# NCERT Solutions for Class 6 Math Chapter 1 Knowing Our Numbers

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ncert solutions for class 6 maths chapter 1 Knowing Our Numbers for English Medium CBSE class 6 maths. Here in class 6 maths solutions learn to solve Knowing Our Numbers problem’s solution and how to solve questions with easiest learning tricks. ncert class 6 maths. ncert solutions for class 6 maths chapter 1 by our experts team solved all questions for students.
NCERT Solutions for Class 6 Math chapter 1, 6th standard maths chapter 1 are part of NCERT 6th maths book. Here we have given NCERT Solutions for class 6 maths solutions.

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# 6th standard maths chapter 1

## NCERT Solutions for Class 6 Maths chapter 1ncert class 6 maths chapter 1Ex . 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.

 (a) 1 lakh = 1,00,000 ten thousand. (b) 1 million = 1,000,000 hundred thousand. (c) 1 crore = 1,00,00,000 ten lakh. (d) 1 crore = 1,00,00,000 million. (e) 1 million = 1,000,000 lakh.

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.
(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 Indian System of Numeration :

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

(a) 87595762
Sol – 8,7595,762
Eight crore seventy-five lakh ninety-five thousand seven hundred sixty two.

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

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

(d) 98432701
Sol – 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) 78921092 (b) 7452283 (c) 99985102 (d) 48049831

(a) 78921092
Sol – 78,921,092

Seventy eight million nine hundred twenty one thousand ninety two

(b) 7452283
Sol – 7,452,283

Seven million four hundred fifty two thousand two hundred eighty three

(c) 99985102
Sol – 99,985,102

Ninety nine million nine hundred eighty five thousand one hundred two

(d) 48049831
Sol – 48, 049,831

Forty eight million forty nine thousand eight hundred thirty one

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.
Sol – Tickets sold on first day = 1094
Tickets sold on second day = 1812
Tickets sold on third day = 2050
Tickets sold on four day = 2751
Total tickets sold = 1094 + 1812 + 2050 + 2751
$$\displaystyle \begin{array}{l}\,\,\,\,1094\\\,\,\,\,1812\\\,\,\,\,2050\\+\,2751\\\overline{\begin{array}{l}\,\,\,\,7707\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}$$

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?
Sol – Runs scored so far = 6980
Runs Shekhar wants to score = 10,000
More runs required = 10,000 − 6980
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {10000} \end{array}\\\begin{array}{*{20}{c}} – & {\,6980} \end{array}\\\overline{{\,\,\,\,\,\,\,\,\,3020}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$
So, Shekhar requires 3,020 more 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?
Sol – Votes secured by successful candidate = 5,77,500
Votes secured by rival = 3,48,700
Margin = 5,77,500 − 3,48,700
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,577500} \end{array}\\\begin{array}{*{20}{c}} – & {348700} \end{array}\\\overline{{\,\,\,\,\,\,\,\,\,228800}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$
So, the margin is 2,28,800

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?
Sol – Value of Books sold in 1 week = Rs 2,85,891
Value of books sold in 2 week = Rs 4,00,768
Total sale = Sale in 1 week + Sale in 2 week
= 2,85,891 + 4,00,768
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,285891} \end{array}\\\begin{array}{*{20}{c}} + & {400768} \end{array}\\\overline{{\,\,\,\,\,\,\,\,\,686659}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$
The sale for the two weeks together was 6,86,659.
Since 4,00,768 > 2,85,891, sale in 2 week was greater than 1 week.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,400768} \end{array}\\\begin{array}{*{20}{c}} – & {285891} \end{array}\\\overline{{\,\,\,\,\,\,\,\,\,114877}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$
∴ The sale in 2 week was larger than the sale in 1 week by 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, 3 each only once.
Sol – Greatest number = 76432
Smallest number = 23467
Difference = 76432 − 23467
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,76432} \end{array}\\\begin{array}{*{20}{c}} – & {23467} \end{array}\\\overline{{\,\,\,\,\,\,\,\,\,52965}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$
So, the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once is 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?
Sol – Screws produced in one day = 2,825
Days in January = 31
Screws produced in 31 days = 2825 × 31
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,\,\,825} \end{array}\\\begin{array}{*{20}{c}} {} & {\,\times \,\,\,31} \end{array}\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,2825\\+\,\,\,8475\times \end{array}}\\\overline{\begin{array}{l}\,\,\,\,\,\,87575\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}$$

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?
Sol – Cost of one radio set = Rs 1200
Cost of 40 radio sets = 1200 × 40 = Rs 48000
Money with Merchant = Rs 78,592
Money spent = Rs 48,000
Money left = 78592 − 48000
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,78592} \end{array}\\\begin{array}{*{20}{c}} – & {48000} \end{array}\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,30592\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}$$
Therefore, Rs 30,592 will remain with her after the purchase.

8. 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?)
Sol – Difference between 65 and 56 = 9
Difference in the answer = 7236 × 9
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,7236} \end{array}\\\begin{array}{*{20}{c}} {} & {\times \,\,\,\,\,\,\,9} \end{array}\\\overline{\begin{array}{l}\,\,\,\,\,65124\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}$$

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?
(Hint: convert data in cm.)
Sol – 2 m 15 cm = 215 cm (1 m = 100 cm)
40 m = 40 × 100
= 4000 cm
Cloth required for one shirt = 215 cm
Number of shirts that can be stitched out of 4000 cm = 4000 ÷ 215
$$\displaystyle \begin{array}{l}\left. {215} \right)4000\left( {18} \right.\\\,\,\,\,\,-215\\\,\,\,\,\,\,\,\,\overline{{1850\,}}\\\,\,\,\,-1720\\\,\,\,\,\,\,\,\,\overline{{\,\,\,130}}\end{array}$$
Therefore, 18 shirts can be made. 130 cm, i.e. 1 m 30 cm, 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?
Sol – 1 kg = 1000 g
4 kg 500 g = 4500 g
800 kg = 800 × 1000 = 800000 g
Number of boxes that can be loaded in the van = 800000 ÷ 4500
$$\displaystyle \begin{array}{l}\left. {4500} \right)800000\left( {177} \right.\\\,\,\,\,\,\,\,\,\,-4500\\\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{{35000\,}}\\\,\,\,\,\,\,\,\,\,-31500\\\,\,\,\,\,\,\,\,\,\,\,\overline{{\,\,\,\,\,35000}}\\\,\,\,\,\,\,\,\,\,\,\,\,-31500\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{{\,\,\,\,3500}}\end{array}$$
Hence, 177 boxes at maximum can be loaded in the van.

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.
Sol – Distance between school and house = 1 km 875 m
Now, 1 km = 1000 m
1 km 875 m = 1875 m
Distance covered each day = 1875 × 2 = 3750 m
Distance covered in 6 days = 3750 × 6
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,3750} \end{array}\\\begin{array}{*{20}{c}} {} & {\times \,\,\,\,\,\,\,6} \end{array}\\\overline{\begin{array}{l}\,\,\,\,\,22500\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}$$
Therefore, distance covered in 6 days = 22,500 m
= 22.5 km or 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?
Sol – Capacity of vessel = 4 l 500 ml
= 4500 ml (1 l = 1000 ml)
Capacity of a glass = 25 ml
Number of glasses that can be filled = 4500 ÷ 25
$$\displaystyle \begin{array}{l}\left. {25} \right)4500\left( {180} \right.\\\,\,\,-25\\\,\,\,\,\,\,\,\overline{{200\,}}\\\,\,\,-200\\\,\,\,\,\,\overline{{\,\,\,\times \,\,\,\,\,\,}}\end{array}$$
∴ 180 glasses can be filled.

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.

(a) 730 + 998
Sol – By rounding off to hundreds, 730 rounds off to 700 and 998 rounds off to 1000.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {1000} \end{array}\\\begin{array}{*{20}{c}} – & {\,\,700} \end{array}\\\,\,\,\,\,\,\overline{{\,\,1700\,}}\\\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(b) 796 – 314
Sol – By rounding off to hundreds, 796 rounds off to 800 and 314 rounds off to 300.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,800} \end{array}\\\begin{array}{*{20}{c}} – & {300} \end{array}\\\,\,\,\,\,\,\overline{{\,\,500\,}}\\\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(c) 12,904 + 2,888
Sol – By rounding off to thousands, 12904 rounds off to 13000 and 2822 rounds off to 3000.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {13000} \end{array}\\\begin{array}{*{20}{c}} + & {\,\,3000} \end{array}\\\,\,\,\,\,\,\overline{{\,\,16000\,}}\\\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(d) 28,292 – 21,496
Sol – By rounding off to nearest thousands, 28296 rounds off to 28000 and 21496 rounds off to 21000.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,28000} \end{array}\\\begin{array}{*{20}{c}} + & {21000} \end{array}\\\,\,\,\,\,\,\overline{{\,\,\,\,\,7000\,}}\\\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

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) 8325 – 491 (d) 4,89,348 – 48,365

Make four more such examples.

(a) 439 + 334 + 4,317
Sol – Rounding off to nearest hundreds, 439, 334,and 4317 may be rounded off to 400,  300,and 4300 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,\,\,\,400} \end{array}\\\begin{array}{*{20}{c}} {} & {\,\,\,\,300} \end{array}\\\begin{array}{*{20}{c}} + & {4300} \end{array}\\\,\,\,\,\,\overline{{\,\,\,5000\,}}\\\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

Rounding off to nearest tens, 439, 334,and 4317 may be rounded off to 440, 330,and 4320 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {\,\,\,\,440} \end{array}\\\begin{array}{*{20}{c}} {} & {\,\,\,\,330} \end{array}\\\begin{array}{*{20}{c}} + & {4320} \end{array}\\\,\,\,\,\,\overline{{\,\,\,5090\,}}\\\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(b) 1,08,734 – 47,599
Sol – Rounding off to hundreds, 1,08,734 and 47,599 may be rounded off to 1,08,700 and 47,600 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {108700} \end{array}\\\begin{array}{*{20}{c}} – & {\,47600} \end{array}\\\,\,\,\,\,\,\overline{{\,\,\,61100\,}}\\\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

Rounding off to tens, 1,08,734 and 47,599 may be rounded off to 1,08,730 and 47,600 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {108730} \end{array}\\\begin{array}{*{20}{c}} – & {\,47600} \end{array}\\\,\,\,\,\,\,\overline{{\,\,\,61130\,}}\\\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(c) 8325 – 491
Sol – Rounding off to hundreds, 8325 and 491 may be rounded off to 8300 and 500 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {8300} \end{array}\\\begin{array}{*{20}{c}} – & {\,\,500} \end{array}\\\,\,\,\,\,\,\overline{{\,\,7800\,}}\\\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

Rounding off to tens, 8325 and 491 may be rounded off to 8330 and 490 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {8330} \end{array}\\\begin{array}{*{20}{c}} – & {\,\,490} \end{array}\\\,\,\,\,\,\,\overline{{\,\,7840\,}}\\\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(d) 4,89,348 – 48,365
Sol – Rounding off to hundreds, 489348 and 48365 may be rounded off to 489300 and 48400 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {489300} \end{array}\\\begin{array}{*{20}{c}} – & {\,\,48400} \end{array}\\\,\,\,\,\,\,\overline{{\,\,440900\,}}\\\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

Rounding off to tens, 489348 and 48365 may be rounded off to 489350 and 48370 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {489350} \end{array}\\\begin{array}{*{20}{c}} – & {\,\,48370} \end{array}\\\,\,\,\,\,\,\overline{{\,\,440980\,}}\\\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

3. Estimate the following products using general rule :

 (a) 5281 × 3491 (b) 5281 × 3491 (c) 1291 × 592 (d) 9250 × 29

Make four more such examples.
(a) 578 × 161
Sol – Rounding off by general rule, 598 and 161 may be rounded off to 600 and 200 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {600} \end{array}\\\begin{array}{*{20}{c}} \times & {200} \end{array}\\\overline{{120000\,}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(b) 5281 × 3491
Sol – Rounding off by general rule, 5281 and 3491 may be rounded off to 5000 and 3000 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {5000} \end{array}\\\begin{array}{*{20}{c}} \times & {3000} \end{array}\\\overline{{1500000\,}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(c) 1291 × 592
Sol – Rounding off by general rule, 1291 and 592 may be rounded off to 1000 and 600 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {1000} \end{array}\\\begin{array}{*{20}{c}} \times & {\,\,600} \end{array}\\\overline{{\,\,600000\,}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

(d) 9250 × 29
Sol – Rounding off by general rule, 9250 and 29 may be rounded off to 9000 and 30 respectively.
$$\displaystyle \begin{array}{l}\begin{array}{*{20}{c}} {} & {9000} \end{array}\\\begin{array}{*{20}{c}} \times & {\,\,\,\,\,30} \end{array}\\\overline{{\,\,270000\,}}\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}$$

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