If only one number be 1 then the product cannot be 1. Examples: 5 x 1 = 5, 4 x 1 = 4, 8 x 1 = 8 If both number are 1, then the product is 1 1 x 1 = 1 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
If only one number be 1 then the product cannot be 1.
Examples: 5 x 1 = 5, 4 x 1 = 4, 8 x 1 = 8
If both number are 1, then the product is 1
1 x 1 = 1
Yes, if we multiply any number with zero the resultant product will be zero. Example: 2 x 0 = 0, 5 x 0 = 0, 9 x 0 = 0 If both numbers are zero, then the result also be zero. 0 x 0 = 0 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
Yes, if we multiply any number with zero the resultant product will be zero.
Example: 2 x 0 = 0, 5 x 0 = 0, 9 x 0 = 0
If both numbers are zero, then the result also be zero.
0 x 0 = 0
Supply of milk in morning = 32 litres Supply of milk in evening = 68 litres Total supply = 32 + 68 = 100 litres Now Cost of 1 litre milk = ₹15 Cost of 100 litres milk = 15 x 100 = ₹1500 Therefore, ₹1500 is due to the vendor per day. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapterRead more
Supply of milk in morning = 32 litres
Supply of milk in evening = 68 litres
Total supply = 32 + 68 = 100 litres
Now
Cost of 1 litre milk = ₹15
Cost of 100 litres milk = 15 x 100 = ₹1500
Therefore, ₹1500 is due to the vendor per day.
Petrol filled on Monday = 40 litres Petrol filled on next day = 50 litres Total petrol filled = 90 litres Now, Cost of 1 litre petrol = ₹ 44 Cost of 90 litres petrol = 44 x 90 = 44 x (100 – 10) = 44 x 100 – 44 x 10 = 4400 – 440 = ₹ 3960 Therefore, he spent ₹ 3960 on petrol. https://www.tiwariacademyRead more
Petrol filled on Monday = 40 litres
Petrol filled on next day = 50 litres
Total petrol filled = 90 litres
Now,
Cost of 1 litre petrol = ₹ 44
Cost of 90 litres petrol = 44 x 90
= 44 x (100 – 10)
= 44 x 100 – 44 x 10
= 4400 – 440
= ₹ 3960
Therefore, he spent ₹ 3960 on petrol.
(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 100 x 1768 = 176800 (b) 4 x 166 x 25 = (4 x 25) x 166 = 100 x 166 = 16600 (c) 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000 (b) 625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000 (e) 285 x 5 x 60 = 284 x (5 x 60) = 284 x 300 = 85500 (f) 125 x 40 xRead more
(a) 2 x 1768 x 50
= (2 x 50) x 1768
= 100 x 1768
= 176800
(b) 4 x 166 x 25
= (4 x 25) x 166
= 100 x 166
= 16600
(c) 8 x 291 x 125
= (8 x 125) x 291
= 1000 x 291
= 291000
(b) 625 x 279 x 16
= (625 x 16) x 279
= 10000 x 279
= 2790000
(e) 285 x 5 x 60
= 284 x (5 x 60)
= 284 x 300
= 85500
(f) 125 x 40 x 8 x 25
= (125 x 8) x (40 x 25)
= 1000 x 1000
= 1000000
(a) 530 > 503; So 503 appear on left side of 530 on number line. (b) 370 > 307; So 307 appear on left side of 370 on number line. (c) 98765 > 56789; So 56789 appear on left side of 98765 on number line. (d) 9830415 < 10023001; So 9830415 appear on left side of 10023001 on number line. htRead more
(a) 530 > 503;
So 503 appear on left side of 530 on number line.
(b) 370 > 307;
So 307 appear on left side of 370 on number line.
(c) 98765 > 56789;
So 56789 appear on left side of 98765 on number line.
(d) 9830415 < 10023001;
So 9830415 appear on left side of 10023001 on number line.
Study the pattern: 1 x 8 + 1 = 9; 12 x 8 + 2 = 98; 123 x 8 + 3 = 987, 1234 x 8 + 4 = 9876; 12345 x 8 + 5 = 98765 Write the next two steps. Can you say how the pattern works?
123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9876543 Pattern works like this: 1 x 8 + 1 = 9 12 x 8 + 2 = 98 123 x 8 + 3 = 987 1234 x 8 + 4 = 9876 12345 x 8 + 5 = 98765 123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9875643 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
Pattern works like this:
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9875643
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessFind using distributive property: (a) 728 x 101 (b) 5437 x 1001 (c) 824 x 25 (d) 4275 x 125 (e) 504 x 35
(a) 728 x 101 = 728 x (100 + 1) = 728 x 100 + 728 x 1 = 72800 + 728 = 73528 (b) 5437 x 1001 = 5437 x (1000 + 1) = 5437 x 1000 + 5437 x 1 = 5437000 + 5437 = 5442437 (c) 824 x 25 = 824 x (20 + 5) = 824 x 20 + 824 x 5 = 16480 + 4120 = 20600 (d) 4275 x 125 = 4275 x (100 + 20 + 5) = 4275 x 100 + 4275 x 2Read more
(a) 728 x 101
= 728 x (100 + 1)
= 728 x 100 + 728 x 1
= 72800 + 728
= 73528
(b) 5437 x 1001
= 5437 x (1000 + 1)
= 5437 x 1000 + 5437 x 1
= 5437000 + 5437
= 5442437
(c) 824 x 25
= 824 x (20 + 5)
= 824 x 20 + 824 x 5
= 16480 + 4120
= 20600
(d) 4275 x 125
= 4275 x (100 + 20 + 5)
= 4275 x 100 + 4275 x 20 + 4275 x5
= 427500 + 85500 + 21375
= 534375
(e) 504 x 35
= (500 + 4) x 35
= 500 x 35 + 4 x 35
= 17500 + 140
= 17640
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessIf the product of two whole number is 1, can we say that one or both of them will be 1? Justify through examples.
If only one number be 1 then the product cannot be 1. Examples: 5 x 1 = 5, 4 x 1 = 4, 8 x 1 = 8 If both number are 1, then the product is 1 1 x 1 = 1 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
If only one number be 1 then the product cannot be 1.
Examples: 5 x 1 = 5, 4 x 1 = 4, 8 x 1 = 8
If both number are 1, then the product is 1
1 x 1 = 1
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessIf the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Yes, if we multiply any number with zero the resultant product will be zero. Example: 2 x 0 = 0, 5 x 0 = 0, 9 x 0 = 0 If both numbers are zero, then the result also be zero. 0 x 0 = 0 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
Yes, if we multiply any number with zero the resultant product will be zero.
Example: 2 x 0 = 0, 5 x 0 = 0, 9 x 0 = 0
If both numbers are zero, then the result also be zero.
0 x 0 = 0
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessA vendor supplies 32 litres of milk to a hotel in a morning and 68 litres of milk in the evening. If the milk costs ₹15 per litre, how much money is due to the vendor per day?
Supply of milk in morning = 32 litres Supply of milk in evening = 68 litres Total supply = 32 + 68 = 100 litres Now Cost of 1 litre milk = ₹15 Cost of 100 litres milk = 15 x 100 = ₹1500 Therefore, ₹1500 is due to the vendor per day. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapterRead more
Supply of milk in morning = 32 litres
Supply of milk in evening = 68 litres
Total supply = 32 + 68 = 100 litres
Now
Cost of 1 litre milk = ₹15
Cost of 100 litres milk = 15 x 100 = ₹1500
Therefore, ₹1500 is due to the vendor per day.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessA taxi-driver, filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs ₹ 44 per litre, how much did he spend in all on petrol?
Petrol filled on Monday = 40 litres Petrol filled on next day = 50 litres Total petrol filled = 90 litres Now, Cost of 1 litre petrol = ₹ 44 Cost of 90 litres petrol = 44 x 90 = 44 x (100 – 10) = 44 x 100 – 44 x 10 = 4400 – 440 = ₹ 3960 Therefore, he spent ₹ 3960 on petrol. https://www.tiwariacademyRead more
Petrol filled on Monday = 40 litres
Petrol filled on next day = 50 litres
Total petrol filled = 90 litres
Now,
Cost of 1 litre petrol = ₹ 44
Cost of 90 litres petrol = 44 x 90
= 44 x (100 – 10)
= 44 x 100 – 44 x 10
= 4400 – 440
= ₹ 3960
Therefore, he spent ₹ 3960 on petrol.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessFind the product using suitable properties: (a) 738 x 103 (b) 854 x 102 (c) 258 x 1008 (d) 1005 x 168
(a) 738 x 103 = 738 x (100 + 3) = 738 x 100 + 738 x 3 = 73800 + 2214 = 76014 (b) 854 x 102 = 854 x (100 + 2) = 854 x 100 + 854 x 2 = 85400 + 1708 = 87108 (c) 258 x 1008 = 258 x (1000 + 8) = 258 x 1000 + 258 x 8 = 258000 + 2064 = 260064 (d) 1005 x 168 = (1000 + 5) x 168 = 1000 x 168 + 5 x 168 = 16800Read more
(a) 738 x 103
= 738 x (100 + 3)
= 738 x 100 + 738 x 3
= 73800 + 2214
= 76014
(b) 854 x 102
= 854 x (100 + 2)
= 854 x 100 + 854 x 2
= 85400 + 1708
= 87108
(c) 258 x 1008
= 258 x (1000 + 8)
= 258 x 1000 + 258 x 8
= 258000 + 2064
= 260064
(d) 1005 x 168
= (1000 + 5) x 168
= 1000 x 168 + 5 x 168
= 168000 + 840
= 168840
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessFind the product by suitable arrangement: (a) 2 x 1768 x 50 (b) 4 x 166 x 25 (c) 8 x 291 x 125 (d) 625 x 279 x 16 (e) 285 x 5 x 60 (f) 125 x 40 x 8 x 25
(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 100 x 1768 = 176800 (b) 4 x 166 x 25 = (4 x 25) x 166 = 100 x 166 = 16600 (c) 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000 (b) 625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000 (e) 285 x 5 x 60 = 284 x (5 x 60) = 284 x 300 = 85500 (f) 125 x 40 xRead more
(a) 2 x 1768 x 50
= (2 x 50) x 1768
= 100 x 1768
= 176800
(b) 4 x 166 x 25
= (4 x 25) x 166
= 100 x 166
= 16600
(c) 8 x 291 x 125
= (8 x 125) x 291
= 1000 x 291
= 291000
(b) 625 x 279 x 16
= (625 x 16) x 279
= 10000 x 279
= 2790000
(e) 285 x 5 x 60
= 284 x (5 x 60)
= 284 x 300
= 85500
(f) 125 x 40 x 8 x 25
= (125 x 8) x (40 x 25)
= 1000 x 1000
= 1000000
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessFind the sum by suitable rearrangement: (a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647
(a) 837 + 208 + 363 = (837 + 363) + 208 = 1200 + 208 = 1408 (b) 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
(a) 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See lessIn each of the following pairs of numbers, state which 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
(a) 530 > 503; So 503 appear on left side of 530 on number line. (b) 370 > 307; So 307 appear on left side of 370 on number line. (c) 98765 > 56789; So 56789 appear on left side of 98765 on number line. (d) 9830415 < 10023001; So 9830415 appear on left side of 10023001 on number line. htRead more
(a) 530 > 503;
So 503 appear on left side of 530 on number line.
(b) 370 > 307;
So 307 appear on left side of 370 on number line.
(c) 98765 > 56789;
So 56789 appear on left side of 98765 on number line.
(d) 9830415 < 10023001;
So 9830415 appear on left side of 10023001 on number line.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-2/
See less