a concrete ix requires 3 parts sand and 5 parts cement.how many pounds of cement are needed for 75 pounds of sand?
2006-11-13 07:11:29 · 7 answers · asked by ls1boi 1 in Science & Mathematics ➔ Mathematics
Cement : sand = 5 : 3
So 3 parts = 75 pounds
So 1 part = 25 pounds
Thus cement needed = 5 parts = 125 pounds
2006-11-13 07:16:38 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
The cement/sand ratio is 5 to 3 or 5/3 which equals a ratio of 1.67/1. Multiply the mass of the sand by 1.67. You need 125 lbs of cement
2006-11-13 07:31:22 · answer #2 · answered by Top Gun 3 · 0⤊ 0⤋
3 parts sand:5 parts cement
75 # sand
75#/3=x/5
x=75*5/3=125# of cement.
2006-11-13 07:40:34 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
125 pounds of cement
2006-11-13 07:17:49 · answer #4 · answered by Johnnyboy0809 1 · 0⤊ 0⤋
...the conventional blend for concrete is 3 parts gravel 2 parts sand and a million area cement....all of this by quantity not weight purely as Russell states...my answer replaced into incorrect and that i replaced it because i remerbered weight had not something to do with the blend...i had the combination instructions good purely had the burden pasrt incorrect..
2016-11-29 02:43:36 · answer #5 · answered by ? 4 · 0⤊ 0⤋
put it in fraction form
3/5 = 75/ x
3 goes into 75 how many times
multiply that answer by 5 and you will get x
2006-11-13 07:14:32 · answer #6 · answered by tomkat1528 5 · 0⤊ 0⤋