Use equations to solve problems involving percentages, discounts, and saltwater concentration. It takes some practice to convert “○%” and “○ tenths off” into mathematical expressions, but these problems can be solved by applying standard patterns.
Expressing Percentages
- x % is x/100. For example, 30% = 30/100 = 0.3
- x tenths is x/10. For example, 2 tenths = 2/10 = 0.2
- x% of a is a × x/100 ＝ ax/100
- Increasing a by x% gives a × (1 + x/100). Example: a increased by 20% ＝ 1.2a
- Decreasing a by x% gives a × (1 − x/100). Example: a discounted by 20% ＝ 0.8a
Example ①
An item was bought at a 20% discount from its list price, and the price paid was 1200 yen. Find the list price.
Solution:
Let the list price be x yen. The discounted price = 0.8x yen.
Equality: 0.8x = 1200 → x = 1500.
The list price is 1500 yen.
An item was bought at a 20% discount from its list price, and the price paid was 1200 yen. Find the list price.
Solution:
Let the list price be x yen. The discounted price = 0.8x yen.
Equality: 0.8x = 1200 → x = 1500.
The list price is 1500 yen.
Problems Involving Increases and Decreases
This type includes “increased by ○% from last year” and “the number of people decreased by ○%.”
・This year ＝ Last year × (1 ＋ rate of change) (when it increases)
・This year ＝ Last year × (1 − rate of change) (when it decreases)
Example ②
In a class, there are 5 more boys than girls, and there are 35 students in total. Find the numbers of boys and girls.
Solution:
If the number of girls is x, the number of boys is (x + 5).
Equality: x + (x + 5) = 35 → 2x = 30 → x = 15.
There are 15 girls and 20 boys.
In a class, there are 5 more boys than girls, and there are 35 students in total. Find the numbers of boys and girls.
Solution:
If the number of girls is x, the number of boys is (x + 5).
Equality: x + (x + 5) = 35 → 2x = 30 → x = 15.
There are 15 girls and 20 boys.
Saltwater Concentration
Equations can also be used in saltwater problems.
Mass of salt ＝ Mass of saltwater × concentration / 100
The important basis for the equality is that “the total mass of salt remains unchanged before and after mixing.”
Example ③
You want to add water to 200 g of 5% saltwater to make 4% saltwater. How many grams of water should you add?
Solution:
Suppose x g of water is added.
Mass of salt: in the original saltwater = 200 × 5/100 = 10 g (the amount of salt does not change when water is added)
Saltwater after adding water: (200 + x) g, concentration 4 %
Equality: (200 + x) × 4/100 = 10
4(200 + x) = 1000 → 800 + 4x = 1000 → 4x = 200 → x = 50 (g)
You want to add water to 200 g of 5% saltwater to make 4% saltwater. How many grams of water should you add?
Solution:
Suppose x g of water is added.
Mass of salt: in the original saltwater = 200 × 5/100 = 10 g (the amount of salt does not change when water is added)
Saltwater after adding water: (200 + x) g, concentration 4 %
Equality: (200 + x) × 4/100 = 10
4(200 + x) = 1000 → 800 + 4x = 1000 → 4x = 200 → x = 50 (g)
Key Points
- Convert ○% to ○/100 and ○ tenths to ○/10.
- ○% increase = ×(1+○/100); ○% decrease = ×(1−○/100).
- Mass of salt = mass of saltwater × concentration/100.
- The mass of salt remains unchanged before and after mixing (when only water is added).
Practice Problems
- An item with a list price of 2500 yen was sold for 2000 yen after a discount of x yen. Find the discount x.
- This year, a middle school has 462 students, which is 5% more than last year. Find the number of students last year.
- How many grams of water must be added to 300 g of 8% saltwater to make 6% saltwater?
Answers and Explanations
- Answer: 500 yen
Explanation: 2500 − x = 2000 → x = 500. - Answer: 440 students
Explanation: Let last year’s number be x. x × 1.05 = 462 → x = 462 ÷ 1.05 = 440. - Answer: 100 g
Explanation: Salt = 300×8/100 = 24 g. (300+x)×6/100 = 24 → 6(300+x) = 2400 → 300+x = 400 → x = 100.