Solve and coach, then show and review
Pairs Check works inside a group of four, split into two pairs. In each pair, one partner solves a problem aloud (the doer) while the other coaches and checks (the coach); then they swap. After every two problems the two pairs compare answers. A Gallery Walk then has each group display one worked solution on the wall, while the class circulates to review and leave feedback.
They pass the PIES test:
The coach is responsible for the doer's success, and the gallery depends on every group's poster.
Each learner solves while being coached, and may be asked in review.
Partners alternate doer and coach, and every group displays.
All pairs work, and then all groups walk, at once.
Three things to prepare
- Print the problem set (below), one per learner or per pair.
- Give each group a sheet of chart paper (or a large whiteboard) and a few sticky notes for the Gallery Walk.
- Arrange groups of four as two pairs. A group of three works as a pair plus a rotating coach.
About 24 minutes, in two phases
In each pair, decide who is the doer and who is the coach for Problem 1.
The doer works the problem aloud, drawing the large triangle or splitting the area; the coach watches, prompts with questions, and checks the answer and the unit.
Change roles for the next problem, so each partner both solves and coaches.
The two pairs compare answers. If they agree, move on; if not, they reconcile before continuing.
Coaching stems for the coach
Each group writes a clear worked solution to its assigned problem on chart paper, showing the graph, the triangle or the split, and the calculation with units.
Groups rotate clockwise; at each poster they leave one star (a strength) and one question (something to clarify) on sticky notes.
Groups return to their own poster, read the feedback, and agree one improvement.
The teacher's role during the activity
During Pairs Check, circulate and make sure the coach is coaching, not taking over the pen, and listen for the key decision: gradient for an acceleration, area for a distance. During the Gallery Walk, read the sticky notes and choose one or two points to discuss with the whole class.
When the room does not behave like the plan
The coach solves instead of coaching: remind them their job is to question and check, not to answer.
A pair races ahead without explaining: ask the doer to talk through the next step aloud.
A poster is unclear: that is exactly what the gallery questions are for; use them in the review.
Uneven group: a three works as a pair plus a coach who rotates each problem.
- Support: start with the constant-speed and single-triangle problems, and give a part-split area to complete.
- Challenge: assign the three-shape area and ask the group to produce the model solution for the gallery.
Take turns: the doer solves aloud, the coach checks
Give a unit with every answer.
A cyclist speeds up from rest to 12 m/s in 6 s. Find the acceleration.
A car slows from 30 m/s to 18 m/s in 4 s. Find the deceleration.
A runner moves at a constant 8 m/s for 15 s. Use the area to find the distance travelled.
A speed-time graph shows a rise from rest to 20 m/s in 5 s, then a constant 20 m/s for 10 s. Find the total distance travelled.
A train speeds up from 10 m/s to 34 m/s in 8 s. Find the acceleration.
A speed-time graph shows a rise from rest to 15 m/s in 3 s, a constant 15 m/s for 7 s, then a fall to rest in 5 s. Find the total distance travelled.
Gallery task
Your group will be given one problem. On the chart paper, produce a clear model solution: sketch the speed-time graph, show the large triangle or the way you split the area, and write the calculation with units. Then walk the gallery and leave a star and a question on each poster.
Acceleration is a gradient; distance is an area
| Problem | Worked answer |
|---|---|
| Problem 1 | a = Δv ÷ Δt = 12 ÷ 6 = 2 m/s² |
| Problem 2 | a = (18 − 30) ÷ 4 = −3 m/s², so the deceleration is 3 m/s² |
| Problem 3 | distance = area = 8 × 15 = 120 m |
| Problem 4 | triangle ½ × 5 × 20 = 50 m; rectangle 10 × 20 = 200 m; total = 250 m |
| Problem 5 | a = (34 − 10) ÷ 8 = 24 ÷ 8 = 3 m/s² |
| Problem 6 | ½ × 3 × 15 = 22.5 m; 7 × 15 = 105 m; ½ × 5 × 15 = 37.5 m; total = 165 m |