The length of metallic rod F is
45 of the length of metallic rod J and the length of metallic rod J is
13 of the length of metallic rod G and
19 of the length of metallic rod H. What fraction of the total length of the 4 metallic rods is metallic rod H?
F |
J |
G |
H |
Total |
4x1 |
5x1 |
|
|
|
|
1x5 |
3x5 |
|
|
|
1x5 |
|
9x5 |
|
4 u |
5 u |
15 u |
45 u |
69 u |
The length of metallic rod J is repeated. Make the length of metallic rod J the same. LCM of 5 and 1 is 5.
Total length of all the 4 metallic rods
= 4 u + 5 u + 15 u + 45 u
= 69 u
Metallic Rod HTotal Length =
4569 =
1523Answer(s):
1523