The ATEDO Method

Have you ever seen a question with bicycles and cars, or adults and children, and felt stuck because there are two unknown numbers? You do not need to guess both numbers at once. ATEDO lets you pretend first, then repair the pretend total one item at a time.

Parents: ask your child to predict the opposite count, then let them adjust one replacement before reading the explanation together.

By the end you will be able to recognise an ATEDO problem, keep the item count fixed, find the big and small gaps, and explain why the final division gives the opposite type.

Main question

A car park has 30 vehicles. Every vehicle is either a bicycle with 2 wheels or a car with 4 wheels. There are 86 wheels altogether. How many bicycles and cars are there?

A — Assume: keep every position

ATEDO fits when you know the total number of items, each item is one of two types, and each type contributes a fixed amount. A car park with 30 vehicles fits: every position is either a bicycle or a car, and each has a known number of wheels.

The first move is Assume. Choose one type and imagine that every position has that type. The number of positions must stay fixed. You are changing labels in your imagination, not adding vehicles.

T — Total: count the pretend world

Now calculate the Total under the assumption. A bicycle has 2 wheels, so 30 assumed bicycles would have:

Working

$$30 \times 2 = 60 \text{ wheels}$$

This is a hypothetical total. It is useful even though the real total is different, because it gives us a clean starting line.

E — Excess: measure the big gap

The question says there are 86 wheels altogether. Our pretend total of 60 is short. Subtract the two totals:

Working

$$86 - 60 = 26 \text{ wheels}$$

ATEDO calls this positive gap the Excess. The word does not mean the assumption must be too high. Here it is a shortage of 26; if your assumption had produced too many wheels, it would be an excess of 26. Use the positive size of the gap and name its direction.

D — Difference: test one swap

Do not divide by the number of wheels on a car. One position already had 2 bicycle wheels in the assumption. Replace one assumed bicycle with one car and compare:

Working

$$4 - 2 = 2 \text{ wheels per swap}$$

That is the Difference per swap, the small gap. The vehicle count stays 30, but the wheel total rises by 2 each time. This is the heart of ATEDO: every replacement repairs the same small amount.

🔬

Replace one assumed vehicle at a time

🤔 Predict first: How many cars are needed to move from 60 wheels to 86 wheels?

13 cars17 cars26 cars

Working

$$60 + (13 \times 2) = 86 \text{ wheels}$$

The grid still has 30 positions. Thirteen bicycles have changed into cars; no thirty-first vehicle appeared.

O — Opposite: divide the big gap by the small gap

Now the final letter is ready. The 26-wheel shortage needs 2-wheel repairs:

Working

$$26 \div 2 = 13 \text{ car replacements}$$

The Opposite count is 13. We assumed bicycles, so 13 of those assumed bicycles are actually cars. The remaining vehicles are:

Working

$$30 - 13 = 17 \text{ bicycles}$$

There are 17 bicycles and 13 cars. Check both facts.

Working

$$17 + 13 = 30 \text{ vehicles}$$

$$17 \times 2 + 13 \times 4 = 34 + 52 = 86 \text{ wheels}$$

When ATEDO fits — and when to pause

ATEDO is a good choice when there are a known number of items, exactly two types, and a fixed contribution for each type. If the question gives pairs or groups, first convert the group contribution to one-person or one-item contribution. In a fruit question, two children sharing 2 apples means each child represents 1 apple.

The two contributions must be different. If both types contribute 3 each, replacing one with the other changes nothing, so no division can identify the types. Also check that the gap is divisible by the difference; a leftover part of a swap means the given numbers do not describe a whole-number solution.

A mistake worth catching

In the car-park example, dividing the big gap by four is not the right working. The four-wheel car does not add four new wheels to a position: that position already contributed two wheels as a bicycle. The replacement adds only the difference shown in the working panel.

Another common slip is to call 26 the number of cars. It is the wheel gap. Each car replacement repairs two wheels, so the quotient gives the replacement count.

Graduated practice

Practice 1 — identify the opposite

Seven seats are made from 3-legged stools and 4-legged chairs. There are 25 legs altogether. How many chairs are there?

Show solution

Working

$$7 \times 3 = 21 \text{ legs}$$

$$25 - 21 = 4 \text{ legs short}$$

$$4 - 3 = 1 \text{ leg per swap}$$

$$4 \div 1 = 4 \text{ chairs}$$

$$7 - 4 = 3 \text{ stools}$$

$$4 \times 4 + 3 \times 3 = 25 \text{ legs}$$

Practice 2 — the assumption gives too much

A shop sold 28 tickets. Adult tickets cost $12 and child tickets cost $7. The shop collected $256. How many adult tickets were sold?

Show solution

Working

$$28 \times \$12 = \$336 \text{ assumed total}$$

$$\$336 - \$256 = \$80 \text{ excess}$$

$$\$12 - \$7 = \$5 \text{ removed per swap}$$

$$\$80 \div \$5 = 16 \text{ child tickets}$$

$$28 - 16 = 12 \text{ adult tickets}$$

$$12 \times \$12 + 16 \times \$7 = \$256$$

Practice 3 — convert a group rate first

Eighteen people share fruit. Each adult receives 3 apples, while every 2 children share 2 apples. Altogether, 38 apples are given out. How many children are there?

Show solution

Working

$$2\text{ children share }2\text{ apples}$$

$$1\text{ child}=2\div2=1\text{ apple}$$

$$18 \times 3 = 54 \text{ assumed apples}$$

$$54 - 38 = 16 \text{ apples excess}$$

$$3 - 1 = 2 \text{ apples per swap}$$

$$16 \div 2 = 8 \text{ children}$$

$$18 - 8 = 10 \text{ adults}$$

$$10 \times 3 + 8 \times 1 = 38 \text{ apples}$$

Watch out — easily mixed up

Quick recap

🎯 Mastery check

Answer all 8 — your progress is saved on this device.

1. In ATEDO, what must stay fixed when you make the assumption?

   The number of item positionsThe actual contribution totalThe number of opposite items

2. Thirty assumed bicycles have 2 wheels each. What is the assumed Total?

   32 wheels60 wheels86 wheels

3. The assumed total is 60 and the actual total is 86. What is the positive Excess gap?

   21326

4. Why is the Difference in the bicycle and car problem 4 − 2 instead of 4?

   A bicycle already contributed 2 wheels in the assumptionCars have only 2 wheelsThe car park has 4 vehicles

5. What does 26 ÷ 2 = 13 count in the main example?

   Wheels in one carCar replacements, so the opposite carsAll vehicles in the car park

6. Seven seats have 3-legged stools and 4-legged chairs, with 25 legs. How many chairs are there?

   347

7. A shop sold 28 adult or child tickets. Adult tickets cost $12, child tickets $7, and revenue is $256. Which assumption makes the first total larger?

   Assume all tickets are adultsAssume all tickets are childrenNeither assumption can make a total

8. Two children share 2 apples. In an ATEDO model for 18 people, what contribution should one child represent?

   2 apples1 apple$\frac{1}{2}$ a person

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