The document NCERT Solution - Knowing Your Number Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.

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**Exercise 1.1**

**Place Value Chart****Q1. 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 ****Ans: ****(a)** 10**(b) **10**(c)** 10**(d)** 10**(e)** 10**Q2. 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. ****Ans:** **(a)** 73,75,307**(b)** 9,05,00,041**(c)** 7,52,21,302**(d)** 58,423,202**(e)** 23,30,010**Q3. Insert commas suitable and write the names according to Indian system of numeration:****(a) 87595762 ****(b) 8546283****(c) 99900046 ****(d) 98432701****Ans:** **(a)** 8,75,95,762: Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.**(b)** 85,46,283: Eight-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.**Q4. Insert commas suitable and write the names according to International system of numeration: ****(a) 78921092 ****(b) 7452283 ****(c) 99985102****(d) 48049831****Ans:** **(a) **78,921,092: Seventy-eight million nine hundred twenty-one thousand ninety-two.**(b) **7,452,483: 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**

**Q1. 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. ****Ans:** Number of tickets sold on first day = 1,094

Number of tickets sold on second day = 1,812

Number of tickets sold on third day = 2,050

Number of tickets sold on fourth day = 2,751

Total tickets sold = 7,707

Therefore,** 7,707 tickets **were sold on all the four days.**Q2. 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? ****Ans: **Total runs scored by Shekhar = 6,980

Number of run he wishes to complete = 10,000

Number of runs he needs

= Total runs he wishes to get - Runs Scored

= 10,000 - 6,980

= **3020****Q3. 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?****Ans:** Number of votes secured by successful candidates = 5,77,500

Number of votes secured by his nearest rival = 3,48,700

Margin by which candidate wins

= Successful candidate votes - Rival votes

= 5,77,500 - 3, 48,700 = **2,28,800****Q4. 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? ****Ans: **Books sold in first week = 2,85,891

Books sold in second week = 4,00,768

Total books sold = 6,86,659

Since 4,00,768,> 2,85,891

Therefore, Sale of the second week is greater than that of the first week.

Books sold in second week = 4,00,768

Books sold in first week = 2,85,891

Difference in sales of week 1 and week 2 = 4,00,768 - 2,85,891

= **1,14,877****Q5. 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.****Ans:** Greatest five-digit number using digits 6,2,7,4,3 = 76432

Smallest five-digit number using digits 6,2,7,4,3 = 23467

Difference = 52965

Therefore, the difference is **52965.****Q6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006? ****Ans:** Number of screws manufactured in one day = 2,825

Number of days in the month of January (31 days) = 2,825 x 31 = 87,575

Therefore, the machine produced **87,575** screws in the month of January.**Q7. A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1,200 each. How much money will remain with her after the purchase?****Ans:** Cost of one radio = Rs 1200

Cost of 40 radios = 1200 x 40 = Rs 48,000

Now, Total money with merchant = Rs 78,592

Money spent by her = Rs 48,000

Money left with her = Rs 30,592

Therefore, Rs **30,592** will remain with her after the purchase.**Q8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? ****Ans:** Wrong answer = 7236 x 65

Correct answer = 7236 x 56

Difference in answers = 470340 – 405216 = **65,124.****Q9. 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? ****Ans: **Cloth required to stitch one shirt = 2 m 15 cm = (2 x 100) cm + 15 cm = 215 cm

Length of cloth = 40 m = 40 x 100 cm = 4000 cm

Number of shirts that can be stitched = 4000/215

Therefore,** 18** shirts can be stitched and **130 cm (1 m 30 cm)** cloth will remain.**Q10. Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a can which cannot carry beyond 800 kg? ****Ans:** The weight of one box = 4 kg 500 g = 4 x 1000 g + 500 g = 4500 g

Maximum load can be loaded in van = 800 kg = 800 x 1000 g = 800000 g

Number of boxes = 800000/4500

Therefore, **177 boxes **can be loaded.**Q11. 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. ****Ans:** Distance between school and home = 1.875 km

Distance between home and school = + 1.875 km

Total distance covered in one day = 3.750 km

Distance covered in six days = 3.750 x 6 = 22.500 km.

Therefore, **22 km 500 m** distance covered in six days.**Q12. A vessel has 4 liters and 500 ml of curd. In how many glasses each of 25 ml capacity, can it be filled? ****Ans: **Capacity of curd in a vessel = 4 litres 500 ml = 4 x 1000 ml + 500 ml = 4500 ml

Capacity of one glass = 25 ml

Number of glasses can be filled = 4500/25

Therefore, **180 glasses** can be filled by curd.

**Exercise 1.3**

**Q1. Estimate each of the following using general rule: ****(a) 730 + 998****(b) 796 – 314 ****(c) 12,904 + 2,888****(d) 28,292 – 21,496****Ans: ****(a) **730 round off to 700

998 round off to 1000

Estimated sum = 1700**(b)** 796 round off to 800

314 round off to 300

Estimated sum = 500**(c) **12904 round off to 13000

2888 round off to 3000

Estimated sum = 16000**(d)** 28292 round off to 28000

21496 round off to 21000

Estimated difference = 7000**Q2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):****(a) 439 + 334 + 4317 ****(b) 1,08,737 – 47,599****(c) 8325 – 491****(d) 4,89,348 – 48,365 ****Ans: ****(a)** 439 round off to 400

334 round off to 300

4317 round off to 4300

Estimated sum = 5000**(b)** 108737 round off to 108700

47599 round off to 47600

Estimated difference = 61100**(c) **8325 round off to 8300

491 round off to 500

Estimated difference = 7800**(d)** 489348 round off to 489300

48365 round off to 48400

Estimated difference = 440900**Q3. Estimate the following products using general rule: ****(a) 578 x 161****(b) 5281 x 3491****(c) 1291 x 592****(d) 9250 x 29 ****Ans:** **(a)** 578 x 161

578 round off to 600

161 round off to 200

The estimated product = 600 x 200 = 1,20,000**(b)** 5281 x 3491

5281 round of to 5,000

3491 round off to 3,500

The estimated product = 5,000 x 3,500 = 1,75,00,000**(c)** 1291 x 592

1291 round off to 1300

592 round off to 600

The estimated product = 1300 x 600 = 7,80,000**(d)** 9250 x 29

9250 round off to 9,000

29 round off to 30

The estimated product = 9,000 x 30 = 2,70,000

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