145 Oasis School Mathematics-8 145 Worked Out Examples Example: 1 Add: 10112 and 11112 Solution: Hence, 10112 + 11112 = 110102 Example: 2 Add: 1110112 and 1111112 Solution: ∴ 1110112 + 1111112 = 11110102 Example: 3 Add the binary numbers: 11011012 , 10101012 and 10110102 Solution: Hence, 1011012 +10101012 +10110102 = 1000111002 Steps • 1 + 1 = 2 = 102 : set down 0, carry 1 • 1 + 1 + 1 = 3 = 112 : set down 1, carry 1 • 1 + 0 + 1 = 2 = 102 : set down 0, carry 1 • 1 + 1 + 1 = 3 = 112 : set down 1, carry1 Carry overs → First number → 1 0 1 12 Second number → + 1 1 1 12 1 1 0 1 02 1 1 1 Steps • 1 + 1 = 2 and 210 = 102 . : set down 0 and carry 1. • (1) + 1 + 1 = 3 and 310 = 112 . : set down 1 and carry 1. • (1) + 0 + 1 = 2 and 210 = 102 .: set down 0 and carry 1. • (1) + 1 + 1 = 3 and 310 = 112 . : set down 1 and carry 1. • (1) + 1 + 1 = 3 and 310= 112 . : set down 1 and carry 1. • (1) + 1 + 1= 3 and 310 = 112 . : set down 1 and carry 1. • We finally set down 1. Carry overs → 111011 2 + 1 1 1 1 1 1 2 11 1 1 0 1 0 2 1 1 1 1 1 1 Carry overs → 1 1 0 1 1 0 12 1 0 1 0 1 0 12 + 1 0 1 1 0 1 02 1 0 0 0 1 1 0 02 1 1 1 1 1 1
146 Oasis School Mathematics-8 Example: 4 Subtract: 111012 from 11100112 . Solution: 1 1 1 0 1 12 – 1 0 1 0 12 1 0 0 1 1 02 0 10 ∴ 1110112 – 101012 = 1001102 Example: 5 Simplify: 110112 + 10012 – 11112 Solution: ∴ 110112 + 10012 = 1001002 Again, ∴ 1001002 – 11112 = 101012 Hence, 11011+10012 – 11112 = 101012 Exercise 9.2 1. Carry out the addition of the following binary numbers. (a) 112 + 112 (b) 1012 + 1112 (c) 11002 + 102 (d) 111112 + 110112 (e) 1000012 + 1011012 (f) 10101012 + 10000012 2. Carry out the addition of the following binary numbers. (a) 111102 + 101112 + 100012 (b) 10101012 + 10110102 + 11011012 Steps • 1 – 1 = 0, set 0 down. • 1 – 0 = 1, set 1 down. • As 0 < 1, borrow 1 from higher places then 10-1=1, set 1 down. • 0 – 0=0, set 0 down. • 1–1=0, set 0 down. • set 1 down. 1 1 0 1 12 + 1 0 0 12 1 0 0 1 0 02 1 0 1 1 1 0 0 1 0 02 – 1 1 1 12 1 0 1 0 12 1 1 10 1 10
147 Oasis School Mathematics-8 147 3. Carry out the subtraction of the following binary numbers: (a) 1112 – 102 (b) 11002 – 112 (c) 100002 – 12 (d) 1111112 – 1000012 (e) 1001012 - 110112 4. Simplify: (a) 11102 + 10102 (b) 111012 – 11002 + 110012 (c) 1111012 – 10012 + 11002 (d) 1111112 +101112 – 101012 Answer 1. (a) 1102 (b) 11002 (c) 11102 (d) 1110102 (e) 10011102 (f) 100101102 2. (a) 10001102 (b) 1000111002 3. (a) 1012 (b) 10012 (c) 11112 (d) 111102 (e) 10102 4. (a) 110002 (b) 1010102 (c) 10000002 (d) 10000012 9.5 Addition and Subtraction of Quinary Numbers a. Addition of Quinary Numbers In quinary system, when the sum of the digits becomes 5 or more, we begin to take carry 1 five, 2 fives, 3 fives, etc. to the higher places. As there are five numerals 0, 1, 2, 3 and 4 in quinary system, the following addition table gives us the rules of addition of quinary numbers. + 0 1 2 3 4 1 1 2 3 4 105 2 2 3 4 105 115 3 3 4 105 115 125 4 4 105 115 125 135 For example, Add: 344105 + 323435 Remember ! 1 + 0 = 1 3 + 0 = 3 1 + 1 = 2 3 + 1 = 4 1 + 2 = 3 3 + 2 = 105 1 + 3 = 4 3 + 3 = 115 1 + 4 = 105 3 + 4 = 125 2 + 0 = 2 4 + 0 = 4 2 + 1 = 3 4 + 1 = 105 2 + 2 = 4 4 + 2 = 115 2 + 3 = 105 4 + 3 = 125 2 + 4 = 115 4 + 4 = 135 Steps: • 0 + 3 = 3. : set 3 down. • 1 + 4 = 5 and 510 = 105 . : set 0 down and carry 1. • 4 + 3 +( 1) = 8 and 810 = 135 . : set 3 down and carry 1. • 4 + 2 + (1) = 7 and 710 = 125 . : set 2 down and carry 1. • 3 + 3 + (1) = 7 and 710 = 125 . : set 2 down and carry 1. Carry over → 111 3 4 4 1 0 5 + 3 2 3 4 3 5 1223035 ⇒ 344105 + 323435 = 1223035
148 Oasis School Mathematics-8 b. Subtraction of Quinary Numbers In quinary system, when the greater quinary numeral is subtracted from the smaller one, we borrow 5 from the previous digit at higher place value to the digit at lower place value. Let's see the following example, For example, Subtract: 32105 – 21145 Worked Out Examples Example: 1 Simplify: 2 3 4 1 25 + 102145 Solution: ∴ 234125 + 102145 = 341315 Example: 2 Carry out the subtraction of the following quinary numbers: 231435 – 204145 . Solution: ∴ Required difference = 22245 105 -1 = 4 115 -1=105 125 -1=115 135 -1=125 105 -2=3 115 -2=4 125 -2=105 135 -2=115 105 -3=2 115 -3=3 125 -3=4 135 -3=105 105 -4=1 115 -4=2 125 -4=3 135 -4=4 3 2 1 05 – 2 1 1 45 1 0 4 15 ∴ Required difference = 10415 Steps 1 10 10 • As 0 < 4, we borrow 1 from higher place then 105 – 4=1, set 1 down. • As 0 < 1, borrow 1 from higher place105 –1=4, set 4 down. • 1 – 1 = 0. So, set 0 down • 3 – 2 = 1. So, set 1 down. Remember ! 1 1 2 3 4 1 25 1 0 2 1 45 3 4 1 3 15 + Steps • 2 + 4 = 115 , set 1 down, carry 1. • 1 + 1 + 1 = 3, set 3 down. • 4 + 2 = 115 , set 1 down, carry 1. • 1 + 3 = 4, set 4 down. • 2 + 1 = 3, set 3 down. Steps • 3 < 4, borrow 1 from higher place then 135 – 4 = 4, set 4 down. • 3 – 1 = 2, set 2 down. • 1 < 4, borrow, 1 from higher place 115 – 4 = 2, set 2 down • 2 – 0 = 2, set 2 down. 2 3 1 4 35 2 0 4 1 45 2 2 2 45 2 11 3 13 –
149 Oasis School Mathematics-8 149 Example: 3 Carry out the addition of the following quinary numbers (i.e. base-5 numbers). 324105 + 123445 + 310015 Solution: Exercise 9.3 1. Carry out the addition of the following quinary numbers (base 5 numbers). (a) 13425 + 4135 (b) 100045 + 4325 (c) 444445 + 111115 (d) 123445 + 343335 (e) 33335 + 10045 2. Carry out the subtraction of the following quinary numbers (base 5 numbers). (a) 3315 – 1315 (b) 43205 – 32235 (c) 20325 – 3435 (d) 40005 – 235 3. Simplify: (a) 203145 + 112035 – 21135 (b) 32015 – 21105 + 314215 (c) 421305 + 220335 – 100345 (d) 1102145 – 20125 – 312035 Answer 1. (a) 23105 (b) 104415 (c) 1111105 (d) 1022325 (e) 43425 2. (a) 2005 (b) 10425 (c) 11345 (d) 34225 3. (a) 244045 (b) 330125 (c) 1041245 (d) 214445 Steps (i) 0 + 4 + 1 = 105 . So, set 0 down and carry 1. (ii) 1 + 4 + 0 + (1)=115 . So, set 0 down and carry 1. (iii) 4 + 3 + 0 + (1)=135 . So, set 3 down and carry 1. (iv) 2 + 2 + 1 + (1)=115 . So, set 1 down and carry 1. (v) 3 + 1 + 3 + (1)=135 . So, set 3 down and carry 1. (vi) Set down 1. Carry overs → 1 1 1 1 3 2 4 1 05 + 1 2 3 4 45 + 3 1 0 0 15 1 3 1 3 1 05 ∴ Required sum = 1313105
150 Oasis School Mathematics-8 Integers 10.1 Integers Introduction: We have already learnt about natural numbers. Let us consider the set of natural numbers. N = {1, 2, 3, 4, 5, ......... } Natural numbers are closed under addition and multiplication. i.e. sum and product of two natural numbers is also a natural number. Let us be clear with an example. 3 and 5 are natural numbers. 3 + 5 = 8 which is a natural number. 3 × 5 = 15 which is also a natural number. Let us see the case of subtraction. (3 – 5) = – 2 which is not a natural number. i.e. natural numbers may not always be closed under subtraction. So, in order to make the operation of subtraction meaningful, negative numbers and zero are introduced. The set of natural numbers together with their negatives including zero are called integers. The set of integers is denoted by Z. ∴ Z = { ..........., –4, –3, –2, –1, 0, 1, 2, 3, 4, ......... } Z+ = {1, 2, 3, 4, ........... } = a set of positive integers. Z- = {–1, –2, –3, –4,........... } = a set of negative integers. Integers are closed under addition, subtraction and multiplication. Unit 10 Integers
151 Oasis School Mathematics-8 151 Laws of addition on integers -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 Take any three integers – 3, 4 and 1 –3 + 4 = 1, which is an integer. 4 + 1 = 5, which is an integer. (–3 + 1) + 4 = 2, which is an integer. Hence, we conclude that the sum of integers is again an integer. If a, b and c be any three integers then a + b is also an integer and a + b + c is also an integer. This property is called closure property. Again, – 3 + 4 = 1, 4 + (–3) = 1 i.e, – 3 + 4 = 4 + (–3) ∴ a + b = b + a This property is called commutative property. Again, (–3 + 4) + 1 = 2 – 3 + (4 + 1) = 2 i.e. (–3 + 4) + 1 = – 3 + (4 +1) ∴ (a + b) + c = a + (b + c) This property is called associative property. Again, – 3 + 3 = 0 3 + (–3) = 0 i.e. a + (–a) = 0 This property is the inverse property. Here, –a is the additive inverse of a. + ve + ve = + ve + ve + – ve = + ve [If greater number is+ve] + ve + – ve = – ve [If greater number is–ve] – ve + – ve = – ve Remember ! If a, b and c be any three integers, then i. a + b and a + b + c are also integers. [Closure property] ii. a + b = b + a [Commutative property] iii. (a + b) + c = a + (b + c) [Associative property] iv. a+ 0 = 0 + a = a [Identity property] v. a + (–a) = (–a) + a = 0 [Inverse property] Remember !
152 Oasis School Mathematics-8 Laws of multiplication on integers Let's take any three integers 2, –3, 5. Here, 2 × (–3) = – 6 is an integer. 2 × 5 = 10 is an integer. ∴ If a and b be any two integers then a × b is also an integer. It is called closure property. Again, 2 × (–3) = – 6 –3 × 2 = – 6 ∴ 2 × (–3) = –3 × 2 i.e. a × b = b × a It is called commutative property. Again, (2×(–3)) × 5 = – 6 × 5 = – 30 2 × (–3 × 5) = 2 × (–15) = – 30 ∴ (2 × (–3)) × 5 = 2 × (–3 × 5) i.e. a × (b × c) = (a × b) × c This property is called associative property. Again, 2 × (– 3 + 5) = 2 × 2 = 4 2 × (–3) + 2 × 5 = – 6 + 10 = 4 ∴ 2 × (–3 + 5) = 2 × (–3) + 2 × 5 i.e. a × (b + c) = a × b + a × c This property is called distributive property. Again, 2 × 1 = 2, 5 × 1 = 5, –3 × 1 = 3 i.e a × 1 = a This property is called identity property. + ve × + ve = + ve, + ve × – ve = – ve – ve × – ve = +ve +ve ÷ + ve = + ve, + ve ÷ – ve = – ve – ve ÷ + ve = –ve, – ve, ÷ – ve = + ve Remember ! Hence, if a, b and c be any three integers then i. a × b and a × b × c are also integers. (Closure property) ii. a × b = b × a (Commutative property) iii. (a × b) × c = a × (b × c) (Associative property) iv. a × (b + c) = a × b + a × c (Distributive property) v. a × 1 = 1 × a = a (Identity property) vi. a × 0 = 0 × a = 0 Note : Division of any integers by zero is not defined.
153 Oasis School Mathematics-8 153 Exercise 10.1 1. Write the additive inverse of the following integers. (a) +6 (b) –9 (c) +7 (d) –2 2. Evaluate: (a) –5 + 4 (b) + 8 +(–3) (c) +7 + (–11) (d) (–8) + (–3) (e) (–6) + (–10) (f) (–7) – (–8) (g) (–12) + (–4) (h) +18 – (–12) 3. Evaluate: (a) (–3) × (+4) (b) (–5)×(–3) (c) (–7) × (–4) (d) (+2) × (–8) (e) (+5) × (+6) (f) (–8) × (+4) 4. Evaluate : (a) 12 ÷ (–3) (b) –24 ÷ (–3) (c) (–15) ÷ (+5) (d) (+30) ÷ (–6) (e) (+48) ÷ (+8) (f) (–16) ÷ (–4) 5. Simplify: (a) (+5) × (–3) × (+2) (b) (+15) × (–3) × (–5) (c) (+12) × (–3) × (–1) (d) (+18) × (–2) × (–3) Answer 1. (a) – 6 (b) + 9 (c) – 7 (d) + 2 2. (a) – 1 (b) + 5 (c) – 4 (d) – 11 (e) – 16 (f) + 1 (g) – 16 (h) + 30 3. (a) – 12 (b) + 15 (c) + 28 (d) – 16 (e) + 30 (f) – 32 4. (a) – 4 (b) + 8 (c) – 3 (d) – 5 (e) + 6 (f) + 4 5. (a) – 30 (b) + 225 (c) + 36 (d) + 108 10.2 Simplification of Integers Order of Operations for Simplification Some mathematical problems contain mixed operations: addition, subtraction, multiplication,divisionanddifferentbrackets.Whenmore thanoneoperationis involved, we have to use the following rules forthe simplification of given expression of integers. This rule is known as 'BODMAS' Steps Remove bar — , brackets ( ), { }, [ ] in order. By simplifying all the operations within it or by simplifying, all the operations within it. (B) Perform the operation involving 'of' (O) Perform the operation involving division (D) Perform the operation involving multiplication (M) Perform the operation involving addition (A) Perform the operation involving subtraction (S)
154 Oasis School Mathematics-8 Worked Out Examples Example: 1 Simplify: 115 + 51 ÷ 17 × 5 – 80 Solution: 115 + 51 ÷ 17 × 5 – 80 = 115 + 3 × 5 – 80 [Operation of division] = 115 + 15 – 80 [Operation of multiplication] = 130 – 80 [Operation of addition] = 50 [Operation of subtraction] Example: 2 Simplify: 50 ÷ {18 – (4 ×10 ÷ 2)} Solution: 50 ÷ {18 – (4 ×10 ÷ 2)} = 50 ÷ {18 – (4 × 5)} = 50 ÷ {18 – 20} = 50 ÷ – 2 = – 25 Example: 3 Simplify: 64 ÷ 8 – 2 [3 + {7 – 3(3 + 4 – 2 )}] Solution: 64 ÷ 8 – 2 [3 + {7 – 3(3 + 4 – 2 )}] = 64 ÷ 8 – 2 [3 + {7 – 3 (3 + 2)}] = 64÷8 – 2[3 + {7 – 3(5)] = 64÷8 – 2 [3 + {7 – 15}] = 64÷8 – 2 [3 + (–8)] = 64÷8 – 2[–5] = 64÷8 + 10 = 8 + 10 = 18 Example: 4 Convert the given statement into mathematical expression and simplify. 4 times 20 is divided by 5 and is added to 3 times 5. Solution: 4 × 20 ÷ 5 + 3 × 5 = 4 × 4 + 3 × 5 = 16 + 15 = 31
155 Oasis School Mathematics-8 155 Exercise 10.2 Simplify 1. (a) 5 × 6 + 10 ÷ 2 (b) 6 × 8 ÷ 4 – 2 × 5 + 7 (c) 40 ÷ 8 + 5 × 4 – 10 (d) 16 ÷ 8 – 4 × 3 + 12 ÷ 3 (e) 115 + 51 ÷ 17 × 5 – 80 (f) (–3) × 6 – (–18) ÷ 6 + 28 ÷ 7 2. (a) 12 + [12 – {4 + (15 ÷ 5 × 3)}] (b) 10 – [20 ÷ {20 – 5(4 – 1)}] (c) 28 – {24 – 3(4 + 2)} (d) 150 ÷ {30 – (4 × 10 ÷ 2} (e) 42 ÷ [3 + 16 ÷ {2 + 8 ÷ (6 – 2)}] (f) 3 [{84 – (6 × 8)} ÷ 12] – 8 (g) 5{ 24 + 8 ÷ (6 – 4)} ÷ (16 × 5 – 10) (h) 5[125 – {27 – 3(6 – 2)}] (i) 102 ÷ [15 ×7 – 15 ÷ 5] (j) 39 – 7 {28 ÷ (17 – 10) + 1} (k) 20 + [122 ÷ {6 + (16 × 3) + 7)] (l) {(4 × 5) – (12 ÷ 3)} × 7 (m) (15 + 60 – 42) ÷ [(–33) ÷ 3] 3. (a) 25 + (35 – 9 + 6 ) (b) 70 + [18 × 3 – 16 – 6 + 9 × 4] (c) 20 + [16 – {4 + (3 + 6 – 4 )}] (d) 60 ÷ [28 – {24 – (20 – 8 – 4 )}] (e) 5 [40 ÷ {8 × 4 – 5 × 4 – (8 – 5 + 1 )}] (f) 70 [460 ÷ {12 × 5 – 30 ÷ 6 – (15 – 8+2 )}] 4. Convert the following statements into mathematical expression and simplify. (a) 25 is added to the product of 6 and 3. (b) 18 is subtracted from 3 times the difference of 13 and 5. (c) 6 is multiplied to one-fourth of 20 and 12 is added to the result. (d) 15 is added to the difference of 12 and 7 and the result is divided by 4. (e) 6 times the sum of 7 and 13 is divided by the sum of 5 and 7. (f) One third of 30 is added to the one tenth of 70 and the result is divided by 17. Answer 1. (a) 35 (b) 9 (c) 15 (d) –6 (e) 50 (f) –11 2. (a) 11 (b) 6 (c) 22 (d) 15 (e) 6 (f) 1 (g) 2 (h) 550 (i) 1 (j) 4 (k) 22 (l) 112 (m) -3 3. (a) 45 (b) 48 (c) 27 (d) 3 (e) 20 (f) 700 4. (a) 43 (b) 6 (c) 42 (d) 5 (e) 10 (f) 1
156 Oasis School Mathematics-8 11.1 Rational Numbers As we know that, N = The set of natural numbers = {1, 2, 3, 4, .......................} W = The set of whole numbers = {0, 1, 2, 3, ....................} Z = The set of integers = {............, -3, -2, -1, 0, 1, 2, 3, ............} Let's take any two integers, 2 and 5 2 + 5 = 7 is an integer. 2 – 5 = –3 is an integer. 2 × 5 = 10 is an integer. 2 ÷ 5 = 2 5 is not an integer. To make the division of integers meaningful rational number is introduced. 2 5 is a rational number who 2 and 5 both are integers. Hence, any number in the form of p q where p ∈ Z, q ∈ Z and q ≠ 0, is a rational number. Terminating and non-terminating recurring decimals Let's convert some rational numbers into decimal, 1 2 = 0.5 (terminating decimals) 1 4 = 0.25 (terminating decimals) 1 3 = 0.3333 ............ (non-terminating recurring decimals) 3 11 = 0.272727 ........ (non terminating recurring block of digits) ∴ Therefore, a rational number is either a terminating or non-terminating recurring decimals. Thus, all integers and fractions, whether positive or negative, including zero are Note: • Every integer is a rational number. • Rational number is either a terminating or non terminating recuring decimals. Unit 11 Rational and Irrational Numbers
157 Oasis School Mathematics-8 157 called rational numbers. 11.2 Irrational Number A set of numbers which have non-terminating and non-recurring decimals, i.e. they cannot be expressed in the form p q where p and q both are integers and q ≠ 0, such numbers are irrational numbers. 2 = 1.4142135 ........ (non-terminating, non recurring decimals) 3 = 1.7320507… (non-terminating, non recurring decimals) ∴ π = 3.14159 … (non-terminating, non recurring decimals) Hence, an irrational number can be expressed as a non-terminating, non-recurring (non-repeating) decimal. Irrational number in number line Draw a number line. Take a point P on the number line, where OP = 1 unit. Draw OPQ = 900 . Cut OP = OQ = 1 unit. Join OQ. Now, OQ2 = OP2 + PQ2 = 1 + 1 = 2. Taking 'O' as the centre and OQ as radius draw an arc which meets numberline at R, which represents 2 as OQ = OR. 11.3 Real Numbers The set of all rational and irrational numbers taken together is called the set of real numbers. The set of real numbers is denoted by R and is written as R = Q ∪ Q' = {x : x ∈ Q or x ∈ Q' } .…(i) where Q = the set of rational numbers Q' = the set of irrational numbers This shows that every rational number is a real number and every irrational number is a real number. Thus, natural numbers constitute a proper subset of integers and the integers constitute a proper subset of rational numbers and then latter constitute a proper subset of real numbers. i.e., N ⊂ W ⊂ Z ⊂ Q ⊂ R. The relation between above set of numbers is shown in the above diagram. Here, Z+ = N = the set of positive integers or natural numbers Z– = the set of negative integers W = the set of whole numbers F = the set of fractional numbers. -3 -2 -1 0 Q P 1 R 2 3 2 Z F W Z– N = Z O R Q Q'
158 Oasis School Mathematics-8 The classification of the set of real numbers is shown in the following diagram. Worked Out Examples Example: 1 Examine whether the following numbers are rational or irrational. (i) 2 (ii) 0.5 (iii) 2 3 (iv) 49 Solution: (i) Here, 2 = 1.4142134…. This is a non-terminating, non-recurring decimal. So, 2 is an irrational number. (ii) This is a terminating decimal. So, 0.5 is a rational number. (iii) 2 3 = 0.6666. This is a non-terminating recurring decimal. So, it is a rational number. (iv) 49 = 7. As the square root of 49 gives perfect whole number, it is a rational number. Example: 2 Find rational number p q, if, (i) p = 4, q = 10 (ii) p = 0.3, q = 0.6 Solution: (i) Rational number (ii) Rational number = p q = 4 10 = 2 5 = p q = 0.3 0.6 = 1 2
159 Oasis School Mathematics-8 159 Exercise 11.1 1. State whether the following numbers are rational or irrational? (a) 2 (b) 3 5 (c) 1.317432... (d) 16 (e) 1 4 (f) 1.3333... (g) 22 7 (h) – 25 (i) 3.272727... 2. Convert the following numbers into decimal and state whether they are terminating or non-terminating decimal. (a) 1 5 (b) 4 5 (c) 5 3 (d) 2 3 (e) 31 3 3. Write the rational number in the form p q if, (a) p = 3, q = 5 (b) p = 0.5, q = 0.6 (c) p = 1, q = 3 (d) p = 1.5, q = 2.5 4. State whether the following statements are true or false. (a) Every integer is a rational number. (b) Every rational number is an integer. (c) 1 is the smallest natural number. (d) An rational number is a non-terminating and recurring decimal. (e) A rational number is a non-terminating and non-recurring decimal. 5. From the adjoining figure, write the elements of the following sets. (a) the set of natural numbers (N) (b) the set of whole numbers (W) (c) the set of negative integers (Z– ) (d) the set of integers (Z) (e) the set of rational numbers (Q) (f) the set of irrational numbers (g) the set of real numbers (R) 6. Answer the following questions. (a) Is 0 a rational number? Why? (b) Why 1.3333... is a rational number? (c) Why 2.4527314... is an irrational number? (d) Is π a rational number? Why? (e) Is 3 an irrational number? Why?
160 Oasis School Mathematics-8 Answer 1. (a) rational (b) rational (c) irrational (d) rational (e) rational (f) rational (g) irrational (h) rational (i) rational 2. (a) 0.5, terminating (b) 0.8, terminating (c) 1.6666..., non-terminating (d) 0.6666..., non-terminating (e) 3.3333..., non-terminating 3. (a) 3/5 (b) 5/6 (c) 1/3 (d) 3/5 4. (a) true (b) false (c) true (d) false (e) false 5. (a) N = {1,2,3,4,10} (b) W = {0, 1, 2, 3, 4, 10} (c) Z = {-1, -2, -3, -4, -10} (d) Z = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10} (e) Q = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10, 4/5, -1/3 2/3, 4/3} (f) = { 3, 2, 13, 125 } (g) R = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10, 4/5, -1/3, 2/3, 4/3, 3, 2, 13, 125 } 6. Consult your teacher. 11.4 Operations on Irrational Numbers 3 is an irrational number 3 + 3 = 2 3 is an irrational number but 3 + (– 3 ) = 0 which is a rational number. ∴ sum of two irrational numbers may not be an irrational 3 × 3 = 3 which is a rational number. 3 × 2 = 6 which is an irrational number. ∴ Product of two irrational numbers may not be irrational. Let's discuss the operations on irrational number. I. Rationalisation Let's take a fraction 3 2 . Here, 2 is an irrational number. To convert 2 into rational number Multiply both numerator and denominator by 2 . Then 3 2 × 2 2 = 2 3 2 This is the rationalisation of denominator. II. Conjugate of Binomial Irrational Numbers Let's take a binomial irrational number 4 + 3 and its conjugate is 4 – 3 . Conjugate of 5 – 3 is 5 + 3 . Hence, the conjugate of a + b is a – b and vice versa.
161 Oasis School Mathematics-8 161 III. Rationalising factor Take an irrational number 3 Now 3 × 3 = 3 which is rational. ∴ 3 is the rationalising factor of 3 . Again, take an irrational number 3 – 1. Now ( 3 – 1) × ( 3 + 1) = ( 3 )2 – 12 = 3 – 1 = 2, which is a rational number. ∴ 3 + 1 is the rationalising factor of 3 –1. IV. Addition and Subtraction of Numbers Involving Radicals It is same as the simplification of the algebraic expressions. Under the distributive law, similar irrational numbers are added and subtracted. Rational numbers which are not similar cannot be added or subtracted. Example, 2 5 + 5 = 3 5 5 3 – 2 3 = 3 3 , etc. Worked Out Examples Example: 1 What are the rationalising factor for each of the following. (i) 2 (ii) 2 +1 Solution: The rationalising factor of 2 is 2 The rationalising factor of 2 + 1 is 2 – 1. Example: 2 Rationalise the denominator of 1 3 . Solution: 1 3 = 1 3 × 3 3 = 3 3 Example: 3 Simplify : (3 2 +1) (3 2 –1) Solution: (3 2 + 1) (3 2 – 1) = (3 2 )2 – (1)2 = 18 – 1 = 17
162 Oasis School Mathematics-8 Example: 4 Simplify : 48 – 6 3 +5 3 Solution: 48- 6 3 5 3 48 6 3 5 3 4 3 6 3 3 3 5 3 4 3 6 3 3 5 3 4 3 2 3 5 3 4 2 5 2 + = − + = × − × + = − + = − + = − ( ) + 3 = 7 3. Example: 5 Simplify: 3 – 1 3 + 1 Solution: 3 – 1 3 + 1 = 3 – 1 3 + 1× 3 + 1 3 + 1 Exercise 11.3 1. Write the rationalising factor for each of the following expressions. (a) 2 3 (b) 2 + 3 (c) 3 2 – 2 3 (d) 5 + 2 2. Rationalise the denominators of the following. (a) 1 3 (b) 4 2 (c) 5 3 2 (d) 1 2 1 + (e) 1 3 1 + (f) 1 7 3 − 3. Simplify the following expressions: (a) 2 3 + 5 3 (b) 6 2 – 5 2 (c) –3 7 – 5 7 (d) 4 8 + 5 8 – 3 8 (e) 3 + 2 5 – 3 3 + 3 5 (f) 4 10 – 3 10 + 5 10 (g) 5 7 + 3 8 – 4 7 + 2 8 [Multiplying the numerator and the denominator with the rationalising factor of 3 + 1.] = ( ) + − = + + − = + + = + = + = + 3 1 3 1 3 2 3 1 3 1 3 2 3 1 2 4 2 3 2 2 2 3 2 2 3 2 2 2 2 ( ) ( ) ( ) ( ) .
163 Oasis School Mathematics-8 163 4. Simplify: (a) 5 × 3 (b) 2 2 × 4 3 (c) 2 3 × 2 × 5 3 (d) 2 5 × 3 × 15 (e) 5 3 × 2 2 × 3 6 (f) 5 10 ÷ 5 2 (g) 12 6 ÷ 2 6 (h) 6 15 ÷ 3 5 5. Simplify: (a) ( 3 + 2 ) ( 3 – 2) (b) ( 5 – 3 ) ( 5 + 3 ) (c) (2 2 + 3 ) (2 2 – 3 ) (d) (5 3 + 2 2 ) (5 3 – 2 2) (e) (2 3 – 2)2 (f) ( 5 + 2)2 (g) (2 3 + 2) (2 3 – 2 ) (h) (2 5 – 3 ) (2 5 + 3 ) 6. Simplify the following expressions: (a) 48 (b) 3 7 + 343 (c) 3 6 + 216 (d) 2 3 + 27 (e) 8 + 32 – 2 (f) 45 – 3 20 + 4 5 (g) 4 12 – 50 – 7 48 (h) 12 – 147 + 192 (i) 4 3 – 3 12 + 2 75 7. Rationalise, (if necessary) and simplify: (a) 3 + 4 3 (b) 4 5 + 5 2 (c) 21 7 – 2 5 7 (d) 5 3 + 2 27 + 1 3 (e) 4 3 + 3 48 – 5 2 3 (f) 216 –5 6 + 294 – 3 6 (g) 3 147 – 7 3 3 + 7 3 8. Rationalise the denominators and simplify: ( ) a b( ) ( ) c d( ) 1 3 1 3 5 2 5 2 2 3 2 3 1 5 3 + − − + + − − Answer 1. a. 3 b. 2– 3 c. 3 2 + 2 3 d. 5 5 – 2 3 2. a. b. 2 2 c. 6 5 2 d. 2–1 e. f. 3. a. 7 3 b. 2 c. –8 7 d. 6 2 e. 5 2 – 2 3 f. 6 10 g. 7 + 5 8 4. a. 15 b. 8 6 c. 30 2 d. 30 e. 180 f. 5 g. 6 h. 2 3 5. a. 1 b. 2 c. 5 d. 67 e. 14–4 6 f. 7+2 10 g. 10 h. 17 6. a. 4 3 b. 10 7 c. 9 6 d. 5 3 e. 5 2 f. 5 g. – 5 2 – 20 3 h. 3 3 i. 8 3 7. a. 3 7 3 b. 10 8 5 + 25 2 c. 2 7 d. 3 34 3 e. 6 91 3 f. 2 15 6 g. 9 203 3 3 3 2 3 -1 3 8. (a) –2– 3 b. 7 –2 10 c. 7 + 4 3 d. 2 3 + 5 4 3 + 7
164 Oasis School Mathematics-8 11.5 Scientific Notation of Numbers Large numbers which are expressed as a × 10n where, (1 ≤ a < 10) and n=0, 1, 2, … are known as the numbers written in scientific notation or in standard form. Worked Out Examples Example: 1 Express in scientific notation. (i) 35,00,00,000 (ii) 53,000,000,000 Solution: (i) Here, 350,000,000 = 35 × 10000000 = 3.5 × 100000000 = 3.5 × 108 (ii) Here, 53,000,000,000 = 53 × 1000,000,000 = 5.3 × 10000000000 = 5.3 × 1010 Example: 2 Express the standard form into usual forms 9.3 × 107 Solution: Here, 9.3 × 10000000 = 93 × 1000000 = 93000000. Example: 3 Hair of a man grows by 0.000004 cm per second. Write it in scientific notation. Solution: Here, 0.000004 cm = 4 1000000 cm = 4 106 cm = 4 × 10–6 cm Example: 4 Simplify: (i) (6.8 × 105 ) + (3.7 × 108 ) (ii) (4 × 10–5) (9 × 104 ) 6 × 107 Solution: (i) (6.8 × 105 ) + (3.7 × 108 ) = (6.8 × 100000) + (3.7 × 100000000) = (68 × 10000) + (37 × 10000000) = 680000 + 370000000 = 370680000 = 3.7068 × 108 Example: 5 Convert the following scientific notation in usual form. (i) 2.5 × 10–4 (ii) 3.75 × 10–5 (iii) 3.567 × 10–7 (iv) 7.003 × 10–10 (ii) (4 × 10–5) (9 × 104 ) 6 × 107 4×9 6 [ 10-5×104 107 [ = 6[10-5 × 104 × 10-7] = 6 × 10-8 =
165 Oasis School Mathematics-8 165 Solution: (i) 2.5 × 10–4 = 2.5 104 = 25 105 = 25 100000 = 0.00025 (ii) 3.75 × 10–5 = 3.75 105 = 375 107 = 375 10000000 = 0.0000375 (iii) 3.567 × 10–7 = 3.567 107 = 3567 1010 = 0.0000003567 (iv) 7.003 × 10–10 = 7.003 1010 = 7003 1013 = 0.0000000007003 Exercise 11.4 1. Express the following in scientific notation. (a) 3,700 (b) 39,000 (c) 3,000000 (d) 99000000 2. Express the following in scientific notation. (a) 3.5 (b) 0.0057 (c) 0.000098 (d) 0.00307 3. Express the following standard form in usual forms. (a) 2.5 × 105 (b) 3.6 × 107 (c) 0.5 × 103 (d) 2 × 107 4. Express the numbers in the statements below in scientific notation. (a) There are 2250000 children in Nepal. (b) A cold drink industry made a profit of Rs. 76200000 in a year. (c) Approximately 257800000 litre of water is contained in Ranipokhari. (d) The average distance of the sun from the earth is 149760000 km. approximately. 5. Convert the numberin scientific notation in the statements below into the general form. (a) The velocity of light is approximately 2.98 × 108 m/sec. (b) The diameter of the sun has been approximated as 1.39 × 1012 metre. (c) The volume of the moon is approximated to be 3.7 × 1013 cubic metre. 6. Express the numbers used in the following statements in scientific notation. (a) The diameter of the hole of needle is 0.0007 cm. (b) The height of a man grows by 0.000000035 cm. per second. (c) The mass of an electron used to be approximated 0.000000375 grams.
166 Oasis School Mathematics-8 7. Simplify and express in scientific notation. (a) (7 × 10–7) (2.5 × 1011) 5 × 1010 (b) (3.5×10–6) (6.3×10-7) 4.9 × 10-15 (c) 0.275×0.005 0.00000625×0.00125 (d) 85000 × 0.00002 0.0000034 8. Simplify and express in general form. (a) (3.75 × 10 – 7 ) + (4.25 × 10 – 3 ) (b) (8 × 10-7)+(1.6 × 10-10) 3.2 × 10-7 (c) (5.25 × 10-6) – (2.25 × 10-7) 3 × 10-9 Answer 1. a. 3.7 × 103 b. 3.9 × 104 c. 3 × 106 d. 9.9×107 2. a. 3.5×100 b. 5.7×10-3 c. 9.8×10-5 d. 3.07×10-3 3. a. 250000 b. 36000000 c. 500 d. 20000000 4. a 2.25×106 b. 7.6×107 c. 2.578 × 108 litres d. 1.4976 × 108 km. 5. a. 298000000 m/second b. 1.390000000000 meter c. 37000000000000 m 6. a. 7 × 10-4 cm. b. 3.5 × 10-8cm/second c. 3.75 × 10-7gm 7. a. 3.5 × 10-6 b. 4.5 × 102 c. 1.76 × 10-5 d. 5 × 105 8. a. 0.004250375 b. 2.5005 c. 1825
167 Oasis School Mathematics-8 167 12.1 Ratio Let us suppose that marks obtained by Lakpa in Mathematics is 80 and the marks obtained by Dorje is 60. Let's compare their marks, Lakpa got (80-60) = 20 marks more than Dorje. Let's compare their marks by division. Marks obtained by Lakpa Marks obtained by Dorje = 80 60 = 80 60 = 4 3 = 4:3 Hence the ratio is the comparison of two quantities having same unit by division. We use the symbol : to express ratio. Write Say a : b 'a' is to 'b' or 'a' to 'b' Terms of ratio: In the ratio 4 : 5 4 is antecedent and 5 is consequent The first term of the ratio is antecedent and the second term is consequent. • In a ratio, both the quantities should be of the same kind and should have the same unit. • A ratio is a pure number and has no unit. Remember ! I. Properties of ratio Take a ratio a : b i.e. a b . Let's multiply both terms by 3, a × 3 : b × 3 = 3a : 3b = a : b Unit 12 Ratio and Proportion
168 Oasis School Mathematics-8 Hence the ratio remains same if both of its terms are multiplied by a non zero con- stant number. Again, let's divide both terms of a ratio a : b by 2 a÷2 : b÷2 = b/2 a/2 = a 2 ×2 b = a : b. Hence the ratio remains same if each term of a ratio is divided by a non zero constant number. II. Compound Ratio A new ratio formed by taking the product of two or more ratios is called compound ratio. Thus, if there are any two ratios a b and c d , then the compound ratio = a b × c d = ac bd = ac : bd. For example, if 2:3 and 4:5 are any two ratios, their compound ratio = 2 3 × 4 5 = 8 15 = 8 : 15 Worked Out Examples Example: 1 Find the ratio of 75 cm and 1 m. Solution: Here 1 m = 100 cm. ∴ Ratio of 75 cm and 1 m = 75cm 100cm = 3 4 Example: 2 Convert the ratio 1 2 : 2 3 in the form of a : b. Solution: 1 2 : 2 3 = 1 2 × 6 : 2 3 × 6 = 3 : 4 Example: 3 Find the compound ratio of 3:4, 4:5 and 5:6 Solution: The compound ratio of 3:4, 4:5 and 5:6 = 3 4 × 4 5 × 5 6 = 3 6 = 1 2 = 1:2 Multiply both terms by the L.C.M of denominators
169 Oasis School Mathematics-8 169 Example: 4 If a:b = 4:5, b:c = 3:4, find a : c. Solution: Given, a : b = 4 : 5 b : c = 3 : 4 Taking compound ratio a b × b c = 4 5 × 3 4 or, a c = 3 5 ∴ a : c = 3:5. Example: 5 The ratio of the number of boys and girls in a school is 3 : 4. If there are 120 boys, find the number of girls. Solution: Let the number of girls be x, Number of boys Number of girls = 3 4 120 x = 3 4 3x = 120 × 4 x = 120 × 4 3 x = 160 ∴ Number of girls = 160. Example: 6 Divide Rs. 7,200 in the ratio 3:4 : 5:6. Solution: Let four parts of the sum be 3x, 4x, 5x and 6x Here, 3x + 4x + 5x + 6x = 7200 or, 18x = 7200 or, x = 400 ∴ 3x = 3 × 400 = Rs. 1200 Alternative method Let no. of boys be 3x and no. of girls be 4x. Then, we have or, 3x = 120 or, x = 120/3 = 40 ∴Number of girls = 4x = 4 × 40 = 160
170 Oasis School Mathematics-8 ∴ 4x = 4 × 400 = Rs. 1600 or, 5x = 5×400 = Rs. 2000 ∴ 6x = 6×400 = Rs. 2400 Hence four parts of money are Rs. 1200, Rs. 1600, Rs. 2000 and Rs. 2400. Example: 7 Two numbers are in the ratio 3:4. If 4 is added to each of them, their ratio becomes 4:5, find the numbers. Solution: Let, the required two numbers be 3x and 4x. From the given condition, 3x + 4 4x + 4 = 4 5 or, 5(3x + 4) = 4(4x + 4) or, 15x + 20 = 16x + 16 or, 16x – 15x = 20 – 16 ∴ x = 4 ∴ Required numbers are 3x = 3 × 4 = 12 and 4x = 4 × 4 = 16 Example: 8 The ratio of the present ages of a father and his son is 5:2. Five years ago, the ratio of their ages was 3:1. Find their present ages. Solution: Let the present ages of the father and his son be 5x years and 2x years respectively. Then, 5 years ago, age of the father = (5x – 5) years. 5 years ago, age of the son = (2x – 5) years. By the question, 5x – 5 2x – 5 = 3 1 or, 1(5x – 5) = 3(2x – 5) or 5x – 5 = 6x – 15 or, 6x – 5x = 15 – 5 ∴ x = 10 Hence, the present age of the father = 5x = 5 × 10 years = 50 years and the present age of his son = 2x = 2 × 10 years = 20 years
171 Oasis School Mathematics-8 171 Exercise 12.1 1. Convert the given ratios into the lowest terms. (a) 15 25 (b) 32 48 (c) 24 64 2. Find the ratios of the following and reduce them to their lowest terms. (a) 75 cm and 125 cm (b) Rs. 4 and 50 paisa (c) 4 months and 2 years (d) 200 gm and 2 kg. 3. Convert the given ratios in terms of a:b. (a) 1 2 : 1 3 (b) 2 5 : 1 4 (c) 2 3 : 3 5 (d) 1 2 : 2 3 4. Find the compound ratio of the following ratios. (a) 2:3 and 3:5 (b) 4:5 and 9:16 (c) 4:7 and 3:8 (d) 3:4 and 7:12 5. Find the value of x in each of the following. (a) x 4 = 2 3 (b) 4 x = 2 3 (c) 4 7 = x 21 (d) 2 3 = 8 x 6. (a) If a : b = 5 : 8 and b : c = 4 : 3, find: a : c (b) If a : b = 2: 3 and b : c = 4 : 5, find a : c. (c) If a : b = 2:3, b:c = 4:5 and c : d = 1:2, find a : d. 7. (a) A ratio is equal to 4:5. If antecedent is 36, find the consequent. (b) If a ratio is 2:3 and its consequent is 45, find its antecedent. 8. (a) Divide Rs. 300 between A and B in the ratio 2 : 3. (b) Divide Rs. 1,500 in the ratio 1:2. (c) Divide Rs. 9,000 between A, B and C in the ratio 2 : 3 : 4. (d) An alloy contains copper, zinc and tin in the ratio of 2 : 3 : 5. Find the mass of each metal in 120 gm of alloy. 9. (a) If the angles of a triangle are in the ratio 2 : 3 : 5, find the size of each angle. (b) If the ratios of the angles of a quadrilateral are 1:3:5:6, find the size of each angle. 10. (a) The ratio of the number of boys and girls in a class is 5 : 3. If there are 35 boys, find the number of girls. (b) The ratio of the ages of two sisters is 3:5. If the elder sister is 25 years old, find the age of the younger sister.
172 Oasis School Mathematics-8 (c) The ratio of monthly income of Imran and Salman is 4:5. If the monthly income of Salman is Rs. 60,000, find the monthly income of Imran. 11. (a) Two numbers are in the ratio 4:5. When 5 is added to each of them, their ratio becomes 5:6. Find the numbers. (b) Two numbers are in the ratio 3:5. When 5 is subtracted from each of term, their ratio becomes 1:2, find the numbers. (c) Two numbers are in the ratio 3:4. If 10 is added to both numerator and denominator, the new ratio becames 4:5. Find the numbers. 12. (a) The ratio of the present ages of a father and his son is 5:2. Ten years hence, the ratio of the ages will be 2:1. Find their present ages. (b) The ratio of the present ages of a mother and her daughter is 3:1. Six years ago, the ratio of their ages was 6:1. Find their present ages. (c) Ages of two brothers are in the ratio 3:4. 5 years ago, the ratio of their ages was 2:3. Find the their present ages. 1. (a) 3/5 (b) 2/3 (c) 3/8 2. (a) 3:5 (b) 8:1 (c) 1:6 (d) 1:10 3. (a) 3:2 (b) 8:5 (c) 10:9 (d) 3:4 4. (a) 2:5 (b) 9:20 (c) 3:14 (d) 7:16 5. (a) 8/3 (b) 6 (c) 12 (d) 12 6. (a) 5:6 (b) 8:15 (c) 4:15 7. (a) 45 (b) 30 8. (a) Rs. 120, Rs. 180 (b) Rs. 500, Rs. 1000 (c) Rs. 2000, Rs. 3000, Rs. 4000 (d) 24gm, 36gm, 60 gm 9. (a) 360 , 540 , 900 (b) 240 , 720 , 1200 , 1440 10. (a) 21 (b) 15yrs (c) Rs. 48,000 11. (a) 20, 25 (b) 15, 25 (c) 30, 40 12. (a) 50 yrs, 20 yrs, (b) 30 yrs, 10 yrs, (c) 15 yrs, 20 yrs. Answer 12.2 Proportion Let's take two ratios 8:12 and 6:9. Here, 8:12 = 8 12 = 2 3 and 6:9 = 6 9 = 2 3 . i.e. 8:12 = 6:9. Such equality of two ratios is called a proportion. Hence, four quantities a, b, c and d are in proportion, if a b = c d i.e. a : b = c : d. A proportion is also written as a : b : : c : d In a proportion, first and fourth terms are called extremes whereas second and third terms are called means. In a b = c d, a × d = b × c. The product of extremes = the product of means.
173 Oasis School Mathematics-8 173 Continued Proportion Let's take any two ratios 3:6 and 6:12. Here, 3 6 = 1 2 = 1 : 2 and 6 12 = 1 2 = 1 : 2 ∴ 3 : 6 = 6 : 12 Hence, 3, 6, 6 and 12 are in proportion. i.e. 3, 6 and 12 are in proportion. This type of proportion is continued proportion. Three quantities are said to be in continued proportion if the ratio of the first to the second is equal to the ratio of second to the third. i.e. three quantities a, b and c are said to be in continued proportion if a : b = b : c, where b is called as mean proportional. • a, b, c, d are said to be proportional if a:b :: c:d. • 'd' is said to be fourth proportional to a, b, c if a b = c d . • If a, b, c, d are in proportion, first and last are extremes and second and third are means. • In proportion, product of extremes = product of means. Remember ! Worked Out Examples Example: 1 Test whether the numbers 2, 3, 8 and 12 are in proportion or not. Solution: Here, ratio of first two = 2 : 3 ratio of last two = 8:12 = 8 12 = 2 3 = 2:3 Since the ratio of the first two and the last two numbers is equal given from terms are in proportion. Example: 2 The first, second and third terms of proportion are 4, 22 and 6 respectively. Find the fourth proportional. Solution: Let fourth proportional be x. Then, 4, 22, 6 and x are in proportion.
174 Oasis School Mathematics-8 Thus, we have, 4 : 22 = 6 : x or, 4 22 = 6 x or, 4x = 22 × 6 or, x = 22×6 4 = 33 Hence, the required fourth proportional is 33. Example: 3 Find the mean proportional between 2 and 18. Solution: Let the mean proportional between 2 and 18 be x. Then, 2, x, 18 are in continued proportion. ∴ 2 x = x 18 or, x² = 36 or, x = 6 ∴ Mean proportional is 6. Example: 4 What number must be added to each of the terms 4, 9, 17 and 35 to be in proportion? Solution: Let, the required number be x. then, (4 + x), (9 + x) (17 + x) and (35 + x) are in proportion. i.e. 4 + x 9 + x = 17 + x 35 + x (4 + x) (35 + x) = (17 + x) (9 + x) or, 140 + 4x + 35 x + x² = 153 + 17x + 9 x + x² or, 140 + 39x = 153 + 26x or, 39x – 26 x = 153 – 140 or, 13x = 13 or, x = 1 ∴ The required number is 1.
175 Oasis School Mathematics-8 175 Exercise 12.2 1. Identify whether the given numbers are in proportion or not? (a) 1, 2, 6, 8 (b) 6, 8, 10, 20 (c) 25, 35, 50, 70 (d) 4 cm, 6cm, 12, cm, 18 cm (e) 5kg, 10kg, 20kg, 40 kg 2. Find the values of x in the following proportions. (a) 2 : 5 = 6 : x (b) 7 : 5 = x : 35 (c) 4 : x = 16 : 20 (d)x : 7 = 16 : 28 3. Find the fourth proportional in the following: (a) 1, 3, 5 (b) 2, 4, 8 (c) 15, 18, 30 (d) 10, 8, 5 4. Find the mean proportional between the following: (a) 4 and 9 (b) 4 and 16 (c) 9 and 16 (d) 5 and 20 5. (a) If 3, x, 12 are in continued proportion, find the value of x. (b) If 4, 12, x, 15 are in proportion, find the value of x. (c) Two extremes of a proportion are 10 and 60. If the first mean is 15, find the second mean. (d) Two means of a proportion are 12, 8. If the first extreme is 6, find the last extreme. 1. (a) Not in proportion (b) Not in proportion (c) Proportion (d) Proportion (e) Proportion 2. (a) 15 (b) 49 (c) 5 (d) 4 3. (a) 15 (b) 16 (c) 36 (d) 4 4. (a) 6 (b) 8 (c) 12 (d) 10 5. (a) 6 (b) 5 (c) 40 (d) 16 Answer 12.3 Types of proportion I. Direct proportion Let the cost of 1 copy = Rs. 10, then the cost of 2 copies = Rs. 2 × 10 = 20 Now, Number of copy Cost (Rs.) 1 2 10 20
176 Oasis School Mathematics-8 Here, more the number of copies, more the cost. Again, ratio of copies = 1 : 2 ratio of cost = 10:20 = 1:2 such type of proportion is direct proportion II. Indirect Proportion (Inverse proportion) Suppose 1 man can do a work in 20 days. then 2 men can do the same work in 10 days. Now, Men Days 1 2 20 10 Here, more the number of men, less the number of days. Ratio of men = 1:2 Inverse ratio of days = 20:10 = 2:1 Such proportion is indirect proportion. • In direct proportion, if one quantity increases another also increases in the same ratio. • In indirect proportion, if one quantity increases another decreases in the same ratio. Remember ! Worked Out Examples Example: 1 If two ratios 4:5 and x:20 are in direct proportion, find the value of x. Solution: Here, 4:5 and x : 20 are in direct proportion, 4 5 = x 20 or, 5x = 80 or, x = 80 5 = 16 ∴ x = 16.
177 Oasis School Mathematics-8 177 Example: 2 If two ratios 2 : 3 and a : 50 are in indirect proportion, find the value of 'a'. Solution: Here, 2:3 and a : 50 are in indirect proportion, Hence, 2 3 = 50 a or, 2a = 50 × 3 or, a = 50 × 3 2 ∴ a = 75. Example: 3 Find the value of x in the given case. Weight (in kg.) Cost (in Rs.) 5 8 400 x Solution: Since weight and cost are in direct proportion, 5 8 = 400 x or, 5x = 8 × 400 or, x = 8×400 5 x = 8×80 ∴ x = Rs. 640. Example: 4 If 40 metres of a cloth costs Rs. 1200, how many metres can be bought for Rs. 870 ? Solution: Let, cloth bought for Rs. 870 be x metres' Length of cloth(m) Cost in (Rs.) 40 1200 x 870 Here, length of cloths and cost are in direct proportion, so, 40 : x = 1200 : 870 or 40 x = 1200 870 Weight (in kg.) Cost (in Rs.) 5 8 400 x
178 Oasis School Mathematics-8 or, 1200 × x = 40 × 870 or x = 40 × 870 1200 = 87 3 = 29 meters Hence, cloth bought for Rs. 870 is 29 metres. Example: 5 A garrison of 600 men had provisions for 36 days. However, a reinforcement of 300 men arrived. For how many days will the food last now? Solution: Let's suppose the food will now last for x days, Number of men Number of days 600 36 600 + 300 = 900 x Clearly, more men require less days for provisions to last. So, it is a case of inverse proportion. ∴ 600 : 900 = x : 36 or, 600 900 = x 36 or, 9x = 6 × 36 or, x = 6 × 36 9 = 24 days. Hence, the food will now last for 24 days. Exercise 12.3 1. (a) If the following ratios are in direct proportion, find the value of x. (i) 3 : 5 and x : 25 (ii) 12 : x and 24 : 36 (iii) 15 : 20 and x : 40 (b) If the following ratios are in indirect proportion, find the value of x. (i) 4 : 5 and x : 16 (ii) x : 10 and 20 : 50 (iii) 5 : 10 and x : 25 2. Identify whether the given variables are in direct or indirect proportion. (a) The quantity of rice and cost (b) The number of book and cost (c) The number of workers and the time taken to finish a certain work (d) The number of days worked and the quantity of work (e) Interest on money and the amount invested
179 Oasis School Mathematics-8 179 3. Find the value of x in each of the following cases. (a) Weight (in kg) Cost (in Rs.) (b) Number of workers Days (c) Number of workers Work 5 120 30 5 5 12 x 50 x 10 x 4. (a) If 8 pens cost Rs. 64, what is the cost of 12 pens ? (b) The cost of 5 books is Rs. 625. How many books can be purchased for Rs. 1000? (c) The cost of 20 books or 30 pens is Rs. 360. What is the total cost of 16 books and 24 pens? 5. (a) A bus covers a distance of 120 km in 2 hours. How long will it take to run 600 km with the same speed? (b) If the cost of painting 4 walls having area 250 m2 is Rs. 3500, what will be the cost of painting the walls having an area of 1000 m2 ? 6. (a) If 18 men can do a piece of work in 36 days, how many men can do it in 54 days ? (b) If 36 men can do a piece of work in 25 days, in how many days will 15 men do it? 7. (a) A garrison has provisions for 240 persons for 30 days. If 60 more persons join the garrison, how long will the provisions last ? (b) A hostel has enough food for 150 boys for 30 days. After 10 days 50 boys leave the hostel. Find how long will the remaining food last. (c) A garrison had provision for 300 men for 90 days. After 20 days 50 more men joined the garrison. How long will the remaining food last at the same rate? 1. (a) (i) 15 (ii) 18 (iii) 30 (b) (i) 20 (ii) 25 (iii) 50 2. (a) Direct variation (b) Direct variation (c) Indirect variation (d) Direct variation (e) Direct variation 3. (a) Rs. 288 (b) 3 days (c) 2/3 4. (a) Rs. 96 (b) 8 books (c) Rs. 576 5. (a) 10 hrs. (b) Rs. 14000 6. (a) 12 men (b) 60 days 7. (a) 24 days (b) 30 days (c) 60 days Answers 1 3
180 Oasis School Mathematics-8 13.1 Introduction: Per cent means out of every hundred or per hundred. For example, a boy got 80 percent marks in Maths means he got 80 marks out of 100. The symbol for percent is % or p.c. A fraction with 100 as its denominator is called percentage. e.g. 35 100 = 35%, 65 100 = 65% and so on. Some operations related to the percentage (i) Conversion of percentage into fraction If the given percentage is divided by 100, then, it converts into fraction. Percentage Fraction 20% 20 100 = 1 5 (ii) Conversion of fraction into percentage If a given fraction is multiplied by 100, then it converts into percentage. Fraction Percentage 1 5 1 5 × 100% = 20% (iii) Conversion of decimal into percentage If a given decimal is multiplied by 100, then it converts into percentage. Decimal Percentage 0.25 0.25×100% = 25% (iv) Calculation of certain percentage of a number Let a number be x. If we have to find the value of 10% of x. 10% of x means 10 out of 100 of x. ∴ 10 % of x = 10 100 of x = 10 100 x = x 10 . Unit 13 Percentage
181 Oasis School Mathematics-8 181 (v) Conversion of one quantity as the percentage of another quantity If the unit of two quantities is the same, then one quantity can be expressed as the percentage of another quantity. If the rate of the first quantity is to be expressed as the percentage of another, the ratio of the first to the second quantity is multiplied by 100 to convert one quantity as the percentage of another. For example: What per cent of 60 is 15? Here, Required percent = 15 60 × 100 % = 1 4 × 100% = 25% Worked Out Examples Example: 1 If 1 5 of the total students of a school are girls, find the percentages of boys and girls. Solution: Let, Total students of a school = 100%. Then, The percentage of girls = 1 5 × 100% = 20% The percentage of boys = 100% – 20% = 80%. Example: 2 What per cent of Rs. 5 is 25 paisa? Solution: Here, Rs. 5 = 500 Paisa. ∴ Required percentage = 25 500 × 100 % = 5 % Example: 3 20% of a number is 60. Find the number. Solution: Let the required number be x. Then 20% of x = 60 or 20 100 × x = 60 or x = 60×100 20 = 300 ∴ Hence, required number is 300.
182 Oasis School Mathematics-8 Example: 4 The monthly income of Kalpana is Rs.15,000. If she spends 30% of her income on food, 15% on house rent, 10% on education and 20% on miscellaneous, find her (a) expenditure on each item. (b) saving in a month. Solution: (a) Here, the monthly income of Kalpana = Rs.15,000. Her expenditure on food = 30% of Rs. 15,000 = 30 100 × 15000 = Rs. 4,500 Her expenditure on house rent = 15% of Rs.15,000 = 15 100 × 15,000 = Rs. 2,250 Her expenditure on education = 10% of Rs.15000 = 10 100 × 15,000 = Rs.1,500 Her expenditure on miscellaneous = 20% of Rs. 15000 = 20 100 × 15,000 = Rs. 3,000 ∴ Total expenditure in a month = Rs. 4,500 + Rs. 2,250 + Rs. 1,500 + Rs. 3,000 = Rs. 11,250 (b) Now, her monthly saving = Rs. 15,000 – Rs. 11,250 = Rs. 3,750 Example: 5 In an election, a candidate secured 40% of the votes polled and lost the election by 6000 votes. Find the total number of votes cast. Solution: Let, the total number of votes cast be x. Then, votes secured by losing candidate = 40% of x = 40 100 × x = 2x 5 ∴ Votes secured by winning candidate = (x – 2x 5 ) = 3x 5 Difference of votes = 3x 5 – 2x 5 = x 5 But, by question, x 5 = 6000 or x = 6000 × 5 = 30000. Hence, the total number of votes cast is 30000.
183 Oasis School Mathematics-8 183 Example: 6 The population of Tamghas Bazaar in 2070 was 1,20,000. In 2071 it was increased by 10% and one year later it was decreased by 5%. What is the population of the town in 2072? Solution: Here, Population of Tamghas in 2070 = 1,20,000. Increment in the population = 10 % of 1,20,000 = 10 100 × 1,20,000 = 12,000 ∴ Population of the town in 2071 B.S. = 1,20,000 + 12,000 = 1,32,000 Again, Reduction in the population = 5 % of 1,32,000 = 5 100 × 1,32,000 = 6600 ∴ Population of the town in 2072 B.S. = 1,32,000 – 6,600 = 1,25,400 Example: 7 A milkman mixes 20 litres of milk containing 25% water with 50 litres of milk containing 10 % of water. What is the percentage of water in the mixture? Solution: Here, Amount of water in the first lot of milk = 25 % of 20 = 25 100 × 20 = 5 litres Amount of water in the second lot of milk = 10 % of 50 = 10 100 × 50 = 5 litres. Total amount of water in the mixture = (5 + 5) litres = 10 litres. Total amount of mixture = (20 + 50) litres = 70 litres. ∴ Percentage of water in the mixture = 10 70 × 100 % = 14 2 7 %
184 Oasis School Mathematics-8 Exercise 13.1 1. Express the following fractions or decimals in percentage. (a) 1 5 (b) 3 10 (c) 0.05 (d) 2.5 2. Express the following percentages in fraction and reduce them to their lowest terms. (a) 35% (b) 30% (c) 50% (d) 85% 3. Convert the following percentages into decimals. (a) 35% (b) 40% (c) 45% (d) 0.5% 4. Convert the following decimals into percentage. (a) 0.75 (b) 0.01 (c) 0.2 (d) 0.5 5. Find the value of each of the following. (a) 25% of 300 (b) 20% of 1000 (c) 2 1 2 % of Rs. 800 (d) 5% of 100 6. (a) What percent of 150 is 75? (b) What percent of Rs. 5 is 50 paisa? (c) What percent of 8 kg is 500 gm? (d) What percent of 2m is 5cm? 7. (a) If 25% of a number is 800, find the number. (b) If 20% of a sum is Rs. 600, find the sum. (c) If 15% of a number is 240, find the number. 8. (a) A number becomes 190 when it is reduced by 5%. What is the number? (b) A number becomes 55 when it is increased by 10%. What is the number? (c) If a number is increased by its 20%,it becomes 360. Find the number. 9. (a) 7 10 of the total number of students in a school are girls. Find the percentages of girls and boys (b) If 1 4 of examinee failed in an examination, calculate the percentage of students who passed? (c) 3 5 of the population of a village is female. Find the percentage of male. 10. (a) 60% of the students in a school are boys and the number of girls is 300. Find the number of boys. (b) In an examination, 60% of the total examinees passed. If 120 students failed, find the total number of examinees. (c) A man spends 60% of his income and saves Rs. 4,000, find his income.
185 Oasis School Mathematics-8 185 (d) A man spends 80% of his salary. If he saves Rs. 1500 per month, find his salary per month. 11. (a) Gunpowder contains 70% nitre, 15% sulphur and rest charcoal. Find the amount of charcoal in 12 kg of Gunpowder. (b) A man earns Rs. 8000 in a month and spends 30% on food 15% on cloth, 20%, on education, 10% on house rent and 15% on others, (i) find the expenditure on each item. (ii) the saving in a month. 12. (a) Among 2,500, only 1000 voters cast their votes in an election. (i) What per cent of the total voters cast their votes? (ii) What per cent of the voters did not cast their votes? (b) Among 1,200 students in a school, 200 were absent, find the (i) percentage of absent students. (ii) percentage of present students. 13. (a) The price of petrol is increased from Rs.800 to Rs. 1,000. Find the percentage increase in the price. (b) The price of an article is decreased from Rs. 500 to Rs. 400. Find the percentage decrease in the price. 14. (a) The population of Beni Bazaar increases by 10% annually. If the present population is 2,000, what will be its population after 2 years? (b) In 2012 A.D., the number of tourists visiting Nepal was 150,000. In 2013 A.D., it increased by 10% and in 2015 A.D., it is decreased by 10%. How many tourists visited Nepal in 2015 A.D? 15. A milkman mixes 300 litres of milk containing 10 % water with 200 litres of milk containing 5 % water, what is the percentage of water in that mixture? 1. (a) 20% (b) 30% (c) 5% (d) 250% 2. (a) 7/20 (b) 3/10 (c) 1/2 (d) 17/20 3. (a) 0.35 (b) 0.4 (c) 0.45 (d) 0.005 4. (a) 75% (b) 1% (c) 20% (d) 50% 5. (a) 75 (b) 200 (c) Rs. 20 (d) 5 6. (a) 50% (b) 10% (c) 6 1 4% (d) 2.5% 7. (a) 3200 (b) Rs. 3000 (c) 1600 8. (a) 200 (b) 50 (c) 300 9. (a) 70%, 30% (b) 75% (c) 40% 10. (a) 450 (b) 300 (c) Rs. 10,000 (d) Rs. 7500 11. (a) 1800gm (b) (i) Food-Rs.2400, Cloth-Rs. 1200, Education-Rs.1600, Other item-Rs. 1200, House rent-Rs. 800 (ii) Rs. 800 12. (a) (i) 40% (ii) 60% (b) (i) 16 2 3% (ii) 83 1 3% 13.(a) 25% (b) 20% 14. (a) 2420 (b) 1,48,500 15. 8% Answer
186 Oasis School Mathematics-8 14.1 Introduction The method of finding the value of articles using unit value is called unitary method. In unitary method, we come across two types of variations namely direct variation and indirect variation. We have already discussed the direct and indirect variation in the chapter of Ratio and Proportion. Worked Out Examples Example: 1 The cost of 8 articles is Rs. 640. What is the cost of 20 articles? Solution: Cost of 8 articles is Rs. 640 Cost of 1 article is Rs.640 8 Cost of 20 articles is Rs.640 8 × 20 ∴ Required cost = Rs. 640 8 × 20 = Rs. 1600 ∴ Cost of 20 articles is Rs. 1600. Example: 2 If the cost of 2 5 of a piece of land is Rs. 5,00,000, what is the cost of 3 4 piece of the same land? Solution: The cost of 2/5 of a piece of land = Rs. 5,00,000 The cost of 1 piece of land = Rs. 500000 2 / 5 Alternative method Articles Cost 8 20 640 x Since the number of articles and cost are direct variations 20 8 = x 640 or, 8x = 20×640 or, x = 20×640 8 = Rs. 1600 Cost of 1 article is less than the cost of 8 articles Cost of 20 articles is more than the cost of 1 article Unit 14 Unitary Method
187 Oasis School Mathematics-8 187 = Rs. 500000×5 2 = Rs. 1250000 The cost of 3 4 of a piece of land = Rs. 3 4 × 12,50,000 = Rs.9,37,500 Hence, the required cost of 3 4 of a piece of the same land is Rs. 9,37,500. Example: 3 The cost of 3 chairs and 2 tables is Rs. 1050. If the cost of one chair is Rs. 150, find the cost of one table. Solution: Here, Cost of 1 chair is Rs. 150 Cost of 3 chairs is Rs. 3×150 = Rs. 450 Now, Cost of 3 chairs + Cost of 2 tables = Rs. 1050 Rs. 450 + Cost of 2 tables = Rs. 1050 Cost of 2 tables = Rs. 1050 – 450 = Rs. 600 Cost of 1 table = Rs. 600 2 = Rs. 300. Example: 4 48 men are employed to construct a building in 75 days. They wanted the building to be ready in 60 days. How many more men should be employed to construct the building in 60 days? Solution: In 75 days, 48 men can construct a building In 1 day, 48 × 75 men can construct the building In 60 days, 48 × 75 60 men can construct the building ∴ Men required to construct the building = 48 × 75 60 = 60. Number of additional men = (60 – 48)men = 12 men More men are required to construct the building in 1 day. Less men are required to construct the building in 60 days. Alternative method Land Cost 500000 x Since the land and cost are direct variations, x 500000 = x = 3 4 × 5 2 × 500000 x = Rs. 937500 3 4 2 5 3 4 2 5
188 Oasis School Mathematics-8 Alternative method Days Men 75 60 48 48+x Since, the men and days are indirect proportion, 48 + x 48 = 75 60 or, 48 + x 48 = 5 4 or, 4(48 + x) = 240 or, 192 + 4x = 240 or, 4x = 240 – 192 or, 4x = 48 ∴ x = 48 4 = 12 Example: 5 A hostel has provisions for 45 students for 30 days. If 15 more students joined the hostel after 6 days, how long will the provision last? Solution: Here, Total number of students = 45 + 15 = 60 Remaining number of days = 30 – 6 = 24 For 45 students remaining provision lasts for 24 days. For 1 student remaining provision lasts for 24 × 45 days. 60 students have provision for 24 × 45 60 days = 6 × 3 days = 18 days. Hence, the remaining provision will last for 18 days. Alternative method Here, the total number of students = 45 + 15 = 60 Remaining number of days = 30 – 6 = 24 Let us suppose the remaining provision will last for x days. Now, Student Days 45 24 60 x As the number of students and days are indirect variations, x 24 = 45 60 or, x = 45 60 × 24 = 18 days.
189 Oasis School Mathematics-8 189 Example: 6 20 men can do a piece of work in 12 days working 9 hours a day. In how many days 30 men can do the same work working 6 hours a day. Solution: Working 9 hours a day 20 men can do a work in 12 days. Working 1 hour a day 20 men can do a work in 12 × 9 days. Working 1 hour a day 1 man can do a work in 12 × 9 × 20 days. Working 6 hours a day 1 man can do a work in 12 × 9 × 20 6 days. Working 6 hours a day 30 men can do a work in 12 × 9 × 20 6 × 30 days. ∴ Required number of days = 12 × 9 × 20 6 × 30 days = 12 days. Alternative method Let's suppose 30 men can do a work in x days working 9 hours a day. Men Hours Days 20 9 12 30 6 x Men and days are inverse variations. Similarly, hours and days are also inverse variations. x 12 = 9 6 × 20 30 x = 9×20×12 6×30 = 12 ∴ Required number of days = 12 days. Exercise 14.1 1. (a) If the cost of 20 pens is Rs 640, what is the cost of 12 pens? (b) If the cost of 5m cloth is Rs. 1,200, what is the cost of 15 m cloth? (c) If a dozen oranges cost Rs. 120, find the cost of 4 oranges. (d) If the cost of 50 kg of sugar is Rs. 5,000, what is the cost of 2 quintals of sugar? 2. (a) Lochan earns Rs. 2,16,000 in 3 months! How much does he earn in 1 year? (b) A worker earns Rs. 54,000 in 6 months. How much does he earn in 9 months? (c) Cost of 18 kg rice is Rs. 900. How much rice can be bought with Rs. 500? (d) If the cost of 25 pens is Rs. 246, how many pens can be bought for Rs. 984? (e) A car consumes 8 litres of petrol in covering a distance of 120 km. How many kilometers will it go with 15 litres of petrol?
190 Oasis School Mathematics-8 3. (a) If 1 4 part of a property is Rs. 48600, find the 2 5 part of it. (b) If the cost of 1 4 part of the land is Rs. 80,000, find the cost of 1 2 part of the land. (c) If 1 3 part of a money is Rs. 40,000, what is 1 4 part of it? 4. (a) Cost of 5 pens and 7 books is Rs. 350. If the cost of one pen is Rs. 28, find the cost of 1 book. (b) The cost of 3 tables and 8 chairs is Rs. 7000. If the cost of one table is Rs. 1000, find the cost of 10 chairs. 5. (a) If 4 men can do a piece of work in 30 days, in how many days could 12 men do the same work? (b) 12 workers can do a piece of work in 20 days. How many workers should be added to complete the work in 16 days? (c) 15 men can do a piece of work in 80 days. How long will it take to finish the work if 5 men are added? (d) 6 taps take 15 minutes to fill up a water tank. How many minutes will 10 taps take to fill up the same tank? 6. (a) A garrison of 1500 men has provisions for 60 days. How long would the food last if the garrison is reduced to 900 men ? (b) 150 students of a hostel have food enough for 45 days. 25 students leave the hostel after 10 days. How long will the remaining food last ? (c) A garrison of 1200 men has food for 60 days. How long does it last if 300 men were added after 10 days? 7. (a) 15 men reap 45 hectares of field in 6 days. How many hectares will 9 men reap in 4 days? (b) 40 men can do a piece of work in 72 days if they work 8 hours a day. How many men are required to finish the same work in 40 days working 9 hours a day? (c) 28 men can dig a field in 18 days working 8 hours a day. In how many days can 16 men dig the same field working 9 hours a day? 1. (a) Rs. 384 (b) Rs. 3600 (c) Rs. 40 (d) Rs. 20,000 2. (a) Rs. 8,64,000 (b) Rs. 81,000 (c) 10 kg (d) 100 pens (e) 225 km 3. (a) Rs. 77760 (b) Rs. 1,60,000 (c) Rs. 30,000 4. (a) Rs. 30 (b) Rs.5,000 5. (a) 10 days (b) 3 (c) 60 days (d) 9 mins. 6. (a) 100 days (b) 42 days (c) 40 days 7. (a) 18 hectares (b) 64 men (c) 28 days. Answer
191 Oasis School Mathematics-8 191 14.2 Time and Work While solving problems on time and work, the following facts should be kept in mind. Let's discuss the relation of time and work. 20 men can do a work in 15 days. 20 men can do 1 15 work in 1 day. i.e. less the number of days, less the work and more the number of days, more the work. Worked Out Examples Example: 1 A can do a piece of work in 10 days and B alone can do it in 15 days. How much time will both take to finish the work ? Solution: A can do 1 work in 10 days. A can do 1 10 work in 1 day. B can do 1 work in 15 days. B can do 1 15 work in 1 day. (A + B) can do ( 1 10 + 1 15) work in 1 day. (A + B) can do 3 + 2 30 work in 1 day. (A + B) can do 5 30 work in 1 day. (A + B) can do 1 work in 30 5 days. ∴ Required number of days = 30 5 days = 6 days. Example: 2 A and B together can do a piece of work in 4 days and A alone can do it in 6 days. In how many days can B alone do it? Solution: (A + B) can do 1 work in 4 days. (A + B) can do 1 4 work in 1 day. A can do 1 work in 6 days. Focus on: • Number of workers to do the work • Duration of time to do the work • Amount of work done A can do 1 6 work in 1 day. B can do ( 1 4 – 1 6 ) work in 1 day. B can do (3 – 2 12 ) work in 1 day. B can do 1 12 work in 1 day. B can do 1 work in 12 days.
192 Oasis School Mathematics-8 Example: 3 A can do 1 3 of a certain work in 6 days and B can do 1 2 of the same work in 12 days. In how many days will they together complete the work? Solution: A can do 1 3 work in 6 days. A can do 1 3×6 work in 1 day. A can do 1 18 work in 1 day. B can do 1 2 work in 12 days. B can do 1 24 work in 1 day. Exercise 14.2 1. (a) A can do a piece of work in 10 days and B can do it in 15 days. In how many days will they do it together? (b) Aadhya can do a piece of work in 20 days and Anasuya can do it in 30 days. How long will both take to do it together? 2. (a) A and B can finish a work in 15 days. A alone would take 20 days to finish it. How many days will be taken by B to finish it alone? (b) A and B can finish a work in 12 days. B alone takes 30 days to finish it. In how many days can A finish it alone? (c) A and B can complete a piece of work in 8 days and A alone completes it in 12 days. In how many days would B alone complete the work? 3. (a) A can do 1 4 of a work in 10 days and B can do 1 3 of the same work in 20 days. In how many days will they complete the work together? (b) A can do 1 4 of a work in 2 days and B can do 2 5 of it in 2 days. In how many days will they do the whole work together? (A + B) can do ( 1 18 + 1 24) work in 1 day. (A + B) can do (4 + 3 72 ) work in 1 day. (A + B) can do 7 72 work in 1 day. (A + B) can do 1 work in 72 7 days. ∴ Required number of days = 72 7 days = 10 2 7 days. 1. (a) 6 days (b) 12 days 2. (a) 60 days (b) 20 days (c) 24 days 3. (a) 24 days (b) 3 1 13 days Answer
193 Oasis School Mathematics-8 193 15.1 Profit and Loss (Introduction) The term profit and loss is generally used by the businessmen in the transaction of goods while buying or selling them. A person buys goods at a certain cost which is called its cost price, in short C.P., and the cost at which s/he sells it is called its selling price, in short S.P. If the cost price is less than the selling price, there is profit and if the cost price is more than the selling price, there is loss in the transaction. Hence, we can relate the terms profit, loss, C.P. and S.P. as follows: (i) If S.P. > C.P. then profit = S.P. – C.P. (ii) If S.P. < C.P. then loss = C.P. – S.P. Note: (i) Overhead expenses like transportation, maintenance, repair and other expenditures are always added with the cost price to get net cost price. (ii) Profit or loss is calculated with the help of S.P. and C.P. of the same number of articles. Profit and Loss Percentage A businessman always desires to expand the business by studying the market value of goods in terms of his/her profit percentage. Profit or loss percentage always gives an exact idea about the nature of demand and supply. Just by knowing the profit or loss, one can't predict for the future plan of the business. For example, if a calculator is bought for Rs. 500 and sold for Rs. 700, the profit is Rs. 200. Also if a geometrical box is bought at Rs. 100 and sold for Rs. 300, the profit is again Rs. 200. But, profit percent in first case = Rs. 200×100 Rs. 500 = 40% While, profit percent in second case = Rs. 200×100 Rs. 100 = 200% Hence, a businessman would better go for the business of the second type. Note: Profit percent or loss percent is always calculated out of the given cost price unless and until it is stated otherwise. Unit 15 Profit and Loss
194 Oasis School Mathematics-8 Hence, we can write, Profit % = Profit×100 C.P. % ........ (i) Loss % = Loss×100 C.P. % ........ (ii) Again, if profit % is given, S.P. = C.P. + Profit % of C.P. ........ (iii) If loss % is given, S.P. = C.P. – Loss % of C.P. ........ (iv) From (iii) and (iv) we can derive: (i) S. P = (100 + profit per cent) × C.P. 100 (iii) C. P = (iv) C. P = S.P × 100 (100+profit per cent) S.P × 100 (100–loss per cent) (ii) S. P = (100 – loss per cent) × C.P. 100 Worked Out Examples Example: 1 A man buys a motorbike for Rs. 1,50,000 and after some days he sells it for Rs. 1,30,000. What is his gain or loss percentage? Solution: Here, C.P. = Rs. 1,50,000 S.P. = Rs. 1, 30,000 Since, C.P. > S.P. Loss = C.P. – S.P. = 1,50,000 – 1,30,000 = 20,000 We have, loss per cent = Loss C.P. × 100% = 20,000 1,50,000 × 100% = 2 15 × 100% = 13 1 3 %