5 × 8 = 40, and 1 + 1 = 2 decimal places → 0.40 = 0.4. Yes, the 0 may go: 0.40 and 0.4 are the same number (40/100 = 4/10). 0.04 or 4.0 would be a miscount of places or a forgotten point.

24 × 35 = 840. Ten times and a hundred times is 1000 times, so divide by 1000 → 0.84. Estimate: 2.4 × 0.35 is about 2 × 0.4 = 0.8 ✓

Diagram: split the 1 m × 1 m square into a 10 × 10 grid of 100 small squares. Colour 6 squares down and 7 across → 6 × 7 = 42 squares. Each is 0.01 m², so 0.42 m² (the same as 6 × 7 ÷ 100).

36 × 25 = 900, 2 places → 9.00 = 9. Estimate: 4 × 2.5 = 10 or so ✓

6 × 15 = 90, and 2 + 2 = 4 decimal places → 0.0090 = 0.009. 0.09 or 0.9 has lost one or two zeros.

A and C (multiplier below 1). B is 12.12 and D is 18, both bigger than 12. B (1.01) is only just above 1, but it still does not make the answer smaller.

Estimate: 5 × 3 = 15, so the point goes to give 14.88. (1.488 or 148.8 are far from 15.)

(25 + 40 + 0 + 35 + 30) ÷ 5 = 130 ÷ 5 = 26 minutes. Watch for 130 ÷ 4 = 32.5, which leaves the 0 day out.

4-test total 6.5 × 4 = 26; needed for 5 tests 7 × 5 = 35 → 35 − 26 = 9 points. (That is within the 10 available.)

(1) 12 + 14 + 13 + 15 + 61 = 115, 115 ÷ 5 = 23 minutes　(2) 54 ÷ 4 = 13.5 minutes　(3) (2). The one 61-minute day drags the average from 13.5 up to 23, and 23 is close to none of the five days. An average is pulled towards one far-out value. A value may be set aside only if you can give a reason.

Diagram: 1 m is 0.45 kg, so 2.8 m is 2.8 times that (as an area: 0.45 by 2.8). 45 × 28 = 1260, and 2 + 1 = 3 places → 1.260 = 1.26 kg. Estimate: 0.5 × 3 = 1.5 or so ✓

(1) 2.4 + 3.1 + 2.3 = 7.8, 7.8 ÷ 3 = 2.6 L　(2) 2.6 × 12.5 = 32.5 km. The other route: total 7.8 × 12.5 = 97.5 km, 97.5 ÷ 3 = 32.5 km. That both routes agree is what "levelling out" means.

82.4 + 91.1 + 68.7 + 88.2 + 79.6 = 410.0, and 410.0 ÷ 5 = 82.0 Mbps

+0.4, +9.1, −13.3, +6.2, −2.4. They add up to 0. What sticks up (0.4 + 9.1 + 6.2 = 15.7) equals what dips down (13.3 + 2.4 = 15.7). An average is "the height after levelling out the bumps".

82.4 + 91.1 + 88.2 + 79.6 = 341.3, 341.3 ÷ 4 = 85.325 → the second decimal is 2, so round down: 85.3 Mbps. It went up (82.0 → 85.3). The value thrown away was slower than the average, so the rest average higher.

The six-run total must be 83 × 6 = 498, and so far it is 410 → 498 − 410 = 88 Mbps.

(1) 205 ÷ 82.0 = 2050 ÷ 820 = 2.5 s　(2) 205 ÷ 68.7 = 2050 ÷ 687 = 2.984… → 3.0 s (the second decimal is 8, so round up). When the divisor is multiplied by 10, was the dividend multiplied by 10 too?

(1) 369 ÷ 4.5 = 3690 ÷ 45 = 82 Mbps (82 Mb each second)　(2) 82 × 1.8 = 147.6 Mb (82 × 18 = 1476, 1 decimal place).

(1) 45 × 8 = 360 Mb　(2) 360 ÷ 90 = 4 s　(3) The MB were never turned into Mb before dividing by Mbps (MB ÷ Mbps does not match units). 45 ÷ 90 = 0.5 s is one eighth of the true 4 s. The M is the same, but b (bit) and B (byte) differ by a factor of 8.

82.0 ÷ 100 = 0.82 times, that is 82%. It is less than 1 (the usual speed is below the advertised "up to").

(1) total 471.0, 471.0 ÷ 6 = 78.5 Mbps　(2) 466.5 ÷ 5 = 93.3 Mbps　(3) (2). 78.5 is a value none of the six runs reached, and the single 4.5 pulled the average down by about 15. The 4.5 run should be measured again, or its cause (a brief drop-out) looked into. If it is thrown away, write the reason first.

(1) Because the counts differ. An average is "total ÷ count", so the five runs of A weigh more in the whole than the four of B. The "average of averages" works only when the counts are equal.　(2) A total 82.0 × 5 = 410, B total 84.5 × 4 = 338, (410 + 338) ÷ 9 = 748 ÷ 9 = 83.11… → 83.1 Mbps. (It sits nearer A's 82.0 than 83.25 does.)

Diagram: top (price) 0 · 270 · ?, bottom (length) 0 · 1.5 · 1. 270 ÷ 1.5 = 2700 ÷ 15 = 180 yen. Estimate: 1 m is shorter than 1.5 m → less than 270 yen ✓

1. Make it 1 kg: 1080 ÷ 1.2 = 10800 ÷ 12 = 900 yen　2. For 0.75 kg: 900 × 0.75 = 675 yen. Estimate: 0.75 kg is less than 1 kg → under 900 yen ✓

7.9 ÷ 5.2 = 79 ÷ 52 = 1.519… → the third decimal is 9, so round up: 1.52 kg

(1) 15.8 ÷ 2.5 = 158 ÷ 25 = 6 remainder 8 → 6 bottles, 0.8 L left (check 2.5 × 6 + 0.8 = 15.8)　(2) 7 bottles. The leftover 0.8 L still needs a bottle, so round up — the meaning of a remainder depends on the situation.

(1) taking B as 1: 2.4 ÷ 1.5 = 24 ÷ 15 = 1.6 times　(2) taking A as 1: 1.5 ÷ 2.4 = 15 ÷ 24 = 0.625 times. What is taken as 1 decides the calculation. (2) is below 1 because B is shorter than A.

□ × 1.1 = 1.32 → □ = 1.32 ÷ 1.1 = 13.2 ÷ 11 = 1.2 m. Check 1.2 × 1.1 = 1.32 ✓. Last year is 1 and this year is 1.1. Going backwards from "how many times" is also ÷.

(1) 118.4 + 124.6 + 109.2 + 131.8 + 116.0 = 600.0, 600.0 ÷ 5 = 120.0 L　(2) 120 ÷ 1.5 = 1200 ÷ 15 = 80 bottles

Per litre 45 ÷ 3.6 = 450 ÷ 36 = 12.5 km → per day 2.4 × 12.5 = 30 km → 5 days 30 × 5 = 150 km. The other route: petrol for 5 days 2.4 × 5 = 12 L, 12 × 12.5 = 150 km.

A: 94 ÷ 0.4 = 940 ÷ 4 = 235 yen　B: 84 ÷ 0.35 = 8400 ÷ 35 = 240 yen → A is cheaper, by 5 yen per kg. Compare per kg, not the price on the label.

Weight of 1 m: 1.8 ÷ 0.75 = 180 ÷ 75 = 2.4 kg → 6 ÷ 2.4 = 60 ÷ 24 = 2.5 m. Check 2.4 × 2.5 = 6 ✓. The weight of 1 m is the "unit price"; weight ÷ unit price = length.

Most children write 50 (= (40 + 60) ÷ 2). Whatever is written, do not mark it right or wrong. The aim is to have a guess on paper; the next questions test it.

(1) 120 ÷ 40 = 3 hours　(2) 120 ÷ 60 = 2 hours (the boxes in the figure: 3 and 2)

distance 120 + 120 = 240 km, time 3 + 2 = 5 hours, 240 ÷ 5 = 48 km/h. Not 50.

The time spent at the slow 40 km/h (3 hours) is longer than at the fast 60 km/h (2 hours). The average is pulled towards the slow speed that lasted longer, so it is 48, below the halfway 50. (40 + 60) ÷ 2 works only when the times are equal (question 5). Mark it right if the explanation uses "time".

Out 40 × 2 = 80 km, back 60 × 2 = 120 km → 200 km ÷ 4 hours = 50 km/h. This time it matches (40 + 60) ÷ 2 = 50. Question 3 has the same distances; question 5 has the same times.

(1) climb 4.5 ÷ 9 = 0.5 hours, descent 4.5 ÷ 18 = 0.25 hours　(2) distance 9 km ÷ 0.75 hours = 12 km/h　(3) the climb took longer, so the slow speed counts for more (the answer comes out below 13.5).

(1) 30 ÷ 6 = 5 m/s, 30 ÷ 4 = 7.5 m/s　(2) 60 ÷ (6 + 4) = 60 ÷ 10 = 6 m/s　(3) It is below the halfway 6.25 (it is 6). The first half took longer (6 s against 4 s), so it sits nearer the slow 5.

The whole trip is 90 × 2 = 180 km. At 40 km/h that takes 180 ÷ 40 = 4.5 hours. The way out took 90 ÷ 30 = 3 hours, so 4.5 − 3 = 1.5 hours are left for the way back. 90 ÷ 1.5 = 60 km/h. Check 180 ÷ (3 + 1.5) = 40 ✓. 30 and 60 give 40, not the halfway 45 — today in one line.

Check against the notebook. Look for: a diagram; an honest circle on the reason; a correct redo; and a check in (d) that uses the answer to get back to the start. Solved the second time means finished.

Diagram: P ●――――――――――――――● Q (whole 2.1 km = 2100 m). Each minute they close 80 + 60 = 140 m. (1) 2100 ÷ 140 = 15 minutes　(2) the one from P walks 80 × 15 = 1200 m (the one from Q walks 60 × 15 = 900 m, and 1200 + 900 = 2100 ✓). Were the km turned into m? For people meeting, you add the two speeds.