How much salt is in a mixture containing 10 gal of an 11% salt solution and 6 gal of a 7% salt solution
1 answer:
In a 10 gal of an 11% salt solution,
11% of 10 gal are sal.
11% *10 gal=(11/100)*10 gal =0.11*10 gal = 0.1 gal is salt.
Similarly, 0.07*6 gal= 0.42 gal is salt of the 6 gal in the second solution.
The mixture is 10 gal +6 gal = 16 gal, and contains a total of
0.1 gal +0.42 gal = 0.52 gal of salt.
0.52 = x% of 16
0.52=x%*16
x=(0.52/16)*100=0.0325*100=3.25
Answer: there is 0.52 gal of salt in the mixture, that is 3.25%
You might be interested in
Hello,
multiple of 9 betheen 50 and 99 are :54,63,72,81,99
and 54,72 are even.
5-4=1 and not 5
7-2=5 ok
72 is the number searched.
Answer:
Step-by-step explanation:
let, the length is z
thus the ratio will be 5z : 4z
perimeter = 2 (a + b)
360 = 2 ( 5z + 4z )
180 = 9z
z = 20 cm
length = 100 cm and width = 80cm
area = length * width
area = 100*80
area = 8000cm²
-12ab+7a power of 2 +a 11
Answer:
the radius is 8.5 mm
U+xy
On replacing for u, x and y.
=9+9*7
=9+63
=72
