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

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.

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

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.

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

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.

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

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.

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

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.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• Playing With Numbers Class 6 Ex 3.3
• Playing With Numbers Class 6 Ex 3.5
• Playing With Numbers Class 6 Ex 3.6
• Playing With Numbers Class 6 Ex 3.7

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4

Question 1.
Find the common factors of:
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120.
Solution :
(a) 20 and 28
Factors of 20 are 1, 2,4, 5, 10 and 20.
Factors of 28 are 1, 2,4, 7, 14 and 28.
Hence, the common factors of 20 and 28 are 1, 2 and 4.

(b) 15 and 25
Factors of 15 are 1, 3, 5 and 15.
Factors of 25 are 1, 5 and 25.
Hence, the common factors of 15 and 25 are 1 and 5.

(c) 35 and 50
Factors of 35 are 1, 5, 7 and 35.
Factors of 50 are 1, 5, 10, 25 and 50.
Hence, the common factors of 35 and 50 are 1 and 5.

(d) 56 and 120
Factors of 56 are 1, 2,4, 7, 8, 14, 28 and 56. Factors of 120 are 1, 2, 3,4, 5, 6, 8,10, 12,15, 20, 24, 30,40, 60 and 120. ,
Hence, the common factors of 56 and 120 are 1, 2,4 and 8.

Question 2.
Find the common factors of:
(a) 4, 8 and 12
(b) 5, 15 and 25.
Solution :
(a) 4,8 and 12 Factors of 4 are 1,2 and 4.
Factors of 8 are 1, 2,4 and 8.
Factors of 12 are 1, 2, 3,4,6 and 12.
Hence, the common factors of 4,8 and 12 are 1, 2 and 4.

(b) 5,15 and 25
Factors of 5 are 1 and 5.
Factors of 15 are 1, 3 and 5.
Factors of 25 are 1,5 and 25.
Hence, the common factors of 5,15 and 25 are 1 and 5.

Question 3.
Find the first three common multiples of:
(a) 6 and 8
(b) 12 and 18.
Solution :
(a) 6 and 8
Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,78, 84,90,96,
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Common multiples of 6 and 8 are 24, 48, 72, 96,
∴ First three common multiples of 6 and 8 are 24, 48 and 72.

(b) 12 and 18
Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, *
Multiples of 18 are 18,36,54,72,90,108,126, 144,
∴ Common multiples of 12 and 18 are 36, 72,108, 144,
∴ First three common multiples of 12 and 18 are 36, 72 and 108.

Question 4.
Write all the numbers less than 100 which are common multiples of 3 and 4.
Solution :
Multiples of 3 are : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108,
Multiples of 4 are : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108,
Common multiples of 3 and 4 are : 12, 24, 36, 48, 60, 72, 84, 96, 108,
∴ All the numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84 and 96.

Question 5.
Which of the following numbers are co-prime :
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16.
Solution :
(a) 18 and 35
Factors of 18 are 1, 2, 3, 6, 9 and 18.
Factors of 35 are 1, 5, 7 and 35.
∴ A common factor of 18 and 35 is 1.
∴ 18 and 3 5 have only 1 as the common factor
∴ 18 and 35 are co-prime numbers.

(b) 15 and 37
Factors of 15 are 1, 3, 5 and 15.
Factors of 37 are 1 and 37.
∴ A common factor of 15 and 37 is 1. v 15 and 37 have only 1 as the common factor ∴ 15 and 37 are co-prime numbers.

(c) 30 and 415
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 415 are 1, 5, 83 and 415.
∴ Common factors of 30 and 415 are 1 and 5. v
∴ 30 and 415 have two common factors
∴ 30 and 45 are not co-prime numbers.

(d) 17 and 68 Factors of 17 are 1 and 17.
Factors of 68 are 1, 2, 4, 17, 34 and 68.
∴ Common factors of 17 and 68 are 1 and 17.
∴ 17 and 68 have two common factors
∴ 17 and 68 are not co-prime numbers.

(e) 216 and 215
Factors of 216 are 1,2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216.
Factors of 215 are 1, 5 and 43
∴ A common factor of 216 and 215 is 1.
∴ 216 and 215 have only 1 as the common factor
∴ 216 and 215 are co-prime numbers.

(f) 81 and 16
Factors of 81 are 1, 3, 9, 27 and 81.
Factors of 16 are 1, 2, 4, 8 and 16.
∴ A common factor of 81 and 16 is 1.
∴ 81 and 16 have only 1 as the common factor
∴ 81 and 16 are co-prime numbers.

Question 6.
A number is divisible by both 5 and 12. By which other numbers will that number be always divisible?
Solution :
The number will be always divisible by 5 × 12 = 60.

Question 7.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution :
Factors of 12 are 1, 2, 3, 4, 6 and 12. So, that number will be divisible by other numbers 1, 2, 3, 4 and 6.

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4, drop a comment below and we will get back to you at the earliest.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• Playing With Numbers Class 6 Ex 3.4
• Playing With Numbers Class 6 Ex 3.5
• Playing With Numbers Class 6 Ex 3.6
• Playing With Numbers Class 6 Ex 3.7

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3

Question 1.
Using divisibility tests, determine which the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11; (say yes or no):

Solution :

Question 2.
Using divisibility tests, determine which of the following numbers are divisible by 4: by 8:
(a) 572
(b) 726352
(C) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150.
Solution :
(a) 572
(i) Divisibility by 4
The number formed by last two digits = 72

∵ Remainder is 0
∴ 72 is divisible by 4
∴ 572 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 572

∵ Remainder = 4 ≠ 0
∴ 572 is not divisible by 8.

(b) 726352
(i) Divisibility by 4
The number formed by last two digits = 52

∵ Remainder is 0
∴ 52 is divisible by 4
∴ 726352 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 352

∵ Remainder is 0
∴ 352 is divisible by 8
∴ 726352 is divisible by 8.

(c) 5500
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4 5500 is divisible by 4.
∴ 500 is Divisibility by 4

(ii)
The number formed by last three digits = 500

∵ Remainder 4 0
∴ 500 is not divisible by 8
∴ 5500 is not divisible by 8.

(d) 6000
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4
6000 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 000, which is divisible by 8 6000 is divisible by 8.
6000 is divisible by 8

(e) 12159
(i) Divisibility by 4
The number formed by last two digits = 59

∵ Remainder = 3 ≠ 0
∴ 59 is not divisible by 4.
∴ 12159 is not divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits =159

∵ Remainder 7 ≠ 0
∴ 159 is not divisible by 8
∴ 12159 is not divisible by 8.

(f) 14560
(i) Divisibility by 4
The number formed by last two digits = 60

∵ Remainder is 0
∴ 60 is divisible by 4
∴ 14560 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 560

∵ Remainder is 0
∴ 560 is divisible by 8
∴ 14560 is divisible by 8.

(g) 21084
(i) Divisibility by 4
The number formed by last two digits = 84

∵ Remainder is 0
∴ 484 is divisible by 4
∴ 21084 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 84

∵ The remainder is not 0
∴ 84 is not divisible by 8
∴ 21084 is not divisible by 8.

(h) 31795072
(i) Divisibility by 4
The number formed by last two digits = 72

∵ Remainder is 0
∴ 72 is divisible by 4
∴ 31795072 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 72

∵ Remainder is 0
∴ 72 is divisible by 8
∴ 31795072 is divisible by 8.

(i) 1700
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4
∴ 1700 is divisible by

(ii) Divisibility by 8
The number formed by last three digits = 700

∵ Remainder 4 ≠ 0
∴ 700 is not divisible by 8
∴ 1700 is not divisible by 8.

(j) 2150
(i) Divisibility by 4
The number formed by last two digits = 50

∵ Remainder = 2 ≠ 0
∴ 50 is not divisible by 4
∴ 2150 is not divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits =150

∵ Remainder 6 ≠ 0
∴ 150 is not divisible by 8
∴ 2150 is not divisible by 8.

Question 3.
Using divisibility tests, determine which of following numbers are divisible by 6 :
(a)297144
(b) 1258
(c) 4335
(d) 61233
(e) 901352
(f) 438750
(g) 1790184
(h) 12583
(i) 639210
(j) 17852.
Solution :
We know that number is divisible by 6 if it is divisible by 2 and 3 both.
(a) 297144
(i) Divisibility by 2
Unit’s digit = 4
297144 is divisible by 2.

(ii) Divisibility by 3
Sum of the digits = 2 + 9 + 7 + 1+ 4 + 4 = 27, which is divisible by 3 297144 is divisible by 3 Since, 297144 is divisible by 2 and 3 both, so it is divisible by 6.

(b) 1258
(i) Divisibility by 2 Unit’s digit = 8 1258 is divisible by 2.
(ii) Divisibility by 3 Sum of the digits =1+2 + 5 + 8=16 which is not divisible by 3 1258 is not divisible by 3. Since 1258 is divisible by 2 but not by 3. so 1258 is not divisible by 6.

(c) 4335
(i) Divisibility by 2
Unit’s digit = 5, which is not any of the digits 0, 2, 4, 6 or 8 4335 is not divisible by 2 4335 is not divisible by 6.

(d) 61233
(i) Divisibility by 2
Unit’s digit = 3, which is not any of the digits 0, 2, 4, 6 or 8

61233 is not divisible by 2
61233 is not divisible by 6.

(e) 901352
(i) Divisibility by 2 v Unit’s digit = 2
901352 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits = 9 + 0+1+3 + 5 + 2 = 20, which is not divisible by 3
901352 is not divisible by 3 Since, 901352 is divisible by 2 but not by 3, so it is not divisible by 6.

(f) 438750
(i) Divisibility by 2 v Unit’s digit = 0
438750 is divisible by 2.

(ii) Divisibility by 3
Sum of the digits = 4 + 3 + 8 + 7 + 5+ 0 = 27, which is divisible by 3
438750 is divisible by 3 Since, 438750 is divisible by 2 and 3 both, so it is divisible by 6.

(g) 1790184
(i) Divisibility by 2 v Unit’s digit = 4
1790184 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits =1+7 + 9 + 0+1 + 8 + 4=30, which is divisible by 3
1790184 is divisible by 3 Since, 1790184 is divisible by 2 and 3 both, so it is divisible by 6.

(h) 12583,
(i) Divisibility by 2
Unit’s digit = 3, which is not any of the digits 0, 2, 4, 6 or 8
12583 is not divisible by 2
12583 is not divisible by 6.

(i) 639210
(i) Divisibility by 2
Unit’s digit = 0
639210 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits = 6 + 3 + 9 + 2+1+0 = 21, which is divisible by 3
639210 is divisible by 3 Since, 639210 is divisible by 2 and 3 both, so it is divisible by 6.

(j) 17852
(i) Divisibility by 2
Unit’s digit = 2
∴ 17852 is divisible by 2.
(ii) Divisibility by 3 Sum of the digits =1+7 + 8 + 5 + 2 = 23, which is not divisible by 3
17852 is not divisible by 3
Since 17852 is divisible by 2 but not by 3, so it is not divisible by 6.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445
(b) 10824
(c) 7138965
(d) 70169308
(e) 10000001
(f) 901153.
Solution :
(a) 5445
Sum of the digits (at odd places) from the right =5+4=9
Sum of the digits (at even places) from the right =4+5=9
Difference of these sums = 9-9 = 0 v 0 is divisible by 11
5445 is divisible by 11.

(b) 10824
Sum of the digits (at odd places) from the right = 4 + 8+1 = 13
Sum of the digits (at even places) from the right = 2 + 0 = 2
Difference of these sums = 13 – 2 = 11
11 is divisible by 11
10824 is divisible by 11.

(c) 7138965
Sum of the digits (at odd places) from the right =5+9+3+7=24
Sum of the digits (at even places) from the right = 6 + 8 + 1 = 15
Difference of these sums = 24 – 15 = 9
Y 9 is not divisible by 11
∴ 7138965 is not divisible by 11.

(d) 70169308
Sum of the digits (at odd places) from the right = 8 + 3+6 + 0=17
Sum of the digits (at even places) from the right = 0 + 9 + 1+7=17 Difference of these sums = 17-17=0
Y 0 is divisible by 11
∴ 70169308 is divisible by 11.

(e) 10000001
Sum of the digits (at odd places) from the right =l+0+0+0=l
Sum of the digits (at even places) from the right =0+0+0+1=1 Difference of these sums =1-1=0 0 is divisible by 11
∴ 10000001 is divisible by 11.

(f) 901153
Sum of the digits (at odd places) from the right = 3+ 1 +0 = 4
Sum of the digits (at even places) from the right = 5+1+9=15
Difference of these sums =15-4=11
Y 11 is divisible by 11
∴ 901153 is divisible by 11.

Question 5.
Write the smallest digit and the largest digit in the blank space of each of the following numbers so that the number is divisible by 3:
(a) __ 6724
(b) 4765 __ 2.
Solution :
(a) __ 6724
(i) Smallest digit
Sum of the given digits = 6 + 7 + 2 + 4=19
19 is not divisible by 3
∴ Smallest digit (non-zero) is 2.

(ii) Largest digit The largest digit is 8.

(b) 4765 _ 2
(i) Smallest digit
Sum of the given digits = 4 + 7 + 6 + 5 + 2 = 24 Y 24 is divisible by 3
∴ Smallest digit is 0.

(ii) Largest digit The largest digit is 9.

Question 6.
Write digit in the blank space of each of the following numbers so that the number is divisible by 11:
(a) 92 __ 389
(b) 8 __ 9484.
Solution :
(a) 92 __ 389
Sum of the given digits (at odd places) from the right = 9 + 3 + 2= 14
Sum of the given digits (at even places) from the right = 8 + required digit + 9 = required digit + 17
Difference of these sums = required digit + 3 For the above difference to be divisible by 11, required digit = 8.
Hence, the required number is 92 8 389.

(b) 8 __ 9484
Sum of the given digits (at odd places) from the right = 4 + 4 + required digit = 8 + required digit
Sum of the given digits (at even places) from the right = 8 + 9 + 8 = 25
Difference of the sums = 17 – required digit For the above difference to be divisible by 11, required digit = 6.
Hence, the required number is 8 6 9484.

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3, drop a comment below and we will get back to you at the earliest.

• Whole Numbers Class 6 Ex 2.1
• Whole Numbers Class 6 Ex 2.2

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3

Question 1.
Which of the following will not represent
(a) 1 + 0
(b) 0 × 0
(c) \(\frac { 0 }{ 2 }\)
(d) \(\frac { 10\quad 10 }{ 2 }\)
Solution :
(a) 1 + 0 = 1 ≠ 0
(b) 0 × 0 = 0
(c) \(\frac { 0 }{ 2 }\)
(d) \(\frac { 10-10 }{ 2 } =\frac { 0 }{ 2 } =0\)
Hence, (a) 1 + 0 will not represent zero.

Question 2.
If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Solution :
Yes! For example :
2 × 0 = 0
0 × 3 = 0
0 × 0 = 0.

Question 3.
If the product of two whole numbers is 1. can we say that one or both of them will be 1? Justify through examples.
Solution :
Both of them must be ‘ 1’ as 1 × 1 = 1.

Question 4.
Find using distributive property :
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35
Solution :

Question 5.
Study the pattern :

Write the next two steps. Can you say how the pattern works ?
(Hint : 12345 =11111 + 1 1 11 + 111 + 11 + 1). Sol. Next two steps are as follows :
Solution :
Next two steps are as follows :
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543.
Working of the pattern

We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3, drop a comment below and we will get back to you at the earliest.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• Playing With Numbers Class 6 Ex 3.3
• Playing With Numbers Class 6 Ex 3.4
• Playing With Numbers Class 6 Ex 3.5
• Playing With Numbers Class 6 Ex 3.7

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6

Question 1.
Find the H.C.F. of the following numbers,
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12 45 75
Solution :
(a) 18,48
Factors of 18 are 1, 2, 3, 6, 9 and 18.
Factors of 48 are 1, 2, 3,4, 6, 8, 12, 16, 24 and
∴ Common factors of 18 and 48 are 1, 2, 3
Highest of these common factors is 6. ∴ H.C.F. of 18 and 48 is 6.

(b) 30,42
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 42 are 1, 2, 3. 6. 7, 14, 21 and 42. ,
∴ Common factors of 30 and 42 are 1,2, 3 and 6.
Highest of these common factors is 6.
∴ H.C.F. of 30 and 42 is 6.

(c) 18,60
Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 60 are 1,2, 3,4, 5,6,10 12,15, 20, 30 and 60.
∴ Common factors of 18 and 60 are 1, 2, 3 and 6.
Highest of these common factors is 6.
∴ H.C.F. of 18 and 60 is 6.

(d) 27,63
Factors of 27 are 1, 3, 9 and 27.
Factors of 63 are 1, 3, 7, 9, 21 and 63.
Common factors of 27 and 63 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 27 and 63 is 9.

(e) 36,84
Factors of 36 are 1, 2, 3,4, 6, 9,12, 18 and 36.
Factors of 84 are 1,2, 3,4,6,7, 12, 14,21, 28, 42 and 84.
Common factors of 36 and 84 are 1,2, 3,4, 6 and 12.
Highest of these common factors is 12.
∴ H.C.F. of 36 and 84 is 12. if) 34,102 • ‘
Factors of 34 are 1,2, 17 and 34.
Factors of 102 are 1, 2, 3,6,17, 34. 51 and 102.
∴ Common factors of 34 and 102 are 1, 2, 17 and 34.
Highest of these common factors is 34.
∴ H.C.F. of 34 and 102 is 34.

(g) 70,105,175
Factors of 70 are 1. 2, 5. 7, 10. 14, 35 and 70.
Factors of 105 are 1, 3, 5. 7. 15. 21. 35 and 105.
Factors of 175 are 1. 5, 7. 25. 35 and 175. .’. Common factors of 70, 105 and 175 are 1, 5 and 35.
Highest of these common factors is 35.
∴ H.C.F. of 70. 105 and 175 are 35.

(h) 91,112,49
Factors of 91 are 1,7, 13 and 91.
Factors of 112 are 1,2. 4. 7, 8, 14. 16. 28, 56 and 112.
Factors of 49 are 1.7 and 49.
Common factors of 91,112 and 49 are 1 and 7.
Highest of these common factors is 7.
∴ H.C.F. of 91, 112 and 49 is 7.

(i) 18,54,81
Factors of 18 are 1. 2, 3. 6, 9 and 18. Factors of 54 are 1, 2. 3, 6. 9, 18. 27 and 54.
Factors of 81 are 1. 3, 9, 27 and 81.
∴ Common factors of 18,54 and 81 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 18, 54 and 81 is 9.
(j) 12, 45, 75
Factors of 12 are 1, 2, 3, 4, 6 and 12. Factors of 45 are 1, 3, 5, 9, 15 and 45.
: Factors of 75 are 1, 3, 5, 15, 25 and 75.
∴ Common factors of 12,45 and 75 are 1 and 3.
Highest of these common factors is 3.
H.C.F. of 12. 45 and 75 is 3.

Question 2.
What is the H.C.F. of two consecutive :
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution :
(a) The H.C.F. of two consecutive numbers is 1.
(b) The H.C.F. of two consecutive even numbers is 2.
(c) The H.C.F. of two consecutive odd numbers is 1.

Question 3.
H. C.F. of co-prime numbers 4 and 15 was found as follows factorization: 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F?
Solution :
No, the answer is not correct. The correct answer is as follows :
H.C.F. of 4 and 15 is 1.

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6, drop a comment below and we will get back to you at the earliest.

• Basic Geometrical Ideas Class 6 Ex 4.2
• Basic Geometrical Ideas Class 6 Ex 4.3
• Basic Geometrical Ideas Class 6 Ex 4.4
• Basic Geometrical Ideas Class 6 Ex 4.5
• Basic Geometrical Ideas Class 6 Ex 4.6

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1

Question 1.
Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments.
Solution :
(a) O, B, C. D, E

Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solution :

Question 3.
Use the figure to name:
(a) The line containing point E.
(b) The line passing through A.
(c) The line on which O lies
(d) Two pairs of intersecting lines.

Solution :
(a) \(\overleftrightarrow { AE }\), etc.
(b) \(\overleftrightarrow { AE }\), etc.
(c) \(\overleftrightarrow { CO } or\overleftrightarrow { OC }\)
(d) \(\overleftrightarrow { CO, } \overleftrightarrow { AE } ;\overleftrightarrow { AE } ,\overleftrightarrow { EF }.\)

Question 4.
How many lines can pass through
(a) one given point?
(b) two given points?
Solution :
(a) Countless lines can pass through one given point.
(b) One and only one line can pass through two given points.

Question 5.
Draw a rough figure and label suitably in each of the following cases :
(a) Point P lies on \(\bar { AB } \)
(b) \(\overleftrightarrow { XY }\) and \(\overleftrightarrow { PQ }\) intersect at M.
(c) Line contains E and F but not D.
(d) \(\bar { Op } \) and \(\bar { OQ } \) meet at O.
Solution :
(a)

(b)

(c)

(d)

Question 6.
Consider the following figure of line \(\bar { MN } \) Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line \(\bar { MN } \)
(b) M, O, N are points on a line segment \(\bar { MN } \).
(c) M and N are end points of line segment \(\bar { MN } \).

(d) O and N are end points of line segment \(\bar { OP } \).
(e) M is one of the end points of line segment \(\bar { QO } \).
(f) M is point on ray \(\overrightarrow { OP }\).
(g) Ray \(\overrightarrow { OP }\) is different from ray \(\overrightarrow { QP }\).
(h) Ray \(\overrightarrow { OP }\) is same as ray \(\overrightarrow { OM }\).
(i) Ray \(\overrightarrow { OM }\) is not opposite to ray \(\overrightarrow { OP }\).
(j) O is not an initial point of \(\overrightarrow { OP }\).
(k) N is the initial point of \(\overrightarrow { NP }\) and \(\overrightarrow { NM }\).
Solution :
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) False
(j) False
(k) True.

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1, drop a comment below and we will get back to you at the earliest.