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NCERT Solutions for class 6 Maths chapter 1 – Knowing Our Numbers

Back Exercise

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 = ten ten thousand.
(b) 1 million = ten hundred thousand.
(c) 1 crore = ten ten lakh
(d) 1 crore = ten million
(e) 1 million = ten 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 crores 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 :
(a) 73,75,307
(b) 9.05.00.041
(c) 7,52.21,302
(d) 58.423.202
(e) 23.30,010.

Question 3.
Insert commas suitably and write the names according to Indian System of Numeration :
(a) 87595762
(b) 8546283
(c)99900046
(d) 98432701.
Solution :
(a) 8, 75, 95, 762. Eight crores seventy-five lakh ninety-five thousand seven hundred and sixty-two.
(b) 85, 46, 283. Eighty-five lakh forty-six thousand two hundred and eighty-three.
(c) i 9, 99, 00, 046. Nine crore ninety-nine lakh and forty-six.
(d) 9, 84, 32, 701. Nine crore eighty-four lakh thirty-two thousand seven hundred and one.

Question 4.
Insert commas suitably and write the names according to International System of Numeration :
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831.
Solution :
(a) 78,921,092. Seventy-eight million nine hundred twenty-one thousand and ninety-two.
(b) 7.452,283. Seven million four hundred fifty-two thousand two hundred and eighty-three.
(c) 99. 985. 102. Ninety-nine million nine hundred eighty-five thousand one hundred and two.
(d) 48. 049. 831. Forty-eight million forty-nine thousand eight hundred and thirty-one.

Exercise 1.2

Question 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 four days.
Solution :
Number of tickets sold on the first day = 1094
Number of tickets sold on the second day =1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the final day = 2751
Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707.

Question 2.
Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10000 runs. How many more runs does he need?
Solution :
Runs scored so far = 6980
Runs wished to be scored = 10000
Runs needed more = 10000 – 6980 = 3020.

Question 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?
Solution :
Votes registered by the successful candidate = 5,77,500
Votes secured by the nearest rival = 3,48.700
Margin by which the successful candidate won the election = 5,77,500 – 3,48,700 = 2,28,800.

Question 4.
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?
Solution :
Sale of books in the first week = ₹ 2,85,891
Sale of books in the second week = ₹ 4,00,768
∴ Sale for the two weeks together = ₹ 2,85,891 + ₹ 4,00,768 = ₹ 6,86,659.
The sale was greater in the second week by ₹ 4,00,768 – ₹ 2,85,891 i.e., by ₹ 1,14,877.

Question 5.
Find the difference between the greatest and least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once.
Sol.
Greatest number that can be written using the digits 6, 2, 7, 4, 3 each only once = 76,432
Least number that can be written using the digits 6, 2, 7, 4, 3 each only once = 23,467
∴ Difference between the greatest and least numbers that can be written using the digits 6,2,7,4, 3 each only once = 76,432 – 23,467 = 52,965.

Question 6.
A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Solution :
The number of screws manufactured by the machine a day on an average = 2,825.
Number of days in the month of January 2006 = 31
The number of screws produced by the machine in the month of January 2006 = 2,825 × 31 =87,575.

Question 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.
Money which the merchant had = ₹ 78,592
Cost of 40 radio sets at ? 1200 each = ₹ 1200 × 40 = ₹ 48,000
Money that will remain with the merchant after the purchase = ₹ 78,592 – ₹ 48,000 = ₹ 30,592.

Question 8.
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?]
Solution :
The wrong answer was greater than the correct answer by
= 7236 × 65 – 7236 × 56
= 7236 × (65 – 56)
= 7236 × 9 = 65.124

Question 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.]
Solution :
2 m 15 cm = 2 m + 15 cm = 2 × 100 cm + 15 cm = 200 cm + 15 cm = 215 cm
40 m = 40 × 100 cm = 4000 cm
4000−−−−√21518
Undefined control sequence \cfrac
Undefined control sequence \cfrac
Hence, 18 shirts can be stitched and 130 cm,
i. e., 1 m 30 cm cloth will remain.

Question 10.
Medicine is packed in boxes, each weighing 4 kg 500 g. Mow many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solution :
4 kg 500 g = 4 kg + 500 g
= 4 × 1000 g + 500 g
= 4000 g + 500 g
= 4500 g 800 kg
= 800 × 1000 g = 800000 g
800000−−−−−−√4500177
Undefined control sequence \cfrac
Undefined control sequence \cfrac
Undefined control sequence \cfrac
Hence, 177 such boxes can be loaded.

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 :
1 km 875 m = 1 km + 875 m
= 1 × 1000 m + 875 m
= 1000 m + 875 m = 1875 m
Distance covered by her in a day in walking both ways between school and home = 1875 × 2 m = 3750 m
∴ Total distance covered by her in six days in walking both ways between school and home = 3750 m × 6 = 22500 m
= 22000 m + 500 m = 220001000 km + 500 m
= 22 km + 500 m = 22 km 500 m.

Question 12.
A vessel has 4 liters and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Solution :
41500 ml = 41 + 500 ml
= 4 × 1000 ml + 500 ml
= 4000 ml + 500 ml = 4500 ml
2018-11-19 12_50_53-NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers 01 - Windows Pho 1
Hence, it can be filled in 180 glasses.

Exercise 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
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solution :
(a) 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314 = 800 – 300 = 500
(c) 12,904 + 2,888 = 13,000 + 3,000 = 16,000
(d) 28,292 – 21,496 = 28,000 – 21,000 = 7,000
Ten more such examples of addition, subtraction and estimation of their outcome are as follows:
Ex. 1.720 + 990
Ex. 2.640 + 880
Ex. 3. 749 + 740
Ex. 4.890 – 420
Ex. 5.680 – 370
Ex. 6.585 – 220
Ex. 7.10803 + 3777
Ex. 8.15663 + 2125
Ex. 9. 30990 – 21660
Ex. 10. 40870-19530
Solution :
1. 720 + 990 = 700 + 1000 = 1700
2. 640 + 880 = 600 + 900 = 1500
3. 749 + 740 = 700 + 700 = 1400
4. 890 – 420 = 900 – 400 = 500
5. 680 – 370 = 700 – 400 = 300
6. 585 – 220 = 600 – 200 = 400
7. 10803 + 3777 = 11000 + 4000 = 15000
8. 15663 + 2125 = 16000 + 2000 = 18000
9. 30990 – 21660 = 31000 – 22000 = 9000
10. 40870 – 19530 = 41000 – 20000 = 21000

Question 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) 8,325 – 491
(d) 4,89,348-48,365 Make four more such examples.
Solution :
(a)

(b)

(c)

(d)

Four more such examples are as follows:
1. 538 + 432 + 5326
2. 2,09, 849 – 57,698
3. 9426 – 395
4. 5,98,459 – 36,463 Sol.
solution :
1. 538 + 432 + 5326
= 500 + 400 + 5300
(Rough estimate Rounding off to nearest hundreds)
= 6200
538 + 432 + 5326
= 540 + 430 + 5330
(Closer estimate Rounding off to nearest tens) = 6300

2. 2,09,849 – 57.698
= 2.09.800 – 57,700
(Rough estimate Rounding off to nearest hun-dreds)
=152100 2,09, 849 – 57,698
= 2,09, 850 – 57,700
(Closer estimate Rounding off to nearest tens) =152150

3. 9426-395
= 9400 – 400 ..
(Rough estimate Rounding off to nearest hundreds)
= 9000 9426 – 395
= 9430 – 400
(Closer estimate Rounding off to nearest tens) = 9030

4. 5,98,459 – 36,463
= 5,98,500 – 36,500
(Rough estimate Rounding off to nearest hundreds)
= 5,62,000 5,98,459 – 36,463
= 5,98,460 – 36,460
(Closer estimate Rounding off to nearest tens) = 5,62,000

Question 3.
Estimate the following products using the general rule:
(a) 578 ×161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
Solution :
(a) 578 × 161
Estimated product = 600 × 200 = 1,20,000
(b) 5281 × 3491
Estimated product = 5000 × 3500 = 1,75,00,000
(c) 1291 × 592
Estimated product = 1300 × 600 = 7,80,000
(d) 9250 × 29
Estimated product = 10000 × 30 = 3,00,000

Four more such examples are:
1. 678 × 261
2. 4271 × 4391
3. 2391 × 629
4. 8250 × 39