CBSE Class 6

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-2-ex-2-2/

• Whole Numbers Class 6 Ex 2.1
• Whole Numbers Class 6 Ex 2.3

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 2
Chapter Name	Whole Numbers
Exercise	Ex 2.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

Question 1.
Find the sum by suitable rearrangement :
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647.
Solution :
(a) 837 + 208 + 363
= 837 + 363 + 208 = (837 + 363) + 208 = 1200 + 208 = 1408
(b) 1962+ 453+ 1538 + 647
= 1962 + 1538 + 453 + 647 = (1962 + 1538) + (453 + 647)
= 3500+ 1100 = 4600.

Question 2.
Find the product by a suitable rearrangement :
(a) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25.
Solution :
(a) 2 × 1768 × 50
= 2 × 50 × 1768 = (2 × 50) × 1768 = 100 × 1768= 1,76,800
(b) 4 × 166 × 25 = 4 × 25 × 166 = (4 × 25) × 166 = 100 × 166 = 16,600
(c) 8 × 291 × 125 = 8 × 125 × 291
= (8 × 125) × 291 = 1000 × 291 =2,91,000
(d) 625 × 279 × 16 = 625 × 16 × 279
= (625 × 16) × 279 = 10000 ×279 = 27,90,000
(e) 285 × 5 × 60 = 285 × (5 × 60) = 285 × 300
= 85,500
(f) 125 × 40 × 8 × 25 = (125 × 40) × (8 × 25) = 5000 × 200 = 10,00,000.

Question 3.
Find the value of the following :
(a) 297 × 17+297 × 3
(b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69
(d) 3845 × 5 × 782 + 769 × 25 × 218.
Solution :
(a) 297 × 17 + 297 × 3 = 297 × (17+ 3)
= 297 × 20 = 5940
(b) 54279 × 92 + 8 × 54279
= 54279 × 92 + 54279 × 8 = 54279 × (92 + 8)
= 54279 × 100 = 54,27,900
(c) 81265 × 169-81265 × 69
= 81265 × (169-69)
= 81265 × 100 = 81,26,500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218 = 3845 × 5 × 782 + (769 × 5) × 5 × 218 = 3845 × 5 × 782 + 3845 × 5 × 218 = 3845 × 5 × (782 + 218)
= 3845 × 5 × 1000 = 19225 × 1000 = 1,92,25,000.

Question 4.
Find the product, using suitable properties :
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168.
Solution :
(a) 738 × 103
= 738 × (100+ 3)
= 738 × 100 + 738 × 3 = 73,800 + 2,214 = 76,014
(b) 854 × 102
= 854 × (100 + 2)
= 854 × 100 + 854 × 2 = 85,400+ 1,708 = 87,108
(c) 258 × 1008
= 258 × (1000+ 8)
= 258 × 1000 + 258 × 8 = 2,58,000 + 2,064 = 2,60,064
(d) 1005 × 168
= 168 × 1005
= 168 × (1000+ 5)
= 168 × 1000+ 168 × 5
= 1,68,000 + 840 = 1,68,840.

Question 5.
A taxi driver filled his car petrol tank with 40 liters of petrol on Monday. The next day he filled the tank with 50 liters of petrol. If the petrol costs ₹ 44 per liter, how much did he spend all on petrol?
Solution :
Petrol filled on Monday = 40 litres
Petrol filled the next day = 50 litres
∴ Total petrol filled on the two days = 40 litres + 50 litres = 90 litres
∴ Cost of petrol per litre = ₹ 44
∴ Cost of 90 litres of 7 petrol = ₹ 44 × 90 = ? 3960.

Question 6.
A vendor’supplies 32 liters of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs ₹ 15 per liter, how much money is due to the vendor per day?
Solution :
Milk supplied in the morning = 32 liters
Milk supplied in the evening = 68 liters
∴ Milk supplied per day = 32 litres + 68 litres = 100 litres
Cost of milk per liter = ₹ 15
Money due to the vendor per day = Cost of
100 litres of milk = ₹ 15 × 100 = ₹ 1500.

Question 7.
Match the following :

Solution :

We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-2/

• Whole Numbers Class 6 Ex 2.2
• Whole Numbers Class 6 Ex 2.3

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 2
Chapter Name	Whole Numbers
Exercise	Ex 2.1
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1

Question 1.
Write the next three natural numbers after 10999.
Solution :
The next three natural numbers after 10999 are 11000, 11001 and 11002.

Question 2.
Write the three whole numbers occurring just before 10001.
Solution :
The three whole numbers occurring just before 10001 are 10000, 9999 and 9998.

Question 3.
Which is the smallest whole number?
Solution :
0 is the smallest whole number.

Question 4.
How many whole numbers are there between 32 and 53?
Solution :
There are 20 whole numbers between 32 and 53.
These are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,43, 44, 45, 46, 47,48, 49, 50, 51 and 52.

Question 5.
Write successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Solution :
(a) The successor of 2440701 is 24,40,702.
(b) The successor of 100199 is 1,00,200.
(c) The successor of 1099999 is 11,00,000.
(d) The successor of 2345670 is 23,45,671.

Question 6.
Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321.
Solution :
(a) The predecessor of 94 is 93.
(b) The predecessor of 10000 is 9,999.
(c) The predecessor of 208090 is 2,08,089.
(d) The predecessor of 7654321 is 76,54,320.

Question 7.
In each of the following pairs of numbers, the state which the whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503
(b) 370, 307
(c) 98765, 56789
(d) 9830415, 10023001.
Solution :
(a) The whole number 503 is on the left of the whole number 530 on the number line. So, 503 < 530.
(b) The whole number 307 is on the left of the whole number 370 on the number line. So, 307 < 370.
(c) The whole number 56789 is on the left of the whole number 98765 on the number line. So, 56789 < 98765. ;
(d) The whole number 9830415 is on the left of the whole number 10023001 on the number line. So, 98,30,415 < 100,23,001.

Question 8.
Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers,
(g) The predecessor of a two digit number is never a single digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor,
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two digit number is always a two digit number.
Solution :
(a) This statement is false (F).
(b) This statement is false (F).
(c) This statement is true (T).
(d) This statement is true (T).
(e) This statement is true (T).
(f) This statement is false (F).
(g) This statement is false (F).
(h) This statement is false (F).
(i) This statement is true (T).
(j) This statement is false (F).
(k) This statement is false (F).
(l) This statement is true (T).
(m) This statement is false (F).

 

We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-1-ex-1-3/

• Knowing Our Numbers Class 6 Ex 1.1
• Knowing Our Numbers Class 6 Ex 1.2

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 1
Chapter Name	Knowing Our Numbers
Exercise	Ex 1.3
Number of Questions Solved	3
Category	NCERT Solutions

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
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)

• Rough estimate (Rounding off to nearest hundreds)
  439 + 334 + 4317
  = 400 + 300 + 4,300 = 5000
• Closer estimate (Rounding off to nearest tens) 439 + 334 + 4317
  = 440 + 330 + 4,320 = 5,090.

(b)

• Rough estimate (Rounding off to nearest hundreds) 1,08,734 – 47,599
  = 1,08,700 – 47,600 = 61,100
• Closer estimate (Rounding off to nearest tens) 1,08,734 – 47,599
  = 1,08,730 – 47,600 = 61,130.

(c)

• Rough estimate (Rounding off to nearest hundreds) 8325 – 491
  = 8300 – 500 = 7800
• Closer estimate (Rounding off to nearest tens)
  8325 – 491
  = 8330 – 490 = 7840.

(d)

• Rough estimate (Rounding off to nearest hundreds)
  4,89,348 – 48,365
  = 4,89,300 – 48,400 = 4,40,900
• Closer estimate (Rounding off to nearest tens)
  4,89,348 – 48,365
  = 4,89,350 – 48,370 = 4,40,980

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

We hope the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-1-ex-1-2/

• Knowing Our Numbers Class 6 Ex 1.1
• Knowing Our Numbers Class 6 Ex 1.3

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 1
Chapter Name	Knowing Our Numbers
Exercise	Ex 1.2
Number of Questions Solved	12
Category	NCERT Solutions

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

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

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 = \(\frac { 22000 }{ 1000 }\) 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

Hence, it can be filled in 180 glasses.

We hope the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-1/

• Knowing Our Numbers Class 6 Ex 1.2
• Knowing Our Numbers Class 6 Ex 1.3

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 1
Chapter Name	Knowing Our Numbers
Exercise	Ex 1.1
Number of Questions Solved	4
Category	NCERT Solutions

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

 

We hope the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1, drop a comment below and we will get back to you at the earliest.