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Knowing Our Numbers, Exercise 1.1, Question 1., Fill in the blanks:, (a) 1 lakh = ………….. ten thousand., (b) 1 million = ………… hundred thousand., (c) 1 crore = ………… ten lakh., (d) 1 crore = ………… million., (e) 1 million = ………… lakh., Solution (a) 1 lakh = ten ten thousand., (b) 1 million = ten hundred thousand., (c) 1 crore = ten ten lakh, (d) 1 crore = ten million, (e) 1 million = ten lakh, Question 2., Place commas correctly and write the numerals:, (a) Seventy-three lakh seventy-five thousand three hundred seven., Ans. 73,75,307, (b) Nine crore five lakh forty-one., Ans. 9,05,00,041, (c) Seven crore fifty-two lakh twenty-one thousand three hundred two., Ans. 7,52,21,302, (d) Fifty-eight million four hundred twenty- three thousand two hundred two., Ans. 5,84,23,202, (e) Twenty-three lakh thirty thousand ten., Ans. 23,30,010, Question 3., Insert commas suitably and write the names according to Indian System of Numeration:, , (a) 87595762, Ans. 8,75,95,762 (Eight crore seventy-five lakh ninety-five thousand seven hundred, sixty- two), (b) 8546283
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Ans., , 85,46,283 (Eighty-five lakh forty-six thousand two hundred eighty-three), , (c) 99900046, Ans. 9,99,00,046 (Nine crore ninety-nine lakh forty-six), (d) 98432701, Ans. 9,84,32,701 (Nine crore eighty-four lakh thirty-two thousand seven hundred one), Question 4., Insert commas suitably and write the names according to International System of, Numeration:, , (a) 78921092, Ans. 78,921,092 (Seventy-eight million nine hundred twenty-one thousand ninetytwo), (b) 7452283, Ans. 7,452,283 (Seven million four hundred fifty- two thousand two hundred eightythree), (c) 99985102, Ans. 99,985,102 (Ninety-nine million nine hundred eighty-five thousand one hundred, two), (d) 48049831, Ans. 48,049,831 (Forty-eight million forty-nine thousand eight hundred thirty-one), , Exercise 1.2, Question 1., A book exhibition was held for four days in a school. The number of, tickets sold at the counter on the first, second, third and final day, was respectively 1094, 1812, 2050 and 2751. Find the total number, of tickets sold on all the four days., Solution:, Number, Number, Number, Number, , of, of, of, of, , tickets, tickets, tickets, tickets, , sold, sold, sold, sold, , on, on, on, on, , the, the, the, the, , first day, second day, third day, final day, , = 1094, = 1812, = 2050, = 2751, , ∴ Total number of tickets sold on all the four days = 1094 + 1812 + 2050 +, 2751, = 7,707.
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Question 2., Shekhar is a famous cricket player. He has so far scored 6980 runs in, test matches. He wishes to complete 10,000 runs. How many more, runs does he need?, Solution:, Shekhar has so far scored 6980 runs, He wishes to complete 10,000 runs., Therefore total number of runs needed by him, = 10,000 – 6980, = 3020 runs, , Question 3., In an election, the successful candidate registered 5,77,500 votes, and his nearest rival secured 3,48,700 votes. By what margin did the, successful candidate win the election?, Solution:, Number of votes secured by the successful candidate, = 5,77,500, Number of votes secured by his nearest rival, = 3,48,700, Therefore, margin of votes to win the election, 3,48,700, , = 5,77,500 –, = 2,28,800, , Question 4., Kirti bookstore sold books worth Rs 2,85,891 in the first week of, June and books worth Rs 4,00,768 in the second week of the month., How much was the sale for the two weeks together? In which week, was the sale greater and by how much?, Solution:, Books sold in first week of June worth Rs 2,85,891, Books sold in second week of the month worth Rs 4,00,768, Therefore, total sale of books in the two weeks together = Rs 2,85,891 + Rs, 4,00,768, = Rs 6,86,659, In the second week of the month, the sale of books was greater.
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Difference of the sale of books, 2,85,891, , `, , = Rs 4,00,768 – Rs, = Rs 1,14,877, , Hence, in second week of june, the sale of books was more by Rs 1,14,877., , Question 5., Find the difference between the greatest and the least numbers that, can be written using the digits 6, 2, 7, 4, 3 each only once., Solution:, Given digits are 6, 2, 7, 4, 3, Greatest number, Least number, , = 76432, = 23467, , Therefore, difference, = 76432 – 23467, = 52,965, , Question 6., A machine, on an average, manufactures 2,825 screws a day. How, many screws did it produce in the month of January, 2006?, Solution:, Number of screws manufactured in a day, = 2,825., Number of screws manufactured in month of January, = 31 x 2825, = 87,575, , Question 7., A merchant had Rs 78,592 with her. She placed an order for, purchasing 40 radio sets at Rs 1200 each. How much money will, remain with her after the purchase?, Solution:, Amount of money with the merchant, Number of radio sets, Price of one radio set, Therefore, cost of 40 radio sets, , = Rs 78,592, = 40, = Rs 1200, , = Rs 1200 x 40, = Rs 48,000
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Remaining money with the merchant = Rs 78,592 – Rs 48000, = Rs 30,592, Hence, amount of Rs 30,592 will remain with her after purchasing the radio, sets., , Question 8., A student multiplied 7236 by 65 instead of multiplying by 56. By how, much was his answer greater than the correct answer?, Solution:, Student has multiplied 7236 by 65 instead of multiplying by 56., Difference between the two multiplications = (65 – 56) x 7236, = 9 x 7236, = 65124, (We don’t need to do both the multiplied), Hence, the answer greater than the correct answer is 65,124., , Question 9., To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how, many shirts can be stitched and how much cloth will remain?, Solution:, Total length of the cloth, , = 40 m, = 40 x 100 cm, = 4000 cm., , Cloth needed to stitch a shirt, = 2 m 15 cm, = 2 x 100 + 15 cm, = 215 cm, Therefore, number of shirts stitched = 4000215, So, the number of shirts stitched = 18, and the remaining cloth = 130 cm, = 1 m 30 cm, Question 10., Medicine is packed in boxes, each weighing 4 kg 500 g. How many, such boxes can be loaded in a van which cannot carry beyond 800
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kg?, Solution:, Weight of one box = 4 kg 500 g, = 4 x 1000 + 500, = 4500 g, and 800 kg = 800 x 1000, = 800000 g, Therefore, 177 boxes can only be loaded in the van., , Question 11., The distance between the school and the house of a student is 1 km, 875 m. Everyday she walks both ways. Find the total distance, covered by her in six days., Solution:, Distance between school and house, , = 1 km 875 m, = (1000 + 875) m, = 1875 m., , Distance travelled by the student in both ways = 2 x 1875, = 3750 m, Distance travelled in 6 days, , = 3750 m x 6 – 22500 m, = 22 km 500 m., , Hence, total distance covered in six days = 22 km 500 m., , Question 12., A vessel has 4 litres and 500 ml of curd. In how many glasses, each, of 25 mL capacity, can it be filled?, Solution:, Quantity of curd in a vessel, , Capacity of 1 glass, , = 4 1 500 mL, = (4 x 1000 + 500) mL, = 4500 mL., , = 25 mL, , Therefore number of glasses, , = 4500 / 25
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= 180, , Exercise 1.3, Question 1., Estimate each of the following using general rule:, (a) 730 + 998, Rounding off 730 nearest to hundreds = 700, Rounding off 998 nearest to hundreds = 1,000, ∴ 700 + 1000 = 1700, , (b) 796 – 314, Rounding off 796 nearest to hundreds = 800, Rounding off 314 nearest to hundreds = 300, ∴ 800 – 300 = 500, , (c) 12,904 + 2,888, Rounding off 12,904 nearest to thousands = 13000, Rounding off 2888 nearest to thousands, = 3000, ∴ 13000 + 3000 = 16000, , (d) 28,292 – 21,496, Rounding off 28,292 nearest to thousands = 28,000, Rounding off 21,496 nearest to thousands = 21,000, ∴ 28,000 – 21,000 = 7,000, , Question 2., Give a rough estimate (by rounding off to nearest hundreds) and, also a closer estimate (by rounding off to nearest tens):, (a)439 + 334 + 4,317
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Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 439 + 334 + 4,317, = 400 + 300 + 4300, = 5,000, (ii) Closer estimate (Rounding off to nearest tens), 439 + 334 + 4317, = 440 + 330 + 4320, = 5090., , (b) 1,08,734 – 47,599, Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 1,08,734 – 47,599, = 1,08,700 – 47,600, = 61,100, (ii) Closer estimate (Rounding off to nearest tens), 1,08,734 – 47,599, = 1,08,730 – 47,600, = 61,130., , (c) 8325 – 491, Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 8325 – 491, = 8300 – 500, = 7800, (ii) Closer estimate (Rounding off to nearest tens), 8325 – 491, = 8330 – 490, = 7840., , (d) 4,89,348 – 48,365, Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 4,89,348 – 48,365
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= 4,89,300 – 48,400, = 4,40,900, (ii) Closer estimate (Rounding off to nearest tens), 4,89,348 – 48,365, = 4,89,350 – 48,370, = 4,40,980, , Example 1:, 384 + 562, Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 384 + 562 = 400 + 600, = 1,000, (ii) Closer estimate (Rounding off to nearest tens), 384 + 562 = 380 + 560, = 940, , Example 2:, 8765 – 3820, Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 8765 – 3820 = 8800 – 3900, = 4900, (ii) Closer estimate (Rounding off to nearest tens), 8765 – 3820 = 8770 – 3820, = 4950, , Example 3:
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6653 – 8265, Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 6653 + 8265 = 6700 + 8300, = 15,000, (ii) Closer estimate (Rounding off to nearest tens), 6653 + 8265 = 6650 + 8270, = 14920, , Example 4:, 3826 – 1262, Solution:, (i) Rough estimate (Rounding off to nearest hundreds), 3826 – 1262 = 3800 – 1300, = 2500, (ii) Closer estimate (Rounding off to nearest tens), 3826 – 1262 = 3830 – 1260, = 2570, , Question 3., Estimate the following products using general rule:, (a) 578 x 161, Solution:, 578 x 161, = 600 x 200, = 1,20,000
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(b)5281 x 3491, Solution:, 5281 x 3491, = 5000 x 3000, = 1,50,00,000, , (c) 1291 x 592, Solution:, 1291 x 592, = 1300 x 600, = 7,80,000, , (d) 9250 x 29, Solution:, 9250 x 29, = 9000 x 30, = 2,70,000, , Make four more such examples., Example 1., 382 x 1062, Solution:, 382 x 1062, = 400 x 1000, = 4,00,000, , Example 2., 6821 x 1291, Solution:, 6821 x 1291, = 7000 x 1000, = 70,00,000
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Example 3., 3858 x 9350, Solution:, 3858 x 9350, = 4000 x 9000, = 3,60,00,000, , Example 4., 3405 x 7502, Solution:, 3405 x 7502, = 3000 x 8000, = 2,40,00,000, , By : Kishore, Sir