# NCERT Solutions – Class 6 Maths Chapter 1: Knowing Our Numbers

Knowing Our Numbers is the first chapter in the NCERT textbook for Class 6 Maths. This chapter introduces students to the Indian and International systems of numeration and teaches them how to write and read large numbers. The NCERT Solutions for chapter 1 of the class 6 maths provide step-by-step explanations of the concepts, along with solved examples. The solutions are clear and concise, and they are written in simple language that is easy for students to understand.

#### Exercise 1.1 – NCERT Class 6 Maths Chapter 1 (With Solutions)

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.

Here are the solutions:

(a) 1 lakh = 10 ten thousand

In simple terms, 1 lakh is equal to 100,000. This can also be written as 10 x 10,000, or 10 x ten thousand.

(b) 1 million = 10 hundred thousand

In simple terms, 1 million is equal to 1,000,000. This can also be written as 10 x 100,000, or 10 x hundred thousand.

(c) 1 crore = 10 ten lakh

In simple terms, 1 crore is equal to 10,000,000. This can also be written as 10 x 1,000,000, or 10 x ten lakh.

(d) 1 crore = 10 million

In simple terms, 1 crore is equal to 10 million. This can also be written as 10 x 1,000,000.

(e) 1 million = 10 lakh

In simple terms, 1 million is equal to 10 lakh. This can also be written as 10 x 100,000.

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

Here are the Solutions:

(a) Seventy three lakh seventy five thousand three hundred seven = 73,75,307

In simple terms, this is equal to 73 lakhs, 75 thousand, and 307.

(b) Nine crore five lakh forty one = 9,05,00,041

In simple terms, this is equal to 9 crores, 5 lakhs, and 41.

(c) Seven crore fifty two lakh twenty one thousand three hundred two = 7,52,21,302

In simple terms, this is equal to 7 crores, 52 lakhs, 21 thousand, and 302.

(d) Fifty eight million four hundred twenty three thousand two hundred two = 58,423,202

In simple terms, this is equal to 58 million, 423 thousand, and 202.

(e) Twenty three lakh thirty thousand ten = 23,30,010

In simple terms, this is equal to 23 lakhs, 30 thousand, and 10.

3. Insert commas suitably and write the names according to the Indian System of Numeration:

(a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701

Here are the solutions:

(a) 87,59,5762 = Eighty-seven lakh fifty-nine thousand five hundred seventy-six two

In simple terms, this is equal to 87 lakhs, 59 thousand, 5762.

(b) 85,46,283 = Eighty-five lakh forty-six thousand two hundred eighty-three

In simple terms, this is equal to 85 lakhs, 46 thousand, 283.

(c) 99,90,0046 = Ninety-nine lakh ninety thousand four hundred sixty-two

In simple terms, this is equal to 99 lakhs, 90 thousand, 4062.

(d) 98,43,2701 = Ninety-eight lakh forty-three thousand two hundred seventy-one

In simple terms, this is equal to 98 lakhs, 43 thousand, 2701.

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

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

Here are the solutions:

(a) 78,921,092 = Seventy-eight million nine hundred twenty-one thousand ninety-two

In simple terms, this is equal to 78 million, 921 thousand, and 92.

(b) 7,452,283 = Seven million four hundred fifty-two thousand two hundred eighty-three

In simple terms, this is equal to 7 million, 452 thousand, and 283.

(c) 99,985,102 = Ninety-nine million nine hundred eighty-five thousand one hundred two

In simple terms, this is equal to 99 million, 985 thousand, and 102.

(d) 48,049,831 = Forty-eight million forty-nine thousand eight hundred thirty-one

In simple terms, this is equal to 48 million, 49 thousand, and 831.

#### Exercise 1.2 – NCERT Class 6 Maths Chapter 1 (With Solutions)

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:

The total number of tickets sold on all four days is 7707.

Here’s how we can find the total number of tickets sold:

On the first day, 1094 tickets were sold.

On the second day, 1812 tickets were sold.

On the third day, 2050 tickets were sold.

On the fourth day, 2751 tickets were sold.

Therefore, the total number of tickets sold on all four days is 1094 + 1812 + 2050 + 2751 = 7707.

We can find the total number of tickets sold by adding the number of tickets sold on each day.

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 needs 3020 more runs to complete 10,000 runs.

Here’s how I calculated it:

Shekhar has scored 6980 runs so far.

He wants to complete 10,000 runs.

Therefore, he needs 10,000 – 6980 = 3020 more runs.

Shekhar needs to score an additional 3020 runs to reach his goal of 10,000 runs. This is because he is currently 3020 runs short of his goal.

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:

The margin of victory is the difference between the number of votes won by the successful candidate and the number of votes won by the nearest rival. In this case, the successful candidate won by a margin of 5,77,500 – 3,48,700 = 2,28,800 votes.

This means that the successful candidate won by more than 2 lakh votes. This is a significant margin of victory, and it suggests that the successful candidate was very popular with the electorate.

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:

The total sale for the two weeks together is Rs 6,86,659. The sale was greater in the second week by Rs 1,14,877.

Here is the calculation:

Sale in the first week = Rs 2,85,891

Sale in the second week = Rs 4,00,768

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

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

The bookstore sold Rs 2,85,891 worth of books in the first week of June and Rs 4,00,768 worth of books in the second week of June. The total sale for the two weeks together is Rs 6,86,659. The sale was greater in the second week by Rs 1,14,877. This means that the bookstore sold Rs 1,14,877 more books in the second week than in the first week.

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:

The greatest 5-digit number that can be written using the digits 6, 2, 7, 4, and 3 is 76432. This is because the digits are arranged in descending order, with the largest digit 7 in the thousands place, followed by 6, 4, 3, and 2.

The least 5-digit number that can be written using the digits 6, 2, 7, 4, and 3 is 23467. This is because the digits are arranged in ascending order, with the smallest digit 2 in the thousands place, followed by 3, 4, 6, and 7.

The difference between these two numbers is 76432 – 23467 = 52965.

In simple terms, to find the difference between the greatest and least 5-digit numbers, we can simply subtract the least number from the greatest number. The answer will be the difference between the two numbers.

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

Solution:

The machine produced 87,575 screws in the month of January 2006.

Here’s the explanation:

The machine produces an average of 2,825 screws per day.

January 2006 has 31 days.

So, the machine produced 2,825 * 31 = 87,575 screws in January 2006.

In other words, if you start with 2,825 screws, and add 2,825 screws 30 more times, you’ll end up with 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:

The cost of 40 radio sets at Rs 1200 each is 40 * 1200 = Rs 48000.

So, the merchant will have Rs 78592 – Rs 48000 = Rs 30592 remaining after the purchase.

In simple terms, the merchant had Rs 78592 to start with. She spent Rs 48000 on the radio sets, so she has Rs 30592 remaining.

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

Solution:

Here’s how we can calculate it:

The correct answer is 7236 * 56 = 405,216

The student’s answer is 7236 * 65 = 470,340

The difference between the two answers is 470,340 – 405,216 = 65,124

In simple terms, the student’s answer was greater than the correct answer because he multiplied 7236 by 9 more than he should have. The difference between 65 and 56 is 9, so the student’s answer was 9 times greater than it should have been.

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:

First, let’s convert the length of cloth needed to stitch a shirt into centimeters. 2 m 15 cm = 2m*100 + 15cm = 215 cm.

Now, we can divide the total length of cloth by the length needed to stitch a shirt. 4000 cm / 215 cm = This gives us 18.60.

Since we can’t stitch part of a shirt, we round 18.60 down to 18. This means that 18 shirts can be stitched from 40 m of cloth.

Total cloth required to stitch 18 shirts = 215 cm (per shirt) * 18 = 3870 cm.

Remaining cloth = 4000 cm – 3870 cm = 130 cm. 130 cm of cloth will remain, which is equal to 1 m 30 cm.

In simple terms, the length of cloth needed to stitch a shirt is 215 cm. Out of 40 m of cloth, there are 4000 cm of cloth. Therefore, 18 shirts can be stitched and 130 cm of cloth will remain.

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:

The weight of one box is 4 kg 500 g = 4500 g.

The maximum load that can be loaded in the van is 800 kg = 800000 g.

The number of boxes that can be loaded in the van is:

number_of_boxes = 800000 / 4500 = 177.77

Since the number of boxes cannot be in decimal form, the van can carry 177 boxes.

In simple terms, the van can carry a maximum of 800000 grams of medicine. Each box weighs 4500 grams, so the van can carry 800000 / 4500 = 177 boxes.

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:

The distance between the school and the student’s house is 1 km 875 m, which is equal to 1875 m. So, the total distance covered by the student every day is 1875 m x 2 = 3750 m.

In six days, the student will cover a total distance of 3750 m x 6 = 22500 m.

In simple terms, the student covers 3750 m every day by walking to school and back home. In six days, she will cover a total distance of 22500 m, which is equal to 22.5 km or 22 km and 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:

First, we need to convert the amount of curd in the vessel to milliliters. 4 litres is equal to 4000 ml, and 500 ml is equal to 500 ml, so the total amount of curd is 4000 + 500 = 4500 ml.

Next, we need to divide the total amount of curd by the capacity of each glass. The capacity of each glass is 25 ml, so we get 4500 / 25 = 180.

Therefore, the curd in the vessel can be filled in 180 glasses, each of 25 ml capacity.

In simple terms, we can think of the problem as follows. If we have a total of 4500 ml of curd and each glass can hold 25 ml of curd, then we can fill 4500 / 25 = 180 glasses with the curd.

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