In this article we are provide you the full ** NCERT Solutions for Class 6 Maths Chapter 1**. Mathematics is a fascinating subject that lays the foundation for logical thinking and problem-solving skills. In Class 8, one of the important topics is rational numbers. In this article, we will explore the **NCERT Solutions for Class 6 Maths Chapter 1** Knowing Our Numbers Exercise 1.1 ,Exercise 1.2 solutions that will help you develop a strong understanding of this topic. So, let’s embark on a journey to master the world of numbers!

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##### If you’re looking for the Hindi and English Medium** NCERT Solutions for Class 6 Maths Chapter 1** Knowing Our Numbers, they are provided here and were created for the CBSE session 2023–2024. There are only two problems in chapter 1 of the class 6 mathematics textbooks for the CBSE 2023–24, according to the rationalised syllabus. Lets start the** NCERT Solutions for Class 6 Maths Chapter 1.**

** NCERT Solutions for Class 6 Maths Chapter 1**

** 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.

** Solutions:**

(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.

**Q 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.

** Solutions:**

###### (a) The numeral of seventy three lakh seventy five thousand three hundred seven is 73,75,307

###### (b) The numeral of nine crore five lakh forty one is 9,05,00,041

###### (c) The numeral of seven crore fifty two lakh twenty one thousand three hundred two is 7,52,21,302 (d) The numeral of fifty eight million four hundred twenty three thousand two hundred two is 5,84,23,202

###### (e) The numeral of twenty three lakh thirty thousand ten is 23,30,010

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

###### Solutions:

###### (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.

**Q 4. Insert commas suitably and write the names according to International System of Numeration:**

**(a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831 **

###### Solutions:

###### (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.

** NCERT Solutions for Class 6 Maths Chapter 1**

**Exercise 1.2**

__ __**Q 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. **

###### Solutions:

###### Number of tickets sold on 1st day = 1094

###### Number of tickets sold on 2nd day = 1812

###### Number of tickets sold on 3rd day = 2050

###### Number of tickets sold on 4th day = 2751

###### Hence, number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707 tickets

**Q 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?**

Solutions:

###### Shekhar scored = 6980 runs

###### He want to complete = 10000 runs

###### Runs need to score more = 10000 – 6980 = 3020

###### Hence, he need 3020 more runs to score

** Q 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? **

##### Solutions:

##### No. of votes secured by the successful candidate = 5,77,500

##### No. of votes secured by his rival = 3,48,,700

##### Margin by which he won the election = 5,77,500 – 3,48,700 = 2,28,800 votes

##### ∴ Successful candidate won the election by 2,28,800 votes

**Q 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?**

##### Solutions:

##### Price of books sold in June first week = Rs 2,85,891

##### Price of books sold in June second week = Rs 4,00,768

##### No. of books sold in both weeks together = Rs 2,85,891 + Rs 4,00,768 = Rs 6,86,659

##### The sale of books is the highest in the second week

##### Difference in the sale in both weeks = Rs 4,00,768 – Rs 2,85,891 = Rs 1,14,877

##### ∴ Sale in second week was greater by Rs 1,14,877 than in the first week.

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

Solutions:

Greatest 5-digit number = 76,432

Least 5-digit number = 23,467

Difference between the two numbers = 76,432 – 23,467 = 52,965

∴ The difference between the two numbers is 52,965

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

###### Solutions:

###### Number of screws manufactured in a day = 2825

###### Since January month has 31 days

###### Hence, number of screws manufactured in January = 31 × 2825 = 87,575

###### Hence, machine produce 87575 screws in the month of January 2006

**Q 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?**

Solutions:

Total money the merchant had = Rs 78592

Number of radio sets she placed an order for purchasing = 40 radio sets

Cost of each radio set = Rs 1200

So, cost of 40 radio sets = Rs 1200 × 40 = Rs 48000

Money left with the merchant = Rs 78592 – Rs 48000 = Rs 30592

Hence, money left with the merchant after purchasing radio sets is Rs 30592

** Q8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?**

Solutions:

Difference between 65 and 56 i.e (65 – 56) = 9

The difference between the correct and incorrect answer = 7236 × 9 = 65124

Hence, by 65124, the answer was greater than the correct answer

**Q 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?**

Solutions:

Given

Total length of the cloth = 40 m

= 40 × 100 cm = 4000 cm

Cloth required to stitch one shirt = 2 m 15 cm

= 2 × 100 + 15 cm = 215 cm

Number of shirts that can be stitched out of 4000 cm = 4000 / 215 = 18 shirts

Hence 18 shirts can be stitched out of 40 m and 1m 30 cm of cloth is left out

**Q 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? **

Solutions:

Weight of one box = 4 kg 500 g = 4 × 1000 + 500 = 4500 g

Maximum weight carried by the van = 800 kg = 800 × 1000 = 800000 g

Hence, number of boxes that can be loaded in the van = 800000 / 4500 = 177 boxes

**Q 11. The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.**

Solutions:

Distance covered between school and house = 1 km 875 m = 1000 + 875 = 1875 m

Since, the student walk both ways.

Hence, distance travelled by the student in one day = 2 × 1875 = 3750 m

Distance travelled by the student in 6 days = 3750 m × 6 = 22500 m = 22 km 500 m

∴ Total distance covered by the student in six days is 22 km and 500 m

** Q12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?**

Solutions:

Quantity of curd in the vessel = 4 l 500 ml = 4 × 1000 + 500 = 4500 ml

Capacity of 1 glass = 25 ml

∴ Number of glasses that can be filled with curd = 4500 / 25 = 180 glasses

Hence, 180 glasses can be filled with curd.