Class 9 · MathematicsSet 2 · Answer key

Answer Key – Set 2 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 2, for teachers and parents to value the paper. Accept any correct answer in the student’s own words; give method marks even when the final answer is wrong.

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ANSWER KEY & MARKING SCHEME · SET 2
Standard IX · Mathematics · Annual Examination
Total score: 60 · Time: 2 hours
Section AAnswer all questions · 10 × 1 = 10 marks
1.
A box has 80 beads, some black and some white. There are 12 more black beads than white beads. If the number of white beads is x, which equation gives x?
(a) x + 12 = 80(b) x + (x − 12) = 80(c) 2x + 12 = 80(d) 12x = 80
[1]

Step-by-step solution

  1. White beads = x, so black beads = x + 12.
  2. Together: x + (x + 12) = 80, that is 2x + 12 = 80.
  3. (Solving: 2x = 68, x = 34 white and 46 black beads; 34 + 46 = 80 ✓.)
✅ Answer: (c) 2x + 12 = 80

Marking: 1 mark for (c).

📖 Revise: Ch 1 · Mental arithmetic and algebra · Chapter 1 – Pairs of Equations

2.
√279 =
(a) 1⅓(b) 1⅔(c) 2⅓(d) 259
[1]

Step-by-step solution

  1. Write the mixed number as a fraction: 2 7/9 = (18 + 7)/9 = 25/9.
  2. 25 = 5² and 9 = 3², so (5/3)² = 25/9.
  3. Hence √279 = 5/3 = 1⅔. (Check: (1⅔)² = 1 + 2 × ⅔ + 4/9 = 1 + 1⅓ + 4/9 = 2 7/9 ✓.)
✅ Answer: (b) 1⅔

Marking: 1 mark for (b).

📖 Revise: Ch 2 · Square roots of fractions · Chapter 2 – New Numbers

3.
Fill in the blanks:   The side of a square of area 1.69 sq cm is cm. The side of a square of area 3 sq cm is written as cm.
[1]

Step-by-step solution

  1. 1.3² = 1.3 × 1.3 = 1.69, so the side is 1.3 cm.
  2. No fraction has square 3, so the side of a square of area 3 sq cm is written √3 cm (വർഗമൂലം); √3 = 1.73205…
✅ Answer: 1.3 cm; √3 cm

Marking: ½ mark for each blank.

📖 Revise: Ch 2 · Measures and numbers · Chapter 2 – New Numbers

4.
The numbers 1 to 36 are written in rows of 6 (1–6 in the first row, 7–12 in the second, and so on). In any 2 × 2 square of these numbers, the products of the numbers at opposite corners always differ by
(a) 1(b) 5(c) 6(d) 7
[1]

Step-by-step solution

  1. A 2 × 2 square looks like x, x + 1 (top row) and x + 6, x + 7 (next row), since each row has 6 numbers.
  2. x(x + 7) = x² + 7x and (x + 1)(x + 6) = x² + 6x + x + 6 = x² + 7x + 6.
  3. The difference is always 6. (Example: 8, 9, 14, 15: 9 × 14 = 126 and 8 × 15 = 120.)
✅ Answer: (c) 6

Marking: 1 mark for (c).

📖 Revise: Ch 4 · Number curiosities · Chapter 4 – Multiplication Identities

5.
√29 =
(a) √23(b) 23(c) √29(d) 29
[1]

Step-by-step solution

  1. For positive numbers, √xy = √x√y.
  2. So √29 = √2√9 = √23.
✅ Answer: (a) √23

Marking: 1 mark for (a).

📖 Revise: Ch 5 · Division of square roots · Chapter 5 – Irrational Numbers

6.
A triangle has sides 2 cm, 3 cm and 4 cm. Which of these triangles has the same angles as it?
(a) 4 cm, 6 cm, 7 cm(b) 3 cm, 4.5 cm, 6 cm(c) 3 cm, 4 cm, 5 cm(d) 4 cm, 5 cm, 6 cm
[1]

Step-by-step solution

  1. 3 = 1½ × 2, 4.5 = 1½ × 3 and 6 = 1½ × 4: all the sides change by the same scale 1½.
  2. If all the sides of two triangles change by the same scale, the triangles have the same angles (സദൃശത്രികോണങ്ങൾ).
  3. In the others the scale is not the same for all sides: (a) 4/2 = 2 but 7/4 = 1¾; (c) 3/2 = 1½ but 4/3 = 1⅓; (d) 4/2 = 2 but 5/3 = 1⅔.
✅ Answer: (b) 3 cm, 4.5 cm, 6 cm

Marking: 1 mark for (b).

📖 Revise: Ch 6 · Sides and angles · Chapter 6 – Similar Triangles

7.
A point moves on the number line from −3 to −8. Its change of position is
(a) 5(b) −5(c) 11(d) −11
[1]

Step-by-step solution

  1. Change of position = final position − starting position = −8 − (−3).
  2. −8 − (−3) = −8 + 3 = −5: the point moved 5 units to the left.
✅ Answer: (b) −5

Marking: 1 mark for (b).

📖 Revise: Ch 7 · Change of position · Chapter 7 – Negative Numbers

8.
Which of these statements about π is correct?
(a) π is exactly equal to 22/7.(b) π is the circumference of the circle of diameter 1, and it cannot be written as a fraction.(c) π is exactly equal to 3.14.(d) π is larger for larger circles.
[1]

Step-by-step solution

  1. The circumferences of all circles change in the same proportion as their diameters, so circumference = π × diameter, where π is the circumference of the circle of diameter 1. So π is the same for every circle.
  2. π cannot be written as a fraction; 22/7 and 3.14 are only approximations (π = 3.14159…).
✅ Answer: (b) π is the circumference of the circle of diameter 1, and it cannot be written as a fraction.

Marking: 1 mark for (b).

📖 Revise: Ch 8 · π, a new number · Chapter 8 – Measures of Circles

9.
|−4| − |3 − 7| =
(a) 0(b) −8(c) 8(d) 14
[1]

Step-by-step solution

  1. |−4| = 4.
  2. 3 − 7 = −4, so |3 − 7| = |−4| = 4.
  3. 4 − 4 = 0.
✅ Answer: (a) 0

Marking: 1 mark for (a).

📖 Revise: Ch 9 · Absolute value · Chapter 9 – Real Numbers

10.
True or false? Give a reason:   “For every real number x, |x| = x.”
[1]

Step-by-step solution

  1. |x| (കേവലമൂല്യം) is x when x is positive or 0, but it is −x when x is negative.
  2. Example: x = −3 gives |−3| = 3, which is not −3. So the statement is false (it is true only for x ≥ 0).
✅ Answer: False. For example |−3| = 3, which is not −3.

Marking: ½ mark for “false”, ½ mark for the reason or a counter-example.

📖 Revise: Ch 9 · Absolute value · Chapter 9 – Real Numbers

Section BAnswer all questions · 8 × 2 = 16 marks
11.
A pen and a notebook together cost ₹70. The notebook costs ₹4 more than twice the price of the pen. Taking the price of the pen as ₹x, form an equation and find the price of each.
[2]

Step-by-step solution

  1. Pen = x rupees, so notebook = 2x + 4 rupees.
  2. Together: x + (2x + 4) = 70, so 3x + 4 = 70 (സമവാക്യം).
  3. 3x = 66, so x = 22. The pen costs ₹22.
  4. Notebook = 2 × 22 + 4 = ₹48. Check: 22 + 48 = 70 ✓.
  5. (Mentally: remove the extra ₹4; the remaining ₹66 is the price of three pens.)
✅ Answer: Pen ₹22, notebook ₹48

Marking: Equation 1 mark; price of the pen ½; price of the notebook ½.

📖 Revise: Ch 1 · Mental arithmetic and algebra · Ch 1 · Solving one equation · Chapter 1 – Pairs of Equations

12.
In the figure, the lines l₁, l₂, l₃ are parallel. They cut one line at A, B, C and another line at P, Q, R. AB = 4 cm, BC = 6 cm and PQ = 5 cm. Find QR and PR.ABCPQR4 cm6 cm5 cm?l₁l₂l₃
[2]

Step-by-step solution

  1. Three or more parallel lines cut any two lines in the same ratio (സമാന്തരവരകൾ). So PQ/QR = AB/BC.
  2. AB/BC = 4/6 = 2/3, so 5/QR = 2/3.
  3. QR = 5 × 3/2 = 7.5 cm.
  4. PR = PQ + QR = 5 + 7.5 = 12.5 cm.
✅ Answer: QR = 7.5 cm, PR = 12.5 cm

Marking: Same-ratio statement 1 mark; QR ½; PR ½.

📖 Revise: Ch 3 · Parallel lines and unequal parts · Chapter 3 – Parallel Lines

13.
Two natural numbers each leave remainder 2 on division by 3. Prove that their product leaves remainder 1 on division by 3. Check with one example.
[2]

Step-by-step solution

  1. Numbers leaving remainder 2 on division by 3 have the form 3m + 2 and 3n + 2 (m, n = 0, 1, 2, …).
  2. (3m + 2)(3n + 2) = 9mn + 6m + 6n + 4 (multiply each number of the first sum by each number of the second).
  3. = 9mn + 6m + 6n + 3 + 1 = 3(3mn + 2m + 2n + 1) + 1.
  4. This is one more than a multiple of 3, so the remainder is 1.
  5. Example: 5 × 8 = 40 = 3 × 13 + 1 ✓.
✅ Answer: (3m + 2)(3n + 2) = 3(3mn + 2m + 2n + 1) + 1, which leaves remainder 1.

Marking: Correct general forms ½; expansion ½; writing as 3( … ) + 1 with the conclusion ½; example ½.

📖 Revise: Ch 4 · Properties of operations · Chapter 4 – Multiplication Identities

14.
(a) Write 6√3 with no square root in the denominator.   (b) Find √45√5.
[2]

Step-by-step solution

  1. (a) Multiply the numerator and the denominator by √3: 6√3 = 6√3√3 × √3 = 6√33 = 2√3.
  2. (b) √45√5 = √455 = √9 = 3.
✅ Answer: (a) 2√3   (b) 3

Marking: 1 mark each.

📖 Revise: Ch 5 · Division of square roots · Chapter 5 – Irrational Numbers

15.
A triangle has sides 6 cm, 8 cm and 10 cm and area 24 sq cm. Another triangle has the same angles, and its longest side is 15 cm. Find (a) its other two sides   (b) its area.
[2]

Step-by-step solution

  1. Same angles, so the sides change by the same scale. Longest side 10 cm → 15 cm: scale factor = 15/10 = 1½.
  2. (a) Other sides: 6 × 1½ = 9 cm and 8 × 1½ = 12 cm.
  3. (b) The area changes by the square of the scale: (1½)² = 2¼. Area = 24 × 2¼ = 54 sq cm.
  4. (Check: 9, 12, 15 is a right triangle, ½ × 9 × 12 = 54 ✓.)
✅ Answer: (a) 9 cm and 12 cm   (b) 54 sq cm

Marking: (a) scale factor ½, both sides ½; (b) 1 mark.

📖 Revise: Ch 6 · Scale factor and area · Ch 6 · Angles and sides · Chapter 6 – Similar Triangles

16.
A point moves along the number line. Complete the table.
starting positionchange of positionfinal position
(i)−69…
(ii)4−10…
(iii)−2…−9
(iv)…−51
[2]

Step-by-step solution

  1. Final position = starting position + change; change = final − start.
  2. (i) −6 + 9 = 9 − 6 = 3.
  3. (ii) 4 + (−10) = 4 − 10 = −6.
  4. (iii) change = −9 − (−2) = −9 + 2 = −7 (7 to the left).
  5. (iv) start + (−5) = 1, so start = 1 + 5 = 6.
✅ Answer: (i) 3   (ii) −6   (iii) −7   (iv) 6

Marking: ½ mark for each entry.

📖 Revise: Ch 7 · Position and number · Ch 7 · Change of position · Chapter 7 – Negative Numbers

17.
The wheel of a bicycle has diameter 70 cm. How far does the bicycle go when the wheel turns once? How many turns does the wheel make to cover 1.1 km? (Take π = 22/7.)
[2]

Step-by-step solution

  1. In one turn the bicycle goes the circumference of the wheel = π × diameter.
  2. = 22/7 × 70 = 22 × 10 = 220 cm = 2.2 m.
  3. 1.1 km = 1100 m = 110000 cm.
  4. Number of turns = 110000 ÷ 220 = 500.
✅ Answer: 220 cm in one turn; 500 turns

Marking: Circumference 1 mark; number of turns 1 mark.

📖 Revise: Ch 8 · Circumference = πd · Chapter 8 – Measures of Circles

18.
For the polynomial p(x) = 7 − 3x² + 5x³ − x, write (a) its degree (b) the coefficient of x² (c) the coefficient of x (d) the constant term.
[2]

Step-by-step solution

  1. Arrange the terms by powers, highest first: p(x) = 5x³ − 3x² − x + 7.
  2. (a) The highest power of x is 3, so the degree (കൃത്യങ്കം) is 3.
  3. (b) The coefficient (ഗുണകം) of x² is −3 (the sign goes with it).
  4. (c) −x = (−1) × x, so the coefficient of x is −1.
  5. (d) The constant term (the term without x) is 7.
✅ Answer: (a) 3   (b) −3   (c) −1   (d) 7

Marking: ½ mark each.

📖 Revise: Ch 10 · Polynomials and degree · Chapter 10 – Polynomials

Section CAnswer all questions · 6 × 3 = 18 marks
19.
Ammu’s mother is 26 years older than Ammu. After 6 years, the mother’s age will be 3 times Ammu’s age then. Taking Ammu’s present age as x years: (a) write the ages of both after 6 years in terms of x; (b) form an equation and solve it; (c) find their present ages and check.
[3]

Step-by-step solution

  1. (a) Now: Ammu x, mother x + 26. After 6 years: Ammu x + 6, mother x + 32.
  2. (b) Mother then = 3 × Ammu then: x + 32 = 3(x + 6) = 3x + 18.
  3. 32 − 18 = 3x − x, so 2x = 14 and x = 7.
  4. (c) Ammu is 7 years old and her mother is 7 + 26 = 33 years old.
  5. Check: after 6 years they are 13 and 39, and 39 = 3 × 13 ✓.
  6. (Mental way: the difference of their ages is always 26. When the mother is 3 times Ammu’s age, the difference is 2 times Ammu’s age, so Ammu will be 13 then, and is 7 now.)
✅ Answer: Ammu 7 years, mother 33 years

Marking: (a) 1 mark; (b) equation 1, solving ½; (c) ages and check ½.

📖 Revise: Ch 1 · Ages · Ch 1 · Mental arithmetic and algebra · Chapter 1 – Pairs of Equations

20.
In the calendar a 3 × 3 square of dates is marked. (a) Find the products of the dates at opposite corners of the marked square, 4 × 20 and 6 × 18, and their difference. (b) Prove that for every 3 × 3 square of dates in a calendar this difference is the same. (c) What is the difference for a 4 × 4 square? Give the reason.SunMonTueWedThuFriSat12345678910111213141516171819202122232425262728293031
[3]

Step-by-step solution

  1. (a) 4 × 20 = 80 and 6 × 18 = 108; difference 108 − 80 = 28.
  2. (b) A row has 7 days, so if the top-left date is x, the corners are x, x + 2 (top) and x + 14, x + 16 (bottom).
  3. x(x + 16) = x² + 16x.
  4. (x + 2)(x + 14) = x² + 14x + 2x + 28 = x² + 16x + 28.
  5. The difference is always 28, whatever x is.
  6. (c) Corners of a 4 × 4 square: x, x + 3, x + 21, x + 24. (x + 3)(x + 21) = x² + 24x + 63 and x(x + 24) = x² + 24x, so the difference is 63. (Check: 4 × 22 − 1 × 25 = 88 − 25 = 63.)
✅ Answer: (a) 80 and 108, difference 28   (b) always 28   (c) 63

Marking: (a) ½; (b) corners ½, both products and the conclusion 1; (c) 1.

📖 Revise: Ch 4 · Calendar curiosities · Chapter 4 – Multiplication Identities

21.
(a) Show that (√7 + √5)(√7 − √5) = 2.   (b) Using this, write 2√7 − √5 with no square root in the denominator.   (c) Taking √7 ≈ 2.646 and √5 ≈ 2.236, find its value correct to two decimal places.
[3]

Step-by-step solution

  1. (a) The identity (x + y)(x − y) = x² − y² holds for irrational numbers too: (√7 + √5)(√7 − √5) = (√7)² − (√5)² = 7 − 5 = 2.
  2. (b) Multiply the numerator and the denominator by √7 + √5: 2√7 − √5 = 2(√7 + √5)(√7 − √5)(√7 + √5) = 2(√7 + √5)2 = √7 + √5.
  3. (c) √7 + √5 ≈ 2.646 + 2.236 = 4.882 ≈ 4.88. (Adding is much easier than dividing 2 by 0.41.)
✅ Answer: (a) 7 − 5 = 2   (b) √7 + √5   (c) ≈ 4.88

Marking: 1 mark for each part.

📖 Revise: Ch 5 · Removing roots from the denominator · Chapter 5 – Irrational Numbers

22.
In the figure, ΔABC has sides 4 cm, 5 cm, 6 cm and angles 83°, 56°, 41° (to the nearest degree). ΔPQR has sides PR = 6 cm, QR = 7.5 cm and PQ = 9 cm. (a) Show that the sides of the two triangles are in the same ratio. (b) Find ∠P, ∠Q and ∠R, giving the reason. (c) How many times the area of ΔABC is the area of ΔPQR?ABCRPQ4 cm6 cm5 cm6 cm9 cm7.5 cm56°41°83°
[3]

Step-by-step solution

  1. (a) Take the sides in order of size: 4, 5, 6 and 6, 7.5, 9. 6/4 = 1½, 7.5/5 = 1½, 9/6 = 1½. All the sides change by the same scale 1½.
  2. (b) If all the sides change by the same scale, the triangles have the same angles (സദൃശം); equal angles are opposite matching sides.
  3. Opposite the smallest sides: ∠C (opposite AB = 4) and ∠Q (opposite PR = 6), so ∠Q = 41°.
  4. Opposite the middle sides: ∠B (opposite CA = 5) and ∠P (opposite QR = 7.5), so ∠P = 56°.
  5. Opposite the largest sides: ∠A (opposite BC = 6) and ∠R (opposite PQ = 9), so ∠R = 83°. (56 + 41 + 83 = 180 ✓)
  6. (c) The area changes by the square of the scale: (1½)² = 2¼ times.
✅ Answer: (a) each side × 1½   (b) ∠P = 56°, ∠Q = 41°, ∠R = 83°   (c) 2¼ times

Marking: (a) 1 mark; (b) reason ½, the three angles 1 (deduct ½ for each wrong angle); (c) ½.

📖 Revise: Ch 6 · Sides and angles · Chapter 6 – Similar Triangles

23.
(a) Using the meaning of subtraction (“how much must be added?”), explain why 4 − (−6) = 10. (b) In a building the basement floors are numbered −1, −2, −3, …. A lift goes from floor −3 up to floor 7 and then down to floor −2. Write the change of floor for each trip as (final floor − starting floor). (c) Find the change from floor −3 to floor −2 directly and show that it is the sum of the two changes in (b).
[3]

Step-by-step solution

  1. (a) 4 − (−6) asks: how much must be added to −6 to get 4?
  2. From −6, adding 6 reaches 0; adding 4 more reaches 4. So we add 6 + 4 = 10, and 4 − (−6) = 10.
  3. (b) First trip: 7 − (−3) = 7 + 3 = 10 (10 floors up).
  4. Second trip: −2 − 7 = −(2 + 7) = −9 (9 floors down).
  5. (c) Directly: −2 − (−3) = −2 + 3 = 1 (1 floor up). And 10 + (−9) = 10 − 9 = 1 ✓.
  6. This always happens: (z − y) + (y − x) = z − x; the total change is final − start.
✅ Answer: (a) add 6 to reach 0, then 4 more: 10   (b) 10 and −9   (c) 1 = 10 + (−9)

Marking: (a) 1 mark; (b) ½ each; (c) 1 mark.

📖 Revise: Ch 7 · Change of position and subtraction · Chapter 7 – Negative Numbers

24.
A window has the shape shown: a rectangle 1.4 m wide and 1.5 m high with a semicircle on top. (a) Find the length of the semicircular arc. (b) Find the perimeter of the window. (c) Another window of the same shape is twice as wide and twice as high. What is its perimeter? (Take π = 22/7.)1.4 m1.5 msemicircle
[3]

Step-by-step solution

  1. (a) The semicircle has diameter 1.4 m. Its arc is half the circumference: ½ × π × 1.4 = 0.7π m.
  2. 0.7π = 0.7 × 22/7 = 0.1 × 22 = 2.2 m.
  3. (b) The boundary is the arc, the two vertical sides and the bottom side (the dashed line is inside the window): 2.2 + 1.5 + 1.5 + 1.4 = 6.6 m.
  4. (c) All the lengths are doubled; the circumference of a circle changes in the same proportion as its diameter, so the arc doubles too. Perimeter = 2 × 6.6 = 13.2 m. (Directly: arc 1.4π = 4.4 m, sides 3 + 3 + 2.8, total 13.2 m.)
✅ Answer: (a) 2.2 m   (b) 6.6 m   (c) 13.2 m

Marking: (a) 1 mark; (b) 1 mark; (c) 1 mark.

📖 Revise: Ch 8 · Circumference and arcs · Ch 8 · Diameter and circumference · Chapter 8 – Measures of Circles

Section DAnswer all questions · 4 × 4 = 16 marks
25.
Squares in a house.   (a) A square kolam has area 6¼ sq m. Find its side. (b) A square floor tile has area 0.0144 sq m. Find its side. (c) A square courtyard has area 5 sq m. Use the table to find between which two numbers with one decimal place, and between which two numbers with two decimal places, its side lies. (d) Write the side of the courtyard correct to a centimetre (two decimal places of a metre).
x2.22.32.232.242.2362.237
x²4.845.294.97295.01764.9996965.004169
[4]

Step-by-step solution

  1. (a) 6¼ = 25/4. Since 5² = 25 and 2² = 4, √(25/4) = 5/2. Side = 2½ m.
  2. (b) 0.0144 = 144/10000 and 12² = 144, 100² = 10000, so √0.0144 = 12/100 = 0.12 m (12 cm).
  3. (c) The side is √5 m. From the table 2.2² = 4.84 < 5 < 5.29 = 2.3², so 2.2 < √5 < 2.3.
  4. 2.23² = 4.9729 < 5 < 5.0176 = 2.24², so 2.23 < √5 < 2.24.
  5. (d) 2.236² = 4.999696 < 5 < 5.004169 = 2.237², so √5 = 2.236… The third decimal digit is 6 (5 or more), so correct to two decimal places the side is 2.24 m.
✅ Answer: (a) 2½ m   (b) 0.12 m   (c) between 2.2 and 2.3; between 2.23 and 2.24   (d) 2.24 m

Marking: (a) 1 mark; (b) 1 mark; (c) ½ + ½; (d) 1 mark (the reason with 2.236 is needed for full marks).

📖 Revise: Ch 2 · Measures and numbers · Ch 2 · Squares of fractions and decimals · Chapter 2 – New Numbers

26.
On a map, a line AB of length 10 cm shows a straight stretch of a canal between two boat jetties A and B. A new jetty P is to be built on AB so that AP : PB = 3 : 4. (a) Calculate the lengths AP and PB on the map. (b) Draw AB = 10 cm and mark P using a ruler, compass and set squares, without measuring AP. Write the steps of construction. (c) Explain why your point P divides AB in the ratio 3 : 4.
[4]

Step-by-step solution

  1. (a) 3 + 4 = 7 parts. AP = 10 × 3/7 = 30/7 = 4 2/7 ≈ 4.29 cm; PB = 10 × 4/7 = 40/7 = 5 5/7 ≈ 5.71 cm. These are not easy to measure with a ruler, so we construct P.
  2. (b) Step 1: Draw AB = 10 cm.
  3. Step 2: From A draw a ray AX making an acute angle with AB.
  4. Step 3: With the compass mark 7 equal steps along AX from A (any convenient length, e.g. 1 cm): A₁, A₂, …, A₇.
  5. Step 4: Join A₇ to B.
  6. Step 5: Through A₃ draw a line parallel to A₇B (slide a set square along a ruler). It meets AB at P.
  7. A1A2A3A4A5A6A7ABP3 parts4 partsXA₃P ∥ A₇BAP ≈ 4.3 cmPB ≈ 5.7 cm
  8. (c) In ΔAA₇B the line A₃P is parallel to the side A₇B, so it cuts the other two sides in the same ratio (അംശബന്ധം): AP : PB = AA₃ : A₃A₇ = 3 : 4 (three equal steps and four equal steps).
✅ Answer: (a) AP = 30/7 ≈ 4.29 cm, PB = 40/7 ≈ 5.71 cm   (b) see the construction   (c) A₃P ∥ A₇B cuts AB and AX in the same ratio

Marking: (a) 1 mark; (b) construction with steps 2 marks (ray and equal steps ½, joining A₇B ½, parallel through A₃ 1); (c) 1 mark.

📖 Revise: Ch 3 · Dividing a line in a ratio · Ch 3 · Construction toolkit · Chapter 3 – Parallel Lines

27.
A KSRTC bus runs on a straight east–west road. The road is drawn as a number line with the bus stand O at 0; distances are in km, east is positive. The stops are A (−6), J (−1), O (0), C (2) and B (9). (a) Find the distance between A and B using absolute value. (b) Raju’s house is on the road, 4 km from C. Write this as an equation with absolute value and find the possible positions of his house. (c) A fuel station is to be built less than 3 km from the junction J. Write this condition with absolute value, find the possible positions and show them on the number line. (d) Which of the stops are more than 5 km from C?−8−7−6−5−4−3−2−1012345678910AJOCBO = bus stand, positions in km (east → positive)
[4]

Step-by-step solution

  1. (a) The distance between x and y on the number line is |x − y|: |−6 − 9| = |−15| = 15 km (bigger minus smaller: 9 − (−6) = 15).
  2. (b) Position x with distance 4 from 2: |x − 2| = 4.
  3. 4 km east of C: x = 2 + 4 = 6; 4 km west of C: x = 2 − 4 = −2.
  4. (c) Distance from J: |x − (−1)| = |x + 1|. Condition: |x + 1| < 3.
  5. So x lies within 3 km on either side of −1: from −1 − 3 = −4 to −1 + 3 = 2, that is −4 < x < 2.
  6. −8−7−6−5−4−3−2−1012345678910AJOCBO = bus stand, positions in km (east → positive)−4 < x < 2 (open circles: the ends are not included)
  7. (d) Distances from C (2): A: |−6 − 2| = 8, J: |−1 − 2| = 3, O: |0 − 2| = 2, B: |9 − 2| = 7.
  8. More than 5 km: A and B. (In algebra: |x − 2| > 5 means x < −3 or x > 7.)
✅ Answer: (a) 15 km   (b) |x − 2| = 4: 6 or −2   (c) |x + 1| < 3: −4 < x < 2   (d) A and B

Marking: (a) 1 mark; (b) equation ½, both positions ½; (c) condition ½, interval and figure ½; (d) 1 mark.

📖 Revise: Ch 9 · Distances · Ch 9 · Absolute value · Chapter 9 – Real Numbers

28.
A rectangular paddy field is 10 m longer than it is wide. Its width is x m. A bund 1 m wide is made all round it, outside the field. (a) Write the area a(x) of the field as a polynomial. What is its degree? Write its coefficients. (b) Write the area b(x) of the bund as a polynomial and give its degree. (c) Is the length of the diagonal of the field a polynomial in x? Is the perimeter of the field a polynomial? Give reasons. (d) Find a(20) and b(20).paddy field(x + 10) mx mbund, 1 m wide
[4]

Step-by-step solution

  1. (a) a(x) = x(x + 10) = x² + 10x. The highest power is x², so the degree is 2. Coefficients: of x² it is 1, of x it is 10, and the constant term is 0.
  2. (b) With the bund, the outer rectangle is (x + 2) m by (x + 12) m. Its area = (x + 2)(x + 12) = x² + 12x + 2x + 24 = x² + 14x + 24.
  3. b(x) = (x² + 14x + 24) − (x² + 10x) = 4x + 24; a first degree polynomial (degree 1).
  4. (c) Diagonal = √x² + (x + 10)² = √2x² + 20x + 100: it takes the square root of an expression with the changing number x, so it is not a polynomial (ബഹുപദം).
  5. Perimeter = 2(x + x + 10) = 4x + 20: only multiplying x by a fixed number and adding a fixed number, so it is a polynomial (degree 1).
  6. (d) a(20) = 20² + 10 × 20 = 400 + 200 = 600 sq m (check: 20 × 30 = 600).
  7. b(20) = 4 × 20 + 24 = 104 sq m (check: 22 × 32 − 600 = 704 − 600 = 104).
✅ Answer: (a) x² + 10x, degree 2   (b) 4x + 24, degree 1   (c) diagonal: no; perimeter: yes   (d) 600 sq m and 104 sq m

Marking: (a) polynomial ½, degree and coefficients ½; (b) 1 mark; (c) ½ + ½; (d) ½ + ½.

📖 Revise: Ch 10 · Polynomials and degree · Ch 10 · Algebra of measures · Chapter 10 – Polynomials

Question numbers by chapter

ChapterA (1)B (2)C (3)D (4)Total
1. Pairs of Equations11119–6
2. New Numbers2, 3––256
3. Parallel Lines–12–266
4. Multiplication Identities41320–6
5. Irrational Numbers51421–6
6. Similar Triangles61522–6
7. Negative Numbers71623–6
8. Measures of Circles81724–6
9. Real Numbers9, 10––276
10. Polynomials–18–286
Total1016181660

Score to grade

Kerala grading scale: A+ 54–60 (90 %+), A 48–53, B+ 42–47, B 36–41, C+ 30–35, C 24–29, D+ 18–23, D 12–17, E below 12. For every question the student lost marks on, the linked guide page explains the idea again.