Question 7/20: Set 3 - Try Out - Mathematical Reasoning (SNBT)

:00

Mrs. Laras owns a cake shop. Because the cakes sold are very delicious and nutritious, many new buyers arrive every day to buy cakes. So her business is getting more popular, and her profits are increasing with a constant increment value. If the total profit up to the $4$ th day is $\text{Rp}700,000.00$ and the total profit up to the $10$ th day is $\text{Rp}2,200,000.00$ .

Then the total profit up to the $20$ th day is...

$\text{Rp}4,800,000.00$

$\text{Rp}5,200,000.00$

$\text{Rp}5,900,000.00$

$\text{Rp}6,300,000.00$

$\text{Rp}7,200,000.00$

Explanation

Forming Equations

The profit increases by a constant amount, so this is an arithmetic progression. The total profit up to the $n$ -th day is denoted as $S_n$ .

Given:

Total profit up to the $4$ th day ( $S_4$ ) is $\text{Rp}700,000.00$ .
Total profit up to the $10$ th day ( $S_{10}$ ) is $\text{Rp}2,200,000.00$ .

Formula for the sum of $n$ terms:

S_n = \frac{n}{2}(2a + (n-1)b)

From $S_4 = 700,000$ :

\frac{4}{2}(2a + 3b) = 700,000

2(2a + 3b) = 700,000

2a + 3b = 350,000 \quad \dots (1)

From $S_{10} = 2,200,000$ :

\frac{10}{2}(2a + 9b) = 2,200,000

5(2a + 9b) = 2,200,000

2a + 9b = 440,000 \quad \dots (2)

Elimination and Substitution

Eliminating equations $(1)$ and $(2)$ :

\begin{aligned} 2a + 9b &= 440,000 \\ 2a + 3b &= 350,000 \\ \hline 6b &= 90,000 \\ b &= 15,000 \end{aligned}

Substituting $b = 15,000$ into equation $(1)$ :

2a + 3(15,000) = 350,000

2a + 45,000 = 350,000

2a = 305,000

a = 152,500

Calculating Total Profit for Day $20$

We calculate $S_{20}$ :

S_{20} = \frac{20}{2}(2a + 19b)

S_{20} = 10(305,000 + 19(15,000))

S_{20} = 10(305,000 + 285,000)

S_{20} = 10(590,000)

S_{20} = \text{Rp}5,900,000.00

Thus, the total profit up to the $20$ th day is $\text{Rp}5,900,000.00$ .

Command Palette

Number 7

Explanation