two investments were made totaling $15,000. for a certain year these investments yielded $1432 in simple interest. part of the $15,000 was invested at 9% and part at 10%. find the amount invested at each rate.
a total of $1150 was invested, part of it at 12% and part at 11%. the total yield was $133.75. How much was invested at each rate?
first one:
let the amount invested at 9% be x
let the amount invested at 10% be 15000-x
.09x + .10(15000-x) = 1432
multiply each term by 100
9x + 10(15000-x) = 143200
-x = -6800
x = 6800
so $6800 were invested at 9% and $8200 at 10%
do the second question the same way.
90+10(15000-x)=1432000
To solve this problem, we'll use a system of equations.
Let's define:
x = amount invested at 9%
y = amount invested at 10%
According to the information given, the total amount invested is $15,000, so:
x + y = 15,000 ---> Equation 1
The total interest earned is $1,432. Since the interest for each investment is calculated independently, we can write:
0.09x + 0.10y = 1,432 ---> Equation 2
Now, let's solve the system of equations.
From Equation 1, we can express y as:
y = 15,000 - x
Substituting this value of y into Equation 2, we have:
0.09x + 0.10(15,000 - x) = 1,432
0.09x + 1,500 - 0.10x = 1,432
-0.01x = 1,432 - 1,500
-0.01x = -68
Now, let's solve for x by dividing both sides of the equation by -0.01:
x = -68 / (-0.01)
x = 6,800
Substituting the value of x back into Equation 1 to find y:
6,800 + y = 15,000
y = 15,000 - 6,800
y = 8,200
Therefore, $6,800 was invested at a rate of 9%, and $8,200 was invested at a rate of 10%.
-------------------------------------------------
Now let's solve the second problem.
Let's define:
a = amount invested at 12%
b = amount invested at 11%
According to the information given, the total amount invested is $1,150, so:
a + b = 1,150 ---> Equation 3
The total interest earned is $133.75. Using similar logic as before, we have:
0.12a + 0.11b = 133.75 ---> Equation 4
To solve this system of equations, we can use the substitution method.
From Equation 3, we can express a as:
a = 1,150 - b
Substituting this value of a into Equation 4, we have:
0.12(1,150 - b) + 0.11b = 133.75
138 - 0.12b + 0.11b = 133.75
0.11b - 0.12b = 133.75 - 138
-0.01b = -4.25
Now, let's solve for b by dividing both sides of the equation by -0.01:
b = -4.25 / (-0.01)
b = 425
Substituting the value of b back into Equation 3 to find a:
a + 425 = 1,150
a = 1,150 - 425
a = 725
Therefore, $725 was invested at a rate of 12%, and $425 was invested at a rate of 11%.
To solve this problem, we can set up a system of equations.
Let's use the following variables:
Let x be the amount invested at 9%.
Let y be the amount invested at 10%.
For the first problem:
x + y = 15000 (since the total investment is $15000)
0.09x + 0.10y = 1432 (since the total interest yielded is $1432)
For the second problem:
x + y = 1150 (since the total investment is $1150)
0.12x + 0.11y = 133.75 (since the total interest yielded is $133.75)
We can use any method to solve this system of equations. In this explanation, we will use the substitution method.
From the first equation, we have:
x = 15000 - y
Now substitute this value of x into the second equation:
0.09(15000 - y) + 0.10y = 1432
Multiply 0.09 by 15000 and distribute the 0.09:
1350 - 0.09y + 0.10y = 1432
Combine like terms:
0.01y = 82
Divide both sides by 0.01:
y = 8200
Now substitute this value of y back into the first equation:
x = 15000 - 8200
x = 6800
Therefore, $6800 was invested at 9% and $8200 was invested at 10%.
For the second problem, we can use a similar approach.
From the first equation, we have:
x = 1150 - y
Now substitute this value of x into the second equation:
0.12(1150 - y) + 0.11y = 133.75
Multiply 0.12 by 1150 and distribute the 0.12:
138 - 0.12y + 0.11y = 133.75
Combine like terms:
-0.01y = -4.25
Divide both sides by -0.01:
y = 425
Now substitute this value of y back into the first equation:
x = 1150 - 425
x = 725
Therefore, $725 was invested at 12% and $425 was invested at 11%.