rate principal time S interest
investment A rA $8000 4yrs SI A
investment B rB $7000 5yrs SI B
It is known that rA : rB = 2 : 3 and that investment B earned $2000 more interest than investment A. Find the values of rA, rB, SI A and SI B
Investment A:
Rate = r.
Po = $8000.
T = 4 yrs.
Ia = Po*r*T = 8000*r*4 = 32,000r = Int. of Investment A.
Investment B:
Rate = 1.5r.
Po = $7,000.
T = 5 yrs.
Ib = Po*1.5r*T = 7000*1.5r*5 = 52,500r = Int. of investment B.
Ib = Ia+2000
52,500r = 32,000r + 2000.
20,500r = 2000
r = 0.0976 = 9.76% = Int. rate of investment A.
1.5r = 1.5 * 9.76 = 14.64% = Int. rate of investment B.
Pa = 1.0976 * 8000. = Principal after 4 yrs.
Pb = 1.1464 * 7000 = Principal after 5 yrs.
To solve this problem, we need to set up equations based on the given information and then solve for the unknown variables.
Let's start by assigning variables to the unknown values:
- rA: interest rate for investment A
- rB: interest rate for investment B
- SI A: interest earned for investment A
- SI B: interest earned for investment B
From the given information, we know that rA:rB = 2:3. We can write this as an equation:
rA/rB = 2/3 (Equation 1)
We also know that investment B earned $2000 more interest than investment A. This can be written as:
SI B - SI A = $2000 (Equation 2)
Next, let's determine the formulas for calculating the interest earned for each investment. The formula for simple interest is:
Simple Interest = (Principal * Rate * Time) / 100
For investment A, the principal is $8000 and the time is 4 years. So, the formula for SI A becomes:
SI A = (8000 * rA * 4) / 100
Similarly, for investment B, the principal is $7000 and the time is 5 years. So, the formula for SI B becomes:
SI B = (7000 * rB * 5) / 100
Now, we have two equations relating the interest rates and two equations to calculate the interest earned for each investment. We can solve these equations simultaneously to find the values of rA, rB, SI A, and SI B.
To eliminate the fractions, we can multiply both sides of Equation 1 by 3rA and both sides of Equation 2 by 100:
3rA/rB = 2 (Equation 3)
100SI B - 100SI A = 200000 (Equation 4)
Now, let's substitute the formulas for SI A and SI B into Equation 4:
100[(7000 * rB * 5) / 100] - 100[(8000 * rA * 4) / 100] = 200000
Simplifying Equation 4:
35000rB - 32000rA = 200000 (Equation 5)
Next, let's solve for rA in terms of rB from Equation 3 and substitute it into Equation 5:
3rA = 2rB
rA = (2/3)rB
Substituting rA = (2/3)rB into Equation 5:
35000rB - 32000[(2/3)rB] = 200000
Simplifying Equation 5:
35000rB - (64000/3)rB = 200000
Combining like terms:
(105000/3)rB - (64000/3)rB = 200000
Simplifying further:
(41000/3)rB = 200000
Solving for rB:
rB = (200000 * 3) / 41000
rB ≈ 14.634
Substituting the value of rB back into Equation 3 to solve for rA:
3rA = 2rB
3rA = 2 * 14.634
3rA ≈ 29.268
rA ≈ 9.756
Now, we can substitute the values of rA and rB into the formulas for SI A and SI B to find their respective values:
For SI A:
SI A = (8000 * 9.756 * 4) / 100
SI A ≈ $3122.08
For SI B:
SI B = (7000 * 14.634 * 5) / 100
SI B ≈ $5112.69
Therefore, the values of rA, rB, SI A, and SI B are approximately:
rA ≈ 9.756
rB ≈ 14.634
SI A ≈ $3122.08
SI B ≈ $5112.69