10th Maths EM Pre Half Yearly 2025: Original question with Official answer key

 10th Maths EM Pre Half Yearly 2025: Original question with Official answer key | Download PDF

PRE HALF YEARLY EXAM – NOVEMBER 2025

Mathematics
Time: 1.30 Hrs.
Total marks: 50

I. Answer any 6 questions only:(Compulsory Qus. no:7)
6 × 2 = 12

1. Let $X=\{1,2,3,4\}$, $Y=\{2,4,6,8,10\}$ and $R=\{(1,2),(2,4),(3,6),(4,8)\}$. Show that $R$ is a function and find its domain, co-domain and range?

2. Find the number of terms in the A.P. 3, 6, 9, 12, ..., 111.

3. Determine the nature of roots: $2x^2-2x+9=0$.

4. Find the area of the triangle whose vertices are $(-3,5)$, $(5,6)$ and $(5,-2)$.

5. Find the angle of elevation of the top of a tower from a point on the ground, which is $30m$ away from the foot of a tower of height $10\sqrt{3}m$.

6. The sum of the cubes of the first $n$ natural numbers is 2025, then find the value of $n$.

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

II. Answer any 6 questions only:(Compulsory Qus. no:14)
6 × 5 = 30

8. Let $A=\{x\in \mathbb{W}\mathrm{\mid }x<2\}$, $B=\{x\in \mathbb{N}\mathrm{\mid }1<x\le 4\}$ and $C=\{3,5\}$. Verify that

   $$A\times (B\cup C)=(A\times B)\cup (A\times C)\mathrm{.}$$

9. If $f(x)=2x+3$, $g(x)=1-2x$ and $h(x)=3x$. Prove that $f\circ (g\circ h)=(f\circ g)\circ h$.

10. Find the sum of: $9^3+{10}^3+{11}^3+\mathrm{.}\mathrm{.}\mathrm{.}+{21}^3$.

11. Find the square root of $121x^4-198x^3-183x^2+216x+144$.

12. State and prove Angle Bisector Theorem.

13. Find the equation of a straight line through the intersection of lines $7x+3y=10$, $5x-4y=1$ and parallel to the line $13x+5y+12=0$.

14. From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be ${30}^{\circ }$ and ${45}^{\circ }$. If the height of the lighthouse is $700m$, find the distance between the two ships. $(\sqrt{3}=1.732)$

III. Answer the following question:
1 × 8 = 8

15. Construct a triangle similar to a given triangle $PQR$ with its sides equal to $\frac{2}{3}$ of the corresponding sides of the triangle $PQR$ (scale factor $\frac{2}{3}<1$).

(Or)

Construct a triangle similar to a given triangle $PQR$ with its sides equal to $\frac{2}{3}$ of the corresponding sides of the triangle $PQR$ (scale factor $\frac{2}{3}>1$).

10th Maths EM Pre Half Yearly 2025: Original question click here to download

Answer key

Part I

1. Function, Domain, Co-domain, Range

Domain = {1, 2, 3, 4}

Co-domain = {2, 4, 6, 8, 10}

Range = {2, 4, 6, 8}

Each element in X has unique image in Y ⇒ R is a function

2. Number of terms in A.P.

a = 3, d = 3, l = 111

l = a + (n-1)d

111 = 3 + (n-1)3

108 = (n-1)3 ⇒ n-1 = 36 ⇒ n = 37

3. Nature of roots

a = 2, b = -2, c = 9

Δ = b² - 4ac = (-2)² - 4(2)(9)

Δ = 4 - 72 = -68

Δ < 0 ⇒ No real roots

4. Area of triangle

Area = ½|(-3)(6-(-2)) + 5(-2-5) + 5(5-6)|

= ½|(-3)(8) + 5(-7) + 5(-1)|

= ½|-24 -35 -5| = ½|-64| = 32 sq. units

5. Angle of elevation

tan θ = (10√3)/30 = 1/√3

θ = 30°

6. Value of n

[n(n+1)/2]² = 2025

n(n+1)/2 = 45

n² + n - 90 = 0

(n+10)(n-9) = 0 ⇒ n = 9

7. Equation of line

Intercepts equal but opposite ⇒ x/a - y/a = 1

x - y = a

Passes (5,7): 5-7 = a ⇒ a = -2

Equation: x - y + 2 = 0

Part II

8. Verify A × (B ∪ C) = (A × B) ∪ (A × C)

A = {0,1}, B = {2,3,4}, C = {3,5}

B ∪ C = {2,3,4,5}

A × (B ∪ C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}

A × B = {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}

A × C = {(0,3),(0,5),(1,3),(1,5)}

(A × B) ∪ (A × C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}

LHS = RHS ⇒ Verified

9. Prove f ∘ (g ∘ h) = (f ∘ g) ∘ h

(g ∘ h)(x) = g(3x) = 1 - 6x

f ∘ (g ∘ h) = f(1-6x) = 2(1-6x)+3 = 5-12x

(f ∘ g)(x) = f(1-2x) = 2(1-2x)+3 = 5-4x

(f ∘ g) ∘ h = (f ∘ g)(3x) = 5-4(3x) = 5-12x

LHS = RHS ⇒ Proved

10. Sum of cubes

Sum = (1³+2³+...+21³) - (1³+2³+...+8³)

= [21×22/2]² - [8×9/2]²

= 231² - 36² = 53361 - 1296 = 52065

11. Square root

√(121x⁴ - 198x³ - 183x² + 216x + 144) = |11x² - 9x - 12|

12. Angle Bisector Theorem

Statement: Internal angle bisector divides opposite side in ratio of adjacent sides

Proof: Draw CE ∥ DA meeting BA extended at E

In ΔBDA and ΔBCE: ∠BDA = ∠BCE, ∠BDA = ∠BEC ⇒ ΔBDA ~ ΔBCE

AB/AE = BD/DC

But AE = AC (isosceles ΔAEC) ⇒ AB/AC = BD/DC

13. Equation of line

Solve: 7x+3y=10 and 5x-4y=1

Multiply: 28x+12y=40 and 15x-12y=3
Add: 43x=43 ⇒ x=1, y=1 ⇒ Point (1,1)
Slope of given line = -13/5
Equation: y-1 = (-13/5)(x-1)

5y-5 = -13x+13 ⇒ 13x+5y-18=0

14. Distance between ships

tan 30° = 700/d₁ ⇒ d₁ = 700√3 ≈ 1212.4 m

tan 45° = 700/d₂ ⇒ d₂ = 700 m

Distance = d₁ + d₂ = 700√3 + 700 ≈ 1912.4 m

 10th Maths EM Pre Half Yearly 2025: Official key click here to download