Model Test – Solved ഉത്തരസൂചിക
The same 19 questions as the model paper, in the same order, each with a full solution, a marking hint and the page of this guide to revisit. Questions 9–19 have step-by-step animations: press Next ▶.
−x + y = y − x: −9 + 4 = 4 − 9 = −(9 − 4) = −5. On the line: start at −9 and move 4 to the right.
Review: Page 5 · Positions and addition
x − y = −(y − x) when x < y: 3 − 10 = −(10 − 3) = −7.
Review: Page 2 · Class 8 rules
x − (−y) = x + y: 6 − (−4) = 6 + 4 = 10. (How much must be added to −4 to reach 6? 4 to reach 0, then 6 more.)
(−x)(−y) = xy: the product of two negative numbers is positive, 3 × 7 = 21.
Review: Page 9 · Multiplication
6 × (−4) = −24, so (−24) ÷ 6 = −4. In general (−x)/y = −(x/y).
Review: Page 9 · Division
−(−x) = 0 − (−x) = 0 + x = x: the negative of a negative number is the number itself.
Review: Page 7 · −(−x) = x
Each pair of (−1)s multiplies to 1. Nine is odd, so one (−1) is left over: (−1)odd = −1.
Review: Page 9 · Side box: oscillation
Final = start + change = −3 + 5 = 5 − 3 = 2.
Review: Page 5 · Moves to the right
| x | y | x + y |
|---|---|---|
| 7 | −10 | |
| −7 | 10 | |
| −7 | −10 | |
| −4 | 4 |
7 + (−10) = 7 − 10 = −3; −7 + 10 = 10 − 7 = 3; (−7) + (−10) = −(7 + 10) = −17; −4 + 4 = 0.
Marks: ½ mark for each correct entry.
Review: Page 5 · Page 6 · Practice set 1, Q1
From x = −6 to y = 5: y − x = 5 − (−6) = 5 + 6 = 11 (11 to the right). From x = 5 to y = −6: y − x = −6 − 5 = −(6 + 5) = −11 (11 to the left).
Marks: 1 for each change with the correct sign.
Review: Page 7 · Change of position
A subtraction a − b asks "how much must be added to b to make a?" (10 − 6 = 4 because 6 + 4 = 10). So 4 − (−7) asks how much to add to −7 to reach 4: 7 to reach 0, then 4 more, 11 in all. Hence 4 − (−7) = 4 + 7 = 11.
Marks: 1 for the "how much to add" meaning, 1 for the two steps 7 + 4 = 11.
Review: Page 7 · Meaning of 4 − (−7)
(−2)(−3) = 6; 6 × (−4) = −24; (−24) × (−5) = 120. Three negative factors give a negative product, four give a positive one.
Marks: 1 for each product.
Review: Page 10 · Practice set 3, Q6
x = −3: (−3)² + 3 × (−3) + 2 = 9 − 9 + 2 = 2. x = −4: (−4)² + 3 × (−4) + 2 = 16 − 12 + 2 = 6. (Check: y = (x + 1)(x + 2) gives (−2)(−1) = 2 and (−3)(−2) = 6.)
Marks: 1 for each value (½ for the correct square, ½ for the rest).
Review: Page 10 · Practice set 3, Q3
Left: (12 + (−5)) × (−3) = 7 × (−3) = −21. Right: 12 × (−3) + (−5)(−3) = −36 + (−5)(−3). For the two sides to be equal, −36 + (−5)(−3) = −21, so (−5)(−3) must be 15 (−36 + 15 = −21). Keeping the distributive law forces "negative × negative = positive".
Marks: 1 for the left side −21, 1 for −36 + (−5)(−3), 1 for concluding 15.
Review: Page 9 · Why (−10)(−2) = 20
x + y = 2 with y negative means x − (size of y) = 2: x must be 2 more than the size of y, e.g. 5 + (−3) = 2, 10 + (−8) = 2. For x + y = −2 with x negative, y must be 2 less than the size of x: −5 + 3 = −2, −10 + 8 = −2. Many other pairs work.
Marks: ¾ for each correct pair (any correct pairs accepted).
Review: Page 6 · Practice set 1, Q2
(i) 6/(−3) = −2 and (−3)/6 = −½, so z = −2 − (−½) = −2 + ½ = −1½ = −3/2. (ii) (−6)/(−3) = 2 and (−3)/(−6) = ½, so z = 2 − ½ = 3/2.
Marks: 1 for the two divisions with signs, 1 for each z (½ each part may be split).
Review: Page 10 · Practice set 4, Q3
Right speed +8, left speed −8; time after the moment positive, before negative. (i) 8 × 3 = 24. (ii) 8 × (−3) = −24 (it had not reached O yet, so it was 24 m to the left). (iii) (−8) × (−3) = 24 (moving left, 3 s earlier it was 24 m to the right). One equation: s = vt.
Marks: ½ for each position, 1½ for s = vt with the sign convention explained.
Review: Page 9 · Time and speed
- For four consecutive natural numbers a, b, c, d, show using algebra that a − b − c + d = 0.
- What is a + b − c − d?
- Check both with 10, 11, 12, 13 and with −1, −2, −3, −4.
- Find all pairs of integers (x, y) with x² + y² = 25.
(i) b = a + 1, c = a + 2, d = a + 3: a − (a + 1) − (a + 2) + (a + 3) = a − a − 1 − a − 2 + a + 3 = 0.
(ii) a + (a + 1) − (a + 2) − (a + 3) = −4.
(iii) 10 − 11 − 12 + 13 = 0 and 10 + 11 − 12 − 13 = −4. For −1, −2, −3, −4 (each one LESS than the one before): −1 + 2 + 3 − 4 = 0 and −1 − 2 + 3 + 4 = 4.
(iv) Squares up to 25: 0, 1, 4, 9, 16, 25; only 0 + 25 and 9 + 16 make 25. With signs: (±5, 0), (0, ±5), (±3, ±4), (±4, ±3) – 12 pairs.
Marks: 1½ for (i), 1 for (ii), 1 for (iii), 1½ for (iv) (½ for 3, 4 and 0, 5; 1 for including all signs).
Review: Page 8 · Practice set 2, Q3 · Page 10 · Practice set 3, Q5
(i) 29.4 − 9.8 = 19.6 m/s; 29.4 − 19.6 = 9.8 m/s. (ii) 29.4 ÷ 9.8 = 3 seconds. (iii) v = 29.4 − 49 = −19.6: the ball is falling at 19.6 m/s.
(iv) Without signs we need two equations: rising v = 29.4 − 9.8t (t < 3), falling v = 9.8t − 29.4 (t > 3). With downward speed negative, the single equation v = 29.4 − 9.8t covers the whole flight.
(v) n = 7: (−1)⁷ = −1, so ½(1 − 1) = 0. n = 10: (−1)¹⁰ = 1, so ½(1 + 1) = 1.
Marks: 1 for (i), ½ for (ii), 1 for (iii) with meaning, 1½ for (iv), 1 for (v).
Review: Page 9 · Side boxes: oscillation and the thrown ball
Marks summary
| Section | Questions | Marks each | Total |
|---|---|---|---|
| A | 1–8 | 1 | 8 |
| B | 9–13 | 2 | 10 |
| C | 14–17 | 3 | 12 |
| D | 18–19 | 5 | 10 |
| Total | 40 | ||
- Section A (10 min): one line each; picture the number line for every sign.
- Section B (20 min): show each step – write the rule you used (x − (−y) = x + y and so on).
- Section C (25 min): keep brackets around negative numbers to avoid sign slips.
- Section D (25 min): write the algebra neatly; for 18(iv) list squares up to 25 first, then add signs.
- Last 10 min: check every sign once more – most lost marks are sign errors.
If you scored 32 or more, you are ready. Otherwise revisit the pages under each question you missed, then try the scored quiz again. Back to the home page.