Rbse Solutions for Class 6 Chapter 1: Knowing Our Numbers

Last Updated on November 5, 2025 by Aman Singh

You want the detailed solutions for RBSE Class 6 Maths Chapter 1: Knowing Our Numbers. I can provide the solutions for Exercises 1.1, 1.2, and 1.3 as requested in the text you provided.

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Class 6 Chapter 1 Exercise 1.1

1. Fill in the blanks:

(a) 1 lakh = 10 ten thousand.

(b) 1 million = 10 hundred thousand.

(c) 1 crore = 10 ten lakhs.

(d) 1 crore = 10 million.

(e) 1 million = 10 lakhs.

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: 9,05,00,041

(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 the Indian System of Numeration:

(a) 87595762: 8,75,95,762 – Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.

(b) 8546283: 85,46,283 – Eighty-five lakh forty-six thousand two hundred eighty-three.

(c) 99900046: 9,99,00,046 – Nine crore ninety-nine lakh forty-six.

(d) 98432701: 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 the International System of Numeration:

(a) 78921092: 78,921,092 – Seventy-eight million nine hundred twenty-one thousand ninety-two.

(b) 7452283: 7,452,283 – Seven million four hundred fifty-two thousand two hundred eighty-three.

(c) 99985102: 99,985,102 – Ninety-nine million nine hundred eighty-five thousand one hundred two.

(d) 48049831: 48,049,831 – Forty-eight million forty-nine thousand eight hundred thirty-one.

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Class 6 Chapter 1 Exercise 1.2

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 four days.

Solution:

Total tickets sold = 1094 + 1812 + 2050 + 2751 = 7707 tickets.

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:

Runs needed = 10,000 – 6,980 = 3020 runs.

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:

Winning margin = 5,77,500 – 3,48,700 = 2,28,800 votes.

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:

Total sale = Rs 2,85,891 + Rs 4,00,768 = Rs 6,86,659.

The sale was greater in the second week.

Difference in sale = Rs 4,00,768 – Rs 2,85,891 = Rs 1,14,877.

5. Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, and 3 each only once.

Solution:

Given digits: 6, 2, 7, 4, 3

Greatest 5-digit number: 76,432

Least 5-digit number: 23,467

Difference = 76,432 – 23,467 = 52,965.

6. A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Solution:

Number of days in January = 31 days.

Total screws produced = 2,825 screws/day × 31 days = 87,575 screws.

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:

Total cost of radios = 40 sets × Rs 1,200/set = Rs 48,000.

Money remaining = Rs 78,592 – Rs 48,000 = Rs 30,592.

8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?

Solution:

Difference in multipliers = 65 – 56 = 9.

Difference in answers = 7,236 × 9 = 65,124.

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 cloth = 40 m = 4000 cm.

Cloth per shirt = 2 m 15 cm = 215 cm.

Number of shirts = 4000 cm / 215 cm = 18 shirts with a remainder.

Remaining cloth = 4000 – (18 × 215) = 4000 – 3870 = 130 cm = 1 m 30 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 one box = 4 kg 500 g = 4,500 g.

Van’s capacity = 800 kg = 800,000 g.

Number of boxes = 800,000 g / 4,500 g = 177 boxes (with a remainder).

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:

Distance one way = 1 km 875 m = 1875 m.

Distance per day (both ways) = 2 × 1875 m = 3750 m.

Total distance in 6 days = 3750 m × 6 = 22,500 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:

Total volume of curd = 4 L 500 mL = 4500 mL.

Capacity per glass = 25 mL.

Number of glasses = 4500 mL / 25 mL = 180 glasses.

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Class 6 Chapter 1 Exercise 1.3

1. Estimate each of the following using the general rule:

Solution:

(a) 730 + 998

• Rounding to the nearest hundreds: 700 + 1000 = 1700

(b) 796 – 314

• Rounding to the nearest hundreds: 800 – 300 = 500

(c) 12904 + 2888

• Rounding to the nearest thousands: 13000 + 3000 = 16000

(d) 28292 – 21496

• Rounding to the nearest thousands: 28000 – 21000 = 7000

2. Give a rough estimate (by rounding off to the nearest hundreds) and also a closer estimate (by rounding off to the nearest tens):

Solution:

(a) 439 + 334 + 4317

• Rough estimate (hundreds): 400 + 300 + 4300 = 5000
• Closer estimate (tens): 440 + 330 + 4320 = 5090

(b) 108734 – 47599

• Rough estimate (hundreds): 108700 – 47600 = 61100
• Closer estimate (tens): 108730 – 47600 = 61130

(c) 8325 – 491

• Rough estimate (hundreds): 8300 – 500 = 7800
• Closer estimate (tens): 8330 – 490 = 7840

(d) 489348 – 48365

• Rough estimate (hundreds): 489300 – 48400 = 440900
• Closer estimate (tens): 489350 – 48370 = 440980

3. Estimate the following products using the general rule:

Solution:

(a) 578 × 161

• Rounding to the nearest hundreds: 600 × 200 = 120,000

(b) 5281 × 3491

• Rounding to the nearest thousands: 5000 × 3000 = 15,000,000

(c) 1291 × 592

• Rounding to the nearest hundreds: 1300 × 600 = 780,000

(d) 9250 × 29

• Rounding to the nearest thousands and tens: 9000 × 30 = 270,000

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