10th Maths Quarterly Exam 2026 Question Paper & Answer Key | Thoothukudi District (EM)

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10th Maths Quarterly Exam 2026 Question Paper & Answer Key | Thoothukudi District (EM)

T • THOOTHUKUDI DISTRICT Std - X Reg. No.: [____________________]

COMMON QUARTERLY EXAMINATION - 2026

MATHEMATICS (ENGLISH MEDIUM)

Time: 3.00 Hours Marks: 100

Part - 1 • Choose the most appropriate answer: 14 × 1 = 14

1. If the ordered pairs (a + 2, 4) and (5, 2a + b) are equal then (a, b) is
a) (2, −2) b) (5, 1) c) (2, 3) d) (3, −2)

Answer: d) (3, −2)
Solution: Equating corresponding components: a + 2 = 5 ⇒ a = 3. Also 2a + b = 4 ⇒ 2(3) + b = 4 ⇒ 6 + b = 4 ⇒ b = −2. Hence (a, b) = (3, −2).

2. Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f: A → B given by f = {(1, 4), (2, 8), (3, 9), (4, 10)} is a
a) many-one function b) Identity function c) one-to-one function d) into function

Answer: c) one-to-one function
Solution: Distinct elements in the domain A have distinct images in the co-domain B, making f an injective (one-to-one) function.

3. If n(A) = 2, n(B) = 3 and n(A × B) = 6, then n(B × A) is
a) 2 b) 3 c) 5 d) 6

Answer: d) 6
Solution: n(B × A) = n(B) × n(A) = 3 × 2 = 6.

4. The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P.
a) 4551 b) 10091 c) 7881 d) 13531

Answer: c) 7881
Solution: t_n = a + (n − 1)d = 1 + (n − 1)4 = 4n − 3. Therefore, t_n + 3 must be divisible by 4. Testing 7881: 7881 + 3 = 7884 ÷ 4 = 1971 (an integer).

5. The next term of the sequence ³/₁₆, ¹/₈, ¹/₁₂, ¹/₁₈, ... is
a) ¹/₂₄ b) ¹/₂₇ c) ²/₃ d) ¹/₈₁

Answer: b) ¹/₂₇
Solution: Common ratio r = (1/8) / (3/16) = (1/8) × (16/3) = ²/₃. Next term = (1/18) × (2/3) = ²/₅₄ = ¹/₂₇.

6. [(3y − 3) / y] ÷ [(7y − 7) / (3y²)] is
a) 9y / 7 b) 9y³ / (21y − 21) c) (21y² − 42y + 21) / (3y³) d) 7(y² − 2y + 1) / y²

Answer: a) 9y / 7
Solution: [3(y − 1) / y] × [3y² / 7(y − 1)] = (3 × 3y) / 7 = 9y / 7.

7. Which of the following should be added to make x⁴ + 64 a perfect square
a) 4x² b) 16x² c) 8x² d) −8x²

Answer: b) 16x²
Solution: x⁴ + 64 = (x²)² + 8². Adding 2(x²)(8) = 16x² gives (x² + 8)².

8. If α + β = 5 and αβ = 6, then α³ + β³ is
a) 25 b) 35 c) 45 d) 55

Answer: b) 35
Solution: α³ + β³ = (α + β)³ − 3αβ(α + β) = 5³ − 3(6)(5) = 125 − 90 = 35.

9. The perimeters of two similar triangles ΔABC and ΔPQR are 36 cm and 24 cm respectively. If PQ = 10 cm, then the length of AB is
a) 6 ²/₃ cm b) (10√6) / 3 cm c) 66 ²/₃ cm d) 15 cm

Answer: d) 15 cm
Solution: Ratio of perimeters equals ratio of corresponding sides: AB / PQ = Perimeter(ΔABC) / Perimeter(ΔPQR) ⇒ AB / 10 = 36 / 24 = ³/₂ ⇒ AB = 15 cm.

10. In a ΔABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm, the length of the side AC is
a) 6 cm b) 4 cm c) 3 cm d) 8 cm

Answer: b) 4 cm
Solution: By Angle Bisector Theorem: AB / AC = BD / DC ⇒ 8 / AC = 6 / 3 = 2 ⇒ AC = 4 cm.

11. If slope of the line PQ is 1 / √3 then slope of the perpendicular bisector of PQ is
a) √3 b) −√3 c) 1 / √3 d) 0

Answer: b) −√3
Solution: For perpendicular lines, m₁ · m₂ = −1 ⇒ m₂ = −1 / (1/√3) = −√3.

12. The straight line given by the equation x = 11 is
a) parallel to X axis b) parallel to Y axis c) Passing through the origin d) passing through the point (0, 11)

Answer: b) parallel to Y axis
Solution: An equation of the form x = c represents a vertical line parallel to the Y-axis.

13. If 5x = sec θ and 5/y = tan θ, then x² − 1/y² is equal to
a) 25 b) ¹/₂₅ c) 5 d) 1

Answer: b) ¹/₂₅
Solution: x = (sec θ) / 5 and 1/y = (tan θ) / 5. Thus x² − 1/y² = (sec²θ − tan²θ) / 25 = 1 / 25.

14. a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² − q² is equal to
a) a² − b² b) b² − a² c) a² + b² d) b − a

Answer: b) b² − a²
Solution: p² − q² = (a cot θ + b cosec θ)² − (b cot θ + a cosec θ)² = a²(cot²θ − cosec²θ) + b²(cosec²θ − cot²θ) = a²(−1) + b²(1) = b² − a².

Part - II • Answer any 10 questions (Q.No. 28 compulsory): 10 × 2 = 20

15. If A = {1, 3, 5}, B = {2, 3} then find i) A × B ii) B × A.

Solution:
i) A × B = {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}.
ii) B × A = {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)}.

16. Show that the function f: N → N defined by f(x) = 2x − 1 is one-one but not onto.

Solution:
For one-one: Let f(x₁) = f(x₂) ⇒ 2x₁ − 1 = 2x₂ − 1 ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂. Thus, f is one-one.
For onto: Range = {1, 3, 5, 7, ...} (all positive odd integers). Since the co-domain is N, Range ≠ Co-domain (e.g., 2 ∈ N has no pre-image). Thus, f is not onto.

17. Solve: 5x ≡ 4 (mod 6).

Solution:
5x − 4 = 6n for some integer n ⇒ 5x = 6n + 4 ⇒ x = (6n + 4) / 5.
For n = 1, x = 10 / 5 = 2. For n = 6, x = 40 / 5 = 8. For n = 11, x = 70 / 5 = 14.
Hence, x = 2, 8, 14, ...

18. Find the 19th term of an A.P. −11, −15, −19, ...

Solution:
First term a = −11, common difference d = −15 − (−11) = −4.
t₁₉ = a + (19 − 1)d = −11 + 18(−4) = −11 − 72 = −83.

19. Find the sum of 1 + 3 + 5 + ... + 55.

Solution:
Here l = 55. Number of terms n = (l + 1) / 2 = (55 + 1) / 2 = 28.
Sum of first n odd natural numbers = n² = 28² = 784.

20. Find the excluded values of: (x³ − 27) / (x³ + x² − 6x).

Solution:
Set denominator equal to zero: x³ + x² − 6x = 0 ⇒ x(x² + x − 6) = 0.
x(x + 3)(x − 2) = 0 ⇒ x = 0, −3, 2.
The excluded values are 0, −3, 2.

21. Simplify: [x³ / (x − y)] + [y³ / (y − x)].

Solution:
[x³ / (x − y)] − [y³ / (x − y)] = (x³ − y³) / (x − y).
= [(x − y)(x² + xy + y²)] / (x − y) = x² + xy + y².

22. Determine the quadratic equation whose sum and product of roots are −⁷/₂ and ⁵/₂.

Solution:
A quadratic equation with given sum and product is: x² − (Sum of roots)x + (Product of roots) = 0.
x² − (−⁷/₂)x + ⁵/₂ = 0 ⇒ x² + (⁷/₂)x + ⁵/₂ = 0.
Multiplying by 2: 2x² + 7x + 5 = 0.

23. Solve 2x² − 3x − 3 = 0 by formula method.

Solution:
Here a = 2, b = −3, c = −3.
x = [−b ± √(b² − 4ac)] / (2a) = [−(−3) ± √((−3)² − 4(2)(−3))] / (2 × 2)
x = [3 ± √33] / 4.

24. In the figure, AD is the bisector of ∠A. If BD = 4 cm, DC = 3 cm and AB = 6 cm, find AC.
[Diagram: ΔABC with internal angle bisector AD meeting BC at D. AB = 6 cm, BD = 4 cm, DC = 3 cm]

Solution:
By Angle Bisector Theorem: AB / AC = BD / DC.
6 / AC = 4 / 3 ⇒ 4 AC = 18 ⇒ AC = 18 / 4 = 4.5 cm.

25. Find the equation of a line which passes through (5, 7) and makes intercepts on the axes equal in magnitude but opposite in sign.

Solution:
Let the intercepts be a and −a. The intercept equation is (x / a) + (y / −a) = 1 ⇒ x − y = a.
Since it passes through (5, 7): 5 − 7 = a ⇒ a = −2.
Equation: x − y = −2 ⇒ x − y + 2 = 0.

26. If the straight lines 12y = −(p + 3)x + 12, 12x − 7y = 16 are perpendicular then find 'p'.

Solution:
Line 1: (p + 3)x + 12y − 12 = 0 ⇒ slope m₁ = −(p + 3) / 12.
Line 2: 12x − 7y − 16 = 0 ⇒ slope m₂ = −12 / (−7) = ¹²/₇.
Perpendicular condition: m₁ × m₂ = −1 ⇒ [−(p + 3) / 12] × (12 / 7) = −1.
−(p + 3) / 7 = −1 ⇒ p + 3 = 7 ⇒ p = 4.

27. Prove that sec θ − cos θ = tan θ sin θ.

Solution:
LHS = sec θ − cos θ = (1 / cos θ) − cos θ = (1 − cos²θ) / cos θ = sin²θ / cos θ.
= (sin θ / cos θ) × sin θ = tan θ sin θ = RHS. Hence proved.

28. [Compulsory] Find the value of k for which the quadratic equation kx² − 12x + 9 = 0 has real and equal roots.

Solution:
For real and equal roots, the discriminant Δ = b² − 4ac = 0.
(−12)² − 4(k)(9) = 0 ⇒ 144 − 36k = 0 ⇒ 36k = 144 ⇒ k = 4.

Part - III • Answer any 10 questions (Q.No. 42 is compulsory): 10 × 5 = 50

29. Let A = {1, 2, 3, 4} and B = {2, 5, 8, 11, 14} be two sets. Let f: A → B be a function given by f(x) = 3x − 1. Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) in a graphical form.

Solution:
f(1) = 3(1) − 1 = 2 ; f(2) = 3(2) − 1 = 5 ; f(3) = 3(3) − 1 = 8 ; f(4) = 3(4) − 1 = 11.
(i) Arrow diagram: Show set A = {1, 2, 3, 4} and B = {2, 5, 8, 11, 14} with arrows: 1 → 2, 2 → 5, 3 → 8, 4 → 11.
(ii) Table form:

x	1	2	3	4
f(x)	2	5	8	11

(iii) Set of ordered pairs: f = {(1, 2), (2, 5), (3, 8), (4, 11)}.
(iv) Graphical form: The points (1, 2), (2, 5), (3, 8), (4, 11) are plotted on coordinate axes.

30. The function 't' which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C) = F where F = ⁹/₅ C + 32. Find: i) t(0) ii) t(28) iii) t(−10) iv) the value of C when t(C) = 212 v) the temperature when the Celsius value is equal to the Fahrenheit value.

Solution:
i) t(0) = ⁹/₅(0) + 32 = 32°F.
ii) t(28) = ⁹/₅(28) + 32 = 50.4 + 32 = 82.4°F.
iii) t(−10) = ⁹/₅(−10) + 32 = −18 + 32 = 14°F.
iv) t(C) = 212 ⇒ ⁹/₅ C + 32 = 212 ⇒ ⁹/₅ C = 180 ⇒ C = 180 × ⁵/₉ = 100°C.
v) C = F ⇒ C = ⁹/₅ C + 32 ⇒ C − ⁹/₅ C = 32 ⇒ −⁴/₅ C = 32 ⇒ C = −40°.

31. Find x if gff(x) = fgg(x) given f(x) = 3x + 1 and g(x) = x + 3.

Solution:
f(f(x)) = 3(3x + 1) + 1 = 9x + 4.
g(f(f(x))) = (9x + 4) + 3 = 9x + 7.
g(g(x)) = (x + 3) + 3 = x + 6.
f(g(g(x))) = 3(x + 6) + 1 = 3x + 19.
Given gff(x) = fgg(x) ⇒ 9x + 7 = 3x + 19 ⇒ 6x = 12 ⇒ x = 2.

32. If p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄ = 113400 where p₁, p₂, p₃, p₄ are primes in ascending order and x₁, x₂, x₃, x₄ are integers. Find the value of p₁, p₂, p₃, p₄ and x₁, x₂, x₃, x₄.

Solution:
Prime factorisation of 113400:
113400 = 1134 × 100 = (2 × 567) × (2² × 5²) = 2³ × (3⁴ × 7) × 5² = 2³ × 3⁴ × 5² × 7¹.
Since p₁ < p₂ < p₃ < p₄:
p₁ = 2, p₂ = 3, p₃ = 5, p₄ = 7.
x₁ = 3, x₂ = 4, x₃ = 2, x₄ = 1.

33. The product of three consecutive terms of a Geometric Progression is 343 and their sum is ⁹¹/₃. Find the three terms.

Solution:
Let three terms be a/r, a, ar.
Product = (a/r) × a × (ar) = a³ = 343 ⇒ a = 7.
Sum = 7/r + 7 + 7r = ⁹¹/₃ ⇒ 7(1/r + 1 + r) = ⁹¹/₃ ⇒ (r² + r + 1)/r = ¹³/₃.
3r² + 3r + 3 = 13r ⇒ 3r² − 10r + 3 = 0 ⇒ (3r − 1)(r − 3) = 0 ⇒ r = 3 or r = ¹/₃.
When a = 7, r = 3: the terms are ⁷/₃, 7, 21.
When a = 7, r = ¹/₃: the terms are 21, 7, ⁷/₃.

34. Simplify: [1 / (x² − 5x + 6)] + [1 / (x² − 3x + 2)] − [1 / (x² − 8x + 15)].

Solution:
Factorising denominators: (x − 2)(x − 3), (x − 1)(x − 2), (x − 3)(x − 5).
= [1 / ((x − 2)(x − 3))] + [1 / ((x − 1)(x − 2))] − [1 / ((x − 3)(x − 5))]
= [ (x − 1)(x − 5) + (x − 3)(x − 5) − (x − 1)(x − 2) ] / [ (x − 1)(x − 2)(x − 3)(x − 5) ]
= [ (x² − 6x + 5) + (x² − 8x + 15) − (x² − 3x + 2) ] / [ (x − 1)(x − 2)(x − 3)(x − 5) ]
= (x² − 11x + 18) / [ (x − 1)(x − 2)(x − 3)(x − 5) ] = [ (x − 9)(x − 2) ] / [ (x − 1)(x − 2)(x − 3)(x − 5) ]
= (x − 9) / [ (x − 1)(x − 3)(x − 5) ].

35. If 4x⁴ − 12x³ + 37x² + bx + a is a perfect square, find the values of a and b.

Solution:
By polynomial long division method for square root:
Root polynomial starts with 2x². Next term is −12x³ / (4x²) = −3x.
Subtracting (4x² − 3x)(−3x) = −12x³ + 9x² leaves 28x² + bx + a.
Next term in root is 28x² / (4x²) = 7.
Multiplying (4x² − 6x + 7)(7) = 28x² − 42x + 49.
Since remainder is zero: b − (−42) = 0 ⇒ b = −42 and a − 49 = 0 ⇒ a = 49.

36. A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the bus.

Solution:
Let original speed be x km/hr. Time taken = 90 / x.
New speed = (x + 15) km/hr. New time taken = 90 / (x + 15).
Given difference in time = 30 minutes = ¹/₂ hr.
(90 / x) − [90 / (x + 15)] = ¹/₂ ⇒ 90 [ (x + 15 − x) / (x(x + 15)) ] = ¹/₂
1350 / (x² + 15x) = ¹/₂ ⇒ x² + 15x − 2700 = 0.
(x + 60)(x − 45) = 0 ⇒ x = 45 or x = −60.
Since speed cannot be negative, original speed = 45 km/hr.

37. If α, β are the roots of 7x² + ax + 2 = 0 and if β − α = −¹³/₇, find the values of a.

Solution:
From the equation: α + β = −a / 7 and αβ = ²/₇.
Identity: (β − α)² = (α + β)² − 4αβ.
(−¹³/₇)² = (−a / 7)² − 4(²/₇)
¹⁶⁹/₄₉ = (a² / 49) − ⁸/₇ = (a² − 56) / 49.
a² − 56 = 169 ⇒ a² = 225 ⇒ a = ±15.

38. State and prove Angle Bisector theorem.

Solution:
Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Proof: In ΔABC, let AD be the internal bisector of ∠A meeting BC at D.
To prove: AB / AC = BD / CD.
Construction: Draw a line through C parallel to AB. Extend AD to meet this line at E.
In ΔAEC, ∠AEC = ∠BAE (alternate interior angles).
Since AD bisects ∠A, ∠BAE = ∠CAE ⇒ ∠AEC = ∠CAE.
Thus ΔACE is isosceles ⇒ AC = CE — (1).
Now, ΔABD ~ ΔECD (by AA similarity):
AB / CE = BD / CD.
Using (1), substitute CE = AC: AB / AC = BD / CD. Hence proved.

39. Find the area of the quadrilateral whose vertices are at (−9, 0), (−8, 6), (−1, −2) and (−6, −3).

Solution:
Plotting and arranging vertices counter-clockwise: A(−9, 0), B(−6, −3), C(−1, −2), D(−8, 6).
Area = ¹/₂ | (x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) − (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄) |
= ¹/₂ | [(−9(−3) + −6(−2) + −1(6) + −8(0)) − (−6(0) + −1(−3) + −8(−2) + −9(6))] |
= ¹/₂ | [(27 + 12 − 6 + 0) − (0 + 3 + 16 − 54)] | = ¹/₂ | [33 − (−35)] | = ¹/₂ | 33 + 35 | = ¹/₂(68) = 34 sq. units.

40. A line makes positive intercepts on coordinate axes whose sum is 7 and it passes through (−3, 8). Find its equation.

Solution:
Let intercepts be a and b. Given a + b = 7 ⇒ b = 7 − a.
Equation of line: (x / a) + [y / (7 − a)] = 1.
Since it passes through (−3, 8): (−3 / a) + [8 / (7 − a)] = 1.
−3(7 − a) + 8a = a(7 − a) ⇒ −21 + 3a + 8a = 7a − a²
a² + 4a − 21 = 0 ⇒ (a + 7)(a − 3) = 0 ⇒ a = 3 (since intercepts are positive).
Then b = 7 − 3 = 4.
Equation: (x / 3) + (y / 4) = 1 ⇒ 4x + 3y − 12 = 0.

41. Find the equation of a straight line through the intersection of lines 5x − 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x − 7y + 13 = 0.

Solution:
Solving 5x − 6y = 2 — (1) and 3x + 2y = 10 — (2):
Multiply (2) by 3: 9x + 6y = 30.
Adding to (1): 14x = 32 ⇒ wait, 5x − 6y = 2 and 3x + 2y = 10:
(1) + 3 × (2): 5x + 9x = 2 + 30 ⇒ 14x = 32 ⇒ wait, if intersection is point: let's recheck: for x = 2, 5(2) − 6y = 2 ⇒ 6y = 8 ≠ integer. Let's solve directly: 14x = 28 would be if constant were different, but here 14x = 32 ⇒ x = ¹⁶/₇. Then 2y = 10 − 3(16/7) = 22/7 ⇒ y = ¹¹/₇.
(If standard question has 3x + 2y = 10 and 5x − 6y = 2, point is (¹⁶/₇, ¹¹/₇)). Alternatively, equation of family: (5x − 6y − 2) + k(3x + 2y − 10) = 0.
Line perpendicular to 4x − 7y + 13 = 0 has the form 7x + 4y + c = 0.
Substituting (¹⁶/₇, ¹¹/₇): 7(16/7) + 4(11/7) + c = 0 ⇒ 16 + ⁴⁴/₇ + c = 0 ⇒ ¹⁵⁶/₇ + c = 0 ⇒ c = −¹⁵⁶/₇.
Multiplying by 7: 49x + 28y − 156 = 0 (or using standard text values x = 2, y = 2 if 5x − 6y = −2, which gives 7x + 4y − 22 = 0). Following given text values: 7x + 4y − ¹⁵⁶/₇ = 0 ⇒ 49x + 28y − 156 = 0.

42. [Compulsory] Find the sum to n terms of the series 7 + 77 + 777 + ...

Solution:
S_n = 7(1 + 11 + 111 + ... to n terms)
= ⁷/₉ [9 + 99 + 999 + ... to n terms]
= ⁷/₉ [(10 − 1) + (10² − 1) + (10³ − 1) + ... + (10ⁿ − 1)]
= ⁷/₉ [(10 + 10² + 10³ + ... + 10ⁿ) − n]
= ⁷/₉ [ {10(10ⁿ − 1) / (10 − 1)} − n ] = ⁷⁰/₈₁ (10ⁿ − 1) − ⁷ⁿ/₉.

Part - IV • Answer both questions. Each question carries 8 marks: 2 × 8 = 16

43. a) Construct a triangle similar to a given triangle PQR with its sides equal to ⁷/₃ of the corresponding sides of the triangle PQR (scale factor ⁷/₃ > 1).
(OR) b) Draw a triangle ABC of base BC = 8 cm, ∠A = 60° and the bisector of ∠A meets BC at D such that BD = 6 cm.

Solution Steps:
a) 1. Draw any ΔPQR. 2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P. 3. Locate 7 points Q₁, Q₂, ..., Q₇ on QX such that QQ₁ = Q₁Q₂ = ... = Q₆Q₇. 4. Join Q₃R and draw a line through Q₇ parallel to Q₃R intersecting the extended line segment QR at R'. 5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'. 6. ΔP'QR' is the required similar triangle.
b) 1. Draw base BC = 8 cm. 2. Draw ∠CBE = 60° downwards from B and draw ray BF ⊥ BE. 3. Draw perpendicular bisector of BC meeting BF at O and BC at G. 4. With O as centre and OB as radius, draw a circle. 5. Mark point D on BC such that BD = 6 cm. 6. Let the perpendicular bisector intersect the circle arc at I. Join I and D, producing it to intersect the circle at A. 7. Join AB and AC. ΔABC is the required triangle.

44. a) A company initially started with 40 workers to complete the work by 150 days. Later, it decided to fasten up the work with increasing the number of workers as shown below:

Number of workers (x)	40	50	60	75
Number of days (y)	150	120	100	80

i) Graph the above data and identify the type of variation.
ii) From the graph, find the number of days required to complete the work if the company decides to opt for 120 workers?
iii) If the work has to be completed by 200 days, how many workers are required?
(OR) b) A garment shop announces a flat 50% discount on every purchase of items for their customers. Draw the graph for the relation between the Marked price and the discount. Hence find:
i) the marked price when a customer gets a discount of Rs. 3250 (from graph)
ii) the discount when the marked price is Rs. 2500.

Solution:
a)
i) As x increases, y decreases. The product xy = 40 × 150 = 50 × 120 = 60 × 100 = 75 × 80 = 6000 (constant). It is an Indirect / Inverse Variation with equation xy = 6000.
ii) When number of workers x = 120: y = 6000 / 120 = 50 days.
iii) When number of days y = 200: x = 6000 / 200 = 30 workers.
b)
Let marked price be x and discount be y. Since discount is 50%, y = 0.5x or y = ¹/₂ x. This is a Direct Variation passing through origin with k = 0.5.
i) When discount y = Rs. 3250 ⇒ x = 3250 / 0.5 = Rs. 6500 (Marked price).
ii) When marked price x = Rs. 2500 ⇒ y = 0.5 × 2500 = Rs. 1250 (Discount).