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NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 6 Maths Chapter 1 are designed as per the CBSE Class 6 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 6 Chapter 1. NCERT Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers Ex 1.1

Exercise 1.1

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

Solution:
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

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.

Solution:
The solution is as follows,
a 73,75,307
b 9,05,00,041
c 7,52, 21,302
d 58, 423, 202
e 23,30,010

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

Solution:
The Indian system of numeration is crores, lakhs, thousands, hundreds, tens and ones. The solution is as follows:
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

Question 4:Insert commas suitable and write the names according to Internationalsystem of numeration:(a) 78921092(b) 7452283(c) 99985102(d) 48049831

Solution:
In the International System of Numeration, we have ones, tens, hundreds, thousands and then
millions.
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 Ex 1.2

Exercise 1.2

Question 1:A book exhibition was held for four days in a school. The number of ticketssold at the counter on the first, second, third and final day was respectively1094, 1812, 2050 and 2751. Find the total number of tickets sold on all thefour days.

Solution:

Calculate the total number of tickets sold on all the four days: Hence, the total number of tickets sold are 7707 .

Question 2:Shekhar is a famous cricket player. He has so far scored 6980 runs in testmatches. He wishes to complete 10,000 runs. How many more runs does heneed?

Solution:
Until now Shekhar has scored 6980 runs and if he wishes to score 10000 runs then 3020 runs are required to complete 10000 runs.

Question 3:In an election, the successful candidate registered 5,77,500 votes and hisnearest rival secured 3,48,700 votes. By what margin did the successfulcandidate win the election?

Solution:
The winner registered 5,77,500 votes and the other candidate got 3,48,700 votes, so to calculate the margin, 2,28,800 is the margin by which the winner won the election.

Question 4:Kirti Bookstore sold books worth ₹ 2, 85,891 in the first week of June andbooks worth ₹ 4, 00,768 in the second week of the month. How much wasthe sale for the two weeks together? In which week was the sale greater andby how much?

Solution:
The total sale can be calculated by adding the sale of both the weeks,
Sale for the first week 2,85,891
Sale for the second week 4,00,768 It can be seen from the question that the sales in the second week were greater than that in the
first week, so, The sales in the second week were more and by
Rs. 1,14,877 .

Question 5:Find the difference between the greatest and the least number that can bewritten using the digits 6, 2, 7, 4, 3 each only once.

Solution:
The largest 5 digit number using the given digits is
76432 .
The smallest 5 digit number using the given digits is
23467 .
The difference is calculated as, The difference between the largest and the smallest number formed with the digits 6, 2, 7, 4, 3 is 52,965 .

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

Solution:
The number of days in the month of January is 31 and the machine produces 2,825 screws in a day on an average, so the total number of screws produced is, 87,575 screws were produced in January 2006 .

Question 7:A merchant had ₹ 78,592 with her. She placed an order for purchasing 40radio sets at ₹ 1,200 each. How much money will remain with her after thepurchase?

Solution:
A radio costs Rs. 1200 and she placed an order for 40 sets, so, The merchant has spent Rs. 48,000 in buying the radios, the money left with her is, Hence, the money left with her is Rs. 30,592 .

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

Solution:
The answer after multiplying 7236 by 65
is, The actual answer 7236 multiplying by 56
is, Thus, the difference is, 65124 .

Question 9:To stitch a shirt 2 m 15 cm cloth is needed. Out of 40 m cloth, how manyshirts can be stitched and how much cloth will remain?

Solution:
First we will convert the units into a common unit, we are converting meter into centimeters.
1 m =100 cm
40 m =4000 cm

Cloth required for 1 shirt = 2 m 15 cm.
Cloth required for 1 shirt = 215 cm.
Now, number of shirts that can be stitched is 4000 ÷ 215 So, 18 shirts can be stitched from the given cloth and 130 cm of cloth will be left out.

Question 10:Medicine is packed in boxes, each weighing 4 kg 500 g. How many suchboxes can be loaded in a van which cannot carry beyond 800 kg?

Solution:
We will convert the given data in a particular unit,
1 kg =1000 grams
4 kg =4000 grams
4 kg 500 grams= 4500 grams
800 kg =800000 grams
Now, we know a box weighs 4500 grams so number of boxes that can come in a van that can hold only 800000 grams, 800000 ÷ 4500, So, 177 boxes can be loaded in the van.

Question 11:The distance between the school and the house of 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.

Solution:
We will convert the given data in a particular unit,
1 km =1000 m
1 km 875 m =1875 m

Now, the girl daily goes to school and comes back from the school so the total distance covered by the girl in a day, The distance covered in a day by the girl is 3750 m. Now calculating the distance travelled by the girl in 6 days, The distance travelled by the girl in 6 days is 22500 m.

Question 12:A vessel has 4 liters and 500 ml of curd. In how many glasses each of 25 mlcapacity, can it be filled?

Solution:
We will convert the given data in a particular unit,
1 liter= 1000 ml
4 liters= 4000 ml
4 liters 500 ml=4500 ml

The capacity of glasses is 25 ml, so the number of glasses is, 180 glasses of 25 ml can be filled.

NCERT Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers Ex 1.3

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

Solution:

(a) 730 is round off to 700 and 998 is round off to 1000 and the estimated sum is 1700
(b) 796 is round off to 800 and 314 is round off to 300 and the estimated difference is 500
(c) 12,904 is round off to 13000 and 2,888 is round off to 3000 and the estimated sum is 16000
(d) 28,292 is round off to 28,000 and 21,496 is round off to 21,000 and the estimated difference is 7,000.

Question 2:Give a rough estimate (by rounding off to nearest hundreds) and also acloser estimate (by rounding off to nearest tens):(a) 439 + 334 + 4317(b) 1, 08,734 – 47,599(c) 8325 – 491(d) 4, 89,348 – 48,365

Solution:
(a) 439 is round off to 400; 334 is round off to 300 and 4317 is round off to 4300, so the estimated sum is 5000 (rounding off to nearest hundreds)
439 is round off to 440; 334 is round off to 330 and 4317 is round off to 4320, so the estimated sum is 5090 (rounding off to nearest tens).
(b) 108734 is round off to 108700 (for hundreds) and 108730 (for tens) and 47599 is round off to 47600 (for hundreds and tens), so the estimated difference is 61100 (for hundreds)  and 61130 (for tens).
(c) 8325 is round off to 8300 (for hundreds) and 8330 (for tens) and 491 is round off to 500 (for hundreds) and 490 (for tens), so the estimated difference is 7800 and 7840.
(d) 489348 is round off to 489300 (for hundreds) and 489350 (for tens) and 48365 is round off to 48400 (for hundreds) and 48370 (for tens), so the estimated difference is 440900 (for hundreds) and 440980 (for tens).

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

Solution:
(a) 578 is round off to 600 and 161 is round off to 200 and the estimated product is 120000.
(b) 5281 is round off to 5000 and 3491 is round off to 3500 and the estimated product is 17500000.
(c) 1291 is round off to 1300 and 592 is round off to 600 and the estimated product is 780000.
(d) 9250 is round off to 9000 and 29 is round off to 30 and the estimated product is 270000. 