Try. Think. Explain.
Then try something new.

Write one quarter in eighths. What fraction is left for the green counters?

1. One quarter is 2/8, so red and blue together make 3/8 + 2/8 = 5/8.
2. Green counters therefore make the remaining 3/8. If 3/8 is 15 counters, 1/8 is 5 counters.
3. The whole collection is eight lots of 5: 8 × 5 = 40 counters.

Answer: 40 counters

Check: Of 40 counters, 15 are red and 10 are blue, leaving 15 green. The three counts add to 40.

What I would ask next: Which fraction describes the 15 counters? Can you show the whole collection as eight equal parts?

Keep the same colour fractions, but suppose there are 21 green counters. How many counters are there altogether?

The red count has not changed. Draw two equal red parts in both pictures, then compare the blue parts.

1. The two red parts are unchanged, so a part has the same size before and after the removal.
2. The blue count falls from five parts to three parts. Two parts therefore equal 12 counters, making each part 6 counters.
3. Originally there were seven parts: 7 × 6 = 42 counters, made up of 12 red and 30 blue.

Answer: 42 counters

Check: Removing 12 blue from 12 red and 30 blue leaves 12 red and 18 blue, which simplifies to 2 : 3.

What I would ask next: Why are we allowed to use the same-sized parts in both ratios? What stayed fixed?

Starting again with the original box, add 6 red counters instead of removing any blue. What is the new red-to-blue ratio?

A multiple of 5 ends in 0 or 5. Treat these as two separate cases.

1. If the last digit is 0, the first two digits can be 2 then 5, or 5 then 2. This gives 250 and 520.
2. If the last digit is 5, the other digits are 0 and 2. Only 205 is a three-digit number; 025 is not.
3. The complete list is 205, 250 and 520, so there are three numbers.

Answer: 3 numbers: 205, 250 and 520

Check: Every listed number uses all three digits once, has a nonzero first digit and ends in 0 or 5. Both possible final digits have been considered.

What I would ask next: How can you organise the possibilities so that you know you have not missed or counted one twice?

Of these multiples of 5, how many are greater than 300?

Sketch the card. Compare the lengths removed from the outside with the two new edges made by the cut.

1. The original perimeter is 2 × (8 + 6) = 28 cm.
2. The corner cut removes one 3 cm section and one 2 cm section of the original boundary.
3. It creates two new boundary sections of exactly the same lengths: 3 cm and 2 cm. The amount removed equals the amount added.
4. The perimeter therefore stays 28 cm.

Answer: 28 cm

Check: Taking the cut at the top right, with the 3 cm cut running along the 8 cm edge, the six boundary lengths are 8, 4, 3, 2, 5 and 6 cm. Their sum is 28 cm.

What I would ask next: The area has decreased. Does that mean the perimeter must decrease too? Which edges have actually changed?

Instead, cut a rectangular notch 3 cm wide and 2 cm deep from the middle of an 8 cm edge, without reaching either corner. What is the new perimeter?

List the minutes after 09:00 at which each bell rings. Find the first positive number appearing in both lists.

1. Bell A rings after 6, 12, 18, 24, 30, … minutes.
2. Bell B rings after 8, 16, 24, 32, … minutes.
3. The smallest positive common multiple is 24, so the next shared ringing is 24 minutes after 09:00.

Answer: 09:24

Check: Twenty-four minutes contains four intervals of 6 minutes and three intervals of 8 minutes. No earlier positive multiple is shared.

What I would ask next: Why are we looking for a common multiple, and why must it be the smallest positive one?

How many times do the bells ring together strictly after 09:00 and strictly before 10:00?

Imagine replacing every notebook with a pen. How much would the total price fall?

1. Replacing the three notebooks with three pens reduces the total by £3, because each notebook costs £1 more.
2. Five pens would therefore cost £11 − £3 = £8.
3. One pen costs £8 ÷ 5 = £1.60. A notebook costs £2.60.

Answer: £1.60 per pen

Check: Three notebooks cost 3 × £2.60 = £7.80 and two pens cost £3.20. Together they cost £11.

What I would ask next: Can you replace the different items with equal-cost units before dividing?

At the same prices, what do two notebooks and three pens cost?

Try pairs of digits adding to 11, or write the original as 10a + b and the reverse as 10b + a.

1. Let the tens digit be a and the units digit be b. Then a + b = 11.
2. The difference between the original and reversed numbers is (10a + b) − (10b + a) = 9(a − b).
3. Since the difference is 27, a − b = 3. Together with a + b = 11, this gives a = 7 and b = 4.
4. The original number is 74.

Answer: 74

Check: The digits add to 7 + 4 = 11, and 74 − 47 = 27.

What I would ask next: Why does reversing two digits always change the number by a multiple of 9?

Keep the digit sum at 11, but make the reversed number 45 smaller. What is the original number now?

Subtract 2 from the number. What must the result be divisible by?

1. If the number is n, then n − 2 must be divisible by both 3 and 4. It must therefore be a multiple of 12.
2. The positive possibilities for n are 2, 14, 26, 38, 50, 62, … .
3. The first one divisible by 5 is 50.

Answer: 50

Check: 50 = 3 × 16 + 2 and 50 = 4 × 12 + 2. It is also 5 × 10. All smaller candidates in the complete sequence have been checked.

What I would ask next: How can a remainder condition become an ordinary divisibility condition? How do we justify that the answer is the smallest?

Replace “multiple of 5” with “multiple of 7”, keeping the two remainder conditions. What is the smallest positive answer?

A rectangle is fixed when you choose its left and right grid lines, then its top and bottom grid lines.

1. There are four vertical grid lines. Choosing two gives six possibilities: three pairs next to each other, two pairs with one line between them, and one outer pair.
2. There are three horizontal grid lines. Choosing two gives three possibilities.
3. Each vertical pair can be combined with each horizontal pair, giving 6 × 3 = 18 rectangles.

Answer: 18 rectangles

Check: Counting by size gives 6 one-cell, 4 two-cells-wide, 2 three-cells-wide, 3 two-cells-tall, 2 rectangles of size 2 by 2, and 1 full-grid rectangle: 6 + 4 + 2 + 3 + 2 + 1 = 18.

What I would ask next: What choices define exactly one rectangle? Why does this method avoid double-counting?

How many rectangles are there in a grid with 3 columns and 3 rows?

First find the interior angle next to the 124° exterior angle. Then use the equal base angles.

1. Angles on a straight line add to 180°, so that base angle is 180° − 124° = 56°.
2. The other base angle is also 56° because the triangle is isosceles.
3. The three interior angles add to 180°, so the remaining vertex angle is 180° − 56° − 56° = 68°.

Answer: 68°

Check: The interior angles are 56°, 56° and 68°, which sum to 180°. The stated exterior angle is 180° − 56° = 124°.

What I would ask next: Which angle is outside the triangle, and which two angles are equal? Mark those facts before calculating.

Suppose instead that the 124° exterior angle is adjacent to the vertex angle between the equal sides. What is each base angle?

Temporarily label the counters R1, R2, R3, B1 and B2. Count unordered pairs of individual counters, not just colour names.

1. There are 5 × 4 ÷ 2 = 10 unordered pairs of counters. Dividing by 2 avoids counting a pair in both orders.
2. The red-red pairs are R1R2, R1R3 and R2R3: three pairs. There is one blue-blue pair, B1B2.
3. Four of the ten equally likely pairs have the same colour, giving 4/10 = 2/5.

Answer: 2/5

Check: Sequentially, the probability is (3/5 × 2/4) + (2/5 × 1/4) = 6/20 + 2/20 = 8/20 = 2/5.

What I would ask next: Why are “red-red”, “red-blue” and “blue-blue” not three equally likely outcomes?

What is the probability of drawing one red and one blue counter?

List selections in increasing order and organise them by their smallest number.

1. If the smallest number is 1, the other two must add to 9. The possible pairs are 3 and 6, or 4 and 5.
2. If the smallest number is 2, the other two must add to 8 and both exceed 2. Only 3 and 5 work.
3. If the smallest number is at least 3, the smallest possible sum is 3 + 4 + 5 = 12, which is already too large.
4. There are three selections: {1, 3, 6}, {1, 4, 5} and {2, 3, 5}.

Answer: 3 selections: {1, 3, 6}, {1, 4, 5}, {2, 3, 5}

Check: Each listed selection uses three different permitted numbers and sums to 10. The smallest-number cases cover every possible selection exactly once.

What I would ask next: What rule for listing the numbers would stop 1, 3, 6 and 6, 1, 3 being counted as different answers?

Using the same numbers and rules, how many selections have a sum of 11?

Square the given sum carefully. What middle term appears?

1. Squaring gives (x + 1/x)² = x² + 2x(1/x) + 1/x².
2. Since x is nonzero, x(1/x) = 1. Therefore 9 = x² + 2 + 1/x².
3. Subtract 2 to obtain x² + 1/x² = 7.

Answer: 7

Check: The expansion contains a middle term of 2. Omitting it would incorrectly produce 9.

What I would ask next: Which expression are we asked for? Can we reach it directly rather than doing unnecessary algebra to find x?

Under the same condition, find x³ + 1/x³.

Multiply through by 6ab, then try adding the same number to both sides to create a factorisation.

1. Multiplying by 6ab gives 6a + 6b = ab, so ab − 6a − 6b = 0.
2. Add 36 to both sides: (a − 6)(b − 6) = 36.
3. Both a and b must exceed 6: each positive fraction 1/a and 1/b is smaller than their sum 1/6. Thus both factors are positive.
4. The unordered positive factor pairs of 36 are (1, 36), (2, 18), (3, 12), (4, 9) and (6, 6).
5. Adding 6 to each factor gives (7, 42), (8, 24), (9, 18), (10, 15) and (12, 12).

Answer: 5 pairs: (7, 42), (8, 24), (9, 18), (10, 15), (12, 12)

Check: For example, 1/10 + 1/15 = 3/30 + 2/30 = 1/6. The factor-pair list exhausts all positive possibilities without counting reversed pairs twice.

What I would ask next: Why can we exclude zero and negative factors? How does a factorisation help prove that the list is complete?

How many of the unordered solutions have a and b different?

Count all routes first. Then count routes passing through the forbidden point and subtract them.

1. Every route uses four right moves and three up moves, in seven positions. Choosing the three positions for the up moves gives 7 × 6 × 5 ÷ (3 × 2 × 1) = 35 routes.
2. To reach (2, 1), a route needs two right moves and one up move. The up move can be in any of three positions: 3 routes.
3. From (2, 1) to (4, 3), it needs two right moves and two up moves. There are 4 × 3 ÷ 2 = 6 orders.
4. There are 3 × 6 = 18 routes through the forbidden point. Each route through it has exactly one first part and one second part.
5. The number avoiding it is 35 − 18 = 17.

Answer: 17 routes

Check: A second method assigns each allowed grid point the sum of the counts immediately below and to its left, starting with 1 at (0, 0) and 0 at the blocked point. The destination count is 17.

What I would ask next: Why can the two parts of a route through the junction be counted independently and multiplied?

How many of the allowed-direction routes pass through (2, 1) instead of avoiding it?

Call the square’s side s. The small triangle above the square is similar to the whole triangle.

1. Let the square’s side be s cm. The small triangle above it has horizontal width s and vertical height 6 − s.
2. It is similar to the original triangle, whose horizontal width is 8 and vertical height is 6. Corresponding height-to-width ratios give (6 − s)/s = 6/8.
3. Cross-multiply: 8(6 − s) = 6s, so 48 − 8s = 6s.
4. Thus 14s = 48 and s = 24/7 cm.

Answer: 24/7 cm (3 3/7 cm)

Check: The hypotenuse joins (8, 0) to (0, 6), so its points satisfy x/8 + y/6 = 1. At the square’s opposite vertex x = y = 24/7, giving 3/7 + 4/7 = 1.

What I would ask next: Which small triangle is similar to the large one? Can you match horizontal lengths with horizontal lengths before forming a ratio?

If the perpendicular sides become 12 cm horizontally and 9 cm vertically, what is the new square’s side length?

The mass and braking force stay fixed. How does the energy change when speed changes from 6 to 10 m/s?

1. Stopping requires the braking work to equal the initial kinetic energy: Fd = ½mv².
2. With m and F fixed, d is proportional to v². The speed ratio is 10/6 = 5/3, so the distance ratio is (5/3)² = 25/9.
3. The new distance is 9 × 25/9 = 25 m.

Answer: 25 m

Check: The speed is multiplied by about 1.67, but the braking distance is multiplied by about 2.78. Substitution into d/v² gives 9/36 = 25/100 = 1/4.

What I would ask next: Which quantities remain fixed? Why does the distance scale with the square of speed rather than speed itself?

Return to the original speed of 6 m/s, but double the cart’s mass while keeping the same braking force. What is the braking distance?

In series, how is the 12 V shared between identical resistors? In parallel, what voltage does each resistor receive?

1. In series, the identical resistors share the voltage equally: 6 V across each. The current through each, and through the supply, is I = 6/6 = 1 A.
2. Arrangement A therefore takes P = 12 × 1 = 12 W from the supply.
3. In parallel, each resistor receives 12 V, so each branch current is I = 12/6 = 2 A. The total supply current is 2 + 2 = 4 A.
4. Arrangement B therefore takes P = 12 × 4 = 48 W. This is four times the series power.

Answer: A: 12 W; B: 48 W

Check: In series, each resistor dissipates 6 V × 1 A = 6 W, giving 12 W total. In parallel, each dissipates 12 V × 2 A = 24 W, giving 48 W total.

What I would ask next: Does the supply current equal one branch current or the sum of them? What changes when the wiring changes?

Use an ideal 6 V supply instead, with the same resistors. What are the total powers in A and B?