 # NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers

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## Class 6 Maths Chapter 1: Exercise 1.1

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.

Solution.

(a) 1 lakh = 10 ten thousand.

(1 lakh = 1, 00, 000)

(b) 1 million = 10 hundred thousand.

(1 million = 1, 000, 000)

(c) 1 crore = 10 ten lakh

(1 crore = 1, 00, 00, 000)

(d) 1 crore = 10 million

(1 crore = 1, 00, 00, 000 and 10 million = 10, 000, 000)

(e) 1 million = 10 lakh

(1 million = 1, 000, 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.

Solution.

(a) 73,75,307

(b) 9,05,00,041

(c) 7,52,21,302

(d) 5,84,23,202

(e) 23,30,010.

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

(a) 87595762

(b) 8546283

(c) 99900046

(d) 98432701

Solution.

(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

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

(a) 78, 921092

(b) 7452283

(c) 99985102

(d) 48049831

Solution.

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

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

(c) 99,985,102 (Ninetynine million nine hundred eighty five thousand one hundred two)

(d) 48,049,831 (Fortyeight million fortynine thousand eight hundred thirtyone)

## Class 6 Maths 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 the four days.

Solution.

Number of tickets sold on first day = 1094

Number of tickets sold on second day = 1812

Number of tickets sold on third day = 2050

Number of tickets sold on fourth day = 2751

∴ Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7, 707

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 scored by Shekhar = 6980

Runs he want to complete = 10,000

∴Runshe needs to score more = 10,000 – 6980 = 3020

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.

Number of votes secured by the successful candidate = 5,77,500

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

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

4.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?

Solution.

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

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

Therefore, total sale of books in the two weeks together = ₹2,85,891 + ₹4,00,768 = ₹6,86,659

The sale of books was greater in the second week.

Difference in the sale in two weeks =  ₹4, 00, 768 – ₹2, 85, 891 = ₹1, 14, 877

∴ Sale in second week was greater than in first week by ₹1, 14, 877.

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

Solution.

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

Greatest number = 76432

Least number = 23467

Difference between the two numbers = 76432 – 23467 = 52,965

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

Solution.

Number of screws manufactured in one day = 2,825.

Now, January month has 31 days.

∴Number of screws manufactured in January = 31 x 2825 = 87,575

7.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?

Solution.

Merchant had total money = ₹78,592

Number of radio sets = 40

Cost of on radio set = ₹ 1200

So, price of 40 radio sets = ₹1200 x 40 = ₹48,000

Money left with the merchant = ₹78,592 – ₹48000 = ₹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 between 65 and 56 = 65 – 56 = 9

So, difference between the correct and incorrect answer = 7236 x 9 = 65124

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 length of the cloth = 40 m = 40 x 100 cm = 4000 cm.

Cloth required to stitch one shirt = 2 m 15 cm = 2 x 100 + 15 cm = 215 cm

∴Number of shirts that can be stitched out of 4000cm = 4000/215 = 18 and 130 cm of cloth is left out.

10. Medicine is packed in boxes, each weighing 4 kg 500 g. 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 x 1000 + 500 = 4500 g

Maximum weight the van can carry = 800 kg = 800 x 1000 = 800000 g

∴ Number of boxes that can be loaded in the van = 800000/4500 = 177

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

Solution.

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

As the student walks both way.

So, distance travelled by the student in one day = 2 x 1875 = 3750 m

Distance travelled by the student in 6 days = 3750 m x 6 = 22500 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.

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

Capacity of 1 glass = 25 mL

Therefore, number of glasses that can be filled with curd =  4500/25 = 180

## Class 6 Maths Chapter 1: Exercise 1.3

1. Estimate each of the following using 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.

Solution.

(a) 730 + 998

Rounding off 730 to the nearest hundreds = 700

Rounding off 998 to the nearest hundreds = 1,000

∴ 730 + 998 = 700 + 1000 = 1700

(b) 796 – 314

Rounding off 796 to the nearest hundreds = 800

Rounding off 314 to the nearest hundreds = 300

∴ 796 – 314 = 800 – 300 = 500

(c) 12,904 + 2,888

Rounding off 12,904 to the nearest thousands = 13000

Rounding off 2888 to the nearest thousands = 3000

∴ 12,904 + 2,888 = 13000 + 3000 = 16000

(d) 28,292 – 21,496

Rounding off 28,292 to the nearest thousands = 28,000

Rounding off 21,496 to the nearest thousands = 21,000

∴ 28,292 – 21,496 = 28,000 – 21,000 = 7,000

Let’s take ten more examples as follwos:

(i) 320 + 275 = 300 + 300 = 600

(ii) 3927 + 5980 = 4000 + 6000 = 10,000

(iii)6382 – 3672 = 6000 – 4000 = 2000

(iv)5438 – 2872 = 5000 – 3000 = 2000

(v)2185 + 1207 = 2000 + 1000 = 3000

(vi)1105 – 1282 = 1000 – 1000 = 0

(vii)915 + 565 = 900 + 600 = 1500

(viii)6344 – 4800 = 6000 – 5000 = 1000

(ix)3721 + 1400 = 4000 + 1000 = 5000

(x)6385 – 4317 = 6000 – 4000 = 2000

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

(a) 439 + 334 + 4,317

(b) 1,08,734-47,599

(c) 8,325-491

(d) 4,89,348-48,365

Make four such examples:

Solution.

(a)439 + 334 + 4,317

Rounding off to nearest hundreds:

439 + 334 + 4,317 = 400 + 300 + 4300 = 5,000

Rounding off to nearest tens:

439 + 334 + 4317 = 440 + 330 + 4320 = 5090.

(b) 1,08,734 – 47,599

Rounding off to nearest hundreds:

1,08,734 – 47,599 = 1,08,700 – 47,600 = 61,100

Rounding off to nearest tens:

1,08,734 – 47,599 = 1,08,730 – 47,600 = 61,130.

(c) 8325 – 491

Rounding off to nearest hundreds:

8325 – 491 = 8300 – 500 = 7800

Rounding off to nearest tens:

8325 – 491 = 8330 – 490 = 7840.

(d) 4,89,348 – 48,365

Rounding off to nearest hundreds:

4,89,348 – 48,365 = 4,89,300 – 48,400 = 4,40,900

Rounding off to nearest tens:

4,89,348 – 48,365 = 4,89,350 – 48,370 = 4,40,980

Four more such examples are given bellow:

(i)  4843 + 652

Rounding off to nearest hundreds:

4843 + 652 = 4800 + 700 = 5500

Rounding off to nearest tens:

4843 + 652 = 4840 + 650 = 5490

(ii) 765 – 380

Rounding off to nearest hundreds:

765 – 384= 800 – 400 = 400

Rounding off to nearest tens:

765 – 384 = 770 – 380 = 390

(iii) 6365 –2865

Rounding off to nearest hundreds:

6365 – 2865 = 6400 + 2900 = 9300

Rounding off to nearest tens:

6365 – 2865 = 6370 – 2870 = 3500

(iv) 8236 –6212

Rounding off to nearest hundreds:

8236 – 6212 = 8200 – 6200 = 2000

Rounding off to nearest tens:

8236 – 6212 = 8240 – 6210 = 2030

3. Estimate the following products using general rule:

(a) 578 x 161

(b)5281 x 3491

(c) 1291 x 592

(d) 9250 x 29

Make four more such examples.

Solution.

(a) 578 x 161 = 600 x 200 = 1,20,000

(b) 5281 x 3491 = 5000 x 3000 = 1,50,00,000

(c) 1291 x 592 = 1000 x 600 = 6,00,000

(d) 9250 x 29 = 9000 x 30 = 2,70,000

Four more such examples are:

(i) 4638 x 1593 = 5000 x 2000 = 10,00,000

(ii) 3858 x 1291 = 4000 x 1000 = 40,00,000

(iii) 9350 x 8963 = 9000 x 9000 = 8,10,00,000

(iv) 4205 x 7502 = 4000 x 8000 = 3,20,00,000

We are also providing the provision to download the complete solution of this chapter in PDF form which students may keep with them to use whenever required. Click on the following link to download the solutions:

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Check chapter-wise NCERT Solutions for Class 6 Maths from the links given below:

NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers

NCERT Solutions for Class 6 Maths Chapter 3 - Playing with Numbers

NCERT solutions for other chapters will be provided here very soon. Check here for the detailed and appropriate solutions.