A piece of wire is to devided into 20cm or 28cm or 24cm for construction and leave 15cm piece. Find the shortest such wire
4 answers
LCM(15,20,24,28) = 840 cm
Sin2B-cosB=0 find the value of B and hence Tan(B+20)
A piece of wire is to devided into 20cm or 28cm or 24cm for construction and leave 15cm piece. Find the shortest such wire
To find the shortest wire that can be divided into lengths of 20cm, 28cm, 24cm, and leave a 15cm piece, we need to find the least common multiple (LCM) of these lengths.
The prime factorization of these lengths are:
20 = 2^2 * 5
28 = 2^2 * 7
24 = 2^3 * 3
15 = 3 * 5
To find the LCM, we need to take the highest power of each prime factor that appears in any of the lengths. So the LCM would be: 2^3 * 3 * 5 * 7 = 840 cm.
Therefore, the shortest wire that can be divided into lengths of 20cm, 28cm, 24cm, and leave a 15cm piece is 840 cm long.
The prime factorization of these lengths are:
20 = 2^2 * 5
28 = 2^2 * 7
24 = 2^3 * 3
15 = 3 * 5
To find the LCM, we need to take the highest power of each prime factor that appears in any of the lengths. So the LCM would be: 2^3 * 3 * 5 * 7 = 840 cm.
Therefore, the shortest wire that can be divided into lengths of 20cm, 28cm, 24cm, and leave a 15cm piece is 840 cm long.
6 answers