(a) A company requires 3 hours of labor for every $N87.00$ worth of raw materials. Given that the company uses $N30,450.00$ worth of raw materials, what is the labor budget needed if the cost of labor is $N18.25$ per hour? (b) An investor invested $Nx$ at 6% simple interest per annum in bank M and $Ny$ at 8% simple interest per annum in bank N. The total investment is $N8,000,000.00$. The total interest received after 4 years is $N2,320,000.00$. Calculate: (i) The values of $x$ and $y$. (ii) The interest paid by the second bank.

1. Problem Description

(a) A company requires 3 hours of labor for every $N87.00$ worth of raw materials. Given that the company uses $N30,450.00$ worth of raw materials, what is the labor budget needed if the cost of labor is $N18.25$ per hour?

(b) An investor invested $Nx$ at 6% simple interest per annum in bank M and $Ny$ at 8% simple interest per annum in bank N. The total investment is $N8,000,000.00$. The total interest received after 4 years is $N2,320,000.00$. Calculate:

(i) The values of $x$ and $y$.

(ii) The interest paid by the second bank.

2. Solution Steps

(a)

First, determine the number of $N87.00$ units in $N30,450.00$.

Number of units = $\frac{30450}{87} = 350$

Total labor hours required = $350 \times 3 = 1050$ hours

Total labor budget = $1050 \times 18.25 = 19162.5$

(b)

(i)

Let $x$ be the amount invested in bank M and $y$ be the amount invested in bank N.

We are given that the total investment is $N8,000,000.00$. Therefore,

$x + y = 8000000$ (Equation 1)

The simple interest formula is:

$Simple \ Interest = \frac{Principal \times Rate \times Time}{100}$

The interest earned from bank M is $\frac{x \times 6 \times 4}{100} = \frac{24x}{100} = 0.24x$

The interest earned from bank N is $\frac{y \times 8 \times 4}{100} = \frac{32y}{100} = 0.32y$

The total interest earned is $N2,320,000.00$. Therefore,

$0.24x + 0.32y = 2320000$ (Equation 2)

From Equation 1, we have $x = 8000000 - y$. Substituting this into Equation 2:

$0.24(8000000 - y) + 0.32y = 2320000$

$1920000 - 0.24y + 0.32y = 2320000$

$0.08y = 2320000 - 1920000$

$0.08y = 400000$

$y = \frac{400000}{0.08} = 5000000$

Therefore, $y = 5000000$

Now, substitute $y = 5000000$ into Equation 1:

$x + 5000000 = 8000000$

$x = 8000000 - 5000000 = 3000000$

Therefore, $x = 3000000$

(ii)

The interest paid by the second bank (bank N) is $0.32y$.

Interest from bank N = $0.32 \times 5000000 = 1600000$

3. Final Answer

(a) $N19162.50$

(b)

(i) $x = 3000000$, $y = 5000000$

(ii) $N1,600,000.00$