Omuando borrows KSh 90,000 for 5 years at 6 rac12% simple interest p.a. What amount does he have to pay at…

Question

Omuando borrows KSh 90,000 for 5 years at 6 rac12% simple interest p.a. What amount does he have to pay at the end of that time?

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Here are the solutions to the problems:

10. Omuando borrows KSh 90,000 for 5 years at $6 \frac{1}{2}\%$ simple interest p.a. What amount does he have to pay at the end of that time?

Identify the given values.
Principal ( $P$ ) = KSh 90,000
Time ( $T$ ) = 5 years
Rate ( $R$ ) = $6 \frac{1}{2}\%$ p.a. = $6.5\%$ = $\frac{6.5}{100} = 0.065$
Calculate the simple interest ( $SI$ ).
The formula for simple interest is $SI = P \times R \times T$ .
$SI = 90000 \text{ KSh} \times 0.065 \times 5$ $SI = 90000 \times 0.325$ $SI = 29250 \text{ KSh}$
Calculate the total amount to be paid.
The total amount ( $A$ ) is the principal plus the simple interest.
$A = P + SI$ $A = 90000 \text{ KSh} + 29250 \text{ KSh}$ $A = 119250 \text{ KSh}$
The total amount Omuando has to pay is $\boxed{119250 \text{ KSh}}$ .

11. Solve for $x$ in the equation, $9^{x+1} + 3^{2x} = 270$
Express all terms with the same base.
The base 3 can be used since $9 = 3^2$ .
Rewrite $9^{x+1}$ as $(3^2)^{x+1}$ .
$(3^2)^{x+1} + 3^{2x} = 270$ $3^{2(x+1)} + 3^{2x} = 270$ $3^{2x+2} + 3^{2x} = 270$
Factor out the common term $3^{2x}$ .
$3^{2x} \cdot 3^2 + 3^{2x} = 270$ $3^{2x}(3^2 + 1) = 270$ $3^{2x}(9 + 1) = 270$ $3^{2x}(10) = 270$
Isolate the exponential term.
$3^{2x} = \frac{270}{10}$ $3^{2x} = 27$
Express the right side with the same base as the left side.
Since $27 = 3^3$ :
$3^{2x} = 3^3$
Equate the exponents and solve for $x$ .
$2x = 3$ $x = \frac{3}{2}$
The value of $x$ is $\boxed{\frac{3}{2}}$ .

Omuando borrows KSh 90,000 for 5 years at 6 rac12\% simple interest p.a. What amount does he have to pay at the end of that time?