Can anyone solve these two problems?

Question

1. Two sums of money totaling $30,000 earn respectively 8% and 12% interest per year. Together they earn $3,120. How much was invested at 12%?

2. Rose made two investment totaling $25,000. On one investment she made a 12% profit, but on the other she took an 18% loss. If her net loss was $1,200, how much was the investment that gained a 12% profit?

Answer 1

1. Average interest is 10% there by giving you 15k per piece. so taking 10% from the total u ll get 3000 which is 120 short and can be amounted to 4%. so the 4% difference in the two intrests is for a principle of 3000. So the 12% one is 18,000.

2. 11K same as above.

Answer 2

Assuming these problems are based on simple interest as opposed to the compounded interest, here’s one way how you might solve them:

1.Let’s assume that x amount of money was invested at 8% annually and y sum was invested at 12%.

Then you have two equations:

0,08x + 0,12y = 3.120
Andx + y = 30.000

Solving these equations you substitute y though x,
i.e. y = 30.000 – x

0,08x + 0,12*(30.000 – x) = 3.120;
0,08x + 3600 – 0,12x = 3.120;
480 = 0,04x;
x = 12.000.

Thus, you have found the amount of money invested at 8% annually, i.e. 12.000.

Answer: 30.000 – 12.000 = 18.000 was invested at 12% annually.

PS Check:
12.000 x 0,08 + 18.000 x 0,12 = 960 + 2.160 = 3.120

2.Same idea, really.

Let’s assume that Rose’s portion of investment making her 12% on a year was x amount of money and that y part of her investment made a loss of 18% on a year.

Then you can write down 2 following equations:
0,12x + (– 0,18y) = -1.200
Andx + y = 25.000

Now all you have to do is solve the equations by substituting y through x, i.e. x = 25.000 – y

0,12x – 0,18*(25.000 – y) = -1.200;
0,12x – 4.500 + 0,18y = -1.200;
0,3x = 4.500 – 1.200;
0,3x = 3.300;
x = 11.000.

Answer: Rose invested 11.000 at 12% profit.

PS Check
11.000 x 0,12 + 14.000 x(-0,18) = 1.320 – 2.520 = -1.200
or a 1.200 loss on her total investment.

Answer 3

1)
let X = amount invested @ 12%
30000- x = amount invested @ 8%

x*.12 + (30000-x)*.08 = 3120
.12x + 2400 - .08x = 3120
.04x = 720
x = 18000
therefore $18,000 was invested @ 12%

2)
let X = amount invested @ 12%
25000- x = amount invested @ 18%

.12x - (25000-x)*.18 = -1200
.12x - 4500 +.18x = -1200
.3x = 3300
x = 11000
Rose invested $11,000 inthe investment that gained a 12% profit.

Answer 4

1) Assume you invested x for 8% and y for 12%
Now, 0.08x + 0.12y = 3120
x + y = 30000
Solving the simultaneous equation you get,
x =12000 and y = 18000
2) 12 x - 18y=-1200
x+y = 25000
Solving for x and y
x=11000 and y=14000
Check:
0.12x11000+(-0.18x14000)=-1200