Pregunta
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ACTIVITY 25 A man was injured in an accident and receives an income from a disability grant of R48 000 in the \( 1^{1 \text { t }} \) year. The grant increases by a fixed amount each year. me ON 1. What is the annual increase, if over 20 years, he would have received R 1435000 in total from the grant? 2. His initial annual expenditure is R26 000 and increases at a rate of R4000 per year. 3. In which year will his expenses exceed his income?

Ask by Ward Boone. in South Africa
Feb 19,2025

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1. The annual increase in the disability grant is R2500. 2. The man's expenses will exceed his income in the 16th year.

Solución

To solve the problem, we will break it down into three parts based on the questions asked. ### Part 1: Finding the Annual Increase 1. **Known Conditions:** - Initial income in the first year: \( I_1 = 48000 \) - Total income over 20 years: \( T = 1435000 \) - Let the annual increase be \( x \). 2. **Income Calculation:** The income for each year can be expressed as: - Year 1: \( I_1 = 48000 \) - Year 2: \( I_2 = 48000 + x \) - Year 3: \( I_3 = 48000 + 2x \) - ... - Year 20: \( I_{20} = 48000 + 19x \) The total income over 20 years can be expressed as: \[ T = I_1 + I_2 + I_3 + \ldots + I_{20} \] This can be simplified to: \[ T = 48000 \times 20 + (0 + 1 + 2 + \ldots + 19)x \] The sum of the first \( n \) integers is given by \( \frac{n(n + 1)}{2} \). For \( n = 19 \): \[ \text{Sum} = \frac{19 \times 20}{2} = 190 \] Therefore, the total income becomes: \[ T = 48000 \times 20 + 190x \] Setting this equal to the total income: \[ 1435000 = 48000 \times 20 + 190x \] 3. **Calculating \( x \):** Now we can calculate \( x \): \[ 1435000 = 960000 + 190x \] \[ 1435000 - 960000 = 190x \] \[ 475000 = 190x \] \[ x = \frac{475000}{190} \] Now, let's calculate \( x \). Calculate the value by following steps: - step0: Calculate: \(\frac{475000}{190}\) - step1: Reduce the numbers: \(\frac{2500}{1}\) - step2: Calculate: \(2500\) The annual increase \( x \) is \( 2500 \). ### Part 2: Finding When Expenses Exceed Income 1. **Known Conditions:** - Initial annual expenditure: \( E_1 = 26000 \) - Annual increase in expenditure: \( y = 4000 \) - Income in the first year: \( I_1 = 48000 \) - Annual increase in income: \( x = 2500 \) 2. **Expenditure Calculation:** The expenditure for each year can be expressed as: - Year 1: \( E_1 = 26000 \) - Year 2: \( E_2 = 26000 + 4000 \) - Year 3: \( E_3 = 26000 + 2 \times 4000 \) - ... - Year \( n \): \( E_n = 26000 + (n-1) \times 4000 \) 3. **Income Calculation:** The income for each year can be expressed as: - Year 1: \( I_1 = 48000 \) - Year 2: \( I_2 = 48000 + 2500 \) - Year 3: \( I_3 = 48000 + 2 \times 2500 \) - ... - Year \( n \): \( I_n = 48000 + (n-1) \times 2500 \) 4. **Finding the Year When Expenses Exceed Income:** We need to find the smallest \( n \) such that: \[ E_n > I_n \] This translates to: \[ 26000 + (n-1) \times 4000 > 48000 + (n-1) \times 2500 \] Rearranging gives: \[ 26000 + 4000n - 4000 > 48000 + 2500n - 2500 \] \[ 22000 + 4000n > 45500 + 2500n \] \[ 4000n - 2500n > 45500 - 22000 \] \[ 1500n > 23500 \] \[ n > \frac{23500}{1500} \] Now, let's calculate \( n \). Calculate the value by following steps: - step0: Calculate: \(\frac{23500}{1500}\) - step1: Reduce the fraction: \(\frac{47}{3}\) Calculating \( n \) gives us: \[ n > \frac{23500}{1500} \approx 15.67 \] This means that the expenses will exceed the income in the 16th year. ### Summary of Results: 1. The annual increase in the disability grant is \( R2500 \). 2. The expenses will exceed the income in the **16th year**.

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Bonus Knowledge

To find the annual increase in the disability grant, we can denote it as \( x \). Over 20 years, his income would be the sum of an arithmetic series starting at R48,000, increasing by \( x \) each year. The formula for the total income over 20 years is: \[ \text{Total Income} = \text{Number of years} \times \text{Average of first and last term} \] Calculating the last term in the sequence, we find: \[ \text{Last term} = 48000 + 19x \] So the total income equation becomes: \[ 20 \times \left(\frac{48000 + (48000 + 19x)}{2}\right) = 1435000 \] This simplifies to: \[ 20 \times (48000 + 9.5x) = 1435000 \] From this, you can solve for \( x \) (the annual increase) and reach the amount he needs incremented each year. Next, let’s think about when his expenses will exceed his income. His expenditures start at R26,000 and increase by R4,000 each year. We need to identify the year \( n \) when: \[ \text{Expenditures} = 26000 + 4000(n - 1) \] \[ \text{Income} = 48000 + (n - 1)x \] Setting these equal to find \( n \): \[ 26000 + 4000(n - 1) > 48000 + (n - 1)x \] This equation can be solved for \( n \) to see when expenses surpass the income. Plugging your calculated \( x \) into the equation will give you the specific year. Happy calculating!

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