mr. mike invested two sums of money on simple interest. the fist was $2200 invested at 8% per annum and the second $1960 at 10% per annum. the sums were allowed to frow till bothe amounted to the same figure. in how many years did this happaended and what was the final amount?
You want x, where
2200(1+.08x) = 1960(1+.10x)
To find out in how many years the two sums of money will amount to the same figure, we can set up an equation based on the simple interest formula:
Simple Interest = Principal × Rate × Time
Let's denote the time it takes for both sums to amount to the same figure as "t" (in years).
For the first sum:
Principal (P1) = $2200
Rate (R1) = 8% = 0.08 (converted to decimal)
For the second sum:
Principal (P2) = $1960
Rate (R2) = 10% = 0.10 (converted to decimal)
After time t, the first sum will be:
Amount1 = P1 + (P1 × R1 × t)
Similarly, the second sum will be:
Amount2 = P2 + (P2 × R2 × t)
Now, since we want to find the time t when both sums amount to the same figure, we can set up the equation:
Amount1 = Amount2
P1 + (P1 × R1 × t) = P2 + (P2 × R2 × t)
Substituting the given values:
$2200 + ($2200 × 0.08 × t) = $1960 + ($1960 × 0.10 × t)
Now we can solve the equation to find the value of t.
$2200 + (176t) = $1960 + (196t)
$2200 - $1960 = 196t - 176t
$240 = 20t
t = $240 / $20
t = 12 years
Therefore, it takes 12 years for both sums of money to amount to the same figure. To find the final amount, we can substitute t = 12 back into either Amount1 or Amount2:
Amount1 = $2200 + ($2200 × 0.08 × 12)
Amount1 = $2200 + $2112
Amount1 = $4312
So, the final amount will be $4312.