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

NCERT Solutions for Class 6 Math Chapter 1 Knowing Our Numbers are provided here with simple step-by-step explanations. These solutions for Knowing Our Numbers are extremely popular among class 6 students for Math Knowing Our Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 6 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 6 Math are prepared by experts and are 100% accurate.

#### Page No 12:

#### 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.

#### Answer:

(a) lakh =ten thousand

(1 lakh = 1,00,000 and ten thousand = 10,000)

(b) 1 million =hundred thousand

(1 million = 1,000,000 and 1 hundred thousand = 1, 00,000)

(c) 1 crore = ten lakh

(1 crore = 1,00,00,000 and Ten lakh = 10,00,000)

(d) 1 crore = million

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

(e) 1 million = lakh

(1 million = 1,000,000 and 1 lakh = 1,00,000)

#### Page No 12:

#### Question 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.

#### Answer:

(a) 73,75,307

(b) 9,05,00,041

(c) 7,52,21,302

(d) 58,423,202

(e) 23,30,010

#### Page No 12:

#### Question 3:

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

(a). 87595762 (b). 8546283

(c). 99900046 (d). 98432701

#### Answer:

(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

##### Video Solution for knowing-our-numbers (Page: 12 , Q.No.: 3)

NCERT Solution for Class 6 math - knowing-our-numbers 12 , Question 3

#### Page No 12:

#### Question 4:

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

(a). 78921092 (b). 7452283

(c). 99985102 (c). 48049831

#### Answer:

(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

##### Video Solution for knowing-our-numbers (Page: 12 , Q.No.: 4)

NCERT Solution for Class 6 math - knowing-our-numbers 12 , Question 4

#### Page No 16:

#### Question 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.

#### Answer:

Tickets
sold on 1^{st} day = 1094

Tickets
sold on 2^{nd} day = 1812

Tickets
sold on 3^{rd} day = 2050

Tickets
sold on 4^{th} day = 2751

Total tickets sold = 1094 + 1812 + 2050 + 2751

∴ Total tickets sold = 7,707

#### Page No 16:

#### Question 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?

#### Answer:

Runs scored so far = 6980

Runs Shekhar wants to score = 10,000

More runs required = 10,000 − 6980

∴ Shekhar requires 3,020 more runs.

##### Video Solution for knowing-our-numbers (Page: 16 , Q.No.: 2)

NCERT Solution for Class 6 math - knowing-our-numbers 16 , Question 2

#### Page No 16:

#### Question 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?

#### Answer:

Votes secured by successful candidate = 5,77,500

Votes secured by rival = 3,48,700

Margin = 5,77,500 − 3,48,700

∴ Margin = 2,28,800

#### Page No 16:

#### Question 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?

#### Answer:

Value of Books sold in 1^{st} week = Rs 2,85,891

Value of books sold in 2^{nd} week = Rs 4,00,768

Total sale = Sale in 1^{st} week + Sale in 2^{nd} week

= 2,85,891 + 4,00,768

The sale for the two weeks together was 6,86,659.

Since 4,00,768 > 2,85,891, sale in 2^{nd} week was greater than 1^{st} week.

∴ The sale in 2^{nd} week was larger than the sale in 1^{st} week by Rs 1,14,877.

##### Video Solution for knowing-our-numbers (Page: 16 , Q.No.: 4)

NCERT Solution for Class 6 math - knowing-our-numbers 16 , Question 4

#### Page No 17:

#### Question 5:

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

#### Answer:

Greatest number = 76432

Smallest number = 23467

Difference = 76432 − 23467

Therefore, 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.

#### Page No 17:

#### Question 6:

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

#### Answer:

Screws produced in one day = 2,825

Days in January = 31

Screws produced in 31 days = 2825 × 31

Therefore, screws produced during Jan, 06 = 87,575

#### Page No 17:

#### Question 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?

#### Answer:

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

Therefore, Rs 30,592 will remain with her after the purchase.

#### Page No 17:

#### Question 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?)

#### Answer:

Difference between 65 and 56 = 9

Difference in the answer = 7236 × 9

Therefore, his answer was greater than the correct answer by 65,124.

#### Page No 17:

#### Question 9:

To stitch a shirt, 2m 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.)

#### Answer:

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

Therefore, 18 shirts can be made. 130 cm, i.e. 1 m 30 cm, cloth will remain.

#### Page No 17:

#### Question 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?

#### Answer:

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

Hence, 177 boxes at maximum can be loaded in the van.

##### Video Solution for knowing-our-numbers (Page: 17 , Q.No.: 10)

NCERT Solution for Class 6 math - knowing-our-numbers 17 , Question 10

#### Page No 17:

#### Question 11:

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

#### Answer:

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

Therefore, distance covered in 6 days = 22,500 m

= 22.5 km or 22 km 500 m

#### Page No 17:

#### Question 12:

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

#### Answer:

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

∴ 180 glasses can be filled.

##### Video Solution for knowing-our-numbers (Page: 17 , Q.No.: 12)

NCERT Solution for Class 6 math - knowing-our-numbers 17 , Question 12

#### Page No 23:

#### Question 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.

#### Answer:

(a) 730 + 998

By rounding off to hundreds, 730 rounds off to 700 and 998 rounds off to 1000.

(b) 796 − 314

By rounding off to hundreds, 796 rounds off to 800 and 314 rounds off to 300.

(c) 12904 + 2822

By rounding off to thousands, 12904 rounds off to 13000 and 2822 rounds off to 3000.

(d) 28,296 − 21,496

By rounding off to nearest thousands, 28296 rounds off to 28000 and 21496 rounds off to 21000.

#### Page No 23:

#### Question 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.

#### Answer:

(a) 439 + 334 + 4317

Rounding off to nearest hundreds, 439, 334, and 4317 may be rounded off to 400, 300, and 4300 respectively.

Rounding off to nearest tens, 439, 334, and 4317 may be rounded off to 440, 330, and 4320 respectively.

(b) 1,08,734 − 47,599

Rounding off to hundreds, 1,08,734 and 47,599 may be rounded off to 1,08,700 and 47,600 respectively.

Rounding off to tens, 1,08,734 and 47,599 may be rounded off to 1,08,730 and 47,600 respectively.

(c) 8325 − 491

Rounding off to hundreds, 8325 and 491 may be rounded off to 8300 and 500 respectively.

Rounding off to tens, 8325 and 491 may be rounded off to 8330 and 490 respectively.

(d) 4,89,348 − 48,365

Rounding off to hundreds, 489348 and 48365 may be rounded off to 489300 and 48400 respectively.

Rounding off to tens, 489348 and 48365 may be rounded off to 489350 and 48370 respectively.

#### Page No 23:

#### Question 3:

Estimate the following products using general rule:

(a) 578 × 161 (b) 5281 × 3491

(c) 1291 × 592 (d) 9250 × 29

#### Answer:

(a) 578 × 161

Rounding off by general rule, 598 and 161 may be rounded off to 600 and 200 respectively.

(b) 5281 × 3491

Rounding off by general rule, 5281 and 3491 may be rounded off to 5000 and 3000 respectively.

(c) 1291 × 592

Rounding off by general rule, 1291 and 592 may be rounded off to 1000 and 600 respectively.

(d) 9250 × 29

Rounding off by general rule, 9250 and 29 may be rounded off to 9000 and 30 respectively.

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