L.C.M. of 48, 72, 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 sec. After 432 seconds, the lights change simultaneously. 432 second = 7 minutes 12 seconds Therefore the time = 7 a.m. + 7 minutes 12 seconds = 7:07:12 a.m. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
L.C.M. of 48, 72, 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 sec.
After 432 seconds, the lights change simultaneously.
432 second = 7 minutes 12 seconds
Therefore the time = 7 a.m. + 7 minutes 12 seconds
= 7:07:12 a.m.
Hints:- L.C.M. of 8, 10, 12 = 2 x 2 x 2 x 3 x 5 = 120 The largest three digit number = 999 Therefore, the required number = 999 – 39 = 960 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Hints:-
L.C.M. of 8, 10, 12 = 2 x 2 x 2 x 3 x 5 = 120
The largest three digit number = 999
Hints:- L.C.M. of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24 The smallest 3-digit number = 100 To find the number, we have to divide 100 by 24 100 = 24 x 4 + 4 Therefore, the required number = 100 + (24 – 4) = 120. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Hints:-
L.C.M. of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24
The smallest 3-digit number = 100
To find the number, we have to divide 100 by 24
100 = 24 x 4 + 4
Therefore, the required number = 100 + (24 – 4) = 120.
The measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm. Factors of 825 = 3 x 5 x 5 x 11 Factors of 675 = 3 x 5 x 5 x 3 x 3 Factors of 450 = 2 x 3 x 3 x 5 x 5 H.C.F. = 3 x 5 x 5 = 75 cm Therefore, the longest tape is 75 cm. https://www.tiwariacademy.com/ncert-solutions/class-6/mRead more
The measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm.
Factors of 825 = 3 x 5 x 5 x 11
Factors of 675 = 3 x 5 x 5 x 3 x 3
Factors of 450 = 2 x 3 x 3 x 5 x 5
H.C.F. = 3 x 5 x 5 = 75 cm
Therefore, the longest tape is 75 cm.
Hint:- For finding minimum distance, we have to find L.C.M of 63, 70 and 77. L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm. Therefore, the minimum distance is 6930 cm. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Hint:-
For finding minimum distance, we have to find L.C.M of 63, 70 and 77.
L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm.
Therefore, the minimum distance is 6930 cm.
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m. at what time will they change simultaneously again?
L.C.M. of 48, 72, 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 sec. After 432 seconds, the lights change simultaneously. 432 second = 7 minutes 12 seconds Therefore the time = 7 a.m. + 7 minutes 12 seconds = 7:07:12 a.m. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
L.C.M. of 48, 72, 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 sec.
After 432 seconds, the lights change simultaneously.
432 second = 7 minutes 12 seconds
Therefore the time = 7 a.m. + 7 minutes 12 seconds
= 7:07:12 a.m.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessDetermine the largest 3-digit number which is exactly divisible by 8, 10 and 12.
Hints:- L.C.M. of 8, 10, 12 = 2 x 2 x 2 x 3 x 5 = 120 The largest three digit number = 999 Therefore, the required number = 999 – 39 = 960 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Hints:-
L.C.M. of 8, 10, 12 = 2 x 2 x 2 x 3 x 5 = 120
The largest three digit number = 999
Therefore, the required number = 999 – 39 = 960
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessDetermine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Hints:- L.C.M. of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24 The smallest 3-digit number = 100 To find the number, we have to divide 100 by 24 100 = 24 x 4 + 4 Therefore, the required number = 100 + (24 – 4) = 120. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Hints:-
L.C.M. of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24
The smallest 3-digit number = 100
To find the number, we have to divide 100 by 24
100 = 24 x 4 + 4
Therefore, the required number = 100 + (24 – 4) = 120.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessThe length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
The measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm. Factors of 825 = 3 x 5 x 5 x 11 Factors of 675 = 3 x 5 x 5 x 3 x 3 Factors of 450 = 2 x 3 x 3 x 5 x 5 H.C.F. = 3 x 5 x 5 = 75 cm Therefore, the longest tape is 75 cm. https://www.tiwariacademy.com/ncert-solutions/class-6/mRead more
The measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm.
Factors of 825 = 3 x 5 x 5 x 11
Factors of 675 = 3 x 5 x 5 x 3 x 3
Factors of 450 = 2 x 3 x 3 x 5 x 5
H.C.F. = 3 x 5 x 5 = 75 cm
Therefore, the longest tape is 75 cm.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessThree boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the maximum distance each should cover so that all can cover the distance in complete steps?
Hint:- For finding minimum distance, we have to find L.C.M of 63, 70 and 77. L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm. Therefore, the minimum distance is 6930 cm. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Hint:-
For finding minimum distance, we have to find L.C.M of 63, 70 and 77.
L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm.
Therefore, the minimum distance is 6930 cm.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See less