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Exercise 1.1, 1. Fill in the blanks:, (a) 1 lakh = 100 ten thousand., (b) 1 million = 10 hundred thousand., (c) 1 crore = 10 ten lakh., (d) 1 crore = 10 million., (e) 1 million = 100 lakh, 2. Place commas correctly and write the numerals:, (a) Seventy three lakh seventy five thousand three hundred seven.= 73,75,307, (b) Nine crore five lakh forty one.= 9,05,000,41, (c) Seven crore fifty two lakh twenty one thousand three hundred two.= 7,52,21,302, (d) Fifty eight million four hundred twenty three thousand two hundred two.= 58,423,202, (e) Twenty three lakh thirty thousand ten.= 23,30,010, 3. Insert commas suitably and write the names according to Indian System of Numeration:, (a) 8,75,95,762: Eight crore seventy five lakh ninety five seven hundred sixty two., (b) 85,46,283: Eighty five lakh forty six thousand two hundred eighty three, c) 9,99,00,046: Nine crore ninety nine lakh forty six, (d) 9,84,32,701: Nine crore eighty four lakh thirty two thousand seven hundred one., 4. Insert commas suitably and write the names according to International System of Numeration:, (a) 78,921,092: = Seventy eight million nine hundred twenty one thousand ninety two., (b) 7,452,283: = Seven million four hundred fifty two thousand two hundred eighty three., (c) 99,985,102: = Ninety nine million nine hundred eighty five thousand one hundred two., (e) 48,049,831:= Forty eight million forty nine thousand eight hundred thirty one., Exercise 1.2, 1.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 of tickets sold first day = 1094, Number of tickets sold second day = 1812, Number of tickets sold third day = 2050, Number of tickets sold fourth day = 2751, Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707, 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 scored = 6980 runs, Shekhar wants to score = 10,000 runs, He need to score 10,000 – 6980 = 3020, 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:-, Successful candidate registered 5,77,500 votes, Score secured by his rival = 3,48,700 votes, 5,77,500 - 3,48,700 = 22880, Successful candidate need 22880 margin to win the election., 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 the first week = 2,85,891, Books sold in the second week = 4,00,768, The sale of two weeks together =, 2,85,891 + 4,00,768 = 686659, Second week of the month books sale 4,00,768 was greater than first week 2,85,891., 4,00,768 – 2,85,891 = 114877, Book sale was greater than by 114877, 5. Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once., Solution:-, Greatest number = 76432, Smallest number = 23467, The difference between greatest and smallest number = 76432 – 23467 = 52965, 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 screw produced in one days = 2,825, As we know in the month of January there is 31 days., Number of screw produced in 31 days = 2,825 × 31 = 87575., So, the number of screw produced in January 2006 = 87575., 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:-, Merchant had 78,592 with her, Cost of one radio = 1200, Cost of 40 radio = 12000 × 40 = 48000, Money spent by merchant by = 48000, Money left = 78,592 – 48000 = 30592, Answer :- Rs.30592 will left after purchasing., 8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?, (Hint: Do you need to do both the multiplications?), Solution:-, Differences between 65 × 56 = 9, Correct answer is greater than by 7236 × 9 = 65124, 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?, (Hint: convert data in cm.), Solution:-, Cloth available = 40 m, As we know 1 m = 100 cm, 40 m = 40 × 100 = 4000, 2 m 15 cm = 215 cm, Required cloth for one t-shirt = 215 cm, Number of cloths stitched = 4000 ÷ 215, Therefore, 18 shirts can be made., Remaining cloths is 1 m 130 cm., 10. Medicine is packed in boxes, each weighing 4 kg 500g.How many such boxes can be loaded in a van which cannot carry beyond 800 kg?, Solution:-, Weight of each boxes is = 4 kg 500 g, 1 kg = 1000 g, 4 kg 500 g = 4500 g, So, 800 kg = 800 × 1000 = 800000 g, Number of boxes required for 800000 ÷ 4500, So, the number of boxes required for medicine = 177., 11.The distance between the school and the house of a student’s house is 1 km 875 m.Every day she walks both ways.Find the total distance covered by her in six days., Solution:-Distance between the school and the house = 1 km 875 m., As we know 1 km = 1000 m, 1 km 875 m = 1875 m., There are two ways, Distance covered by her in each day was 1875 × 2 = 3750 m, And the distance covered in six days = 3750 × 6 = 22500, ∴ Distance covered in six days = 22500 m = 22 km 500 m ., 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:-, Vessel can store = 4 litres and 500 ml, As we know, 1 l = 1000 ml, We have to find number of classes that can store 25 ml of curd., Number of glasses = 4500 ÷25, ∴ 180 glasses are required for 25 ml of curd., Exercise 1.3, 1. Estimate each of the following using general rule:, (a) 730 + 998, (b) 796 – 314, (c)12,904 +2,888, (d) 28,292 – 21,496, Solution:-, (a)730 + 998, • Rounding off to nearest hundred.730 round off to 700 and 998 round off to 1000., 700 + 1000 = 1700, b) 796 – 314, 800 – 300 = 500, c) 12,904 + 2,888, 13000 + 3000 = 16000, d) 28,292 – 21,496, 28000 – 21000 = 7000, 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, (b) 1,08,734 – 47,599, (c) 8325 – 491, (d) 4,89,348 – 48,365, Solution:-, (a)439 + 334 + 4,317, • Rounding of hundred = 400, 300, 4000, 400 + 300 + 4000 = 5000, • Rounding of ten = 440, 330, 4,320, 440 +330 + 4,320 = 5090., b) 1,08,734 – 47,599, • Rounding of hundred = 1,08,700 and 47,600, 1,08,700 - 47,600 = 61100, • Rounding of ten = 1,08,730 and 47,600, 1,08,730 - 47,600, c) 8325 – 491, • Rounding of hundred 8300 and 500, 8300 - 500 = 7800, • Rounding of ten 8330 and 490, 83330 – 490 = 7840, d) 4,89,348 – 48,365, • Rounding of hundred = 4,89,300 – 48400 = 440900, • Rounding of ten = 489350 – 48370 = 440980, 3. Estimate the following products using general rule:, (a) 578 × 161, (b) 5281 × 3491, (c) 1291 × 592, (d) 9250 × 29, a) Rounding off to nearest hundred ,578 round off to 600 and 161 round off to 200., 600 × 200 = 120000, b) 5000 × 3000 = 15000000, c) 1000 × 600 = 600000, d) 9000 × 30 = 270000