We are given four problems. Problem 1: A laptop originally costs $1500. The price increases by 10%. What is the new price? Problem 2: A bicycle was originally priced at $600, but is now on sale with a 15% discount. What is the sale price? Problem 3: A company's revenue was $5,000,000 in 2010. Each year, the company's revenue increases by 6%. (a) What is the company's revenue in 2011? (b) What is the company's revenue in 2015? Problem 4: A town had a population of 50,000 people. Due to migration, the population decreased by 7% each year. (a) What is the population 1 year later? (b) In 7 years time, will the population be less than 20,000 people? Show working.

1. Problem Description

We are given four problems.

Problem 1: A laptop originally costs $

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5

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0. The price increases by 10%. What is the new price?

Problem 2: A bicycle was originally priced at $600, but is now on sale with a 15% discount. What is the sale price?

Problem 3: A company's revenue was $5,000,000 in

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0. Each year, the company's revenue increases by 6%.

(a) What is the company's revenue in 2011?

(b) What is the company's revenue in 2015?

Problem 4: A town had a population of 50,000 people. Due to migration, the population decreased by 7% each year.

(a) What is the population 1 year later?

(b) In 7 years time, will the population be less than 20,000 people? Show working.

2. Solution Steps

Problem 1:

The original price is $

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0. The price increases by 10%, so the increase is $1500 * 0.10 = $

1

5

0. The new price is $1500 + $150 = $

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Problem 2:

The original price is $

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0. The discount is 15%, so the discount amount is $600 * 0.15 = $

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0. The sale price is $600 - $90 = $

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Problem 3:

(a) The revenue in 2010 is $5,000,

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0. The revenue increases by 6% each year.

In 2011, the revenue is $5,000,000 * (1 + 0.06) =$5,000,000 * 1.06 = $5,300,

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(b) The revenue in 2010 is $5,000,

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0. The revenue increases by 6% each year.

We want to find the revenue in 2015, which is 5 years after

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0. We can use the formula:

$Revenue_{2015} = Revenue_{2010} * (1 + growth\,rate)^{number\,of\,years}$

$Revenue_{2015} = 5,000,000 * (1.06)^5$

$Revenue_{2015} = 5,000,000 * 1.3382255776 = 6,691,127.888$

$Revenue_{2015} \approx 6,691,127.89$

Problem 4:

(a) The original population is 50,

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0. The population decreases by 7% each year.

The population 1 year later is $50,000 * (1 - 0.07) = 50,000 * 0.93 = 46,500$.

(b) The original population is 50,

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0. The population decreases by 7% each year.

We want to find the population in 7 years.

$Population_{7\,years} = Population_{original} * (1 - decrease\,rate)^{number\,of\,years}$

$Population_{7\,years} = 50,000 * (0.93)^7$

$Population_{7\,years} = 50,000 * 0.6017060130566441 \approx 30,085.30$

Since the population after 7 years is approximately 30,085.30, it will not be less than 20,000 people.

In 7 years, the population will be around 30,085, which is greater than 20,

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3. Final Answer

1. New price: $1650

2. Sale price: $510

3. (a) Revenue in 2011: $5,300,000

(b) Revenue in 2015: $6,691,127.89

4. (a) Population 1 year later: 46,500

(b) In 7 years time, no, the population will be approximately 30,085.30, which is not less than 20,

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