NCERT Solutions Class 6 Mathematics Chapter 1 Knowing our Numbers

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 lakhs.
(d) 1 crore = _______ million.
(e) 1 million = _______ lakh.
Sol.1) a) 1 lakh = 1,00,000
= 100 thousand
= 10 ten thousand

b) 1 million = 1,000,000
= 1000 thousand
= 10 hundred thousand

c) 1 crore = 1,00,00,000
= 100 lakh
= 10 ten lakh

d) 1 crore = 1,00,00,000
Adding commas to the number 10000000 according to the International system, we have
10,000,000 = 10 million
1 crore = 10 million

e) 1 million = 10,000,000
Adding commas to the number 10000000 according to the Indian system, we have
1,00,00,000 = 1 crore = 100 lakh
1 million = 100 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.
Sol.2) 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.
58,423,202

e) Twenty three lakh thirty thousand ten.
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
Sol.3) a) 87595762
Inserting commas according to the Indian system: 8,75,95,762
Number name: Eight crore seventy five lakh ninety five thousand seven hundred sixty two.

b) 8546283
Inserting commas according to the Indian system: 85,46,283
Number name: Eighty five lakh forty six thousand two hundred eighty three.

c) 99900046
Inserting commas according to the Indian system: 9,99,00,046
Number name: Nine crore ninety nine lakh forty six.

d) 98432701
Inserting commas according to the Indian system: 9,84,32,701
Number name: Nine crore eighty four lakh thirty thousand seven hundred one

Q.4) Insert commas suitably and write the names according to International System of Numeration:
(a) 78921092 (b) 7452283
(c) 99985102 (c) 48049831
Sol.4) a) 78921092
Inserting commas according to the Indian system: 78,921,092
Number name: Seventy eight million nine twenty one thousand ninety two

b) 7452283
Inserting commas according to the Indian system: 7,452,283
Number name: Seven million four hundred fifty two thousand two hundred
eighty three

c) 99985102
Inserting commas according to the Indian system: 99,985,102
Number name: Ninety nine million nine hundred eighty five thousand one hundred two.

d) 48049831
Inserting commas according to the Indian system: 48,049,831
Number name: Forty eight million forty nine thousand eight hundred thirty one

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.
Sol.1) Number of tickets sold at counter on 1st day = 1,094
Number of tickets sold at counter on 2nd day = 1,812
Number of tickets sold at counter on 3rd day = 2,050
Number of tickets sold at counter on 4th day = 2,751
Total number of tickets sold at counter = sum of all 4 days’ tickets sold
= 1,094 + 1,812 + 2,050 + 2,751 = 7,707
∴ total number of tickets sold at counter 7,707 tickets on all the 4 days

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?
Sol.2) Runs required by Shekhar = 10,000
Runs scored by him = 6,980
Runs needed to be scored by Shekhar = 10,000 − 6980 = 3020
∴ he needs 3,020 more runs.

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?
Sol.3) Number of votes secured by successful candidate = 5,77,500
votes secured by his opponent = 3,48,700
Margin between their votes = 577500 − 348700 = 2,28,800
∴ the successful candidate won by of 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?
Sol.4) Value of books sold in 1st week = 𝑅𝑠 285891
Value of books sold in 2nd week = 𝑅𝑠 400768
Total value of books sold = Books sold in 1st week + Books sold in 2nd week
= 𝑅𝑠 285891 + 𝑅𝑠 400768 = 𝑅𝑠 686659
Since, 𝑅𝑠 400768 > 𝑅𝑠 285891
∴ sale of 2nd week is greater than that of 1st week.
Books sold in 2nd week = 400768
Books sold in 1st week= 285891
More books sold in 2nd week = 400768 − 285891 = 114877
Therefore, 1,14,877 more books were sold in 2nd week.

Q.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.
Sol.5) The Largest five-digit number formed using digits 6,2,7,4,3
= 76432
The Smallest five-digit number formed using digits 6,2,7,4,3
= 23467
Difference between them = 76432 − 23467 = 52965
Therefore, the difference is 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?
Sol.6) Number of screws manufactured in one day= 2,825
Number of days in the month of January (31 days) = 2825 × 31
= 87,575
∴ the factory machine produced 87,575 screws in the month of January 2006.

Q.7) A merchant had `78,592 with her. She placed an order for purchasing 40 radio sets at`1200 each. How much money will remain with her after the purchase?
Sol.7) The cost of one radio set = 𝑅𝑠. 1200
Total cost of 40 radio sets = 1200 × 40 = 𝑅𝑠. 48,000
Now, Total money with merchant = 𝑅𝑠. 78,592
The sum of Money spent by her = 𝑅𝑠. 48,000
Money left remaining with her= 𝑅𝑠. 78,592 − 𝑅𝑠. 48,000 = 𝑅𝑠. 30,592
Therefore, 𝑅𝑠. 30,592 will remain with her after the product is been purchased.

Q.8) A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Sol.8) Wrongly multiplied answer = 7236 × 65 = 470340
Exact answer = 7236 × 56 = 405216
Difference given by the answers
= 470340 – 405216
= 65,124
Therefore, student’s answer was greater than the correct answer by 65,124.

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?
Sol.9) Cloth required to stich one shirt = 2 𝑚 15 𝑐𝑚
Now, we have to convert higher units into lower units i.e. from meters to cm(1𝑚 = 100𝑐𝑚).
= 2 × 100𝑐𝑚 + 15𝑐𝑚
= 215 𝑐𝑚 per shirt
Length of cloth = 40 𝑚
= 40 × 100 𝑐𝑚
= 4000𝑐𝑚
Number of shirts can be stitched out of 4000𝑐𝑚 = 4000 ÷ 215 = 18
Therefore, 18 shirts can be stitched 130𝑐𝑚 reaming cloth (Quotient = 18, Remainder =130).
Other method to find how much cloth is left
215𝑐𝑚 × 18 = 3870𝑐𝑚
4000𝑐𝑚 − 3870𝑐𝑚 = 130𝑐𝑚 cloth left

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?
Sol.10) The weight of one box = 4 𝑘𝑔 500 𝑔
= 4 × 1000𝑔 + 500𝑔 (1 𝑘𝑔 = 1000 𝑔)
= 4500𝑔
Maximum load can be loaded in van = 800 𝑘𝑔
= 800 × 1000𝑔 = 800000𝑔
Number of boxes = 800000 ÷ 4500 = 177.7 approx.
∴ The total number of medicine boxes which the van can carry 177 boxes.

Q.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.
Sol.11) Distance between school and house of a student = 1 𝑘𝑚 875 𝑚
= 1 × 1000 + 875𝑚 = 1875𝑚
Distance Everyday she walks both ways= 1875 × 2 𝑚 = 3750𝑚
Distance covered by her in 6 days = 3750 × 6 = 22500
Therefore, 22 𝑘𝑚 500 𝑚 distance covered by her in six days.

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?
Sol.12) Capacity of curd in a vessel = 4 𝑙
500 𝑚𝑙 = 4 × 1000 𝑚𝑙 + 500 𝑚𝑙 = 4500 𝑚𝑙 (1 liter = 1000 ml)
Capacity of one glass = 25 𝑚𝑙
Number of glasses can be filled = 4500/25 = 180
Therefore, 180 bottles can be filled by curd.

Exercise 1.3

Q.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 outcomes.
Sol.1) (a) 730 + 998
By rounding off to hundreds,
730 rounded off to 700 and
998 rounded off to 1000
700 + 1000 = 1700

(b) 796 – 314
By rounding off to hundreds,
796 rounded off to 800 and
314 rounded off to 300.
800 − 300 = 500

(c) 12,904 + 2822
By rounding off to thousands,
12,904 rounded off to 13,000 and
2,822 rounded off to 3,000.
13,000 + 3,000 = 16,000

(d) 28,296 – 21,496
By rounding off to nearest thousand,
28,296 rounded off to 28,000 and
21,496 rounded off to 21,000.
28,000 – 21,000 = 7,000

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 + 4,317 (b) 1,08,734 – 47,599
(c) 8325 – 491 (d) 4,89,348 – 48,365
Make four more such examples.
Sol.2) (a) 439 + 334 + 4,317
Rounding off to nearest hundreds, 439,334, and 4317 may be rounded off to 400, 300, and 4300 respectively.
400 + 300 + 4,300 = 5,000
Rounding off to nearest tens, 439, 334, and 4,317 will be rounded off to 440, 330, and 4,320 respectively.
440 + 330 + 4,320 = 5,090

(b) 1,08,734 - 47,599
Rounding off to hundreds
1,08,734 may be rounded off to 1,08,700 and
47,599 may be rounded off to 47,600.
1,087,00 − 47,600 = 61,100
Rounding off to tens
1,08,734 may be rounded off to 1,08,730 and
47,599 may be rounded off to 47,600.
1,08,734 − 47600 = 61,130

(c) 8,325 – 491
Rounding off to hundreds
8,325 may be rounded off to 8300 and
and 491 may be rounded off to 500.
8,300 − 500 = 7,800
Rounding off to tens
8,325 may be rounded off to 8,330 and
491 may be rounded off to 490.
8,325 − 490 = 7,840

(d) 4,89,348 – 48,365
Rounding off to hundreds
4,89,348 and 4,83,65 can be rounded off to 4,89,300 and 4,84,00 respectively.
4,89,300 − 48,400 = 4,40,900
Rounding off to tens
489348 and 48365 can be rounded off to 489350 and 48370 respectively.
4,89,350 − 48,370 = 4,40,980

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.
Sol.3) (a) 578 × 161
Rounding off 598 and 161 can be rounded off to 600 and 200 respectively.
600 × 200 = 1,20,000

(b) 5281 × 3491
Rounding off 5281 and 3491 can be rounded off to 5000 and 3000 respectively.
5000 × 3000 = 1,50,00000

(c) 1291 × 592
Rounding off 1291 and 592 can be rounded off to 1000 and 600 respectively.
1000 × 600 = 6,00000

(d) 9250 × 29
Rounding off 9,250 and 29 can be rounded off to 9,000 and 30 respectively.
9,000 × 30 = 2,70,000