pls. mathematicians help me out with these problem and explain in the best and simplest form you can?

Question

i have been battling how to sovle this problem all night long. but i just can break it down. i need you guys help. pls..

1) How many liters of a 17% acid solution should be mixed with 8 liters of an 11% acid solution to
obtain a mixture that is 15%?

2) We have invested a total of $ 5000 in two bank accounts. One account earns 7% interest per year,
the other earns 8% per year. How much did we invest at 8% if the combined interest from the
two account was $ 382?

cut that off. no stupid ambiguios answers

Answer 1

Set up 1) like this:

Let L be the # of liters of 17%
.17(L)+.11(8)=.15(8+L) I'll multiply thru to get rid of those pesky
decomal points.

17L-15L=4(8)
2L=32
L=16

2) B is the dollar amount in the 8% bank so
5000-B is the dollar amount in the 7% bank
.08(B)+.07(5000-B)=382 and multiply thru by 100

B=-35000+38200
B=3200

Check:
.08(3200) +.07(1800)=256+126= 382 OK

Answer 2

1) How many liters of a 17% acid solution should be mixed with 8 liters of an 11% acid solution to
obtain a mixture that is 15%?

x=number of liters added
.15*(8+x) acid in final solution
8*.11=.88 l acid in 11% soln
.17x l acid in soln you are adding
.17x+.88=.15(8+x)
.17x+.88=1.2+.15x
.02x=.32
x=16 l


2) We have invested a total of $ 5000 in two bank accounts. One account earns 7% interest per year,
the other earns 8% per year. How much did we invest at 8% if the combined interest from the
two account was $ 382?


.08x+.07(5000-x)=382
.08x-.07x+350=382
.01x=32
x=3500 at 8%
1500 at 7%

Answer 3

(1) How many liters of a 17% acid solution should be mixed with 8 liters of an 11% solution to obtain a mixture that is 15%?

Let x = the number of liters of 17% acid solution to be used.

A) Number of Liters of the Solution
B) Percent Pure Acid
C) Number of Liters of Pure Acid

~17% Solution~
A) x
B) 17%
C) 0.17x

~11% Solution ~
A) 8
B) 11%
C) (0.11)(8)

~Final 15% Solution ~
A) x + 8
B) 15%
C) 0.15(x + 8)

Now we use our (C) entries to form an equation:
0.17x + (0.11)(8) = 0.15(x + 8) --- Multiply...
0.17x + 0.88 = 0.15x + 1.2 --- Subtract 0.15x from each side...
0.02x + 0.88 = 1.2 --- Subtract 0.88 from each side...
0.02x = 0.32 --- Divide both sides by 0.02...
x = 16

Answer: 16 liters of a 17% acid solution should be mixed with 8 liters of an 11% acid solution to obtain a mixture that is 15%.

(2) We have invested a total of $5,000 in two bank accounts. One account earns 7% interest per year, and the other earns 8% per year. How much did we invest at 8% if the combined interest from the two accounts was $382?

Let x = amount of money invested at 8% interest.
That makes the amount invested at 7% interest (5000 - x).

0.08x + 0.07(5000 - x) = 382 --- Multiply the 0.07 into the parentheses...
0.08x + 350 - 0.07x = 382 --- Combine like terms...
0.01x + 350 = 382 --- Subtract 350 from both sides...
0.01x = 32 --- Divide both sides by 0.01...
x = 3200
5000 - x = 1800

Answer: The amount invested at 8% was $3,200.00.

Answer 4

1) for the first one, we have need x(17acid + 83water)
mixed by 8(11acid + 89water)
so we get 17x + 83x mixed by 88+712 to get a ratio of
15acid/85water

add side-by-side to get

88 + 17x (the acid) / 712 + 83x (the water) = 15/85 = 3/17

so you will get 3(712+83x) = 17(88+17x)
solve and you will end up with 2136+249x = 1496+289x
and then 640 = 40x, so 64=4x , 16.4=4x so x = 16 liters.

2) let the first account be 5000-x dollars. and second will be x.
interest in 1 year would be 7(5000-x) / 100 = 35000-7x / 100

second account will be 8x / 100

add them together to get the total interest

35000 - 7x + 8x / 100 = 35000+x / 100
since this was 382 dollars

35000+x = 38200

x = 3200 dollars.

Answer 5

Answer to Q. No. 1
Let x liters of 17% acid solution be mixed with 8 liters of 8 liters of 11% acid solution.
The total volume of the mixture = (x+8)
This should be 15%
Mathematically
17 x + 11 X 8 = 15(x+8)
Solve this, you get x = 16 liters

Answer to Q. No. 2
Let $ x be ivested in Bank No. 1 and $(5000-x) ivested in Bank No. 2.
Sum of interests = 7 x/100 + (5000-x) 8/100 = 382
Solve, you get x = $1800

Should you need more answers promptly, mail me in
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I am from India

Answer 6

for #2:
You can set up the problem with the following equation:
x*0.08 + (5000-x)*0.07 = 382, where x is the amount invested in the 8% account. Solving for this equation you get.
x*0.08 + 5000*0.07 - x*0.07 = 382
x*0.08 + 350 - x*0.07 = 382
x*(0.08-0.07) = 32
x*0.01 = 32
x = $3,200
You invested $3,200 in the 8% account and $2,800 in the 7% account.

Answer 7

to find it you must break it into parts,

number of liters will be x

.17x (percentage of solution that is 17% acidic)
.11(8) (eight liters of 11% acidic solution)
total is .15x + .15(8) which is number of liters of original variable and eight liters that combine to be 15%

.17x + .11(8)= .15x +.15(8)
.17x + .88 = .15x + 1.2
subtract .88 from left, then right side
subtract .15x from right, then left side
you're then left with
.02x = .32
divide both sides by .02 and you get
x=16

Answer 8

0.17x+8(0.11)=(8+x)(0.15)
17x+88=120+15x
2x=32
x=16

2.0.08x+(5000-x)*0.07=382
8x+35000-7x=38200
x=3200
so he has to invest $3200 at 8% and $1800 at 7%

Answer 9

i suggest "a guide"...im weak in maths too....but i think its -70/17% for that 1st ques...