class 6th NCERT Math Solution Chapter 1

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Class 6th Math Chapter 1 NCERT Solutions

Here are the solutions to all questions in every exercise of Chapter 1 "Knowing Our Numbers" from the NCERT Class 6th Math textbook.

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 1: (a) 1 lakh = 10 ten thousand. (b) 1 million = 10 hundred thousand. (c) 1 crore = 10 ten lakh. (d) 1 crore = 10 million. (e) 1 million = 10 lakh.

Question 2: Place commas correctly and write the numerals: (a) Seventy-three lakh seventy-five thousand three hundred seven. (b) Nine crore five lakh forty-one. (c) Seven crore fifty-two lakh twenty-one thousand three hundred two. (d) Fifty-eight million four hundred twenty-three thousand two hundred two. (e) Twenty-three lakh thirty thousand ten.  

Solution 2: (a) 73,75,307 (b) 9,05,00,041 (c) 7,52,21,302 (d) 58,423,202 (e) 23,30,010

Question 3: Insert commas suitably and write the names according to Indian System of Numeration: (a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701

Solution 3: (a) 8,75,95,762 - Eight crore seventy-five lakh ninety-five thousand 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.

Question 4: Insert commas suitably and write the names according to International System of Numeration: (a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831  

Solution 4: (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. (d) 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 1: Total number of tickets sold = 1094 + 1812 + 2050 + 2751 = 7707 Total number of tickets sold on all four days is 7707.  

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 2: Runs needed more = 10000 - 6980 = 3020 Shekhar needs 3020 more 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 3: Margin of victory = 5,77,500 - 3,48,700 = 2,28,800 The successful candidate won by a margin of 2,28,800 votes.  

Question 4: Kirti bookstore sold books worth ₹ 2,85,891 in the first week of June and books worth ₹ 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 4: Total sale for two weeks = ₹ 2,85,891 + ₹ 4,00,768 = ₹ 6,86,659 Sale in second week is greater. Difference in sales = ₹ 4,00,768 - ₹ 2,85,891 = ₹ 1,14,877 Sale was greater in the second week by ₹ 1,14,877.  

Question 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 5: Greatest number = 76432 Least number = 23467 Difference = 76432 - 23467 = 52965 The difference is 52965.

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 6: Total screws produced in January 2006 = 2825 * 31 = 87575 The machine produced 87575 screws in January 2006.  

Question 7: A merchant had ₹ 78,592 with her. She placed an order for purchasing 39 radio sets at ₹ 1200 each. How much money will remain with her after the purchase?  

Solution 7: Total cost of 39 radio sets = 39 * 1200 = ₹ 46800 Money remaining = ₹ 78,592 - ₹ 46800 = ₹ 31,792 ₹ 31,792 will remain with her.  

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 8: His answer was greater than the correct answer by 7236 * (65 - 56) = 7236 * 9 = 65124 His answer was greater than the correct answer by 65124.

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 9: Number of shirts stitched = 4000 cm / 215 cm = 18 (quotient) Remaining cloth = 130 cm = 1 m 30 cm 18 shirts can be stitched and 1 m 30 cm cloth will remain.

Question 10: Medicine box contains 5,00,000 tablets each weighing 20 mg. What is the total weight of all the tablets in the box in grams and in kilograms?  

Solution 10: Total weight in mg = 5,00,000 * 20 = 1,00,00,000 mg Total weight in grams = 1,00,00,000 / 1000 = 10,000 g Total weight in kilograms = 10,000 / 1000 = 10 kg Total weight is 10,000 grams or 10 kilograms.

Question 11: The distance between the school and 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 11: Distance walked both ways in a day = 1875 m * 2 = 3750 m Distance walked in six days = 3750 m * 6 = 22500 m = 22 km 500 m Total distance covered in six days is 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 12: Number of glasses = 4500 ml / 25 ml = 180 It can be filled in 180 glasses.  

Exercise 1.3

Question 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 1: (a) 730 + 998 ≈ 700 + 1000 = 1700 (b) 796 – 314 ≈ 800 - 300 = 500 (c) 12,904 + 2,888 ≈ 13,000 + 3,000 = 16,000 (d) 28,292 – 21,496 ≈ 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 (b) 1,08,734 – 47,599 (c) 8325 – 491  

Solution 2: (a) 439 + 334 + 4,317 * Rough estimate (nearest hundreds): 400 + 300 + 4300 = 5000 * Closer estimate (nearest tens): 440 + 330 + 4320 = 5090

(b) 1,08,734 – 47,599 * Rough estimate (nearest hundreds): 1,08,700 - 47,600 = 61,100 * Closer estimate (nearest tens): 1,08,730 - 47,600 = 61,130

(c) 8325 – 491 * Rough estimate (nearest hundreds): 8300 - 500 = 7800 * Closer estimate (nearest tens): 8330 - 490 = 7840

Question 3: Estimate the following products using general rule: (a) 578 × 161 (b) 5281 × 3491 (c) 1291 × 592 (d) 9250 × 29

Solution 3: (a) 578 × 161 ≈ 600 × 200 = 1,20,000 (b) 5281 × 3491 ≈ 5000 × 3000 = 1,50,00,000 (c) 1291 × 592 ≈ 1000 × 600 = 6,00,000 (d) 9250 × 29 ≈ 9000 × 30 = 2,70,000

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