 5 km is Rs. 100. i. Find the bus fare for 1 km distance per student. ii. Find the bus fare of 3 students for traveling 10 km. (b) The speed of a car is 60 km per hour. i. What distance does a driver in 5 hours? ii. How long will it take to drive the same distance if his speed is 40 km per hour?

90 The Leading Maths - 8 11. (a) 2 men can do 3 works in 9 days. i. Find in how many days 1 man will do 1 work. ii. Find in how many days 3 men will do 2 works. (b) A man runs at the rate of 8 km per hour. i. What distance does he cover in 3 hours? ii. He changes his speed to 6 km per hour. Find how long he will take to cover the same distance. 12. (a) In a camp, there are provisions for 400 persons for 23 days. i. If it is used for 1 person, how long does it take to complete? ii. If 60 more persons join the camp, find the number of days the provision will last. (b) 10 workers complete a work in 12 days working 4 hours daily. i. In how many days will 1 worker complete it working 1 hour daily? ii. In how many days will 8 workers complete the same work working for 6 hours daily? ANSWERS 3. (a) Rs. 300 (b) Rs. 780 4. (a) 2 days (b) 4 pipes 5. (a) Rs. 50 (b) Rs. 50 6. (a) Rs. 53 (b) Rs. 6 7. (a) 25 km (b) 50 mins 8. (a) 25 days (b) 40 men 9. (a) i. 11.25 kg ii. 60 books (b) i. Rs.180 ii. 7 pens 10. (a) i. Rs. 10 ii. Rs. 300 (b) i. 300 km ii. 7.5 hrs 11. (a) i. 6 days ii. 4 days (b) i. 24 km ii. 4 hrs 12. (a) i. 9200 days ii. 20 days (b) i. 480 days ii. 10 days Project Work 5.1 You can go to a shop neighboring your home and buy five goods that have more than one piece on each goods. If you need more such types of goods, estimate the cost for them. Prepare a report and present it in your classroom. SN Particulars Quantity Cast in Rs. Calculating process 1. Biscuits buying 12 pa. 200 Biscuits needed 15 pa. x 2. ........... buying ....... pcs ........ .......... needed ....... pcs y

ARITHMETIC 91 MRP : Rs. 1000 Discount : Rs. 200 CHAPTER 6 SIMPLE INTEREST Lesson Topics Pages 5.1 Unitary Method …… ” What is percentage? What percent is 2 parts out of 5 parts? ” What is the percent of 9 4 ? ” What is the value of 10% of Rs. 70? ” What percent of Rs. 1000 is Rs. 200? ” What is the meaning of 12%? ” How much is 12% of Rs. 50000 in one year? Similarly, how much count is it for 3 years? WARM-UP MRP : Rs. 70 Discount : 10%

92 The Leading Maths - 8 6.1 Simple Interest At the end of this topic, the student will be able to: ¾ solve the simple problems related to simple interest. Learning Objectives I. Introduction A New Year day is a happy day. On that day, we give gifts and get gifts. We have thus an exchange: a give and take process. In this process, neither the gift is given back nor any money is paid. But, putting (or depositing) in or taking out (or drawing or borrowing) some money from a bank (or merchant) is completely different. In such a case, we shall get or have to pay back the money after some time. At the end of certain time, we get or have to pay some additional money. The amount of additional money depends upon the total money deposited or borrowed or lent and is fixed in advance for every unit or 100 units of money. All these are very common in our life situation. We therefore need to make this workable. ACTIVITY - 1 Anjali gives Rs. 50000 to Bimal as loan by paying 15% interest per year. How much more amount does Bimal bay to Anjali in one year? And how much amount dose he pay back to her in total in this year? How much more amount does Bimal pay to her in two years and three years separately? How much total amount does he pay in 2 years and 3 years separately. For this, we need the following terminology. Principal : The money lent or borrowed is called the principal. It is denoted by P. Interest : The additional money paid or charged for using some-one’s money is called the interest. It is denoted by I.

ARITHMETIC 93 Amount : The sum total of the principal and amount is called the amount. It is denoted by A. Rate : Rate is the interest on every 100 units of money for 1 year. It is denoted by R. Time : Interest on a certain sum of money is reckoned on for how long (i.e., time) the money is lent or borrowed. It is denoted by T. The interest on a given sum of money is said to be simple if it is calculated on the basis of the original sum lent or borrowed only. It may be paid or collected at regular intervals (monthly, half-yearly, yearly or the like), but it is not added to the principal at the end of no interval of time. Simple interest increases at a constant or fixed ratio with each of the following quantities: (a) Principal, P (b) Time, T and (c) Rate, R So, we can calculate the simple interest directly from the definition (or from the first principal) by using the method of unitary method (direct proportion). We may first derive the following formula: I = P × T × R 100 by using the technique of unitary method. To derive the formula, we start with the notations; Principal = P, Time = T, Rate = R and Interest = I. By definition or from the first principle, Interest on 100 units of money for 1 year = R So, interest on 1 unit of money for 1 year = R 100 So, interest on P units of money for 1 year = R 100 × P = P × R 100 So, interest on P units of money for T years = P × R 100 × T = P × T × R 100 . That is, I = P × T × R 100 From this, we first derive the formulae for P, T, and R.

94 The Leading Maths - 8 Since, I = P × T × R 100 So, 100 × I = P × T × R, or, P × T × R = 100 × I Hence, P = 100 × I T × R , T = 100 × I P × R and R = 100 × I P × T . We further know that A = P + I = P + P × T × R 100 = P 1 + T × R 100 = P 100 + TR 100 or, P = 100 × A 100 + T × R. and P = A – I , I = A – P. Remarks (i) Use, as far as possible units such as Rs, $, £, etc. and not their smaller denominations while computing the principal, interest and amount, (ii) Take year as the unit of time or convert the time into year, (iii) Express rate in percent per annum ( p.a.) , and (vi) Take 1 year = 12 months, 1 year = 52 weeks 1 year = 365 days and 1 week = 7 days 1 year = 366 days in the leap year. CLASSWORK EXAMPLES Example : 1 Determine the unknown quantities: (a) P = Rs. 500, T = 3 years, R = 4 %, I = ? (b) T = 5 years, R = 2 %, I = Rs. 50, P = ? (c) R = 3 %, I = Rs. 30 , P = Rs. 200, T = ? (d) I = Rs. 75, P = Rs. 500 , T = 5 years, R = ?