Percentage and Its Applications

PERCENTAGE AND ITS APPLICATIONS CONTENTS • Ratio • Equivalent Ratio • Proportion • Percentage • Profit and Loss • Profit and Loss Percent • Simple Interest Note : 1. A ratio remains unchanged if both of its terms are multiplied by the same non-zero quantity. Let k ≠ 0, then clearly, a ka (i) = and therefore a : b = ka : kb b kb a a/k a b (ii) = and therefore a : b =  :  b b/k k k 2. The ratio a : b is said to be in simplest form if HCF of a and b is 1. Ex.1 Sol. We can compare two quantities by two methods. 1. By finding the differences of their magnitudes : When we want to see how much more or less one quantity is than the other, we find the difference of their magnitudes and such a comparison is known as comparison by difference. 2. By finding the division of their magnitudes: If we want to see how many times more (or less) one quantity is than the other, we find the ratio (or division) of their magnitudes and such a comparison is known as the comparison by division. RATIO Ratio is the comparison by division of same kind of quantities or the ratio of two quantities of same kind and in same units is a fraction that shows how many times the one quantity is of the other. a The ratio a is to b is the fraction , and is written b as a : b. We call 'a' as the first term or antecedent and 'b' the second term or consequent. 8 Ex.2 Sol. EXAMPLES Express 60 : 90 in its simplest form. In order to express the given ratio in its simplest form we divide its first and second term by their HCF. We have 60 = 2 × 2 × 3 × 5 90 = 2 × 3 × 3 × 5 So, HCF of 60 and 90 is 2 × 3 × 5 i.e., 30. 60 60 ÷ 30 2 ∴ 60 : 90 = = = =2:3 90 90 ÷ 30 3 Hence, the simplest form of 60 : 90 is 2 : 3. Comparison of Ratios In order to compare two given ratios, we express each of them in simplest form and then compare these fractions by making their denominators equal. Compare 5 : 12 and 3 : 5 Writing, the given ratio as fractions, we have 5 3 5 : 12 = and 3 : 5 = 12 5 LCM of 12 and 5 is 60. Making the denominator of each fraction equal to 60, we have 5 5×5 25 3 3 × 12 36 = = and = = 12 12 × 5 60 5 5 × 12 60 Clearly, 36 > 25. 36 25 3 5 ∴ > ⇒ > 60 60 5 12 EQUIVALENT RATIO Ex.5 Are 25, 15, 6, 5 in proportion ? Sol. We have A ratio obtained by multiplying or dividing the numerator and denominator of a given ratio by the same non zero number is called an equivalent ratio. a : b = 25 : 15 = 5 : 3 c:d=6:5 as EXAMPLES a:b≠c:d ∴ 25, 15, 6, 5 are not in proportion. Ex.3 Find two equivalent ratio of 12 : 8. Sol. We have Alternative method Product of extemes = ad = 25 × 5 = 125 12 12 ÷ 4 3 = = 8 8÷4 2 Product of means = bc = 15 × 6 = 90. as ∴ 3 : 2 is an equivalent ratio of 12 : 8. 12 12 × 2 24 = = 8 8× 2 16 Also, a = 25, b = 15, c = 6, d = 5 ⇒ 25, 15, 6, 5 are not in proportion. Ex.6 Find the ratio of (i) So, 24 : 16 is an equivalent ratio of 12 : 8. Hence, 3 : 2 and 24 : 16 are two equivalent ratio of 12 : 8. ad ≠ bc. j− 5 to 50 paise (iii) 9 m to 27 cm Sol. (i) j− 5 to 50 paise = 500 : 50 = 10 : 1 If 12 bowls cost 20 such bowl ? (iv) 30 days to 36 hours. = 5 × 100 paise to 50 paise Unitary Method Ex.4 (ii) 15 kg to 210 gm j− 72, What will be the cost of (dividing first and second term by their HCF i.e. by 50) (ii) 15 kg to 210 gm Sol. Q cost of 12 bowl = ∴ cost of 1 bowl = − j 72 j− 72 = 12 Hence, cost of 20 bowl = = 15 × 1000 gm to 210 gm = 15 × 1000 : 210 = 15000 : 210 j− 6 = 500 : 7 (HCF of 15000 and 210 is 30, so dividing I and II terms by j− 6 × 20 = j− 120 30) (iii) 9 m to 27 cm PROPORTION = 9 × 100 cm to 27 cm Four numbers a, b, c, d are said to be in proportion, if a : b = c : d and we write a : b :: c : d or in other words we can say that an equality of two ratios is called a proportion. = 900 : 27 = 100 : 3 (dividing I and II terms by the HCF of 900 and 27 which is 9) (i) The first and fourth terms are called extreme terms, second and third terms are called mean terms. (iv) 30 days to 36 hours = 30 × 24 hours to 36 hours = 30 × 24 : 36 = 720 : 36 If product of means = product of extremes, then given numbers are in proportion. (ii) d is called the fourth proportional to a, b, c. EXAMPLES = 20 : 1 (dividing I and II terms by the HCF of 720 and 36 which is 36) Ex.7 In a computer lab, there are 3 computer for every 6 students. How many computer will be needed for 24 students ? Sol. 6 students have = 3 computers 1 student has = Sol. Let the third term of the proportion be x. 3 computers 6 217 : 112 :: x : 32 3 24 students have = × 24 computers 6 We know that if numbers in proportion, then product of means = product of extremes = 12 computers Hence, 24 students will be needed 12 computers Ex.8 Population of Rajasthan is 570 lakh and population of UP is 1660 lakh. Area of Rajasthan is 3 lakh km2 and area of UP is 2 lakh km2. ⇒ 112 × x = 217 × 32 (i) How many people are there per km2 in both these state ? Hence, the third term of the given proportion is 62. (ii) Which state is less populated ? ⇒ x= Ex.12 Express the ratio (i) 24 to 48 (ii) 12 cm to 1 m in their simplest form. Sol. (i) Population of Rajasthan = 570 lakh Sol. Area of Rajasthan = 3 lakh km2. ∴ Number of people in per km2 = (ii) before comparing 12 cm and 1 m they must be expressed in the same unit. 1660 = 830. 2 (ii) As population of Rajasthan per km2 is less than the population of U.P. per km2 so Rajasthan state is less populated. Sol. The daily pocket expenses of X and Y are j− 45 and j− 90 respectively. What is the ratio of their expenses in simplest form ? ∴ Ex.13 Express the following ratios in their simplest form : (i) 2 : (ii) Required ratio = 45 : 90 45 45 ÷ 45 1 = = 90 90 ÷ 45 2 Sol. (i) 2 : =8:3 Let a = 63, b = 42, c = 33, d = 22. Product of means = 33 × 42 = 1386. So, Product of extremes = Product of means Hence, 63, 42, 33, 22 are in proportion. Ex.11 The first, second and fourth terms of a proporiton are 217, 112, 32. Find the third term. 3 3 =2×4: ×4 4 4 (Multiplying both the numbers by 4) Ex.10 Are 63 42, 33, 22 in proportion ? As product of extremes = 63 × 22 = 1386 3 4 6 15 : 7 14 Hence, required ratio is 1: 2. Sol. 12cm 12cm 12 3 = = = 1m 1 × 100cm 100 25 So 12 cm : 1 m = 3 : 25 HCF of 45 and 90 = 45 = 24 1 = 48 2 (dividing both the numbers by 24) Area of U.P. = 2 lakh km2. Ex.9 (i) 24 to 48 = 570 = 190 3 and population of U.P. = 1660 lakh ∴ Number of people in per km2 = 217 × 32 ; x = 62 112 (ii) 6 15 6 15 6 14 4 : = ÷ = × = 7 14 7 14 7 15 5 ∴ 6 15 4 : = =4:5 7 14 5 Ex.14 Which ratio is greater, 5 : 4 or 7 : 6 ? Sol. To compare 5 : 4 and 7 : 6 we need to compare 5 7 and so that we may express both of them 4 6 with the same denominator. ∴ 5 5× 6 30 7 7×4 28 = = and = = 4 4× 6 24 6 6×4 24 Clearly, 30 28 > or 5 : 4 > 7 : 6. 24 24 Ex.15 A family has 15 pets of which 6 are cats or kittens, 3 are dogs and the rest are birds. Find the ratio of the numbers of (i) birds to dogs (ii) birds to pets Sol. (i) Total no. of pets = 15 No. of cats or kittens = 6 No. of dogs = 3 No. of birds = Total no. of pets – (No. of cats + No. of dogs) = 15 – (6 + 3) ⇒ 15 – 9 = 6 So, the no. of birds = 6 There are 6 birds and 3 dogs. So, the number of birds : number of dogs =6:3=2:1 (ii) There are 6 birds and 15 pets So, the number of birds : number of pets = 6 : 15 = 2 : 5 Ex.16 Find the missing numbers in the following ratios : (i) : 15 = 8 : 10 (ii) 15 = 4 10 Ex.17 Two length are in the ratio 3 : 7. The second length is 42 cm. Find the first length. Sol. Let the first length be x cm. Then we write the ratio of the length as x : 42; but it must be equal to the given ratio 3 : 7 ∴ 3 : 7 = x : 42 ⇒ x= x 3 = 42 7 3 3× 6 × 42 = = 18 7 1 Hence, the first length is 18 cm. Ex.18 In a class of 60 pupils the ratio of the number of boys to the number of girls is 7 : 8. How many boys and girls are there ? Sol. Given that 7 are boys and 8 are girls so they are 15 together. 7 Therefore, boys are 7 out of 15, i.e., of 60. 15 8 and girls are 8 out of 15, i.e. of 60. 15 7 ∴ The number of boys = of 60 15 7 = × 60 = 7 × 4 = 28 15 8 and the number of girls = × 60 = 8 × 4 = 32 15 Check : 28 + 32 = 60 Ex.19 Divide j− 2600 amongst three people so that their shares are in the ratio 4 : 5 : 4. Sol. Given ratio is 4 : 5 : 4 Now sum of the rations = 4 + 5 + 4 = 13 Therefore, the share of first person is 4 out of 13. 4 i.e., × j− 2600 = 4 × j− 200 = j− 800 13 Similarly, the share of the second person is 5 out of 13. 5 i.e., × j− 2600 = 5 × j− 200 = j− 1000 13 and the share of the third person is 4 out of 13 4 i.e., × j− 2600 = 4 × j− 200 = j− 800 13 Check : j− 800 + j− 1000 + j− 800 = j− 2600 Alternative Method : Let the shares be 4x, 5x and 4x Now the sum of shares = 4x + 5x + 4x = 13x According to the questions 13x = j− 2600 − 2600 ⇒x= j = j− 200 13 Hence the share of first person is 4x = 4 × 200 = j− 800 Share of second person = 5x = 5 × j− 200 = j− 1000 and share of the third person = 4x = 4 × j− 200 = j− 800 Check : Sum of shares = j− 800 + j− 1000 + j− 800 = j− 2600 PERCENTAGE When we take 100 as the denominator of fractions, the numerators are called percentages. For convenience, the symbol % is used for percent. Or ‘‘A percentage is simply a ratio in which the second term is arranged to be 100’’. Also percent is an abbreviation of the Latin phrase per centum, meaning per hundred or hundredths. (i) A fraction may be converted into a percentage by multiplying that fraction by 100%. This does not change its value, since 100% is 1. (ii) A decimal may be converted into a percentage by multiplying it by 100%. Ex.25 Convert a percentage into fraction (i) 45% Sol. 7 as a percentage. 20 1 1   100  =  × 100 % =  % = 25% 4 4   4  (b) 22  22  =  × 100 % = 50% 44  44  (c) 4  4  =  × 100 % = 16% 25  25  Ex.23 Out of 50 students in a class, 15 like to play cricket. What is percentage of students who like to play cricket ? Sol. Total students = 50 Students who like to play cricket = 15 So, % age of students who like to play cricket  15  =  × 100 % = 30% .  50  Ex.24 Convert the given decimals to percent : (a) 0.6 (b) 0.75 (c) 0.08 (d) 0.56 Sol. We have (a) 0.6 = (0.6 × 100)% = 60% (b) 0.75 = (0.75 × 100)% = 75% (c) 0.08 = (0.08 × 100)% = 8% (d) 0.56 = (0.56 × 100)% = 56% 45 9 = 100 20 (ii) 65% = 65 13 = 100 20 (a) 53% Sol. 7 7 = × 100% = 35% 20 20 Ex.21 Express 0.625 as a percentage. Sol. 0.625 = 0.625 × 100% = 62.5% 1 22 4 Ex.22 Write (a) (b) (c) as percent. 4 44 25 (a) We have (i) 45% = 42.5 425 85 17 = = = . 100 1000 200 40 Ex.26 Convert each of the following into decimal fraction : Sol. Sol. (iii) 42.5% We have (iii) 42.5% = EXAMPLES Ex.20 Express (ii) 65% (a) 53% = (b) 0.38% 53 = 0.53 100 (b) 0.38% = (c) 4.7% = (c) 4.7% 0.38 = 0.0038 100 4 .7 47 = = 0.047 . 100 1000 Ex.27 What percentage of the adjoining figure is shaded and what percentage is unshaded ? Find it. Sol. First we find the fraction of the figure that is shaded or unshaded. From this fraction we will find the percentage of the shaded and unshaded regions. 1 4 1 4 1 4 1 4 1 1 1 3 So, shaded region =  + +  = 4 4 4 4 Now, percentage of shaded region 3  =  × 100 % = 75% 4   Unshaded region = 1 4 Now, percentage of unshaded region 1  =  × 100 % = 25% 4  Uses of Percentages 1. Interpreting percentages. 2. Converting percentage to ‘How many’. 3. Converting ratio to percentage. 4. Increase or decrease as percent. Eg. : Raju invests 10% of his pocket money in buying toffees means j− 10 out of j− 100 are invested by Raju in buying the toffees. Eg. : A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win ? Here, the total number of matches played are 20. Out of these 25% are won by the team. I method (direct). Out of 100, 25 matches are won by the team. So, out of 20, number of matches won by the team = 25 × 20 100 = 5 matches. Ex.29 Arun bought a car for j− 3,50,000. The next year, the price went upto j− 3,70,000. What was the percentage of price increase? Sol. Original price = j− 3,50,000 Change in price = j− 3,70,000 – j− 3,50,000 = j− 20,000. Percentage increase Amount of change in price = × 100 Original price = 20,000 2 × 100 = × 100 3,50,000 35 2 40 5 × 20 = =5 7 7 7 5 Percentage increase = 5 % 7 Hence, percentage increase in price of car 5 =5 % 7 Ex.30 Estimate what region of the following figures is shaded and hence find percentage of that shaded region. = II method (using percentage). 25% of 20 = 25 × 20 = 5 . 100 (i) (ii) EXAMPLES Ex.28 Convert each of the following ratios into a percentage : (iii) (i) 15 : 45 (ii) 3 : 5 Sol. We have, (i) Shaded region = We have, 15  15  (i) 15 : 45 = =  × 100 % 45  45  3  =  × 100 % 9   1  =  × 100 % 3  = Sol. 100 1 % = 33 % 3 3 3  (ii) 3 : 5 =  × 100 % = 60% 5  1 4 1  % of shaded region =  × 100  % = 25% 4  3 (ii) Shaded region = 5 3  % of shaded region =  × 100  % = 60% 5  3 (iii) Shaded region = 8 3  % of shaded region =  × 100  % 8   3 75   =  × 25  % = % = 37.5% 2 2   Sol. (i) Given, 3 : 1 Ex.31 Find (i) 15% of 250 Total = 3 + 1 = 4 (ii) 1% of 1 hour (iii) 20% of j− 2500 (iv) 75% of 1kg (i) 15% of 250 = Sol. Which shows in fractions : 15 15 × 250 = × 10 = 37.5 100 4 3 3  =  × 100  %= 75% 4 4  Also,  1  (ii) 1% of 1 hour =  × 1 hour  100  1 1  =  × 100  % = 25%. 4 4   1  3 3  = × 60  = min or  × 60  sec  100  5 5  = (ii) Given 2 : 3 : 5 Total = 2 + 3 + 5 = 10 3 min or 36 sec 5 (iii) 20% of j− 2500 = Which shows in fraction : 20 1 × 2500 = ×2500 100 5 Also = j− 500 (iii) 20% (iv) 5% Sol. Ex.34 The population of a city decreased from 25,000 to 24,500. Find the percentage decrease. Sol. S.No. (i) (ii) (iii) Percentage Fraction 25 1 = 100 4 150% 150 3 = 100 2 1.50 20 1 = 100 5 0.20 5 1 = 100 20 0.05 (iv) 5% (ii) 2 : 3 : 5 = Decrease in population ×100 Initial population = 25000 − 24500 25000 0.25  500  = × 100  % = 2% 25000   Ex.35 In a city 30% are females, 40% are males and remaining the children. What % are children ? Ex.33 Convert each of the ratio to percentage : (i) 3 : 1 Percentage decrease Decimal 25% 20% 2  2  =  × 100  % = 20% 10  10  5 5  =  × 100  % = 50% 10  10  Ex.32 Convert given percents to decimal fractions and also to fractions in simplest form : (ii) 150% 2 3 5 , , . 10 10 10 3  3  =  × 100  % = 30% 10  10   75  (iv) 75% of 1 kg =  × 1 kg = 0.75 kg 100   (i) 25% 3 1 and 4 4 Sol. Percentage of females = 30% Percentage of males = 40% Percentage of children = (100 – 30 – 40)% = 30% Ex.36 (i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk. Ex.40 Express 33.6 g as a percentage of 80g. Sol. The first quantity as a percentage of the second quantity is (ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick ? Sol. 33.6 3360 336 × 100% = %= % =42% 80 80 8 (i) As chalk contains, Calcium : Carbon : Oxygen = 10 : 3 : 12 Ex.41 Find the value of Total = 10 + 3 + 12 = 25 Carbon in chalk = 3 25 (i) 44% of 650 km Sol. (i) 44% of 650 km = percentage of carbon in chalk  3  =  × 100  % = 12% 25   Total = 10 + 3 + 12 = 25 If = Chalk = Sol. 3 × 25 gm 3 So, weight of chalk = 25 gm Ex.37 If in a school 45% are girls, what percentage are boys ? Sol. If 45% are girls, then (100 – 45)% are boys i.e., 55% are boys Ex.38 In a particular town if 85% houses have a telephone, what percentage do not have Sol. Sol. = 165 × 16 mm 200 = 165 × 2 33 × 2 66 mm = mm = mm = 13.2 mm 25 5 5 Distance covered by bus = 70% of 10 km = If 85% have a telephone, then (100 – 85)% do not, i.e., 15% do not have telephone Sol. 1 165 82 % of 16 mm = % of 16mm 2 2 Ex.43 William travelled a distance of 10 km. He covered 70% of the distance by bus and the remaining on foot. What distance did he travel by bus ? How much distance did he cover on foot ? All house (i.e. 100% of houses) either have or do not have a telephone. Ex.39 Express 15 cm as a percentage of 3 m. 25 1 200 × × 64 = = 2 kg 8 100 100 1 Ex.42 Find the value of 82 % of 16 mm. 2 Carbon = 3g 3 3gm = × chalk 25 44 44 × 65 × 650 = = 286 km 100 10 1 25 (ii) 3 % of 64 kg = % of 64 kg 8 8 (ii) As ratio of Calcium, Carbon and oxygen in chalk is 10 : 3 : 12 1 (ii) 3 % of 64 kg 8 70 × 10 = 7 km 100 Distance covered on foot = 10 km – 7 km = 3 km First express 3m in cm to bring both quantities to the same unit. Ex.44 55% of the population of a town is male. If the total population of the town is 128200, find the female population of the town. ∴ 3 m = 3 × 100 cm = 300 cm Sol. Then the first quantity as a percentage of the second quantity is 15 1 × 100% = × 100% = 5% 300 20 Male population of the town = 55% of 128200 = 55 × 128200 = 70510 100 Therefore, the female population of the town = 128200 – 70510 = 57690 Ex.45 A person donates 6% of his total savings to the Prime Minister's Relief Fund. He divides the remaining money equally between his one son and one daughter. If the total saving of the person is j− 1500000, find the amount donated to the Prime Minister's Relief Fund. Find the amounts received by his son and daughter respectively Sol. Ex.48 The population of India is 113 crore. If it increases by 1.7% every year, Find India's population after one year. Sol. India's population = 113 crore Increased by 1.7%  1.7  = 113 crore +  × 113  crore 100   192.1 crore 100 = 113 crore + 1.921 crore = 14.921 crore Amount donated to the Prime Minister's Relief Fund = 6% of j− 15,00,000 = = = 113 crore + 6 × j− 15,00,000 100 PROFIT AND LOSS j− 90000 Cost Price The price that a person spends to purchase or manufacture some goods is called the cost price. In short, we write C.P. for cost price. Amount of savings left after donation to P.M.'s Relief Fund = j− 15,00,000 – j− 90,000 = j− 14,10,000 Selling Price The price at which a shopkeeper or a person sells his good is called the selling price. In short, we write S.P. for selling price In case of profit In case of Loss Amount of received by his son = j− 14,10,000 ÷ 2 = j− 7,05,000 because the remaining amount of his savings has been divided between his son and daughter equally. So amount of received by his daughter = j− 7,05,000. • Profit = S.P. – C.P. • Loss = C.P. – S.P. • S.P. = Profit + C.P. • C.P. = Loss + S.P. • C.P. = S.P. – Profit • S.P. = C.P.– Loss Ex.46 There are 800 students in a school, out of which 560 are girls. Find the percentage of girl students in the school. EXAMPLES Ex.49 Find the profit or loss : Sol. Required percentage of girl students = 560 × 100 = 70% 800 Ex.47 Out of an income of j− 15000, Hardik spends j− 10200. What percentage of his income does be save ? Sol. Hardik's total income is Hardik's spending is His saving is = Sol. (i) Here S.P. > C.P., therefore Profit = S.P. – C.P. = j− 15000. j− 10200. j− (15000 – 10200) j− 4800 j− 215.80 – j− 176.50 = j− 39.30 (ii) Here S.P. < C.P., therefore Loss = C.P. – S.P. = j− 499 – j− 357 = j− 142 (iii) Here C.P. > S.P. Therefore, required percentage of his saving = j− 176.50 ; S.P. = j− 215.80 (ii) C.P. = j− 499 ; S.P. = j− 357 (iii) C.P. = j− 44,450 ; S.P. = j− 38,578 (i) C.P. = 4800 ×100 = 32% 15000 So, Loss = C.P. – S.P. j− 44,450 – j− 38,578 = j− 5,872 = Ex.50 A trade purchased 10 quintals of wheat from a farmer for j− 8,750. He sold it at j− 11.50 per kg. Find the amount of profit/ loss of the trader. Sol. We know that 1 quintal = 100 kg ∴ 10 quantials = 10 × 100 kg = 1000 kg So, the cost price of 1000 kg wheat = j− 8,750 Also the selling price of 1kg wheat = 11.50 Therefore, the S.P. of 1000 kg wheat = 1000 × 11.50 = j− 11,500.00 Since S.P. > C.P. So, the profit = S.P. – C.P. = j− 11,500 – j− 8750 = j− 2750 Thus, the profit of the trader is j− 2750. Ex.51 A shopkeeper earns a profit of j− 325.75 on each sewing machine. If the C.P. of a machine is j− 2018.50, what is the selling price ? Sol. Profit = j− 325.75, Cost Price = j− 2018.50 ∴ S.P. = Profit + C.P. = j− 325.75 + j− 2018.50 = j− 2344.25 Ex.52 A milkman buys 20 litres of milk from a dairy for j− 370. He sells it at the rate of j− 21.50 per litre. Find his profit or loss. Sol. C.P. of 20 litre milk = j− 370 S.P. of 1 litre milk = j− 21.50 Therefore, S.P. of 20 litres milk = j− 21.50 × 20 = 430 Clearly, S.P. > C.P., so profit = j− 430 – j− 370 = j− 60 Ex.53 A girl purchased 12 packet for j− 156. Each packet contains 10 pencils. She sold all the pencils at a price of j− 2 per pencil. Find the profit or loss. Sol. 12 packets have 12 × 10 = 120 pencils. C.P. for 120 pencils = j− 156 Selling price for 1 pencil = j− 2 Therefore, the S.P. of 120 pencil = 120 × 2 j− = j− 240 Since S.P. > C.P., therefore, there will be the profit. Profit = j− 240 – j− 156 = j− 84. Ex.54 Bela purchased a second hand car for j− 89,000. She spent j− 21,000 on its repair and sold it to Aman for j− 1,10,000. Find her profit or loss in this transaction. The amount at which the Bela purchased the car = j− 89,000 Sol. The amount he spent on repair = j− 21000 Therefore, the cost price j− 89000 + j− 21,000 = j− 1,10,000 = Note : Total C.P. = actual cost price + overheads. Since S.P. = 1,10,000 ⇒ S.P. = C.P. Therefore Bela neither suffered loss nor gained any profit. PROFIT OR LOSS PERCENT In order to calculate profit or loss in percent, we use the following formulae : 1 (i) Profit % = Amount of profit × 100 C.P. i.e. Profit%= (ii) Loss % = Pr ofit × 100 C.P. Loss × 100 C.P. 2. Profit or loss percent is always calculated on the C.P. Also we can find S.P. = C.P. × (100 + Pr ofit %) 100 In case of profit % S.P. = C.P. × (100 − Loss%) 100 In case of Loss % C.P. = S.P × 100 (100 + Pr ofit %) In case of Profit % S.P. × 100 C.P. = (100 − Loss%) In case of Loss % Ex.57 Neelu bought 2400 bananas at j− 15 a dozen. She sold 1350 of them at j− 4 for 2 and remaining at j− 8 for 5. Find her gain or loss percent. EXAMPLES Ex.55 Find the profit or loss percent if : j− 500; (ii) C.P. = j− 600; j− 600 S.P. = j− 500 (i) C.P. = Sol. (i) We have, C.P. = S.P. = j− 500, S.P. = Sol. j− 600 C.P. of 12 bananas = C.P. of 1 banana = Clearly S.P. > C.P. Therefore profit = 600 – 500 = 100 = 100 × 100 = 20% 500 So, Profit percent = 20% (ii) We have, C.P. = − 600, j S.P. = S.P. of 1 banana = 4 2 j 15 ×2400 = 12 j− 3000 4 × 1350 = 2 j− 2700 Remaining bananas = 2400 – 1350 = 1050 j− 100 Therefore, Loss percent Loss 100 50 2 × 100 = × 100 = = 16 C.P. 600 3 3 So Loss percent = 16 S.P. of 5 remaining bananas = j− 8 S.P. of 1 remaining bananas = 8 5 S.P. of 1050 remaining bananas = 2 % 3 = C.P. of one dozen of pencils = = 20 × 150 = j− 3000 j− 2.50 Gain (%) = 1380 × 100 = 46% 3000 Ex.58 A book wholesaler sold 300 copies of a book at a profit of 15%. If C.P. of a book is j− 48, find the selling price of the books. Sol. S.P. of 1 dozen (i.e. 12) pencils C.P. of 1 copy of the book = j− 48 C.P. of 300 copies of the book = 300 × 48 j− 30 = 2.50 × 12 = = Therefore, S.P. of 150 dozen pencils = 150 × 30 = Profit = S.P. – C.P. = = j− 4500 j− (4500 – 3000) j− 1500 j− 1,680 j− 2700 + j− 1680 = j− 4380 Gain = j− (4380 – 3000) = j− 1380 j− 20 C.P. of 150 dozens of pencils 8 × 1050 5 Total S.P. = Ex.56 Karim bought 150 dozens of pencils at j− 20 a dozen. He sold them at j− 2.50 per pencil. Find the profit or loss percent. Now, S.P. of 1 pencil = j− 4 − 500 Loss = C.P. – S.P. = 600 – 500 = Sol. S.P. of 2 bananas = S.P. of 1350 bananas = Clearly C.P. > S.P. = 15 12 C.P. of 2400 bananas = Profit × 100 C.P. Hence, Profit percent = j− 15 Profit (%) = 15%, Profit = j− 14400 15 × 14400 100 = j− 2160 Therefore, S.P. of books = 14400 + 2160 1500 Profit (%) = ×100 = 50% 3000 = j− 16560 Ex.59 A horse bought for j− 8000 was sold at a loss of 6%. At what price was the horse sold ? Sol. C.P. = j− 8000, Loss (%) = 6% Loss = 6 ×8000 = 100 Ex.62 An old bike bought for j− 2000 is sold for j− 2200. Find the profit and the profit % (or Gain%). Sol. j− 480 Therefore, S.P. = C.P. – Loss = 8000 – 480 = C.P. of a dozen i.e. 12 eggs = Profit = 15 ×1600 = 100 j− 240 1840 1200 S.P. of 100 eggs = = j− Sol. Given S.P. of the hand-cart = − 153 j 1 . 3 j− 420 at a profit of 12%. Profit = S.P. = So, C.P. = S.P. × 100 (100 − Loss%) 720 × 100 720 × 100 = = 100 − 25 75 In this case, S.P. = j− 112, C.P. = j− 100 When S.P. is j− 1, C.P. = j− When S.P. is j− 420, C.P. = j− = Hence, C.P. = − 375. j 100 112 100 × 420 112 − 375 j C.P. × (100 + Gain%) 100 = 960 × (100 + 25) 960 × 125 = 100 100 = j− 1200 Using unitary method, we have When S.P. is j− 960 Desired gain = 25% j− 12 j− 100 + j− 12 = 112. j− 720, Loss = 25% 1840 × 100 1200 Let the cost price of the article be 100 Given Sol. C.P. = 460 = 3 Ex.61 An article is sold for Find the C.P.  Gain   200  = × 100  % =  × 100  % = 10%  C.P.   2000  Ex.63 If a man were to sell his hand cart for j− 720, he would loss 25%. What must be the selling price if he were to gain 25% ? j− 1600 + j− 240 = j− 1840 Therefore, S.P. of 1200 eggs = j− 1840 j− j− 2200 – j− 2000 = j− 200 Therefore, gain % j− 1600 S.P. = S.P. of 1 egg = j− 2200 = 16 Therefore, C.P. of 1200 eggs = ×1200 12 Profit (%) = 15% S.P. of the old bike = So, Profit = S.P. – C.P. j− 16 = j− 2000 Clearly, S.P. > C.P. j− 7520 Ex.60 Shweta bought 1200 eggs at j− 16 a dozen. At what price per hundred must she sell the eggs so as to earn a profit of 15% ? Sol. C.P. of the old bike = Ex.64 Nandan sells a quintal of wheat for thereby, gaining a profit of 12%. j− 308 By selling a quintal of rice for the same amount, he losses 12%. Find the C.P. of both rice and wheat. Also his total gain or loss. Given S.P. of wheat = Sol. We know, C.P. = j− 308, Gain = 12% Note : If the rate of interest is given per annum then the time period must be expressed in terms of year. S.P. × 100 (100 + gain %) For Example 308 × 100 308 × 100 Therefore, C.P. = = 100 + 12 112 − = j 275 Now, S.P. of rice = j− 308 Loss = 12% 100 × S.P. We know, C.P. = (100 − Loss%) Therefore, C.P. of rice = Time period 3 1 = year 12 4 T = 6 month should be written as 6 1 = year 12 2 100 × 308 (100 − 12) 100 × 308 = = 88 Total C.P. of wheat and rice = j− (275 + 350) = j− 625 Total S.P. = j− (308 × 2) = j− 616 We can see that S.P. < C.P. Loss = j− 625 – j− 616 = j− 9 T = 9 month should be written as j− 350 SIMPLE INTEREST 9 3 = year 12 4 EXAMPLES Ex.65 Find the simple interest when; Principal = j− 600, Rate = 2% per annum and Time = 20 months. Sol. Interest : Interest is the amount paid in lieu of using some money which is not owned by us. 4 The amount of money deposited, lent or borrowed is called principal (P). 4 The additional money given at the end of a period for using the principal is called interest. 4 j− 600, R = Rate We have, P = Principal = percent per annum = 2 And T = Time = 20 months = 20 year 12 Therefore, simple interest (S.I.) = The total money we receive or pay is called the amount due at that time. Thus the sum of principal and interest is called amount. i.e. amount = principal + interest 4 T = 3 month should be written as The time for which the money is kept in the bank or for which the loan has been borrowed is called the time period. P×R ×T = 100 Thus S.I. =  600 × 2 × 20    100 × 12  j−  j− 20. Ex.66 Find the principal when Simple Interest = j− 72, Rate = 3% per annum and Time = 3 months. Sol. We have, SI = j− 72, R = 3%, To find the simple interest on a certain amount of money we need to know three quantities. T = 3 months = (i) Amount deposited or borrowed is called principal (P) Therefore, Principal (P) = (ii) Rate of interest (R) (iii) Time period (T) P×R ×T So, simple Interest = 100 P = 3 1 = year 12 4 100 × S.I. R ×T  100 × 72 × 4   = 3 ×1   j−  = j− (100 × 24 × 4) j− 9600 Ex.67 Find the rate when Principal = j− 700, Simple Interest = j− 168 and Time = 16 months Sol. We have, P = j− 700, SI = j− 168, T = 16 months = Therefore, Rate = Rate = = 16 year 12 100 × S.I. % P×T 168 × 100 × 12  168 × 12  %=  % 700 × 16  7 × 16  2016 % = 18% 112 We have, P = R = 12 j− 640, 1 25 %= per annum, SI = 2 2 Therefore, T = Thus, T = j− 40 S.I × 100 1 40 × 100 × 2 = = 640 × 25 P×R 2 1 year or 6 months. 2 1 % per 2 annum from a bank. If he paid j− 1863.75 as 1 interest for 2 years, find the sum. 2 Ex.69 Neeraj borrowed a sum of money at 10 Sol. 1 21 %= , S.I.= 2 2 1 5 and T = 2 years = years. 2 2 We have, R = 10 j− 1863.75 We have to find the sum. Now, Principal (P) = S.I. × 100 R ×T = 1863.75 × 100 × 2 × 2 = 1775 × 4 21 × 5 = j− 7100 Hence, the required sum = j− 7100 7 of itself in 6 4 years at a certain rate of interest. Find the rate of interest. Ex.70 A sum of money becomes Let the Principal be j− P. Then amount = j− We have, principal = Ex.68 Find the time when principal = j− 640, Rate 1 =12 % per annum and Simple Interest = j− 40. 2 Sol. Sol. j− P, Amount = j− 7 P 4 7 P, 4 T = 6 years. We have to find the rate (R) Then, Amount = Principal + S.I. 7P = P + S.I. 4 7P 7 P − 4P 3P S.I. = –P= = 4 4 4 We know that, P×R ×T S.I. = 100 3P P × R × 6 = 4 100 3P × 100 = 4 × P × R × 6 300 P = 24 P × R Therefore, Rate (R) 300P 300P 300 = % or R = = % 24P 24P 24 300 ÷ 12 25 1 ⇒R= %= % = 12 % 24 ÷ 12 2 2 Hence, required rate percent 1 = 12 % per annum 2 Ex.71 If Meena gives an interest of j− 45 for one year at 9% rate p.a., what is the sum she has borrowed ? Sol. S.I. = 45, R = 9%, T = 1 year P×R ×T S.I. = 100 P × 9 ×1 45 = 100 45 × 100 P= = 500 9 Hence, Meena has borrowed j− 500 Ex.72 What rate gives j− 280 as interest on a sum of j− 56,000 in 2 years ? Sol. We have, P = j− 56000, T = 2, R = ? S.I. = j− 280 P×R ×T S.I. = 100 56000 × R × 2 280 × 100 280 = ⇒ R= 100 56000 × 2 Hence Rate (R) = 0.25% Ex.73 Find the amount to be paid at the end of 3 years in each case : (i) Principal = j− 1200 at 12% p.a. (ii) Principal = j− 7500 at 5% p.a. Sol. (i) We have, P = j− 1200, R = 12%, T = 3 years P × R × T 1200 × 12 × 3 S.I. = = 100 100 S.I. = j− 432. A = P + S.I. = j− (1200 + 432) A = j− 1632. (ii) We have, P = j− 7500, R = 5%, T = 3 years P×R ×T 7500 × 5 × 3 S.I. = = 100 100 S.I. = 1125 A = P + S.I. = 7500 + 1125 A = j− 8625 Ex.74 Amina buys a book for j− 275 and sells it at a loss of 15%. How much does she sell it for ? Sol. We have C.P. = j− 275 Loss % = 15% Loss Loss% = × 100 C.P. Loss 15 = × 100 275 15 × 275 Loss = 100 = j− 41.25. S.P. = C.P. – Loss = 275 – 41.25 Hence, S.P. = j− 233.75 Ex.75 Juhi sells a washing machine for j− 13,500. She loses 20% in the bargain. What was the price at which she bought it ? Sol. We have S.P. = j− 13500 Loss %  C.P. − S.P.  Loss = Loss% = × 100  100 C.P.  C.P.  100 × S.P. 100 × 13500 C.P. = = (100 − loss%) 100 – 20 100 × 13500 = Hence C.P. = j− 16,875 80 Ex.76 I bought a T.V. for j− 10,000 and sold it at a profit of 20%. How much money do I get for it? Sol. We have, C.P. = j− 10,000 Profit % = 20% Pr ofit Profit % = × 100 C.P. Pr ofit % × C.P. 20 × 10,000 Profit = = 100 100 Profit = j− 2000 S.P. = C.P. + Profit = j− (10,000 + 2000) = j− 12,000 Hence I got j− 12000 for T.V. Ex.77 An article was bought for j− 400 and sold for j− 350. Find the loss and loss percent. Sol. We have C.P = j− 400 S.P = j− 350 As C.P. > S.P. Loss = C.P. – S.P. = j− (400 – 350) Loss = j− 50 Loss Loss% = × 100 C.P 50 = × 100 400 Loss % = 12.5% Ex.78 An article was purchased for j− 500 and sold for j− 550. Find the gain and gain percent. Sol. We have, C.P. = j− 500 S.P. = j− 550 As S.P. > C.P. ∴ Profit = j− 50 Pr ofit Profit % = × 100 C.P. 50 = × 100 = 10% 500 Hence, Profit % = 10%