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MATHEMATICS IN EVERYDAY LIFE–7, Chapter 10 : Percentage and Its Applications, EXERCISE 10.1, 1., , 75 3, , 100 4, , (i) 75% =, , (ii) 0.25% =, (iii), , 4., , 25, 1, 1, 0.25, , , =, 100 100 400, 100, , (iv) 26% =, , 26 13, , = 13 : 50, 100 50, , 40, 1, 2, 1, 40, , , (v) 13 % =, % =, = 2 : 15, 3, 100, 15, 3, 3, 125, = 1.25, 100, , (i) 125% =, (ii) 45% =, , 45, = 0.45, 100, , (iii) 2.5% =, , 2.5, = 0.025, 100, , (iv), , 1 × 25, 1, 1, 1, 25, % =, =, =, =, 4, 4  100 400, 400 × 25, 10000, , = 0.0025, (v) 6, , 1, 25, 25, % =, % =, 4, 4, 4  100, =, , 25 × 25, 625, =, = 0.0625, 400 × 25, 10000, , Mathematics In Everyday Life-7, , 3 × 100 5, , = 4, , × 87, , 29, , 1, kg, 2, , 27,  9, ,  6  litres =, (vi) 9% of 6 litres = , litre., 50,  100, , , 0.25, 25, 1, , =, = 1 : 400, 100, 100  100 400, , 25 1,  =1:4, 100 4, , 1, , 1, , 4, 29, = 5, 5, 5, 7.5,  75  60 , (v) 7.5% of 60 kg =, × 60 kg = ,  kg, 100,  10  100, , , 0.55, 55, 11, =, =, 100 10000 2000, , (iii) 25% =, , 20, , =, , 12, 3, =, = 3 : 25, 100, 25, , (i) 12% =, , (ii) 0.25% =, , 3., , 12,  800 = 96, 100, , 2, 20, (iv) 6 % of 87 =, % of 87 =, 3, 3, , 140 7, , (iv) 140% =, 100 5, , 2., , 70, (i) 70% of `500 = ` ,  500  = `350,  100, , 25 × 96, 6, 0.25, (ii) 0.25% of 96 =, × 96 =, =, 100, ×, 100, 25, 100, , (iii) 12% of 800 =, , 15, 3, 15, , % =, 2  100 40, 2, , (v) 0.55% =, , ANSWER KEYS, , 5., , (i) x% of 125 is 8., , , , x,  125 = 8, 100, , x=, , 8  100 32, 2, , 6, 125, 5, 5, , Hence, the value of x is 6, , 2, ., 5, , (ii) x% of 1 kg is 50 g., x, × 1 kg = 50 g, 100, x, , × 1000 g = 50 g, (1 kg = 1000 g), 100, 50 × 100, , x=, =5, 1000, Hence, the value of x is 5., (iii) x% of 15200 is 1824., , , , x, × 15200 = 1824, 100, 1824 × 100, , x=, 15200, , x = 12, Hence, the value of x is 12., , , , 1

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(iv) x% of 110 is 11., , , x, × 110 = 11, 100, , 11  100, 110, , x = 10., Hence the value of x is 10., (v) x% of `320 is `272., , , , , , x=, , 7., , 65,  100 % = 65%, (i) 0.65 =, 100, , x=, , 50, 2, = 16, 3, 3, ,  15  75 ,  15  3 , 45,  `, `, = `, , , 4,  100 ,  4 , , = `11.25, Hence, 15% less than `75 = `75 – `11.25, = `63.75., ,  20  90 , 11. 20% of `90 = ` ,  = `18,  100, , , 12., , 1, (i) 12 % of x = 600, 2, , , 3,  100 % = 30%, 10, , , , (i) Let x% of 5 kg be 50 g., Then,, , , , 50  100, , x=, 5000, , x= 1, Hence, 1% of 5 kg is 50 g., (ii) Let x% of 8.5 m is 17 m., Then,, , , x,  8.5 m = 17 m, 100, 17  100, x=, 8.5, x = 200, , , Hence, 200% of 8.5 m is 17 m., 9. Let x% of 2 days be 8 hours. Then,, x, × 2 days = 8 hours, 100, , x, × 48 hours = 8 hours, 100, , 25, % of x = 600, 2, 25,  x = 600, 2  100, , 600  200, 25, , x = 4800, Hence, the value of x is 4800., , x,  5 kg = 50 g, 100, x,  5000 g = 50 g, 100, , , , 2, % of 2 days is 8 hours., 3, , Hence, 20% more than `90 = `90 + `18, = `108., , 7,  100 % = 7%, (ii) 0.07 =, 100, , 2, , , ,  15  75 , 10. 15% of `75 = ` , ,  100, , , x=, , 3, ,  4 × 100 % = (3 × 25)% = 75%, , 8., , 8  100, 48, , x, × `320 = e`272, 100, , 6., , (iii) 0.3 =, , x=, , Hence, 16, , 272  100, 320, , x = 85, Hence, the value of x is 85., , , , , , ( 1 day = 24 hours), , x=, , (ii) 3.7% of x = 74, , , 3.7, of x = 74, 100, , , , 37,  x = 74, 10  100, , 74  1000, = 2000, 37, Hence, the value of x is 2000., 13. 6% of x = 84, , , , , , x =, , 6,  x = 84, 100, , 84  100, 6, , x = 1400, Hence, the value of x is 1400., 14. Let the required number be x., , , , x=, , Then,, 40% of x = 8, Answer Keys

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40, x =8, 100, 8  100, , x=, 40, , x = 20, Hence, the number is 20., 15. Let the distance be x km. Then, , , , , , , , 1, 3 % of x km = 3 km, 3, 10, % of x km = 3 km, 3, 10, x = 3, 3  100, 3  3  100, x=, 10, x = 90, , , Hence, the distance is 90 km., 16. Let the money be `x. Then, 13% of `x = `156, , , , 4. Let population of a town be x., Increase in population every year = 10.5% of x., 10.5, ×x, 100, , =, , 105, 21x, ×x=, 1000, 200, , Increased population = x +, , 21x, 221x, =, 200, 200, , Therefore,, 221x, = 795600, 200, 795600 × 200, 221, , x = 72000, Hence, before a year the population of the town, is 72000., , , , x=, , 5. Let the maximum marks be x., Then, passing marks = 36% of maximum marks, , 13, × `x = `156, 100,  156  100 , x= , , 13, , , x = (12 × 100), , , , x = 1200, Hence, the money is `1200., , =, , 36, x, 100, , 36x, 100,  He scores 109 marks and fails by 35 marks., As per question,, =, , 36x, = 109 + 35, 100, , EXERCISE 10.2, 1. Total number of cycle parts, Number of defective parts, , = 124, =8, 8, Percentage of defective parts =,  100 %, 124, , 200, % = 6.45%, 31, Hence, there are 6.45% defective parts of cycle., 2. Total number of voters = 150, , =, , , , 9,  100 %, =, 150, = 6%, Hence, 6% of voters did not cast their votes., 3. Total number of students in the class = 40, , Number of absent students = 8, Number of present students = 40 – 8 = 32, 32,  100 = 80%, Percentage of present students =, 40, Hence, 80% students were present in the class., , 36x, = 144, 100, , 144  100, 36, , x = 400, Hence, the maximum marks is 400., 6. Let the original price of the book be `x., , , x=, , Then, increase in price = 12% of `x, , Number of voters who did not cast their votes = 9, Percentage of voters who did not cast their votes, , Mathematics In Everyday Life-7, , =, , =`, , 12, x, 100, , =`, , 12 x, 100, , 112 x, , 12 x , Increased price = `  x ,  =`, 100, 100 , , As per question,, 112 x, = `145.60, 100, , , ,  145.60 × 100 , x= , , 112, , 3

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 14560 , x= , ,  112 , , x = 130, Hence, the original price of the book is `130., ,  Saving = `2520, , , , , , 7. Let the original salary be `x., Increase in salary = 20% of `x, , 20 x, 100, , 20x,  120 x  x ,  =`, = `,  100, , 100, Percentage of reduction, 20 x, `, 100  100, =, 120 x, `, 100, , 20,  100 %, 120, , 100, 50, =, %=, %, 6, 3, , 2, = 16 %, 3, , Out of 1, mark obtained by Anil =, , 8. Let the total monthly income be `x., Spent on house rent = 20% of `x, , 20, × `x, 100, 20x, x, =`, =`, 100, 5, =, , , x, 4x, Remainder = `  x   = `, , 5, 5, , 4x, 5, ,  60 4 x , 12 x, ×  =`, = `,  100 5 , 25, , 375, × 100 = 75%, 500, Out of 600, marks obtained by Sunil = 420, , Out of 100, marks obtained by Anil =, , Out of 1, mark obtained by Sunil =, , 420, 600, , 420, × 100 = 70%, 600, Anil secured 75% marks and Sunil secured 70%, marks., , Out of 100, marks obtained by Sunil =, , Hence, Anil’s performance is better than Sunil., 10. Total number of students = 1800, Number of girls = 40% of 1800, , 40,  1800, 100, = 40 × 18, = 720, Thus, the number of boys = 1800 – 720, = 1080, Hence, there are 1080 boys in the school., 11. Percentage marks =, , 45, × 100%, 75, , 3, × 100% = 60%, 5, Hence, Mohnish obtained 60% marks in Mathematics., 12. Total income = `25000, , =, , Expenditure = `16500, Saving = `25000 – `16500 = `8500, Percentage of saving =, , Now,,  4 x 12 x , , Saving = ` , , 25 ,  5, ,  20 x  12 x , 8x, = `,  = ` 25, 25, , 4, , 375, 500, , =, , 2, Hence, the new salary should be reduced by 16 %, 3, to restore the original salary., , Household expenses = 60% of `, , 8x, = 2520, 25, ,  2520  25 , x= , , 8, , , x = 7875, Hence, total monthly income is `7875., 9. Out of 500, marks obtained by Anil = 375, , 20 x , 120 x, , Increased salary = `  x ,  =`, 100 , , 100, Reduction in salary to restore the original salary, , =, ,  8x , `   = `2520,  25 , , , ,  20  x , = `, ,  100, , = `, , , , =, , `8500, × 100, `25000, 8500, × 100%, 25000, , 850, % = 34%, 25, Hence, he saves 34% of his income., =, , Answer Keys

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13. Let the number be 100., Increase in the number = 10%,  10, , Increased number = 100 + , × 100,  100, , , = 100 + 10, = 110, Now, decrease = 10% of 110, =, , 1, Percentage reduction = 6 × 100%, 1, , 100, 50, 2, %=, % = 16 %., 6, 3, 3, Therefore, she must decrease her consumption by, , =, , 16, , 10,  110, 100, , 2, % so as not to increase her expenditure on it., 3, , 16. Let B’s income be `100,  A’s income = `100 + 20% of `100, = `100 + `20 = `120, Difference in salary = `120 – `100 = `20, , = 11, New number = 110 – 11 = 99,  The number is decreased., Net decrease = 100 – 99 = 1, , 20, × 100%, 120, , Required Percentage =, ,  1, , Net decreased per cent = , × 100 % = 1%,  100, , , ,  decrease,  ,  100 % ,  Percentage decrease = ,  original value,  , , 14. Let the original price of a shirt be `x., Reduce in price = 15% of `x, , =, , 100, 2, % = 16 %, 6, 3, , 2, % less than that of A., 3, 17. Let the total value of the land be `x., , Hence, B’s income is 16, , Commission on sale of the land = 6% of `x =, ,  15  x , = `, ,  100, , , = `, ,  3x , = ` ,  20 , According to question,,  3x , `x – `   = `123.25,  20 , , , , 20 x  3x, = 123.25, 20, , 123.25  20, 17, , x = 145, Hence, the original price of a shirt is `145., 15. Let the price of sugar be `100 per kg., Increase in price = 20%, Increased price = `100 + 20% of `100 = `120, In `120, she gets 1 kg of sugar., In `1, she gets, , x=, , Mathematics In Everyday Life-7, , , , x=, , 5688  100, 6, x = 94800, ,  18  100 ,  kg, = ,  100, , = 18 kg, Now,, If copper is 18 kg, then ore = 100 kg,  If copper is 1 kg, then ore =,  If copper is 63 kg, then ore =, , 100, 5, kg =, kg of sugar, 120, 6, , Reduction in consumption = 1 kg –, ,  6x , `, = `5688,  100 , , 100, kg, 18, 50, , 1, kg of sugar, 120, , In `100, she will get =, , , , 6x, 100, , Hence, the total value of land is `94800., 18. Let the total amount of ore be 100 kg., Quantity of copper = 18% of 100 kg, , 17 x, = 123.25, 20, , , , 6, × `x, 100, , 5, 1, kg =, kg, 6, 6, , 100, 7,  63, 18 2, , = 50 × 7, = 350 kg, Hence, 350 kg of ore contained 63 kg of copper., , 5

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19. Original value of car = `3,45,000, Decrease in value for first year = 20% of `3,45,000, , =, , = 6, , 2, %, 3, , Hence, profit per cent is 6, , 2, %., 3, ,  20, ,  3, 45, 000 ,  100, , , = `, , = `69,000, The value of car for next year = `(3,45,000 – 69,000), = `2,76,000, Now, decrease in value of car for second year, = 20 % of `2,76,000, ,  20  2,76,000 , , = `,  100, , = `55,200, Hence, the value of car after 2 years, = `(2, 76,000 – 55,200), = `2,20,800., , 3. Let the cost price of each article be `x., S.P. of 12 articles = C.P. of 16 articles = `16x, C.P. of 12 articles = 12x, Since,, S.P. > C.P., , Gain = S.P. – C.P., = `(16x – 12x), = `4x,  Gain, ,  100 %, Gain per cent = ,  C.P., , ,  4x  100  %, = , ,  12 x, , , EXERCISE 10.3, 1. C.P. of a chair = `700, S.P. of a chair = `570, Since,, S.P. < C.P., , Loss = C.P. – S.P., = `700 – `570, = `130,  Loss, ,  100  %, Now, loss per cent = , C.P., , , ,  130  100  %, , = ,  700, , =, , 130, 4, %  18 %, 7, 7, , 4, Hence, loss per cent is 18 % ., 7, 2. C.P. of a sewing machine = `1250, Overhead expenses = `250, Effective cost price = `(1250 + 250) = `1500, S.P. of sewing machine = `1600, , S.P. > C.P., , Profit = S.P. – C.P., = `1600 – `1500, = `100,  Profit, ,  100  %, Profit per cent = ,  C.P., , ,  100  100  %, = , ,  1500, , 6, , 100, 20, %, %, 15, 3, , =, , 100, %, 3, , 1, = 33 %, 3, 1, Hence, gain is 33 % ., 3, , 4. Let the C.P. of each pencil be `x., Then,, S.P. of 10 pencils = C.P. of 12 pencils = `12x, C.P. of 10 pencils = `10x, Since,, S.P. > C.P., , Profit = S.P. – C.P., = `(12x – 10x), = `2x, ,  Profit, ,  100  %, Profit per cent = ,  C.P., ,  2x  100  %, = , ,  10 x, , = 20%, Hence, profit per cent is 20%., 5. Let the cost price be `100., If it is sold at a loss of 5%,, then, S.P. = `(100 – 5) = `95, If it is sold at a profit of 7%., Then, S.P. = `(100 + 7) = `107, Increase in actual gain = `(107 – 95) = `12, Answer Keys

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If gain is `12, then C.P. = `100, If gain is `1, then C.P. = `, , 100, 12, ,  100  27 , If gain is `27, then C.P. = ` , ,  12, , = `225, Hence, the C.P. of watch is `225., 6. C.P. of an article = `1960,  Profit, ,  100  %, Profit per cent = , C.P., , , 10 =, , Profit,  100, 1960, , 1960  10, 100, Profit = `196, , , , Profit =, , , Thus, S.P. = C.P. + Profit = `1960 + `196 = `2156., 7. C.P. of a table = `1250.,  Loss, ,  100 %, Loss% = ,  C.P., ,  Loss, ,  100, 15 = ,  1250, , , , Loss =, , 15  1250, 100, , Loss = `187.50, Hence, S.P. of the table = C.P. – Loss, = `1250 – `187.50, = `1062.50, Hence, S.P. of a table is `1062.50., 8. S.P. of flower vase = `600, Gain % = 15%, We have,, , ,  S.P.  100 , C.P. = , ,  (100  gain%) ,  600  100 , = `, ,  (100  15) , ,  60000 , = `, ,  115 , C.P. = `521.74, Hence, the cost price of a flower vase is `521.74., 9. Let the cost price be `x., ,  20 x   ` x, Loss = 20% of `x = ` , , 5,  100 , But, it make a loss of `90., Mathematics In Everyday Life-7, , x, = `90, 5, x = `450, S.P. = C.P. – Loss, , , , `, , , , S.P. = `(450 – 90) = `360, (i), S.P. = `360, (ii), C.P. = `450., 10. S.P. of an article = `1600, Loss = 20%, We have,,  S.P.  100 , , C.P. = ,  100  loss% , ,  1600  100 , = `  100  20 , , , ,  160000 , , = `,  80 , C.P. = `2000, Now, required gain = 20%,  C.P. × (100 + Gain %) , ,  S.P. = , 100, , , 2000, , (100, +, 20), , , = `, , 100, , , = `{20 × 120}, = `2400, Hence, he should sell the article for `2400 to make a, gain of 20%., 11. Let C.P. of a watch be `x., , 1, of C.P.., 6, 1, x, =, × `x = `  , 6, 6, S.P. = C.P. + Gain, , Gain =, , Now,, , x, , `1505 = `  x  , 6, , 7x, , = 1505, 6, 1505  6, , x=, 7, , x = 1290, Hence, the C.P. of the watch is `1290., 12. Let the S.P. of one dozen pencils be `x., The S.P. of 8 dozen pencils = `8x,  Gain = S.P. of one dozen pencils,  Gain = `x, Then, C.P. of 8 dozen pencils = S.P. – Gain, C.P. = `(8x – x) = `7x, 7

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 Gain, ,  100  %, Now, Gain % = , C.P., , , ,  x  100  %, = , ,  7x, , , 100, 2, %  14 %, 7, 7, 2, Hence, Gain % = 14 %, 7, 13. C.P. of lemons = `600, =, , , , 3, of lemons sold at a loss of 20%., 4, , C.P. of, , 3, 3, of lemons =, × `600 = `450, 4, 4, ,  20, , × 450, Then, Loss = ` ,  100, , , Loss = `90, And remaining lemons sold at a gain of 20%., Then, C.P. of remaining lemons = `600 – `450 = `150, Gain = `(20% × 150),  20, , × 150, = `,  100, , , = `30, Now, S.P. of lemons = `(600 – 90 + 30), = `540, Now, for whole transaction., Therefore,, , C.P. > S.P., Loss = C.P. – S.P., ,  C.P. × (100 + Gain %) , S.P. = , , 100, , ,  7000 × (100  10) , S.P. = ` , , 100, , ,  7000 × 110 , = `, , 100, , , S.P. = `7700, Hence, new selling price is `7700., 15. C.P. of machine = `3750, Overhead expenses = `250, Effective C.P. = `(3750 + 250), = `4000, Gain = `40,  Gain, ,  100  %, Gain % = , C.P., , , ,  40  100  %, = , ,  4000, , = 1%, Hence, gain per cent is 1%., 16. S.P. = `3600, Profit = 20%, Then,, ,  S.P. × 100 , C.P. = , ,  (100 + Gain%) , , = `(600 – 540), = `60, ,  3600 × 100 , = ` , ,  (100 + 20) , ,  Loss, , Loss % = , × 100,  C.P., , , = ` 360000, 120, C.P. = `3000, If he sold the article for `4000., i.e., S.P. = `4000, Then, Gain = S.P. – C.P., = `(4000 – 3000), = `1000, ,  60  100  %, , = ,  600, , = 10 %, Hence, he lose 10% on the whole transaction., 14. S.P. = `8050, Gain % = 15%, Then,,  S.P. × 100 , C.P. = , ,  100  Gain % ,  8050 × 100 , = `, ,  (100  15) ,  8050 × 100 , = `, , 115, , , C.P. = `7000, 8, , For a profit of 10%., Then,, , , , , ,  1000, ,  100 %,  Gain % = , , 3000,  1000 , %, = ,  30 , 1, = 33 %, 3, 1, Hence, gain% would be 33 % ., 3, Answer Keys

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EXERCISE 10.4, 1. Principal (P) = `1800, Time (T) = 2 years, Rate (R) = 12% per annum, S.I. =, , 5. P = `15000, R = 16% per annum,, T= 3, , P×R×T, 100, , S.I. =, ,  1800 × 12 × 2 , = `, , 100, , , = `(18 × 12 × 2) = `432, Hence, interest is `432., 2. P = `5600, T = 3 years, R = 12% per annum, S.I. =, , P×R×T, 100, ,  5600 × 12 × 3 , = `, , 100, , , = `(56 × 12 × 3), S.I. = `2016, Now,, Amount = Principal + Interest, = `(5600 + 2016), = `7616, Hence, he will pay `7616 after 3 years., 1, 29, % per annum,, 3. P = `3800, R = 14 % =, 2, 2, 1, 5, T = 2 years =, years, 2, 2, P×R×T, S.I. =, 100, , 29 5 , 3800 ×, ×, , 2 2, = `, , 100, , , , , ,  3800 × 29 × 5 , = `,  100 × 4 , , 7, 1, years =, years, 2, 2, P×R×T, 100, ,  15000 × 16 × 7 , = `, , 100 × 2, , , S.I. =, Now, Amount =, =, =, , `8400, Principal + Interest, `(15000 + 8400), `23400, , Hence, he paid `23400 after 3, , 1, years., 2, , 6. P = `7500, R = 15% per annum,, T= 5, , 3, 23, years =, years, 4, 4, , S.I. =, , P×R×T, 100, ,  7500 × 15 × 23 , = `, 100 × 4 , , , S.I. = `6468.75, , , Amount = Principal + Interest, = `(7500 + 6468.75), = `13968.75, , Hence, Surabhi will get an amount of `13968.75., 7. P = `6500, R = 9% per annum,  6500 × 9 × 1, Interest for 1 year = ` , , 100, , , = `(65 × 9) = `585, Amount after 1 year = `(6500 + 585), ,  38 × 29 × 5 , = `, , 4, , , Now,, , S.I. = `1377.50, Amount = Principal + Interest, = `(3800 + 1377.50), = `5177.50, , 1, Hence, he paid `5177.50 after 2 years., 2, , 4.  Principal = Amount – Interest, = `3600 – `200, Principal = `3400, Mathematics In Everyday Life-7, , = `7085,  Priyanka withdrew `1250 after 1 year. Therefore,, , Balance amount = `(7085 – 1250), = `5835, Now, interest for next two years on balance amount,  5835 × 9 × 2 , = `, , 100, , , = `1050.30, Hence, the amount after 3 years = `(5835 + 1050.30), = `6885.30, 9

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8. P = `17000, R = 12% par annum, T = 1 year, S.I. =, , 11. R = 10% per annum, T = 6 years, , P×R×T, 100, , S.I. =, ,  17000 × 12 × 1, = `, , 100, , , S.I. = `2040, Income tax deduction = 30% of S.I., 30, = ` ,  2040 ,  100, , = `612, , Hence, the net annual income, = `(2040 – 612), = `1428, Hence, the net annual income is `1428., 9. Let the principal be `P., R = 10% per annum, T = 5 years, P×R×T, S.I. =, 100, , P  10  5, P, S.I = `, =`, 100, 2, Amount = Principal + Interest, , P, `3600 = `  P  , , 2, , , , 3P, = 3600, 2, , 3600  2, 3, P = 2400, Hence, the principal is `2400., 10. P = `1200, R = 8% per annum, T = ?, , , , , , , Now,, , P=, , Amount = 2P = `2400, Interest = Amount – Principal, = `(2400 – 1200), = `1200, P×R×T, 100, , Interest =, 1200 =, , , Hence, after 12, , 10, , T=, , 2160 =, , , 1, years the amount will double itself., 2, , P × 10 × 6, 100, 2160 × 100, 10 × 6, , P = `3600, Hence, principal is `3600., 12. Let the sum be `P., T = 5 years, R = 10% per annum, P, P × 10 × 5, S.I. =, =`, 2, 100, Now, T = 4 years, R = 16% p.a, P × 16 × 4, 16P, =`, 100, 25, P, 16P, =, – 196, 2, 25, 16P P, , = 196, 25, 2, , S.I. =, , Now,, , , , 32P  25P, = 196, 50, , , , 7P, = 196, 50, 196  50, 7, P = 28 × 50, , , , P=, , , , P = 1400, Hence, the principal amount is `1400., 13. P = `1200, A = `1600, T = 6 years, R = ?,  Interest = Amount – Principal, Interest = `(1600 – 1200) = `400, S.I. =, , P×R×T, 100, , , , 400 =, , 1200 × R × 6, 100, , , , R=, , 1200 × 8 × T, 100, 100, 1, = 12, years, 8, 2, , P=, , P×R×T, 100, , =, , , 00 × 100, 1200 × 6, , 400, 50, =, 9, 72, , 5, R = 5 % per annum, 9, , Hence, rate of interest is 5, , 5, % per annum., 9, Answer Keys

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14. P = `12300, R = 10% per annum, Amount = `18450 T = ?, Interest = Amount – Principal, = `(18450 – 12300), Also,, = `6150, S.I. =, 6150 =, , , T=, ,  3x , = ` ,  4 , , P×R×T, 100, , , , 12300 × 10 × T, 100, , , , 6150 × 100, 615, =, 123, 12300 × 10, , T = 5 years, Hence, `12300 will amount to `18450 at 10% per, annum in 5 years., 15. P = `5000, R = 16% per annum, T = 2 years, S.I. =, , , x, Selling price of the shirt = `  x  , , 4, , P×R×T, 100, ,  5000 × 16 × 2 , = `, , 100, , , = `(50 × 16 × 2), S.I. = `1600,  Amount = Principal + Interest, = `(5000 + 1600) = `6600, , MULTIPLE CHOICE QUESTIONS,  40, , 1. 40% of `750 = ` ,  750  = `300,  100, , Hence, option (c) is correct., 2. Let the total number of students be x., Then, number of girls = 40% of x, 40, x, 100, 0 x, =, 100, , =, , 0 x, = 150, 100, 150  100, , x=, 40, , x = 375, Total number of students is 375., Thus, the number of boys = 375 – 150 = 225, Hence, option (a) is correct., 3. Let the marked price be `x., Discount = 25% of `x, , `, , 3x, = `609, 4, , 609  4, 3, = 203 × 4 = 812, The marked price of the shirt = `812, , x=, , Hence, option (b) is correct., 4. Let the number be x., As per question,, (42% of x) + 29 = x, , , ,  42  x   29 = x, , ,  100, , , x, , 42x, = 29, 100, , 58 x, = 29, 100, 2900, x=, 58, x = 50, , , , , The number is 50., Hence, option (b) is correct., 5. x% of `75 is `9., Then,, ,  x  75 , ,  =9,  100, , , 900, 75, x = 12, Hence, option (a) is correct., , , , x=, , 6. Let the number be x., 12% of x =, , 12, 3x, x, 100, 25, , 8% of x =, , 8, 2x, x, 100, 25, , But,, , x,  25, , = `,  x  = ` , 4,  100, , Mathematics In Everyday Life-7, , As per question,, 3x, 2x,  20, =, 25, 25, , 3x 2x, , = 20, 25 25, x, = 20, 25, x = 500, Hence, option (d) is correct., 11

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7. Let the original price be `x., ,  8, , x, Reduction in value = 8% of `x = ` ,  100, , = `, , 10. Let the number be x. Then, x + (20% of x) = 54, , 8x, 100, , , , 92 x, , 8x , New value = `  x –,  =`, 100, 100 , , , As per question,, , 92 x, = 115, 100, 115  100, , x=, 92, x = 125, Hence, option (a) is correct., , , , 8. Let maximum marks be x., Passing marks = 33% of x, =, , 33, x, 100, , =, , 33 x, 100, , 54  5, =9×5, 6, x = 45, Hence, option (b) is correct., 11. 92% of the total examines passed., , 24  100, 8, × x = 24  x =, = 300, 100, 8, Hence, option (d) is correct., , , , 12. Let the total number of pens be x., As per question,, x = 60% of x + 120, , , 33 x, = 140 + 25, 100, , , , 33x, = 165, 100, , , , , , P×R×T, 100, , 2P =, , P × R × 10, 100, , R, 10, R = 20% per annum, , 2=, , , Hence, option (c) is correct., 12, , x, , 60 x,  120, 100, , 6x, = 120, 10, 4x, = 120, 10, , x=, , MENTAL MATHS CORNER, Fill in the blanks:, 1. The ratio 3 : 5 as percentage is 60%., , = 3P – P, = 2P, S.I. =, , x=, , 1200, = 300, 4, Hence, option (a) is correct., , x=, , 9. Let principal be `P. Then, amount is 3P.,  Interest = Amount – Principal, , x=, , i.e., 8% of the total examines failed,  8% of x = 24, , , , 165  100, 33, , x = 500, Hence, option (b) is correct., , x, = 54, 5, , 6x, = 54, 5, , , , As per question, , , , 20x, = 54, 100, , x, , , ,  92 x , `, = `115,  100 , , , , x, , , ,  3:5=, , 2., , 6, as percentage is 75%., 8, , , , 3., , 3,  100 % = 60%, 5, , 6, 6, =  100 % = 75%, 8, 8, , 1, 2, ., 6 % expressed as fraction is, 15, 3, , , , 20, 1, 1, 2, , 6 % =, =, 3, 100, 3, 15, Answer Keys

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4. If x% of 25 is 8, then x is equal to 32.,  x% of 25 = 8, x,  25 = 8, 100, x, , = 8, 4, , x = 32, 5. The simple interest of `600 at 6% per annum for 5, months is `15., , , 20x  = `  6 x , New salary = `  x ,  , ,  5 , 100 , , , , ,  P = `600, R = 6% p.a., T = 5 months =, , Interest = `, , 600  6  5, 100  12, , 5, year, 12, , P×R×T, ,  S.I. =, , , 100 , , = `15, 6. A number decreased by 35% gives 39, then the, number is 60.,  Let the number be x., Decrease in number = 35% of x, =, , 35 x, 100, , Therefore,, , , , , x, , 35x, = 39, 100, 65x, = 39, 100, 3900, x=, 65, x = 60, , , 7. By selling a book for `100, a man gains `25., , But,, , , ,  6x , `18000 = `  ,  5 , , 18000  5, 6, , x=, , x = 15000, Thus, original salary is `15000., 9. A man buys an article for `40 and sells it as `60, his, gain % is 50%., Gain = S.P. – C.P., = `(60 – 40) = `20,  20, , Gain % =   100  % = 50%,  40, , 10. 5% of a number is 6, then the number is 120., Let the number be x. Then,, 5% of x = 6, , , , 5x, =6, 100, , , , x=, , 6  100, 5, x = 120, , , , REVIEW EXERCISE, 1. Let the money that boy had be `x initially., Money spent = 45% of `x,  45, ,  x, = `,  100, , , 1, His gain % is 33 % ., 3, , , , , , =`, , C.P. = S.P. – Gain, = `(100 – 25), C.P. = `75, , , 9x , Left money = `  x –, , 20 , , ,  Gain, , Gain % = ,  100  %, C.P., , , 25, =   100  %,  75, , , 1, = 33 %, 3, 8. By increasing the salary of a man by 20% it becomes, `18000, his original salary is `15000., Let the original salary be `x. Then, Increase in salary = 20% of `x, 20 x,  20, ,  x = `, = `, 100,  100, , Mathematics In Everyday Life-7, , 9x, 20, ,  20 x  9 x , = `,  20 , , , , =`, , 11x, 20, , Therefore,, , ,  11x , `,  = `176,  20 , , , , 176  20, = 16 × 20, 11, x = 320, , x=, , Hence, he had `320., 13

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2. S.P. for T.V. set = `7200, Loss % = 25%, Then,, , 5. C.P. for a T.V. set = `12000, Profit = 20%, then, ,  C.P. × (100 + Profit%) , S.P. = , , 100, , , ,  S.P. × 100 , C.P. = , ,  (100  Loss%) , , 12000 × (100 + 20) , = `, , 100, , , , 7200 × 100 , = ` , ,  100  25 , , = ` , , , 7200 × 100 , , 75, , , C.P. = `9600, For a gain of 25%.,  C.P. × (100 + Gain %) , S.P. = ` , , 100, , , , 12000 × 120 , = `, , 100, , , S.P. = `14400, Hence, selling price of a T.V set is `14400, 6. I Watch:, S.P. = `780, Gain = 20%,  S.P. × 100 , C.P. = , ,  (100 + Gain %) , ,  9600 × (100 + 25) , S.P. = ` , , 100, , , , = `(96 × 125), S.P. = `12000, Hence, he must sell for `12000 to make a gain of, 25%., 3. Number of boys in a school = 60%, Number of girls in a school = (100 – 60)% = 40%, Let total number of students be x., Therefore,, 40% of x = 360, , , , ,  780  100 , = `,  100  20 , 780  100 , = ` , ,  120 , , C.P. = `650, II Watch:, S.P. = `780, Loss = 20%,  S.P. × 100 , C.P. = , ,  (100  Loss %) , , 40,  x = 360, 100, x=, , 780  100 , = ` , ,  100  20 , , 360  100, 40, , , x = 900, Hence, number of boys = 900 – 360 = 540., 4. Original population = 25000, Decrease percentage in population,  25000  24500, ,  100  %, = , , 25000, , ,  78000 , = `, ,  80 , , Now,, , , , , , ,  Decrease,  ,  100  % ,  Decrease% = ,  Original value,  , , ,  500, , = ,  100  %,  25000, , 50000, =, % = 2%, 25000, , 14, , C.P. = `975, Total C.P. = `(650 + 975), = `1625, Total S.P. = `1560, S.P. < C.P., Loss = C.P. – S.P., = `(1625 – 1560), Loss = `65, , , ,  65, , Loss % = ,  100 ,  1625, , Loss = 4%, , Answer Keys

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7. Let the sum borrowed by Riya be `P., ,  C.P. × (100 + Gain %) , S.P. = ` , , 100, , , , R = 5%, p.a., T = 2 years, , S.I. =, , P×5×2, P, =, 100, 10, , 1360 × (100 + 15) , = `, , 100, , , , Now, Interest + Amount = Principal, P+, , P, = `1732.50, 10, , 11P, = `1732.50, 10, , , ,  1360 × 115 , = `, , , 100, , , S.P. = `1564, 11. P = `6950, S.I. = `347.50, T = 5 months =, , 17325, 11, , , , P =`, , , , P = `1575, , S.I. =, , Hence, she borrowed `1575., 347.50 =, , 8. Let sum be `P. Then, Amount = 2P, , , R=, , Amount = Principal + Interest, Interest = 2P – P = P, , P × 10 × R, S.I. =, 100, , Now,, , P×R, 10, , , , P=, , , , R = 10%, , P×R×T, 100, 6950  R  5, 100  12, , 347.50 × 100 × 12, 6950 × 5, , R = 12%, Hence, the rate of interest is 12% per annum., , HOTS QUESTIONS, 1. Marks obtained by Rohit in Mathematics = 64% of 100, 64,  100, 100, , =, , Hence, the rate of interest is 10% p.a., 9. Let the original salary of a bank clerk be `x., Increase in salary = 10% of `x, , 5, year, R = ?, 12, , = 64, Marks obtained by Mohit in Mathematics = 80% of 100, , 10, x, = ` ,  x  = `,  100, , 10, , 80,  100, 100, , =, , = 80, , 11x, , x ,  New salary = `  x   = `, 10, 10 , , , Let marks obtained by Shobhit be x., , Therefore,, , Average marks of three students = 76% of 100, , , , , , , `, , 11x, = `8503, 10, x=, , 85030, 11, , =, ,  Average marks of three out of 100 = 76, Now,, , x = 7730, , Hence, the original salary is `7730., 10. C.P. of an article = `1315, Overhead expenses = `45, Effective C.P. = `(1315 + 45) = `1360, Gain = 15%, , Mathematics In Everyday Life-7, , 76,  100  76, 100, , , , , Average marks =, , 64  80  x, 3, , 144  x, 3, x = 228 – 144, , 76 =, , x = 84, Hence, Shobhit scored 84 marks out of 100 i.e. 84%., , 15

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2. Let the sum be `x., , VALUE BASED QUESTION, , For 5 years :, R = 8% p.a., T = 5 years, S.I. =, , P × R × T x  8  5 2x, , , 100, 100, 5, , For 3 years and 4 months :, R = 18% p.a., T = 3 years 4 months, = (3 × 12 + 4) months, =, , 40, years, 12, , =, , 10, years, 3, , P × R × T x  18  10 3 x, S.I. =, , , 100, 100  3, 5, , Now, as per question,, 2x, 3x,  200 =, 5, 5, , , , 3x 2 x, , = 200, 5, 5, , , , x, = 200, 5, , , , x = 200 × 5, , , , x = 1000, , Hence, the sum is `1000., , SUMMATIVE ASSESSMENT, Let C.P. be `x.,  Marked price = `x + (25% of `x), , 25, , = ` x , × x, , , 100, , x, = ` x  , , 4,  5x , Marked price = `  ,  4 , , Discount = 10% of marked price, =, , 10,  5x , × ` ,  4 , 100, ,  10 5 x , x, = `, ×, = ` , ,  100 4 , 8, , Therefore, S.P. = Marked price – Discount,  5x x , = `,  ,  4, 8, ,  10 x  x , = `,  8 ,  9x , S.P. = `  ,  8 , ,  S.P. > C.P.,  Profit = S.P. – C.P.,  9x, , x, = `,  x = `  ,  8, , 8,  Profit, , Profit % = , × 100  %,  C.P., , , x, , 8, , =  × 100  %, x, , , , , 1, , 1, =  × 100  % = 12 %, 8, , 2, , Hence, Tradesmen profit 12, , 1, %., 2, , No, the Tradesmen did not showing honesty in his, transaction., He misguide the customer., , 16, , Answer Keys