I invest a sum of money at 10% annual simple rate and twice that amount at 12% annual simple interest.My total yearly in come from both investment was $2890,how much was invested at each rate?
Also What is the speed of each train if one train travels 8 mph faster than the other.In 1.5 hrs the trains are 162 miles apart.
To solve the first problem, let's represent the amount invested at 10% by 'x' and the amount invested at 12% by '2x'.
The formula to calculate simple interest is: Interest = (Principal x Rate x Time).
For the first investment at 10% annual simple interest:
Interest1 = (x x 0.10 x 1), as 1 represents 1 year.
For the second investment at 12% annual simple interest:
Interest2 = ((2x) x 0.12 x 1).
The total income from both investments is given as $2890, so we can write the equation:
Interest1 + Interest2 = 2890,
(x x 0.10 x 1) + ((2x) x 0.12 x 1) = 2890.
Now, we can simplify and solve the equation:
0.10x + 0.24x = 2890,
0.34x = 2890,
x = 2890 / 0.34,
x ≈ $8500.
Therefore, the amount invested at 10% is approximately $8500, and twice that amount, i.e., the amount invested at 12%, is 2x ≈ 2 x $8500 ≈ $17000.
Now let's move on to the second problem, where two trains are traveling at different speeds. Let's assume the speed of the slower train is 'x' mph, and the speed of the faster train is 'x + 8' mph.
In 1.5 hours, the slower train covers a distance of 1.5x miles, and the faster train covers a distance of 1.5(x + 8) miles. Since the total distance covered by both trains is given as 162 miles, we can write the equation:
1.5x + 1.5(x + 8) = 162.
Now, we can simplify and solve the equation:
1.5x + 1.5x + 12 = 162,
3x + 12 = 162,
3x = 162 - 12,
3x = 150,
x = 150 / 3,
x = 50.
Therefore, the speed of the slower train is 50 mph, and the speed of the faster train is x + 8 = 50 + 8 = 58 mph.