Help me answer this. I have already found an answers, so I thought it would be nice to know what others people get for the answer.
How many liters of a 60% acid solution must be added to 20 liters of a 90% acid solution to get a mixture that is 70% acid?
2006-12-02 06:05:29 · 7 answers · asked by yashodauk 2 in Science & Mathematics ➔ Mathematics
P. S show your work.
2006-12-02 06:13:17 · update #1
hmm..
what if we consider this:
volume of acid in the 90% solution + volume of it added from 60% solution = volume of it in 70 % solution
see, we are more more acid, so the amount of new acid must equal to ....
volume of the acid in 90% solution = 20 * 0.9
Volume of it in 60% solution = x * 0.6
we dunn know how much second solution is added, so i considered it as an unknown variable.
and now since x L of second solution is added, so new total volume = x+20
and then volume of acid in it = 0.7 * (x+20)
so
20 * 0.9 +0.6x = 0.7(x+20)
0.6x -0.7x = 0.7*20 - 20*0.9
-0.1x = 20(0.7-0.9)
-0.1x = -20 * 0.2
x= 40 L
2006-12-02 06:18:45 · answer #1 · answered by Anonymous · 0⤊ 0⤋
20L*.9=18 liters of acid
.6*xL=.6x
(18+.6x)=(20+x)*.7
18+.6x=14+.7x subtract 14 from each side
4+.6x=.7x subtract .6x
.1x=4 divide by .1
x=40 L
check
.6*40+.9*20=24+18=42L acid
40+20=60L solution
42/60=.7=70% acid
2006-12-02 14:13:06 · answer #2 · answered by yupchagee 7 · 1⤊ 0⤋
let x be that number
you have as 60% = 0.6 70% = 0.7 90% = 0.9
20*0.9 + 0.6*x = (20+x)0.7= 14 +0.7x
18 + 0.6x = 14+0.7x
4 =0.1 x and x=40
2006-12-02 14:13:10 · answer #3 · answered by maussy 7 · 0⤊ 0⤋
I think the answer is 3
2006-12-02 14:13:55 · answer #4 · answered by ? 3 · 0⤊ 1⤋
it can be 40
2006-12-02 14:10:22 · answer #5 · answered by Ogün E 1 · 0⤊ 2⤋
I think 20.. lol i dont do high school math
2006-12-02 14:09:24 · answer #6 · answered by Anonymous · 0⤊ 2⤋
i don't know...like 18?
2006-12-02 14:08:23 · answer #7 · answered by meisastarya 2 · 0⤊ 2⤋