Answer :
Let's break down and solve each part of this problem step-by-step.
### Part (a)
Given the vectors:
[tex]\[ p = \begin{pmatrix} m + 3 \\ 2 - n \end{pmatrix} \][/tex]
[tex]\[ q = \begin{pmatrix} 3m - 1 \\ n - 8 \end{pmatrix} \][/tex]
and knowing that [tex]\( p = q \)[/tex], we can set the corresponding components equal to each other:
[tex]\[ \begin{pmatrix} m + 3 \\ 2 - n \end{pmatrix} = \begin{pmatrix} 3m - 1 \\ n - 8 \end{pmatrix} \][/tex]
This gives us two scalar equations:
1. [tex]\( m + 3 = 3m - 1 \)[/tex]
2. [tex]\( 2 - n = n - 8 \)[/tex]
Let's solve these equations one by one.
Equation 1:
[tex]\[ m + 3 = 3m - 1 \][/tex]
Subtract [tex]\( m \)[/tex] from both sides:
[tex]\[ 3 = 2m - 1 \][/tex]
Add 1 to both sides:
[tex]\[ 4 = 2m \][/tex]
Divide by 2:
[tex]\[ m = 2 \][/tex]
Equation 2:
[tex]\[ 2 - n = n - 8 \][/tex]
Add [tex]\( n \)[/tex] to both sides:
[tex]\[ 2 = 2n - 8 \][/tex]
Add 8 to both sides:
[tex]\[ 10 = 2n \][/tex]
Divide by 2:
[tex]\[ n = 5 \][/tex]
So, the values are:
[tex]\[ m = 2 \][/tex]
[tex]\[ n = 5 \][/tex]
### Part (b)
A man shared an amount of money between his children Baaba and William in the ratio 6:5. Baaba received GH 1,200.00.
(i) Find the total amount shared.
Given that the ratio of the amounts is 6:5, let the total amount shared be [tex]\( T \)[/tex].
Baaba’s share is 6 parts and William’s share is 5 parts, so:
[tex]\[ \text{Baaba’s share} = \frac{6}{11} \times T \][/tex]
We are given that Baaba received GH 1,200.00:
[tex]\[ \frac{6}{11} \times T = 1200 \][/tex]
Solving for [tex]\( T \)[/tex]:
[tex]\[ T = 1200 \times \frac{11}{6} \][/tex]
[tex]\[ T = 2,200.00 \][/tex]
So, the total amount shared is GH 2,200.00.
(ii) William invested his share in an account at the rate of 20% simple interest per annum for 2 years. Find the total amount in his account at the end of the 2 years.
First, we find William’s share of the total amount:
[tex]\[ \text{William’s share} = \frac{5}{11} \times 2,200.00 \][/tex]
[tex]\[ \text{William’s share} = 1,000.00 \][/tex]
Next, we calculate the simple interest earned over 2 years at the rate of 20% per annum.
Using the simple interest formula:
[tex]\[ I = P \times r \times t \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (GH 1,000.00)
- [tex]\( r \)[/tex] is the annual interest rate (20%, or 0.20)
- [tex]\( t \)[/tex] is the time in years (2 years)
[tex]\[ I = 1,000.00 \times 0.20 \times 2 \][/tex]
[tex]\[ I = 400.00 \][/tex]
So, the interest earned is GH 400.00.
The total amount in William’s account at the end of 2 years:
[tex]\[ \text{Total amount} = \text{Principal} + \text{Interest} \][/tex]
[tex]\[ \text{Total amount} = 1,000.00 + 400.00 \][/tex]
[tex]\[ \text{Total amount} = 1,400.00 \][/tex]
Thus, at the end of 2 years, the total amount in William's account is GH 1,400.00.
### Part (a)
Given the vectors:
[tex]\[ p = \begin{pmatrix} m + 3 \\ 2 - n \end{pmatrix} \][/tex]
[tex]\[ q = \begin{pmatrix} 3m - 1 \\ n - 8 \end{pmatrix} \][/tex]
and knowing that [tex]\( p = q \)[/tex], we can set the corresponding components equal to each other:
[tex]\[ \begin{pmatrix} m + 3 \\ 2 - n \end{pmatrix} = \begin{pmatrix} 3m - 1 \\ n - 8 \end{pmatrix} \][/tex]
This gives us two scalar equations:
1. [tex]\( m + 3 = 3m - 1 \)[/tex]
2. [tex]\( 2 - n = n - 8 \)[/tex]
Let's solve these equations one by one.
Equation 1:
[tex]\[ m + 3 = 3m - 1 \][/tex]
Subtract [tex]\( m \)[/tex] from both sides:
[tex]\[ 3 = 2m - 1 \][/tex]
Add 1 to both sides:
[tex]\[ 4 = 2m \][/tex]
Divide by 2:
[tex]\[ m = 2 \][/tex]
Equation 2:
[tex]\[ 2 - n = n - 8 \][/tex]
Add [tex]\( n \)[/tex] to both sides:
[tex]\[ 2 = 2n - 8 \][/tex]
Add 8 to both sides:
[tex]\[ 10 = 2n \][/tex]
Divide by 2:
[tex]\[ n = 5 \][/tex]
So, the values are:
[tex]\[ m = 2 \][/tex]
[tex]\[ n = 5 \][/tex]
### Part (b)
A man shared an amount of money between his children Baaba and William in the ratio 6:5. Baaba received GH 1,200.00.
(i) Find the total amount shared.
Given that the ratio of the amounts is 6:5, let the total amount shared be [tex]\( T \)[/tex].
Baaba’s share is 6 parts and William’s share is 5 parts, so:
[tex]\[ \text{Baaba’s share} = \frac{6}{11} \times T \][/tex]
We are given that Baaba received GH 1,200.00:
[tex]\[ \frac{6}{11} \times T = 1200 \][/tex]
Solving for [tex]\( T \)[/tex]:
[tex]\[ T = 1200 \times \frac{11}{6} \][/tex]
[tex]\[ T = 2,200.00 \][/tex]
So, the total amount shared is GH 2,200.00.
(ii) William invested his share in an account at the rate of 20% simple interest per annum for 2 years. Find the total amount in his account at the end of the 2 years.
First, we find William’s share of the total amount:
[tex]\[ \text{William’s share} = \frac{5}{11} \times 2,200.00 \][/tex]
[tex]\[ \text{William’s share} = 1,000.00 \][/tex]
Next, we calculate the simple interest earned over 2 years at the rate of 20% per annum.
Using the simple interest formula:
[tex]\[ I = P \times r \times t \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (GH 1,000.00)
- [tex]\( r \)[/tex] is the annual interest rate (20%, or 0.20)
- [tex]\( t \)[/tex] is the time in years (2 years)
[tex]\[ I = 1,000.00 \times 0.20 \times 2 \][/tex]
[tex]\[ I = 400.00 \][/tex]
So, the interest earned is GH 400.00.
The total amount in William’s account at the end of 2 years:
[tex]\[ \text{Total amount} = \text{Principal} + \text{Interest} \][/tex]
[tex]\[ \text{Total amount} = 1,000.00 + 400.00 \][/tex]
[tex]\[ \text{Total amount} = 1,400.00 \][/tex]
Thus, at the end of 2 years, the total amount in William's account is GH 1,400.00.
