NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Knowing Our Numbers Class 6 NCERT Topic-Comparing Numbers

Q1 Can you instantly find the greatest and the smallest numbers in each row?
2. 1473, 89423, 100, 5000, 310
3. 1834, 75284, 111, 2333, 450
4. 2853, 7691, 9999, 12002, 124

Answer: By Observing the number of digits and digit in the leftmost place can straight away tell us about the greatest and smallest number

(2) 1473, 89423, 100, 5000, 310

89423 is the greatest while 100 is the smallest

(3) 1834, 75284, 111, 2333, 450

75284 is the greatest 111 while is the smallest

(4) 2853, 7691, 9999, 12002, 124

12002 is the greatest while 124 is the smallest

Q2 Find the greatest and the smallest numbers.
(a) 4536, 4892, 4370, 4452
(b) 15623, 15073, 15189, 15800
(c) 25286, 25245, 25270, 25210
(d) 6895, 23787, 24569, 24659

Answer:

(a) 4536, 4892, 4370, 4452

4892 is the greatest while 4370 is the smallest

(b) 15623, 15073, 15189, 15800

15800 is the greatest while 15073 is the smallest
(c) 25286, 25245, 25270, 25210

25286 is the greatest while 25245 is the smallest
(d) 6895, 23787, 24569, 24659

24659 is the greatest while 6895 is the smallest

Knowing Our Numbers Class 6 NCERT Topic: Number Formation

Q1 Use the given digits without repetition and make the greatest and smallest 4-digit numbers.
(a) 2, 8, 7, 4
(b) 9, 7, 4, 1
(c) 4, 7, 5, 0
(d) 1, 7, 6, 2
(e) 5, 4, 0, 3

Answer:

(a) 2, 8, 7, 4

8742 is the greatest while 2478 is the smallest

(b) 9, 7, 4, 1

9741 is the greatest 1479 while is the smallest

(c) 4, 7, 5, 0

4750 is the greatest while 4057 is the smallest

(d) 1, 7, 6, 2

7621 is the greatest while 1267 is the smallest

(e) 5, 4, 0, 3

5430 is the greatest while 3045 is the smallest

Q2 Now make the greatest and the smallest 4-digit numbers by using any one digit twice.
(a) 3, 8, 7
(b) 9, 0, 5
(c) 0, 4, 9
(d) 8, 5, 1

Answer:

(a) 3, 8, 7

8873 is the greatest while 3387 is the smallest

(b) 9, 0, 5

9950 is the greatest while 5009 is the smallest

(c) 0, 4, 9

9940 is the greatest while 4009 is the smallest

(d) 8, 5, 1

8851 is the greatest while 1158 is the smallest

Q3 Make the greatest and the smallest 4-digit numbers using any four different digits with conditions as given.

(a) Digit 7 is always at one's place
Greatest - 9 8 6 7
Smallest - 1 0 2 7
(Note, the number cannot begin with the digit 0. Why?)

(b) Digit 4 is always at tens place
Greatest - _ _ 4 _
Smallest - _ _ 4 _

(c) Digit 9 is always at hundreds place
Greatest - _ 9 _ _
Smallest - _ 9 _ _
(d) Digit 1 is always at thousands place
Greatest- 1 _ _ _
Smallest - 1 _ _ _

Answer:

(b) Digit 4 is always at tens place
Greatest - 98 47
Smallest - 10 42

(c) Digit 9 is always at hundreds place
Greatest - 8 976
Smallest - 1 902

(d) Digit 1 is always at thousands place
Greatest- 1987
Smallest - 1023

Q4 Take two digits, say 2 and 3. Make 4-digit numbers using both the digits an equal number of times.
Which is the greatest number?
Which is the smallest number?
How many different numbers can you make in all?

Answer:

Greatest number - 3322

Smallest number - 2233

Different numbers are

2222, 2232, 2233, 2322, 2333, 2332, 2323, 2223, 3222, 3223, 3232, 3233, 3322, 3333, 3332, 3323

There are 16 in total

Solutions For Knowing Our Numbers Class 6 NCERT Topic: Arrangement of Numbers

Q1 Arrange the following numbers in ascending order :
(a) 847, 9754, 8320, 571
(b) 9801, 25751, 36501, 38802

Answer:

(a) 847, 8320, 8320, 571

571 < 847 < 8320 < 8320

(b) 9801, 25751, 25751, 38802

9801 < 25751 < 25751 < 38802

Q2 Arrange the following numbers in descending order :
(a) 5000, 7500, 85400, 7861
(b) 1971, 45321, 88715

Answer:

(a) 5000, 7500, 85400, 7861

85400 > 7861 > 5000 > 75000

(b) 1971, 45321, 88715

88715 > 45321 > 1971

Solutions For Knowing Our Numbers Class 6 NCERT Topic-Revisiting Place Value

Q1 Read and expand the numbers wherever there are blanks.

1. 50000- ________________________________________
2. 41000- ________________________________________
3. 47300- ________________________________________
4. 57630- ________________________________________

5. 29485________________________________________
6. 29085________________________________________

Answer:

1. 50000-Fifty thousand
2. 41000-forty one thousand
3. 47300- forty-seven thousand and three hundred
4. 57630- fifty-seven thousand six hundred

5. 29485- twenty-nine thousand and four hundred eighty-five
6. 29085- twenty-Nine Thousand and Eighty-Five

Q2 Read and expand the numbers wherever there are blanks.

4,57,928 _______________ _______________

4,07,928 _______________ _______________

4,00,829 _______________ _______________

4,00,029 _______________ _______________

Answer:

4,57,928 Four Lakhs Fifty Seven Thousand and Nine Hundred Twenty Eight

4,07,928 Four Lakhs Seven Thousand and Nine Hundred Twenty-Eight

4,00,829 Four Lakhs Eight Hundred Twenty-Nine

4,00,029 Four Lakhs and Twenty-Nine

Q3 You have the following digits 4, 5, 6, 0, 7 and 8.

Using them, make five numbers each with 6 digits.

(a) Put commas for easy reading.

(b) Arrange them in ascending and descending order.

Answer:

Given, digits : $4,5,6,0,7,8$

Five 6-digit numbers are: $456078,45708,548760,750486,876540$

NCERT Solutions for Class 6 Maths Chapter 1 - 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.

Answer:

Let us take this as a physical world problem:

You went to a shop and you bought a toy of Rs. 100. You checked your pocket and found many notes but of only Rs.10

How many Rs.10 notes will you give to shopkeeper:

Well, it’s simple.

You will pay 10 notes.

How?

100/10 =10

The similar easy approach can be done in all the blanks,

Hint: Always remember to keep your eyes on a number of zeros

(a)1 lakh= 100000

1thousand= 1000

Ten thousand= 10*1000 = 10000

Therefore,

1 lakh/ 10 thousand =100000/10000 =10

So, 1lakh= 10 ten thousand

(b) 1 million= 1000000

1thousand= 1000

Hundred thousand= 100*1000 = 100000

Therefore,

1 million/ hundred thousand =1000000/100000 =10

So, 1milion= 10 hundred thousand

(c) 1 crore= 10000000 (7 zeros)

1lakh= 100000 (5 zeros)

Ten lakh= 10*1000 = 1000000( 6 zeros)

Therefore,

1 crore/ Ten lakh=10000000/1000000 =10

So, 1 crore= 10 Ten lakh

(d)1 crore= 10000000 (7 zeros)

1million= 100000 (6 zeros)

Therefore,

1 crore/ 1million=10000000/1000000 =10

So, 1 crore= 10 million

(e) 1 million= 1000000

1lakh= 100000

Therefore,

1million/ 10 lakh = 1000000/100000 =10

So, 1 million= 10 lakhs

See, all answers came out to be 10 but not everytime

Q2 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) 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.

$58,423,302$

(e) Twenty three lakh thirty thousand ten.

$23,30,010$

Q3 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 crores seventy five lakhs ninety five thousand seven hundred sixty two

(b) $85,46,283$ : Eighty five lakhs forty six thousand two hundred eighty three.

(c) $9,99,00,046$ : Nine crores ninety nine lakhs and forty six.

(d) $9,84,32,701$ : Nine crores eighty four lakhs thirty two thousand seven hundred and one.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers: Large Numbers in Practise

Q1 A box contains 2,00,000 medicine tablets each weighing 20 mg. What is the total weight of all the tablets in the box in grams and in kilograms? Answer:

Number of medicine tablets = $2,00,000$

Weight of each medicine tablet = $20mg=\frac{20}{1000}g$

Therefore, total weight = $2,00,000\times \frac{20}{1000}g$

$=4,000g$

$=4kg$

Q2 A bus started its journey and reached different places with a speed of 60 km/hour.
(i) Find the total distance covered by the bus from A to D.
(ii) Find the total distance covered by the bus from D to G.
(iii) Find the total distance covered by the bus, if it starts from A and returns back to A.
(iv) Can you find the difference of distances from C to D and D to E?
Answer:

(i) Total distance from A to D = AB + BC + CD

$=(4170+3410+2160)km=9740km$

(ii) Total distance between D to G = DE + EF + FG

$=(8140+4830+2550)km=15520km$

(iii) Required distance = AD + DG + GA

$=(4170+3410+2160+8140+4830+2550+1290)km=26550km$

(iv) The difference of distances from C to D and D to E $=(8140-2160)km=5980km$

NCERT Solutions for Class 6 Maths Chapter 1 - 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.

Answer: Given,

• Tickets sold on the first day of exhibition = 1094
• Tickets sold on the second day = 1812
• Tickets sold on the third day = 2050
• Tickets sold on the last day = 2751
• Total number of tickets sold =

$$\begin{gathered} =1094+1812+2050+2751 \\ =7707 \end{gathered}$$

Total number of tickets sold = $7,707$

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?

Answer:

The number of runs scored by Shekhar so far = 6980

The number of runs Shekhar wishes to score = 10,000

$\therefore$ The number of runs required = $10,000-6,980$

$=3,020runs$

Therefore, Shekhar requires $3,020runs$ .

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?

Answer: Given,

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

Number of votes registered by the rival candidate = 3, 48,700 votes

The margin $=5,77,500-3,48,700$

$=2,28,800votes$

Therefore, the successful candidate won the election by a margin of $2,28,800$ votes.

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?

Answer: Given,

Worth of books sold in the first week = $Rs2,85,891$

Worth of books sold in the second week = $Rs4,00,768$

Total sale in the two weeks = $Rs.(2,85,891+4,00,768)$

$=Rs.6,86,659$

Clearly, sales in the second week is greater than the first year by

$Rs.(4,00,768-2,85,891)=Rs.1,14,877$

Q5 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.

Answer:

Given, digits: 6, 2, 7, 4 and 3

Since, the digits have to be used only once, arrange them in ascending and descending order to get the minimum and maximum number.

Greatest number = $76,432$

Smallest number = $23,467$

$\therefore$ The difference between the greatest and the least number
$$\begin{gathered} =76,432-23,467 \\ =52,965 \end{gathered}$$

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

Answer:

Given,
Screws produced by machine in one day = $2,825$

We know, there are 31 days in January

$\therefore$ Screws produced in 31 days = $2,825\times 31$

$=87,575screws$

Therefore, screws produced in January 2006 = $87,575$

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

Answer:

Total money merchant has = $Rs.78,592$

Cost of one radio set = $Rs.1200$

$\therefore$ Cost of 40 radio sets = $Rs.(1200\times 40)=Rs.48,000$

$\therefore$ Money left with the merchant = $Rs.(78,592-48,000)=Rs.30,592$

Therefore, money left with her after purchase = $Rs.30,592$

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?)

Answer: Given,

The student multiplied $7236by65$ instead of $56$ .

Wrong answer = $7236\times 65$

Correct answer = $7236\times 56$

$\therefore$ The difference in the answers = $7236\times 65-7236\times 56$

$=7236\times (65-56)=7236\times 9$

$=65124$

Hence, his answer was greater than the correct answer by $65124$

Ques 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.)

Answer: We know, $(1m=100cm)$

Given,

Length of cloth required to stitch a shirt = $2m15cm=215cm$

Also, $40m=(40\times 100)cm=4000cm$

Now, the cloth required for one shirt = $215cm$

Number of shirts that can be stitched out = $4000÷215$

$4000=(18\times 215)+130$

Therefore, 18 shirts can be made from the given cloth.

And, $130cm=1m30cm$ cloth will remain unused.

Ques 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?

Answer- We know, $(1kg=1000g)$

Weight of each medicine box= $4kg500g=4500g$

Weight limit of the van = $800kg=(800\times 1000)g=8,00,000g$

$\therefore$ The number of boxes that can be loaded in the van = $800000÷4500$

$800000=(4500\times 177)+3500$

Hence, 177 boxes can be loaded in the van.

Ques 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.

Answer: We know, $(1km=1000m)$

Given, - Distance between the school and her house = $1km875m=1875m$

The distance she covers each day = $[(1875\times 2)\times 6]m$

$(1875\times 12)m=22500m$

$=22000m500m=22km500m$

Therefore, she will cover $22km500m$ in six days.

Ques 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:

We know, $(1l=1000ml)$

Given, Capacity of vessel = $4l500ml=4500ml$

The capacity of a single glass = $25ml$

$\therefore$ Number of glasses required to fill the vessel = $4500÷25$

$4500=(180\times 25)+0$

Hence, 180 glasses are needed to fill the vessel completely.

NCERT Solutions for Class 6 Maths Chapter 1 - Estimating to the Nearest Thousand Rounding Off

Ques- 1 Round off the given numbers to the nearest tens, hundreds and thousands.

Given Number Approximate to Nearest Rounded Form
75847 Tens ________________
75847 Hundreds ________________
75847 Thousands ________________
75847 Ten thousands ________________

Answer:

NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.3

Ques 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

Answer:

(a) $730+998$
By rounding off to hundreds,

730 rounds off to 700,

998 rounds off to 1000.

Required sum = $700+1000$

$=1700$

(b) $796-314$

By rounding off to hundreds,

796 rounds off to 800,

314 rounds off to 300.

Required difference = $800-300$

$=500$

(c) $12904+2822$

By rounding off to thousands,

2904 rounds off to 13000,

2822 rounds off to 3000.

Required sum = $13000+3000$

$=16000$

(d) $28,296-21,496$

By rounding off to nearest thousands,

28,296 rounds off to 28000,

21,496 rounds off to 21,000

Required difference = $28,000-21,000$

$=7,000$

Q2 Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) :
$$\begin{gathered} (a)439+334+4,317 \\ (b)1,08,734-47,599 \\ (c)8325-491 \\ (d)4,89,348�48,365 \end{gathered}$$

Answer:

(a) $439+334+4,317$

Rounding off to nearest hundreds,

439 rounds off to 400

334 rounds off to 300

4317 rounds off to 4300,

Required sum = $400+300+4300$

$=5000$

Again, by rounding off to nearest tens,

439 rounds off to 440

334 rounds off to 330

4317 rounds off to 4320,

Required sum = $440+330+4320$

$=5090$

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

Rounding off to nearest hundreds,

$1,08,734$ rounds off to $1,08,700$

$47,599$ rounds off to $47,600$

Required difference = $1,08,700-47,600$

$=61100$

Also, rounding off to nearest tens,

$1,08,734$ rounds off to $1,08,730$

$47,599$ rounds off to $47,600$

Required difference = $1,08,730-47,600$

$=61130$

(c) $8325-491$

Rounding off to nearest hundreds,

8325 rounds off to 8300

491 rounds off to 500

Required difference = $8300-500$

$=7800$

Also, rounding off to nearest tens,

8325 rounds off to 8330

491 rounds off to 490

Required difference = $8330-490$

$=7840$

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

Rounding off to nearest hundreds,

$4,89,348$ rounds off to $4,89,300$

$48,365$ rounds off to $48,400$

Required difference = $489300-48400$

$=440900$

Also, rounding off to nearest tens,

$4,89,348$ rounds off to $4,89,350$

$48,365$ rounds off to $48,370$

Required difference = $489350-48370$

$=440980$

Ques 3- Estimate the following products using the general rule:
(a) 578 × 161 (b) 5281 × 3491 (c) 1291 × 592 (d) 9250 × 29

Answer:

(a) $578\times 161$

Rounding off by general rule,

578 rounds off to 600

161 rounds off to 200,

Required product = $600\times 200$

$=120000$

(b) $5281\times 3491$

Rounding off by the general rule,

5281 rounds off to 5000

3491 rounds off to 3500

Required product = $5000\times 3500$

$=1,75,00,000$

(c) $1291\times 592$

Rounding off by general rule,

1291 rounds off to 1300

592 rounds off to 600

Required product = $1300\times 600$

$=780000$

(d) $9250\times 29$

Rounding off by general rule,

9250 rounds off to 10000

29 rounds off to 30Required product = $10000\times 30$

$=300000$

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

Ques 1- Write the expressions for each of the following using brackets.
(a) Four multiplied by the sum of nine and two.
(b) Divide the difference of eighteen and six by four.
(c) Forty five divided by three times the sum of three and two.

Answer:

(a) Four multiplied by the sum of nine and two.

$=4\times (9+2)=4\times 11=44$
(b) Divide the difference of eighteen and six by four.

$=\frac{18-6}{4}=\frac{12}{4}=3$
(c) Forty-five divided by three times the sum of three and two.

$=\frac{45}{3(3+2)}$

$=\frac{45}{15}=3$

Ques 2- Write in Roman numerals.
1. 73
2. 92

Answer: In Roman numerals:
$$\begin{gathered} 1.73-LXXIII \\ 2.92-XCII \end{gathered}$$