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

Written by Team Trustudies
Updated at 2021-02-11

## NCERT solutions for class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.1

Q.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.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

(a) 1 lakh = 10 ten thousand
(b) 1 million = 10 hundred thousand
(c) 1 crore = 10 ten lakh
(d) 1 crore = 10 million
(e) 1 million = 10 lakh

Q.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.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

(a) 73,75,307
(b) 9,05,00,041
(c) 7,52,21,302
(d) 5,84,23,202
(e) 23,30,010

Q 3. Insert commas suitably and write the names according to Indian System of Numeration:
(a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

(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) 78921092 (b) 7452283 (c) 99985102 (d) 48049831

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

(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

## NCERT solutions for class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.2

Q.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.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Number of tickets sold on 1st day = 1094
Number of tickets sold on 2nd day = 1812
Number of tickets sold on 3rd day = 2050
Number of tickets sold on 4th day = 2751
Hence, number of tickets sold on all the four days
= 1094 + 1812 + 2050 + 2751 = 7707 tickets

Q.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?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Shekhar scored = 6980 runs
He want to complete = 10000 runs
Runs need to score more = 10000 – 6980 = 3020

Q.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?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

No. of votes secured by the successful candidate = 5,77,500
No. of votes secured by his nearest rival = 3,48,700
Margin = 5,77,500 – 3,48,700 = 2,28,800 votes

Q.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?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Price of books sold in June first week = Rs 2,85,891
Price of books sold in June second week = Rs 4,00,768
No. of books sold in both weeks together
= Rs 2,85,891 + Rs 4,00,768 = Rs 6,86,659
The sale of books is the highest in the second week
Difference in the sale in both weeks
= Rs 4,00,768 – Rs 2,85,891 = Rs 1,14,877
$?$ Sale in second week was greater by Rs 1,14,877 than in the first week.

Q.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.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Greatest 5-digit number = 76432
Least 5-digit number = 23467
Difference between the two numbers
= 76432 – 23467 = 52965

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

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Number of screws manufactured in a day = 2825
Since January month has 31 days
Hence, number of screws manufactured in January = 31 × 2825 = 87575

Q.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?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Total money the merchant had = Rs 78592
Cost of each radio set = Rs 1200
So, cost of 40 radio sets = Rs 1200 × 40 = Rs 48000
Money left with the merchant = Rs 78592 – Rs 48000 = Rs 30592

Q.8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Difference between 65 and 56 i.e (65 – 56) = 9
The difference between the correct and incorrect answer = 7236 × 9 = 65124

Q.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?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Given
Total length of the cloth = 40 m
= 40 × 100 cm = 4000 I’m
Cloth required to stitch one shirt = 2 m 15 cm
= 2 × 100 + 15 cm = 215 cm
Number of shirts that can be stitched out of 4000 cm = $\frac{4000}{215}$ = 18 shirts

Q.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?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Weight of one box = 4 kg 500 g
= 4 × 1000 + 500 = 4500 g
Maximum weight carried by the van
= 800 kg = 800 × 1000 = 800000 g.
Hence, number of boxes that can be loaded in the van =$\frac{800000}{4500}$= 177 boxes

Q.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.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Distance covered between school and house
= 1 km 875 m = 1000 + 875 = 1875 m
Hence, distance travelled by the student in one day = 2 × 1875 = 3750 m
Distance travelled by the student in 6 days
= 3750 m × 6 = 22500 m = 22 km 500 m

Q.12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Quantity of curd in the vessel
= 4 l 500 ml = 4 × 1000 + 500 = 4500 ml
Capacity of 1 glass = 25 ml
$?$Number of glasses that can be filled with curd = ${\text{\ftac}}450025$ = 180 glasses

## NCERT solutions for class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.3

Q.1 Estimate each of the following using general rule:
(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496
Make ten more such examples of addition, subtraction and estimation of their outcome.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

(a) 730 + 998
Round off to hundreds
730 rounds off to 700
998 rounds off to 1000
Hence, 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314
Round off to hundreds
796 rounds off to 800
314 rounds off to 300
Hence, 796 – 314 = 800 – 300 = 500
(c) 12904 + 2888
Round off to thousands
12904 rounds off to 13000
2888 rounds off to 3000
Hence, 12904 + 2888 = 13000 + 3000 = 16000
(d) 28292 – 21496
Round off to thousands
28292 round off to 28000
21496 round off to 21000
Hence, 28292 – 21496 = 28000 – 21000 = 7000
Ten more such examples are
(i) 540 + 360 = 500 + 400 = 900
(ii) 7657 + 4450 = 8000 + 4000 = 12000
(iii) 9892 – 2372 = 10000 – 2000 = 8000
(iv) 9740 – 2002 = 10000 – 2000 = 8000
(v) 2678 + 1111 = 3000 + 1000 = 4000
(vi) 1110 – 1222 = 1000 – 1000 = 0
(vii) 980 + 525 = 1000 + 500 = 1500
(viii) 7400 + 4300 = 7000 + 4000 = 11000
(ix) 6831 + 2222= 7000 + 2000 = 9000
(x) 9285 + 1119 = 9000 + 1000 = 10000

Q.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 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365
Make four more such examples.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

(a) 439 + 334 + 4317
Rounding off to nearest hundreds
439 + 334 + 4317
= 400 + 300 + 4300 = 5000
Rounding off to nearest tens
439 + 334 + 4317
= 440 + 330 + 4320 = 5090
(b) 108734 – 47599
Rounding off to nearest hundreds
108734 – 47599 = 108700 – 47600
= 61100
Rounding off to nearest tens
108734 – 47599 = 108730 – 47600
= 61130
(c) 8325 – 491
Rounding off to nearest hundreds
8325 – 491 = 8300 – 500
= 7800
Rounding off to nearest tens
8325 – 491 = 8330 – 490
= 7840
(d) 489348 – 48365
Rounding off to nearest hundreds
489348 – 48365 = 489300 – 48400
= 440900
Rounding off to nearest tens
489348 – 48365 = 489350 – 48370
= 440980
Four more examples are as follows
(i) 2343 + 1692
Rounding off to nearest hundreds
2343 + 1692
= 2300 + 1700
= 4000
Rounding off to nearest tens
2343 + 1692 = 2340 + 1690= 4030
(ii) 9086 – 1390
Rounding off to nearest hundreds
10000 – 1400 = 8600
= 8600
Rounding off to nearest tens
9086– 1390 = 9090 – 1390
= 7700
(iii) 1234 – 346
Rounding off to nearest hundreds
1234 – 346= 1200 – 300= 900
Rounding off to nearest tens
1234 – 346= 1230 – 350= 880
(iv) 2251+7654
Rounding off to nearest hundreds
2251+7654 = 2300+7700
= 10000
Rounding off to nearest tens
2251+7654 = 2250+7650
= 9900

Q.3 Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

(a) 578 × 161
Rounding off by general rule
578 and 161 rounded off to 600 and 200 respectively
600×200= 120000
(b) 5281 × 3491
Rounding off by general rule
5281 and 3491 rounded off to 5000 and 3500 respectively
5000 × 3500 = 17500000
(c) 1291 × 592
Rounding off by general rule
1291 and 592 rounded off to 1300 and 600 respectively
1300 × 600 = 780000
(d) 9250 × 29
Rounding off by general rule
9250 and 29 rounded off to 9000 and 30 respectively
9000 × 30 = 270000 Four more examples are as follows
(a) 178 × 351
Rounding off by general rule
178 and 351 rounded off to 200 and 400 respectively
200×400= 80000
(b) 5571 × 871
Rounding off by general rule
5571 and 871 rounded off to 6000 and 900 respectively
6000 × 900 = 5400000
(c) 181 × 92
Rounding off by general rule
181 and 92 rounded off to 200 and 90 respectively
200 × 90 =18000
(d) 4150 × 69
Rounding off by general rule
4150 and 69 rounded off to 4000 and 70 respectively
4000 × 70 = 280000

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