Rate Word Problems
Apply your rate skills to solve real-world problems involving tables and tiered rates
1. Table-Based Rate Problems
Key Idea:
Some problems give you a rate table to look up values. Read the table carefully to find the correct rate.
Example: Postage Charges
The table below shows the amounts a certain country charges for sending letters.
| Mass up to | Charges |
|---|---|
| 20 g | $2.55 |
| 50 g | $2.65 |
| 100 g | $2.85 |
| 250 g | $3.15 |
| 400 g | $5.00 |
Important: "Mass up to 50 g" means the charge is for items with mass greater than 20 g but less than or equal to 50 g.
(a) Devi wants to send a letter with a mass of 400 g. How much does she have to pay?
Look at "Mass up to 400 g" → Charge = $5.00
(b) Andy wants to send a letter that has a mass of 80 g. How much does he have to pay?
80 g is more than 50 g but not more than 100 g
Look at "Mass up to 100 g" → Charge = $2.85
Example: Parcel Charges
The table shows the postal charges for sending parcels to Country A.
| Mass up to | Charges |
|---|---|
| 2 kg | $25 |
| 4 kg | $32 |
| 6 kg | $46 |
| 8 kg | $55 |
Henry needs to send a parcel with a mass of 2 kg 800 g to Country A. How much does Henry need to pay?
2 kg 800 g = 2.8 kg
2.8 kg is more than 2 kg but not more than 4 kg
Look at "Mass up to 4 kg" → Charge = $32
2. Tiered Rate Problems
What are Tiered Rates?
Sometimes rates change after a certain amount. For example, parking charges might be different for the first hour compared to additional hours.
Example: Parking Charges
Parking rates:
First hour: $2.20
Every subsequent ½ hour or part of it: $1.20
Mrs Leng parked her car from 10.00 a.m. to 12.30 p.m. How much did she have to pay?
Charges from 10.00 a.m. to 11.00 a.m. = $2.20 (first hour)
Charges from 11.00 a.m. to 12.30 p.m. = $1.20 × 3 = $3.60 (three ½ hours)
Total charges = $2.20 + $3.60 = $5.80
Example: Water Charges
The table below shows the charge for each cubic metre of water.
| Volume of water | First 40 m³ | After the first 40 m³ |
|---|---|---|
| Water charges | $1.43 per m³ | $1.81 per m³ |
(a) Company A uses 35 m³ of water. How much is Company A charged?
35 m³ is less than 40 m³, so use $1.43 per m³
35 × $1.43 = $50.05
(b) Company B uses 46 m³ of water. How much is Company B charged?
46 m³ is more than 40 m³, so we split it:
First 40 m³: 40 × $1.43 = $57.20
Remaining: 46 − 40 = 6 m³
Additional 6 m³: 6 × $1.81 = $10.86
Total = $57.20 + $10.86 = $68.06
3. Multi-Step Word Problems
Problem-Solving Strategy:
- Read carefully — what information is given?
- Identify — what is the question asking?
- Plan — what steps do you need?
- Calculate — work step by step
- Check — does your answer make sense?
Example: Sending Multiple Parcels
Using the parcel charges table:
| Mass up to | 2 kg | 4 kg | 6 kg | 8 kg |
|---|---|---|---|---|
| Charges | $25 | $32 | $46 | $55 |
Mdm Chan needs to send two different parcels separately to Country A. The masses of the parcels are 3 kg and 5 kg 500 g. How much does she need to pay for the two parcels altogether?
Step 1: Find charge for first parcel (3 kg)
3 kg is more than 2 kg but not more than 4 kg → Charge = $32
Step 2: Find charge for second parcel (5 kg 500 g = 5.5 kg)
5.5 kg is more than 4 kg but not more than 6 kg → Charge = $46
Step 3: Find total
$32 + $46 = $78
Example: Graph Interpretation
The line graph shows the number of jars a machine filled with peanut butter over 10 minutes.
(i) How many jars of peanut butter did the machine fill in 1 minute?
From the graph: 350 jars in 10 minutes
350 ÷ 10 = 35 jars per minute
(ii) At this rate, how many jars can the machine fill in 1 hour?
1 hour = 60 minutes
35 × 60 = 2100 jars
Practice Problems
Practice 1: Table Reading
Using the postage charges table (up to 20 g: $2.55, up to 50 g: $2.65, up to 100 g: $2.85, up to 250 g: $3.15, up to 400 g: $5.00):
How much does it cost to send a letter with a mass of 180 g?
ShowHide solution
180 g is more than 100 g but not more than 250 g
Look at "Mass up to 250 g" → $3.15
Practice 2: Parking Calculation
Parking rates: First hour $2.20, every subsequent ½ hour or part of it $1.20
Dan parked his car from 1.15 p.m. to 4.05 p.m. How much did he have to pay?
ShowHide solution
1.15 p.m. to 2.15 p.m. = 1 hour → $2.20
2.15 p.m. to 4.05 p.m. = 1 h 50 min = 3 half-hours and 20 min. The extra 20 min is charged as another ½ hour, so 4 half-hours → $1.20 × 4 = $4.80
Total = $2.20 + $4.80 = $7.00
Practice 3: Tiered Rates
Water charges: First 40 m³ at $1.43 per m³, after the first 40 m³ at $1.81 per m³
A company uses 55 m³ of water. How much is the company charged?
ShowHide solution
First 40 m³: 40 × $1.43 = $57.20
Remaining: 55 − 40 = 15 m³
Additional 15 m³: 15 × $1.81 = $27.15
Total = $57.20 + $27.15 = $84.35
Practice 4: Multi-Step
A vacuum cleaner cleans a floor area of 10 m² every 4 min. What is the floor area it cleans in 30 min?
ShowHide solution
Method 1: Find rate per minute first
10 m² ÷ 4 = 2.5 m² per minute
2.5 × 30 = 75 m²
Method 2: Find how many times 4 min goes into 30 min
30 ÷ 4 = 7.5 times
10 m² × 7.5 = 75 m²
Key Takeaways
- Table-based problems: Read the table carefully — use the first "up to" mass your item fits into. "Up to 50 g" includes exactly 50 g
- Tiered rates: Different rates apply to different portions — calculate each portion separately, then add
- Multi-step problems: Break down into smaller steps and work systematically
- Check your work: Does your answer make sense in the context of the problem?
- Units matter: Convert units if needed (g to kg, min to hours, etc.)