These NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers InText Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers InText Questions
NCERT In-text Question Page No. 2
Question 1.
Can you instantly find the greatest and the smallest numbers in each row?
1. 382,4972,18, 59785, 750.
Answer:
59785 is the greatest and 18 is the smallest.
2. 1473, 89423, 100, 5000, 310.
Ans. ………………………..
3. 1834, 75284, 111, 2333,450 .
Ans. ………………………..
4. 2853, 7691, 9999,12002,124.
Ans. ………………………..
Was that easy? Why was it easy?
Answer:
2. 89423 is the greatest and 100 is the
smallest.
3. 75284 is the greatest and 111 is the smallest.
4. 12002 is the greatest and 124 is the smallest.
This is not very difficult. The greatest number has the most thousands and the smallest is only in hundreds or in tens.
Question 2.
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.
Ans:
(a) Each of the given numbers is having four digits and their digits at the thousands place are the same.
∴ The greatest number is 4892.
The smallest number is 4370.
(b) Each of the given numbers is having five digits and their two leftmost places have the same digits.
∴ The greatest number is 15800.
The smallest number is 15073.
(c) Each of the given numbers is having five digits and their three leftmost places have the same digits.
On comparing the fourth leftmost digits, we have:
The greatest number is 25286.
The smallest number is 25210.
(d) The given numbers are:
The number 6895 is having four digits, so, it must be the smallest number.
2378 7,2456 9,2465 9, we observe that 4 > 3.
Again, the two leftmost places of 24569 and 24659 are having the same digits (i.e. 2 and 4).
∴ On comparing their third leftmost digits, we have:
∴ The greatest number is 24659.
The smallest number is 6895.
NCERT In text Question Pages Nos 3 & 4
Question 1.
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
(Hint: 0754 is a 3-digit number.)
Answer:
(a) greatest-8742 — smallest-2478
(b) greatest-9741 — smallest-1479
(c) c.greatest-7540 — smallest-4057
(d) greatest-7621 — smallest-1267
(e) greatest-5430 — smallest-3045
Question 2.
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
(Hint: Think in each case which digit will you use twice.)
Answer:
(a) 8873 is the greatest while 3387 is the smallest
(b) 9950 is the greatest while 5009 is the smallest
(c) 9940 is the greatest while 4009 is the smallest
(d) 8851 is the greatest while 1158 is the smallest
Question 3.
Make the greatest and the smallest 4-digit numbers using any four different digits with conditions as given.
(a) Digit 7 is always at ones place
(Note, the number cannot begin with the digit 0. Why?)
(b) Digit 4 is always at tens place
(c) Digit 9 is always at hundreds place
(d) Digit 1 is always at thousands place
Answer:
Keeping the digit ‘7’ at ones place, we have:
∴ The greatest 4-digit number = 9867.
The smallest 4-digit number = 1027.
(b) Keeping the digit 4 at tens place, we have:
∴ The greatest 4-digit number = 9847.
The smallest 4-digit number = 1042.
(c) Keeping the digit 9 at the hundreds place, we have:
∴ The greatest 4-digit number = 8976.
The smallest 4-digit number = 1902.
(d) Keeping the digit 1 at the thousands place, we have:
∴The greatest 4-digit number = 1987.
The smallest 4-digit number = 1023.
Question 4.
Take two digits, say 2 and 3. Make 4-digit numbers using both the digits equal number of times.
Which is the greatest number?
Which is the smallest number?
How many different numbers can you make in all?
Answer:
The given digit are 2 and 3. For making 4-digit numbers, we will repeat the given digits equal number of times.
∴ The possible 4-digit numbers are 3322, 2233, 2323, 3232, 3223 and 2332.
Among these, we have
(i) The greatest number = 3322.
(ii) The smallest number = 2233.
(iii) We can make six different 4-digit numbers.
NCERT In text Question Page No 4
Question 5.
Think of five more situations where you compare three or more quantities.
Answer:
Do it yourself
NCERT In-text Question Page No. 5
Question 1.
Arrange the following numbers in ascending order:
(a) 847, 9754, 8320, 571
(b) 9801,25751,36501,38802
Answer:
(a) 571 < 847 < 8320 < 8320
(b) 9801 <25751 < 25751 <38802
Question 2.
Arrange the following numbers in descending order:
(a) 5000,7500, 85400, 7861
(b) 1971,45321,88715,92547
Make ten such examples of ascending/ descending order and solve them
Answer:
(a) 85400 > 7861 > 7500 > 5000
(b) 92547 > 88715 >45321 > 1971
NCERT In-text Question Pages Nos. 6 & 7
Question 1.
Read and expand the numbers wherever there are blanks.
Answer:
I’m 28
Write five more 5 – digits numbers, read them and expand them.
NCERT In-text Question Pages No. 7
Question 1.
Read and expand the numbers wherever there are blanks.
Answer:
| Number | Number Name | Expansion |
| 3,00.000 | Three lakh | 3 × 1,00,000 |
| 3,50,000 | Three lakh fifty thousand | 3 × 1,00,000 + 5 × 10,000 |
| 3,53,500 | three lakh fifty three thousand five hundred 3 | 3 × 1,00,000 + 5 × 10,000 × 1,00,000 + 5 × 10,000 +7 × 1 + 9 × 100 + 2 × 10 + 8 × 1 |
| 4,57,928 | Four lakh fifty seven thousand nine hundred twenty eight | 4 × 1,00,000 + 5 × 10,000 + 7 × 1 +9 × 100 + 2 × 10 + 8 × 1 |
| 4,07,928 | Four lakh seven thousand nine hundred twenty eight | 4 × 1,00,000 + 7 × 1,000 + 9 × 100 + 2 × 10 + 8 × 1 |
| 4,00,829 | Four lakh eight hundred twenty nine | 4 × 1,00,000 + 8 × 100 + 2 × 10 + 9 × 1 |
| 4,00,029 | Four lakh twenty nine | 4 × 1,00,000 + 2 × 10 + 9 × 1 |
NCERT In-text Question Page No. 8
Question 1.
1. What is 10 – 1 = ?
2. What is 100 – 1 = ?
3. What is 10,000 – 1 = ?
4. What is 1,00,000 – 1 = ?
5. What is 1,00,00,000 – 1 = ?
(Hint: Use the said pattern.)
Answer:
Using the pattern, we have:
10 – 1 = 9 100 – 1 = 99
1.0 – 1 = 999
10.0 – 1 = 9,999
1.0. 000 – 1 = 99,999
10.0. 000 – 1 = 9,99,999
1.0. 00.000 – 1 = 99,99,999
Using this pattern, we can say:
1. 10-1
2. 100 – 1 = 99
3. 10,000 – 1 = 9,99
4. 1,00,000 – 1 = 99,999
5. 1,00,00,000 – 1 = 99,99,999.
Question 2.
Give five examples where the number of things counted would be more than 6-digit number.
Answer:
- The number of stars seen in a clear dark night.
- Number of cars in Delhi.
- Number of children in a big city.
- Number of grains in a sack full of grains.
- Number of pages of note books of all students in a town.
Question 3.
Starting from the greatest 6-digit number, write the previous five numbers in descending order.
Answer:
The greatest 6-digit number is 999999
∴ 1st previous number = 999999 – 1 = 999998
2nd previous number 999999 – 2 = 9999997
3rd previous number = 999999 – 3 = 999996
4th previous number = 999999 – 4 = 999995
5th previous number = 999999 – 5 = 9999994
Writing these numbers in descending order, we have: 999998, 999997, 999996, 999995, 999994.
Question 4.
Starting from the smallest 8-digit number, write the next five numbers in ascending order and read them.
Answer:
The smallest 8-digit number = 1000000
1st next number is: 10000000 + 1 = 10000001
2nd next number is: 10000000 + 2 = 10000002
3rd next number is: 10000000 + 3 – 10000003
4th next number is: 10000000 + 4 = 10000004
5th next number is: 1000000 + 5 = 10000005
Reading these numbers, we have:
10000001: one crore one
10000002: one crore two
10000003: one crore three
10000004: one crore four
10000005: one crore five
NCERT In-text Question Page No. 11
Question 5.
Read these numbers. Write them using placement boxes and then write their expanded forms.
(i) 475320
(ii) 9847215
(iii) 97645310
(iv) 30458094
(a) Which is the smallest number?
(b) Which is the greatest number?
(c) Arrange these numbers in ascending and descending orders.
Answer:
(i) 475320
We read it as: four lakh seventy five thousand three hundred twenty.
Expanded form: 475320 = 4 × 100000 + 7 × 10000 + 5 × 1000 + 3 × 1oo + 2 × 10 + 0
(ii) 9847215
We read it as: ninety eight lakh forty seven thousand two hundred fifteen.
Expanded form: 9847215 = 9 × 1000000 + 8 × 100000 + 4 × 10000 + 7 × 1000 + 2 × 100 + 1 × 10 + 5 × 1
(iii) 97645310
We read it as: nine crore seventy six lakh forty five thousand three hundred ten.
Exxpanded form: 97645310 = 9 × 10000000 + 7 × 1000000 + 6 × 100000 + 4 × 10000 + 5 × 1000 + 3 × 100 + 1 × 10 + 0
(iv) 30458094
We read it as: three crore four lakh fifth eight thousand ninety four.
Expanded form: 30458094 = 3 × 10000000 + 4 × 100000 + 5 × 10000 + 8 × 1000 + 9 × 10 + 4 × 1
(a) The smallest number is 475320.
(b) The greatest number is 97645310.
(c) Ascending order: 475320, 9847215, 30458094, 97645310
Descending order: 97645310, 30458094, 9847215, 475320.
Question 6.
Read these numbers.
(i) 527864 (ii) 95432 (iii) 18950049 (iv) 70002509
(a) Write these numbers using placement boxes and then using commas in Indian as well as International System of Numeration.
(b) Arrange these in ascending and descending order.
Answer:
Reading these numbers, we have:
(i) 5, 27, 864: Five lakh twenty seven thousand eight hundred sixty four
(ii) 95, 432: Ninety five thousand four hundred thirty two
(iii) 1,89,50,049: One crore eighty nine lakh fifty thousand forty nine
(iv) 7,00,02,509: Seven crore two thousand five hundred nine
(a) Using the placement boxes, we have:
International System of Numeration
Using commas, we can re-write these numbers as:
(i) 527864 = 5,27,864 ← (Indian System) = 527,864 ← (International System)
(ii) 95432 = 95,432 ← (Indian System) = 95,432 ← (International System)
(iii) 18950049 = 1,89,50,049 ← (Indian System) = 18,950, 049 ← (International System)
(iv) 70002509 = 7,00,02,509 ← (Indian System) = 70,002,509 ← (International System)
(b) Ascending order: 95, 432; 5,27,864; 1,89,50,049; 7,00,02,509
Descending order: 7,00,02,509; 1,89,50,049; 5,27,864; 95,432
Question 7.
Take three more groups of large numbers and do the exercise given above.
Answer:
Do it yourself.
Question 8.
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:
Five numbers each with 6 digits using the given digits are:
(i) 876540
(ii) 876450
(iii) 867540
(iv) 867405
(v) 876045
(a) Rewriting the above numbers using commas,
(i) 8,76,540
(ii) 8,76,450
(iii) 8,67,540
(iv) 8,67,405
(v) 8,76,045
(b) Ascending order: 8, 67, 405; 8, 67, 540; 8,76, 045; 8,76,450; 8,76,540
Descending order: 8,76,5402; 8,76, 450; 8,76,045; 8,67,540; 8,67,405
Question 9.
Take the digits 4, 5, 6, 7, 8 and 9. Make any three numbers each with 8 digits. Put commas for easy reading.
Answer:
Three numbers each with 8 digits using given digits are:
(i) 9,88,77,456
(ii) 9,88,77,465
(iii) 9,88,77,654
Question 10.
From the digits 3, 0 and 4, make five numbers each with 6 digits. Use commas.
Answer:
Five number each with 6 digits using the digits 3, 0 and 4 are:
(i) 4,44,330
(ii) 3,43,340
(iii) 4,40,340
(iv) 3,00,343
(iv) 3,03,403
NCERT In-text Question Page No. 12
Question 1.
How many centimetres make a kilometre?
Answer:
∵ 1 km = 1000 m and 1 m = 100 cm
∴ 1 km = 1000 × 100 cm =1,00,000 cm
= one lakh cm
There are 1 lakh centimetres in a kilometre.
Question 2.
Name five large cities in India. Find their population. Also, find the distance in kilometres between each pair of these cities.
Answer:
Do it yourself
NCERT In-text Question Page No. 13
Question 1.
How many milligrams make one kilogram?
Answer:
∵ 1 kg = 1000 g and 1 g = 1000 mg
∴ 1 kg = 1000 × 1000 mg = 10,0,000 mg
= 10 lakh mg
Question 2.
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 tablets = 2,00,000
Weight of one tablet = 20 mg
∴ Total weight of all tablets
= 20 x 200000 mg
= 4000000 mg
Since 1 kg = 1000 g and 1 g = 1000 mg
∴ 4000000 mg = \(\frac{4000000}{1000}\)g = 4000 g
Thus, the total weight of all the tablets in grams = 4000 g.
Also, 4000000 mg = \(\frac{4000000}{1000 \times 1000}\) = 4 kg
Thus, the total weight of all the tablets in kilograms = 4 kg
NCERT In-text Question Pages Nos. 13 & 14
Question 1.
A bus started its journey and reached different places with a speed of 60 km/ hour. The journey is shown below.
(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?
(v) Find out the time taken by the bus to reach
(i) A to B
(ii) A to B
(iii) A to B
(iv) Total journey
Answer:
(i) Distance covered by the bust for going from A to D
= 4170 km + 3410 km + 2160 km
= [4170 + 3410+ 2160] km
= 9740 km
(ii) Total distance covered by the bus for going from D to G
= 8140 km + 4830 km + 2550 km
= [8140 + 4830 + 2550] km
= 15520 km
(iii) Total distance covered by the bus for going round from A to A
= [Distance between A and D] + [Distance between D and G] + [Distance between G and A]
= [9740 km] + [15520 km] + [1290 km]
= [9740 + 15520 + 1290] km
= 26550 km
(iv) [Distance between D and E]
– [Distance between C and D]
= [8140 km] – [2160 km]
= [8140 – 2160] km = 5980 km
(v) Since time = \(\frac{\text { dis tan ce }}{\text { speed }}\) and the speed of the bus is 60 km/hour.
(i) Time taken by the bus to each from A to B
= \(\frac{4170 \mathrm{~km}}{60 \mathrm{~km} / \text { hour }}=\frac{4170}{60} \text { hour }=69 \frac{1}{2}\)hours
60km /hour 60 . 2
(ii) Time taken by the bus to reach from C to D = \(\frac{2160}{60}\) hours = 36 hours
(iii) Time taken by the bus to reach from E to G
(iv) Total time taken by the bus for the total journey
Question 2.
Raman’s Shop
| Things | Price |
| Apples | ₹ 40 per kg |
| Oranges | ₹ 30 per kg |
| Combs | ₹ 3 per kg |
| Tooth brushes | ₹ 10 per kg |
| Pencils | ₹ 1 per kg |
| Note books | ₹ 6 per kg |
| Soap cakes | ₹ 8 per kg |
The sales during the last year
| Apples | 2457 kg |
| Oranges | 3004 kg |
| Combs | 22760 |
| Tooth brushes | 25367 |
| Pencils | 38530 |
| Note books | 40002 |
| soap cakes | 20005 |
(a) Can you find the total weight of apples and oranges Raman sold last year?
Weight of apples = ________ kg
Weight of oranges = _______kg
Therefore, total weight = kg + _____ kg= kg
Answer – The total weight of oranges and apples = _______ kg.
(a) Can you find the total money Raman got by selling apples?
(b) Can you find total money Raman got by selling apples and oranges together?
(c) Can you find total money Raman got by selling apples and oranges together?
(d) Make a table showing how much money Raman received from selling each item. Arrange the entries of amount of money received in descending order. Find the item which brought him the highest amout. How much is this amount?
Answer:
(a) Quantity of oranges and apples sold by Raman last year:
Weight of apples = 2457 kg
Weight of oranges = 3004 kg
Therefore, total weight = 2457 kg + 3004 kg
= 5461 kg
The total weight of oranges and apples = 5461 kg.
(b) Selling price of 1 kg of apples = ₹ 40
Weight of apples sold = 2457 kg
∴ Total money Raman got by selling apples
= ₹ 2457 × 40 = ₹ 98,280
(c) Selling price of 1 kg of oranges = ₹ 30
Weight of oranges sold = 3004 kg
Total money Raman got by selling oranges
= ₹ 3004 × 30 = ₹ 90,120
∴ Total money Raman got by selling apples and oranges together = ₹ 98,280 + ₹ 90, 120 = ₹ 1,88,400
Arranging the entries of amount of money received in descending order, we have:
₹ 253670; ₹ 240012; ₹ 160040;
₹ 98120; ₹ 90120; ₹ 68280; 38530
Obviously, the highest amount of money is received against the item “tooth brushes”. This highest amount of money is ₹ 2,53,670.
NCERT In-text Question Page No. 19
Question 1.
Round these numbers to the nearest tens.
Answer:
Number 5 is rounded off to 10.
NCERT In-text Question Page No. 20
Question 2.
Round off the given numbers to the nearest tens, hundreds and thousands.
Answer:
| Given Number | Approximate to Nearest | Rounded Form |
| 75847 | Tens | 75850 |
| 75847 | Hundreds | 75800 |
| 75847 | Thousands | 76000 |
| 75847 | Ten thousands | 80000 |
NCERT In-text Question Page No. 22
Question 1.
Estimate the following products :
(a) 87 × 313
(b) 9 × 795
(c) 898 × 785
(d) 958 × 387
Make five more such problems and solve them
Answer:
(a) 87 × 313
87 → 90 [Rounding to tens]
313 → 300 [Rounding to hundreds]
.’. Estimated product = 90 × 300 = 27000
(b) 9 × 795
9 → 10 [Rounding to tens]
795 → 800 [Rounding to hundreds]
.’. Estimated product = 10 × 800 = 8000
(c) 898 × 900
898 → 900 [Rounding to tens]
785 → 800 [Rounding to hundreds]
.’. Estimated product = 900 × 800 = 720000
(d) 958 × 387
958 → 1000 [Rounding to tens]
387 → 400 [Rounding to hundreds]
∴ Estimated product = 1000 × 400 = 400000
NCERT In-text Question Page No. 23
Question 2.
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 × (9+2)
(b) Divide the difference of eighteen and six by four (18-6) ÷ 4
(c) 45 ÷ [3(3+2)]
Question 3.
Write three different situations for (5 + 8) × 6. (One such situation is : Sohani and Reeta work for 6 days; Sohani works 5 hours a day and Reeta 8 hours a day. How many hours do both of them work in a week?)
Answer:
Rahul pays ? 5 for a morning tea and ? 8 for the evening coffee daily. He takes tea and coffee for six days in a week. What is his weekly expanses for tea and coffee? Prema and Shanti work for six days. Prema earns ? 5 per day and Shanti earns ? 8 per day. What do they earn together in six days?
Prateek read 5 pages of a novel in the morning and 8 pages in the evening. How many pages did Prateek read in 6 days?
Question 4.
Write five situations for the following where brackets would be necessary.
(a) 7(8-3) (b) (7+ 2) (10-3).
Answer:
(a) 7 (8 – 3) = 7 × (8) + 7 × (-3)
= 56 – 21
= 35
(b) (7 + 2) (10-3) = 7 (10-3)+ 2 (10-3)
Using distributive property
= 7 × (7) + 2 × (7)
= 49+14
= 63
NCERT In-text Question Page No. 25
Question 1.
Write in Roman numerals.
1. 73
2. 92
Answer:
1. 73 = LXXIII
2. 92 = XCII