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Maths 9 โ€“ Practical problems

๐Ÿ“˜ Notes

Word Problems: Discounts, Percentage Increase, Speed, Distance and Direct Proportion

In these questions, the main skill is to recognise the type of problem first, choose the correct method, and work step by step.

This topic includes discounts, percentage increase, speed-distance-time, and direct proportion.

1. How to recognise the type of problem

If the question says What to do
discount of \(10\%\), \(20\%\), \(30\%\) Find the discount amount, then subtract it from the original price
price increased by \(20\%\), \(30\%\) Find the increase, then add it to the original price
travelled for 2 hours at 12 km/h Use \(s=vt\)
for 8 people needs 400 g, for 20 people? Use direct proportion
there and back along the same road First find the distance, then find the return speed

2. Main formulae

Discount

\[ \text{discount}=\frac{p}{100}\times \text{original price} \] \[ \text{new price}=\text{original price}-\text{discount} \]

Percentage increase

\[ \text{increase}=\frac{p}{100}\times \text{original price} \] \[ \text{new price}=\text{original price}+\text{increase} \]

Speed, distance and time

\[ s=vt \] \[ v=\frac{s}{t} \] \[ t=\frac{s}{v} \]

Direct proportion

When more people, more metres, or more area need more materials, use:

\[ \text{new amount}= \frac{\text{new value}}{\text{old value}} \times \text{old amount} \]

3. Worked examples

Example 1: Discount

A phone costs 7000000 sum. During a sale, Sardor buys it with a discount of \(10\%\). How much does he pay?

Solution

\[ 10\% \text{ of } 7000000 = \frac{10}{100}\times 7000000 = 700000 \] \[ 7000000-700000=6300000 \]
Answer: \(6300000\) sum.

Example 2: Distance in parts

Amir travels for \(2\) hours on a road at \(12\) km/h and for \(0.5\) hours on a mountain road at \(8\) km/h. What distance does he travel?

Solution

Road:

\[ s=vt=12\times 2=24 \]

Mountain road:

\[ s=8\times 0.5=4 \]

Total distance:

\[ 24+4=28 \]
Answer: \(28\) km.

Example 3: Direct proportion

To make compote for 8 people, Feruza needs 400 g of dried fruit. How much does she need for 20 people?

Solution

\[ \text{new amount}=\frac{20}{8}\times 400 \] \[ =2.5\times 400=1000 \]
Answer: \(1000\) g.

Example 4: Percentage increase

In September, 1 kg of oranges cost 20000 sum. In October, the price increased by \(20\%\). What is the new price?

Solution

\[ 20\% \text{ of } 20000 = \frac{20}{100}\times 20000=4000 \] \[ 20000+4000=24000 \]
Answer: \(24000\) sum.

Example 5: Return speed

A taxi driver travelled from Tashkent to Samarkand for \(3\) hours at \(80\) km/h. On the way back along the same road, he travelled for \(4\) hours. What was his average speed on the way back?

Solution

First find the distance going there:

\[ s=vt=80\times 3=240 \]

The return distance is the same:

\[ v=\frac{s}{t}=\frac{240}{4}=60 \]
Answer: \(60\) km/h.

Example 6: Direct proportion by area

To lay 40 m\(^2\) of tiles, a team needs 8 bags of glue. How many bags are needed for 100 m\(^2\)?

Solution

\[ \text{new amount}=\frac{100}{40}\times 8 \] \[ =2.5\times 8=20 \]
Answer: \(20\) bags.

4. Quick methods

If there is a discount

  1. Find the percentage of the original price.
  2. Subtract it from the original price.

If the price increased

  1. Find the increase.
  2. Add it to the original price.

If the journey has several parts

  1. Find each distance separately using \(s=vt\).
  2. Add the distances together.

If the same rate is used for different amounts

  1. Find how many times bigger or smaller the new value is.
  2. Multiply the old amount by this factor.

5. Common mistakes

  • Adding the percentage in a discount question instead of subtracting it.
  • Subtracting the percentage in an increase question instead of adding it.
  • Forgetting that half an hour is \(0.5\) hours.
  • In return journey questions, forgetting to find the distance first.
  • Using the wrong ratio in direct proportion questions.
  • Forgetting the units: sum, km, g, l.

6. Summary

\[ s=vt \] \[ \text{new price with discount}=\text{old price}-\text{discount} \] \[ \text{new price after increase}=\text{old price}+\text{increase} \] \[ \text{new amount}= \frac{\text{new value}}{\text{old value}} \times \text{old amount} \]

The main idea in these questions is to understand the situation, choose the correct formula, and work carefully step by step.