In this article, we provide comprehensive **NCERT solutions for Class 6 Maths Chapter 1** – Knowing Our Numbers.

This chapter focuses on the concept of numbers and their different forms. It starts by introducing the number system and the place value of digits in a number. The chapter then covers the comparison of numbers and their properties, such as even and odd numbers, prime and composite numbers, etc. The chapter also covers estimation, rounding off numbers, and the use of brackets.

## Class 6 Maths Chapter 1 Exercise 1.1 Solutions

## Exercise 1.1

### 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.

Answers:

(a): 10 ten thousand equals 1 lakh

(b): 1 million equals 10 hundred thousand

(c): 1 crore equals 10 ten lakh

(d): 1 crore equals 10 million

(e): 1 million equals 10 lakh

### Q2. Place commas correctly and write the numerals:

(a) Seventy-three lakh seventy-five thousand three hundred seven.

(b) Nine crores 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.

Answers:

As you learned (Th, H, T, O), you can easily convert English written numbers into digits

First, we put Ones, then Tens to left, and so on like this table

Ten Thousand | Thousands | Hundreds | Tens | Ones |
---|---|---|---|---|

10,000 | 1,000 | 100 | 10 | 1 |

Ten Crores | Crores | Ten Lakhs | Lakhs |
---|---|---|---|

10,00,00,000 | 1,00,00,000 | 10,00,000 | 1,00,000 |

(a): Seventy-three lakh seventy-five thousand three hundred seven.

In this question, we need to take Ten Lakhs, digits and replace each digit according to the question

73,75,307

(b): 9,05,00,041

(c): 7,52,21,302

(d): 5,84,23,202

(e): 23,30,010.

### Q3. Insert commas suitably and write the names according to the Indian System of Numeration :

(a) 87595762

(b) 8546283

(c) 99900046

(d) 98432701

Answers:

(a) 8,75,95,762 (Eight crores 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 crores ninety-nine lakh forty-six)

(d) 9,84,32,701 (Nine crores eighty-four lakh thirty-two thousand seven hundred one)

### Q4. Insert commas suitably and write the names according to the International System of Numeration :

(a) 78921092

(b) 7452283

(c) 99985102

(d) 48049831

Answers:

(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)

## Class 6 Maths Chapter 1 Exercise 1.2 Solutions

## 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. The sum of tickets sold at the counter on the first, second, third, and final day is

1st day: 1094

2nd day: 1812

3rd day: 2050

4th day: 2751

= 1094 + 1812 + 2050 + 2751 = 7707

Make it simple by summing the Th, H, T, O

- Take Thousands first
- Then Hundreds
- Then Tens
- Then Ones
- And sum them in sequence

Thousands are = 1000 + 1000 + 2000 + 2000 = 6000

Hundreds are = 800 + 700 = 1500

Tens are = 90 + 10 + 50 + 50 = 200

Ones are = 4 + 2 + 1 =7

So, sum them together, 6000 + 1500 + 200 + 7 = 7707

### 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. Subtracting 6980 from 10,000 would be the runs he does need

10000 – 6980 = 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?

And. We can get the margin by subtracting the lower number from the higher number

lower number is 348700

higher number is 577500

577500 – 348700 = 228800

### Q4. 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?

Ans. Let’s know the total sale for the two weeks = first-week sales + second-week sales

= 285891 + 400768 = 686659

285891 < 400768, so the second week made higher sales

### Q5. 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.

Ans. Digits are given = 6, 2, 7, 4, and 3

highest number of given digits = 76432

smallest number of given digits = 23467

Difference between 76432 – 23467 = 52965

### Q6. A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Ans. These are given

machine-manufactured screws a day = 2825

January always has 31 days

So, we will multiply 2825 by 31 = 2825 x 31 = 87575

Hence, the 87575 screws manufactured by the machine in January 2006

### Q7. A merchant had 78,592 with her. She placed an order for purchasing 40 radio sets at 1200 each. How much money will remain with her after the purchase?

Ans. What was given

She has 78592

She has to buy 40 radio sets at 1200

First, we will multiply 40 by 1200 and then subtract the multiplication amount from 78592

= 40 x 1200

= 48000

Let’s find out what amount will remain with her

= she has – she spent on radio

= 78592 – 48000

= 30592

Hence, she needs 48000, and the 30592 amount will remain with her

### Q8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)

And. Since, 65 > 56

= 65 – 56 = 9 x 7236 = 65124

### 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? (Hint: convert data in cm.)

Ans.

2 m 15 cm = (2 × 100 cm) + 15 cm = 200 cm + 15 cm = 215 cm

40 m = 40 × 100 cm = 4000 cm

Now that we have converted the measurements to centimeters, we can proceed with the calculation.

To stitch a shirt, 215 cm of cloth is needed. Therefore, the number of shirts that can be stitched from 40 m (4000 cm) of cloth can be found by dividing the total cloth available by the cloth required for each shirt:

Number of shirts = Total cloth available / Cloth required per shirt Number of shirts = 4000 cm / 215 cm ≈ 18.60

Since we cannot have a fraction of a shirt, we will consider the whole number of shirts that can be stitched. In this case, 18 shirts can be stitched.

To determine how much cloth will remain, we need to subtract the total cloth used for stitching the shirts from the total cloth available:

Cloth remaining = Total cloth available – (Cloth required per shirt × Number of shirts)

Cloth remaining = 4000 cm – (215 cm × 18)

Cloth remaining = 4000 cm – 3870 cm = 130 cm

Therefore, after stitching 18 shirts, there will be 130 cm of cloth remaining.

### Q10. 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?

Ans. We will convert everything to grams.

Weight of each box = 4 kg 500 g = (4 × 1000 g) + 500 g = 4000 g + 500 g = 4500 g Maximum weight the van can carry = 800 kg = 800,000 g

Now, we can calculate the number of boxes that can be loaded into the van.

Number of boxes = Maximum weight the van can carry / Weight of each box

Number of boxes = 800,000 g / 4500 g ≈ 177.78

Since we cannot have a fraction of a box, we will consider the whole number of boxes that can be loaded. In this case, 177 boxes can be loaded into the van without exceeding its maximum weight capacity.

### Q11. 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.

Ans.

Given:

Distance between school and house = 1 km 875 m

Round trip distance for one day:

Since the student walks both ways (to school and back home), the round trip distance for one day is twice the distance between the school and the house.

Round trip distance = 2 * Distance between school and house

Convert the distance to the same unit: 1 km = 1000 m

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

Round trip distance = 2 * 1875 m = 3750 m

Total distance covered in six days:

Total distance covered = Round trip distance * Number of days

Total distance covered = 3750 m * 6 = 22500 m

Therefore, the student covers a total distance of 22,500 meters in six days.

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

Ans.

Given:

Quantity of curd = 4 liters 500 ml

The capacity of each glass = 25 ml

Converting the quantity of curd to milliliters: 1 litre = 1000 ml

Quantity of curd = 4 litres * 1000 ml + 500 ml = 4000 ml + 500 ml = 4500 ml

Now, we can calculate the number of glasses that can be filled: Number of glasses = Quantity of curd / Capacity of each glass

Number of glasses = 4500 ml / 25 ml

Dividing 4500 ml by 25 ml, we get: Number of glasses = 180

Therefore, the vessel can be filled with 180 glasses, each with a capacity of 25 ml.

## Class 6 Maths Chapter 1 Exercise 1.3 Solutions

## Exercise 1.3

### Q1. Estimate each of the following using the general rule:

(a) 730 + 998

(b) 796 – 314

(c) 12,904 +2,888

(d) 28,292 – 21,496

Make ten more such examples of addition, subtraction, and estimation of their outcome.

**Ans.**

(a) 730 + 998:

Estimating the sum of 730 and 998, we can round them to the nearest hundred: 700 + 1000 = 1700

(b) 796 – 314:

Estimating the difference between 796 and 314, we can round them to the nearest hundred: 800 – 300 = 500

(c) 12,904 + 2,888:

Estimating the sum of 12,904 and 2,888, we can round them to the nearest thousand: 13,000 + 3,000 = 16,000

(d) 28,292 – 21,496:

Estimating the difference between 28,292 and 21,496, we can round them to the nearest thousand: 28,000 – 21,000 = 7,000

#### Here are ten more examples:

(i) 1,234 + 987:

Estimating the sum, we round to the nearest hundred: 1200 + 1000 = 2200

(ii) 5,678 – 1,234:

Estimating the difference, we round to the nearest thousand: 6000 – 1000 = 5000

(iii) 9,876 + 3,210:

Estimating the sum, we round to the nearest thousand: 10,000 + 3000 = 13,000

(iv) 7,890 – 2,543:

Estimating the difference, we round to the nearest thousand: 8000 – 3000 = 5000

(v) 25,678 + 15,432:

Estimating the sum, we round to the nearest ten thousand: 30,000 + 20,000 = 50,000

(vi) 18,765 – 9,876:

Estimating the difference, we round to the nearest ten thousand: 20,000 – 10,000 = 10,000

(vii) 32,456 + 45,678:

Estimating the sum, we round to the nearest ten thousand: 30,000 + 50,000 = 80,000

(viii) 76,543 – 23,456:

Estimating the difference, we round to the nearest ten thousand: 80,000 – 20,000 = 60,000

(ix) 123,456 + 987,654:

Estimating the sum, we round to the nearest hundred thousand: 100,000 + 1,000,000 = 1,100,000

(x) 987,654 – 123,456:

Estimating the difference, we round to the nearest hundred thousand: 1,000,000 – 100,000 = 900,000

These estimations provide a quick approximation of the results, which can be helpful in mental calculations and getting a rough idea of the outcome.

### Q2. Give a rough estimate (by rounding off to the nearest hundreds) and also a closer estimate (by rounding off to the nearest tens) :

(a) 439 + 334 + 4,317

(b) 1,08,734 – 47,599

(c) 8325 – 491

(d) 4,89,348 – 48,365

Make four more such examples.

**Ans.**

(a) 439 + 334 + 4,317:

Rough estimate (rounding off to the nearest hundreds): 400 + 300 + 4,300 = 5,000

Closer estimate (rounding off to the nearest tens): 440 + 330 + 4,320 = 5,090

(b) 1,08,734 – 47,599:

Rough estimate (rounding off to the nearest hundreds): 1,08,700 – 47,600 = 61,100

Closer estimate (rounding off to the nearest tens): 1,08,730 – 47,600 = 61,130

(c) 8325 – 491:

Rough estimate (rounding off to the nearest hundreds): 8,300 – 500 = 7,800

Closer estimate (rounding off to the nearest tens): 8,320 – 490 = 7,830

(d) 4,89,348 – 48,365:

Rough estimate (rounding off to the nearest hundreds): 4,89,300 – 48,400 = 4,40,900

Closer estimate (rounding off to the nearest tens): 4,89,350 – 48,360 = 4,40,990

#### Here are four more examples:

(i) 1,234 + 567 + 89:

Rough estimate (rounding off to the nearest hundreds): 1,200 + 600 + 100 = 1,900

Closer estimate (rounding off to the nearest tens): 1,230 + 570 + 90 = 1,890

(ii) 12,345 – 6,789:

Rough estimate (rounding off to the nearest hundreds): 12,300 – 6,800 = 5,500

Closer estimate (rounding off to the nearest tens): 12,350 – 6,790 = 5,560

(iii) 987 + 654 + 321:

Rough estimate (rounding off to the nearest hundreds): 1,000 + 700 + 300 = 2,000

Closer estimate (rounding off to the nearest tens): 990 + 650 + 320 = 1,960

(iv) 56,789 – 12,345:

Rough estimate (rounding off to the nearest hundreds): 56,800 – 12,300 = 44,500

Closer estimate (rounding off to the nearest tens): 56,790 – 12,350 = 44,440

These rough and closer estimates provide a quick approximation of the results, giving us a general idea of the outcome while performing mental calculations or making quick assessments.

### Q3. Estimate the following products using general rule:

(a) 578 × 161

(b) 5281 × 3491

(c) 1291 × 592

(d) 9250 × 29

Make four more such examples

**Ans.**

(a) 578 × 161:

Rough estimate: 600 × 200 = 120,000

Closer estimate: 580 × 160 = 92,800

(b) 5281 × 3491:

Rough estimate: 5000 × 3500 = 17,500,000

Closer estimate: 5300 × 3500 = 18,550,000

(c) 1291 × 592:

Rough estimate: 1300 × 600 = 780,000

Closer estimate: 1300 × 590 = 767,000

(d) 9250 × 29:

Rough estimate: 9000 × 30 = 270,000

Closer estimate: 9300 × 30 = 279,000

#### Here are four more examples:

(i) 356 × 897:

Rough estimate: 400 × 900 = 360,000

Closer estimate: 360 × 900 = 324,000

(ii) 2187 × 451:

Rough estimate: 2000 × 500 = 1,000,000

Closer estimate: 2200 × 450 = 990,000

(iii) 672 × 83:

Rough estimate: 700 × 80 = 56,000

Closer estimate: 670 × 80 = 53,600

(iv) 7950 × 126:

Rough estimate: 8000 × 130 = 1,040,000

Closer estimate: 8000 × 120 = 960,000

These rough and closer estimates provide a quick approximation of the products, allowing for easier mental calculations and a general understanding of the magnitude of the results.

## Here are the NCERT Solutions for Class 6 Maths Chapter 1:

**Q: Write the greatest and smallest 6-digit numbers using the digits 6, 0, 5, 2, 1, 9.**

A: The greatest 6-digit number using the given digits is 965210, and the smallest 6-digit number using the given digits is 102569.

**Q: Write the Roman numerals for the given numbers: 50, 100, 500, and 1000.**

A: The Roman numerals for the given numbers are: 50 – L, 100 – C, 500 – D, and 1000 – M.

**Q: Write the successor and predecessor of the following numbers: 59, 155, 2011.**

A: The successor and predecessor of the given numbers are:

59: successor – 60, predecessor – 58

155: successor – 156, predecessor – 154

2011: successor – 2012, predecessor – 2010**Q: Write the smallest composite number and the smallest prime number using the digits 2, 3, 5, 0, 1, 7 (each digit used only once).**

A: The smallest composite number using the given digits is 102, and the smallest prime number using the given digits is 17.

**Q: Estimate the sum of 3224 and 6784.**

A: To estimate the sum of 3224 and 6784, we can round off both numbers to the nearest thousand. So, 3224 ≈ 3000 and 6784 ≈ 7000. Therefore, the estimated sum of 3224 and 6784 is 3000 + 7000 = 10000.