The problem has two parts: (a) Evaluate the expression $(3.69 \times 10^5) \div (1.64 \times 10^{-3})$ and leave the answer in standard form. (b) A man invested N20,000 in bank A at $y\%$ simple interest per annum and N25,000 in bank B at $1.5y\%$ simple interest per annum. His total interest at the end of the year was N4,600. Find the value of $y$.

1. Problem Description

The problem has two parts:

(a) Evaluate the expression $(3.69 \times 10^5) \div (1.64 \times 10^{-3})$ and leave the answer in standard form.

(b) A man invested N20,000 in bank A at $y\%$ simple interest per annum and N25,000 in bank B at $1.5y\%$ simple interest per annum. His total interest at the end of the year was N4,

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0. Find the value of $y$.

2. Solution Steps

(a) We are asked to evaluate $(3.69 \times 10^5) \div (1.64 \times 10^{-3})$ and leave the answer in standard form.

We can rewrite the expression as:

$\frac{3.69 \times 10^5}{1.64 \times 10^{-3}} = \frac{3.69}{1.64} \times \frac{10^5}{10^{-3}}$

We can simplify the expression by dividing $3.69$ by $1.64$:

$\frac{3.69}{1.64} \approx 2.25$

We can simplify the expression by dividing $10^5$ by $10^{-3}$:

$\frac{10^5}{10^{-3}} = 10^{5 - (-3)} = 10^{5+3} = 10^8$

Therefore, we have:

$2.25 \times 10^8$

(b) The simple interest formula is:

$Interest = Principal \times Rate \times Time$

The interest from bank A is:

$I_A = 20000 \times \frac{y}{100} \times 1 = 200y$

The interest from bank B is:

$I_B = 25000 \times \frac{1.5y}{100} \times 1 = 250 \times 1.5y = 375y$

The total interest is:

$I_A + I_B = 200y + 375y = 575y$

We are given that the total interest is N4,600:

$575y = 4600$

Dividing both sides by 575, we get:

$y = \frac{4600}{575} = 8$

3. Final Answer

(a) $2.25 \times 10^8$

(b) $y = 8$