Prepare for your Class 10 Mathematics First Quarter Book Test 2027 with our specially designed Math Test 1 (Units 1 to 3), based on the latest Punjab Board syllabus. This test is part of the Punjab Board 2027 Model Papers series, following the new syllabus issued by PECTAA Punjab. Students can use this Class 10 Math practice test to assess their mathematical concepts, improve problem-solving skills, and prepare effectively for school examinations, monthly tests, and board exams. Download and practice the latest Mathematics test papers to strengthen your preparation for the 2027 annual examinations.
First Quarter Book Test
Unit 1 – 3
T-1
Objective Type
Note: Four possible answer A, B, C and D to each question are given. The choice which you think is correct, fill that circle front of that question with Marker or ink pen in the answer book. Cutting filling two or more circles will result in zero mark in that question.
1.A complex number Z = x + iy is called pure imaginary if
(A) x = 0, y = 0
(B) x ≠ 0, y = 0
(C) x = 0, y ≠ 0
(D) none of these
Answer: C
2.If Z = 1 – i then | Z | =
(A) 0
(B) √2
(C) 2
(D) none of these
Answer: B
3.If Z = x + iy then
(A) x and y both are real numbers
(B) x and y both are imaginary numbers
(C) x is real and y is imaginary
(D) x is imaginary and y is real
Answer: A
4.Im (Z) can be expressed as:
(A) Z + 2
(B) Z – 2
(C) Z + 2i
(D) Z – 2i
Answer: D
5.Z =
(A) Z2
(B) | Z |2
(C) – Z2
(D) None of these
Answer: B
6.For any a and b, which of the following is true?
(A) a2 + b2 ≥ 2ab
(B) a2 + b2 ≤ 2ab
(C) a2 + b2 = 2ab
(D) None of these
Answer: A
7.If (x – 2) (x – 3) = 6, then we can write:
(A) x – 2 = 6
(B) x – 3 = 6
(C) x + 6 = 6
(D) None of these
Answer: D
8.Graph of the function f(x) = x2 + 2x – 3 = (x + 3) (x – 1) is.
(A)
(B)
(C)
(D)
Answer: B (upward parabola, roots at –3 and 1)
9.The expression ax2 + bx + c = 0 is a quadratic equation if:
(A) a ≠ 0
(B) b ≠ 0
(C) c ≠ 0
(D) a ≠ 0, b ≠ 0, c ≠ 0
Answer: A
10.Which of the following expression is a quadratic equation in standard form?
(A) 2x2 = 3x + 1
(B) 1x+1 + 1x+2 = 2
(C) x2 + 1x2 = 2
(D) None of these
Answer: A
11.A = x6515 is singular if x =
(A) 0
(B) 1
(C) 2
(D) not possible
Answer: C (15x – 30 = 0 ⇒ x = 2)
12.If A = 44 and B = 25 then BA =
(A) [28]
(B) [ – 12]
(C) [12]
(D) 820820
Answer: A (2×4 + 5×4 = 28)
13.If A is of order m-by-n and B is of order n-by-f then AB is of order:
(A) n-by-n
(B) f-by-m
(C) m-by-f
(D) None of these
Answer: C
14.If A = 2341 and I = 1001 then AI =
(A) 2341
(B) 2001
(C) 3342
(D) none of these
Answer: A
15.A matrix A is called skew symmetric if:
(A) A′ = A
(B) A′ = – A
(C) A′ = 1A
(D) A′ = –1A
Answer: B
Answer Key: 1-C, 2-B, 3-A, 4-D, 5-B, 6-A, 7-D, 8-B, 9-A, 10-A, 11-C, 12-A, 13-C, 14-A, 15-B
Here is Subjective Part of Model Paper first quarter Chapter No.1 to Chapter No. 3
Subjective Type
Time Allowed: 2:10 Min.
(Part-I)
Max. Marks: 60
2. Write short answers to any six (06) questions:2 × 6 = 12
(i) Simplify the following: i7
i7 = (i4)(i3) = –i
(ii) Define Equality of Complex Numbers.
a + bi = c + di if and only if a = c and b = d.
(iii) Write in terms of i. 3 – √–7
3 – i√7
(iv) Find the values of x and y. (2x + 5) + (y – 1)i = 1 + 2i
2x + 5 = 1 ⇒ x = –2; y – 1 = 2 ⇒ y = 3
(v) Simplify: (i) (8 + 2i) – (5 – 6i).
= 3 + 8i
(vi) If z = –7 – 4i, then, show that z + 0 = z.
z + 0 = (–7 – 4i) + (0 + 0i) = –7 – 4i = z
(vii) Write additive inverse for each complex number: 4 – 3i
Additive inverse = –4 + 3i
(viii) Find multiplicative inverse for each complex number: 7 – 3i
1/(7 – 3i) = (7 + 3i)/58 = 758 + 358i
(ix) If z1 = 2 + 5i, z2 = 1 – 3i and z3 = 2 + i, then verify that: z1 + z2 = z2 + z1
z1 + z2 = 3 + 2i; z2 + z1 = 3 + 2i. Hence verified.
3. Write short answers to any six (06) questions:2 × 6 = 12
(i) If the equation x2 + 2(1 + k) x + k2 = 0 has equal roots, then find the value of k.
D = 0 ⇒ 4(1 + k)2 – 4k2 = 0 ⇒ 4 + 8k = 0 ⇒ k = –12
(ii) Perimeter (P) of a rectangle is P = 2(ℓ + w), make ℓ as the subject of this formula.
ℓ = P2 – w
(iii) The area of a circle is A = πr2. Make r the subject of the formula.
r = √Aπ
(iv) Find the values for m, if the quadratic equation mx2 – 8x + 1 = 0 has real and equal roots.
D = 64 – 4m = 0 ⇒ m = 16
(v) Examine the nature of roots of the following quadratic equations. 3x2 – 9x – 2 = 0
D = 81 + 24 = 105 > 0 (not a perfect square): roots are real, unequal and irrational.
(vi) Find the value of k if the equation (k + 1)x2 + (2k – 1)x + (k – 1) = 0 has equal roots.
D = (2k – 1)2 – 4(k2 – 1) = 5 – 4k = 0 ⇒ k = 54
(vii) Find the value of m in the equation 2x2 + 3x + m = 0 when sum of its roots is equal to double the product of its roots.
Sum = –32, Product = m2. –32 = 2 · m2 ⇒ m = –32
(viii) Find the equation whose roots are double the roots of x2 – px + q = 0.
Roots 2α, 2β: sum = 2p, product = 4q ⇒ x2 – 2px + 4q = 0
(ix) Form a quadratic equation whose roots are given below: – 4, 9
Sum = 5, Product = –36 ⇒ x2 – 5x – 36 = 0
4. Write short answers to any six (06) questions:2 × 6 = 12
(i) Define non-singular matrix with example.
A square matrix A with |A| ≠ 0, e.g. A = 1234, |A| = –2 ≠ 0.
(ii) If A = 7321, B = 5324, then find:(i) | B | (ii) Adj B
|B| = 20 – 6 = 14; Adj B = 4–3–25
(iii) If A = 4276, B = 5588, then find: (A – B)t
A – B = –1–3–1–2; (A – B)t = –1–1–3–2
(iv) Find the adjoint of the following matrices:
A = abcd, B = 5–237,
A = abcd, B = 5–237,
Adj A = d–b–ca; Adj B = 72–35
(v) 4 knives and 6 forks cost Rs. 136, whereas 6 knives and 5 forks cost Rs. 164. Find the cost of a knife and a fork by using matrices.
4x + 6y = 136, 6x + 5y = 164; |A| = –16. x = 19, y = 10. Knife = Rs. 19, Fork = Rs. 10.
(vi) Find the values of each of the determinant.
– 58– 3– 7
– 58– 3– 7
= (–5)(–7) – (8)(–3) = 35 + 24 = 59
(vii) Find the value of x when A = x6515 is a singular matrix.
|A| = 15x – 30 = 0 ⇒ x = 2
(viii) Find the value of the determinant of B = 8243.
|B| = 24 – 8 = 16
(ix) Find AB and BA, if possible. A = 12–10, B = 321–1
AB = 50–3–2; BA = 1622
(Part-II) Note: Attempt any two (02) questions.(2 × 8 = 16)
5:(a) If z1 = 2 + 5i, z2 = 1 – 3i and z3 = 2 + i, then verify that: (z1 + z2) + z3 = z1 + (z2 + z3)
L.H.S = (3 + 2i) + (2 + i) = 5 + 3i. R.H.S = (2 + 5i) + (3 – 2i) = 5 + 3i. L.H.S = R.H.S.
(b) If (1 + i)22 – i = x + iy, then find the values of x and y.
(1 + i)2 = 2i; 2i(2 + i)5 = –25 + 45i. So x = –25, y = 45
6:(a) If z1 = 5 + 4i, z2 = 3 + 2i, then prove that
((z1z2)) = .
z1/z2 = (5 + 4i)(3 – 2i)/13 = (23 + 2i)/13, so its conjugate = (23 – 2i)/13. Also / = (5 – 4i)(3 + 2i)/13 = (23 – 2i)/13. Hence proved.
(b) Find the modulus of the following complex numbers: (i) 35 – 45i (ii) – 5 – 4i
(i) |z| = √(9/25 + 16/25) = 1. (ii) |z| = √25 + 16 = √41
7:(a) If a ball is thrown upward with a velocity v, the maximum height it reaches can be determined by using a formula h = v22g. Rearrange the formula to make v the subject.
v2 = 2gh ⇒ v = √2gh
(b) Form a quadratic equation whose roots are: 6, 32.
Sum = 152, Product = 9 ⇒ x2 – 152x + 9 = 0, i.e. 2x2 – 15x + 18 = 0
(Part-III) Note: Attempt any one (01) question.(1 × 8 = 8)
8:(a) Solve the inequality 2x2 – 5x + 2 ≤ 0
(2x – 1)(x – 2) ≤ 0 ⇒ 12 ≤ x ≤ 2
(b) A shop employs, 5 men and 3 women, pays total daily wages Rs. 3500. If the number of men is reduced to 2 and 3 extra women are taken on, the daily wages amount to Rs. 5000. Find daily wages of a man and a woman by using matrices.
5m + 3w = 3500, 2m + 6w = 5000; |A| = 24. m = 250, w = 750. Man = Rs. 250, Woman = Rs. 750.
9:(a) Three forces act on a particle which must be in equilibrium i.e. F1 + F2 + F3 = 0, where F1 = 52, F2 = –3x, F3 = –2–4. Find the value of x.
2 + x – 4 = 0 ⇒ x = 2 (first row: 5 – 3 – 2 = 0 ✓)
(b) If A = 5–32–1, then find A–1 and prove that AA–1 = A–1A = I.
|A| = –5 + 6 = 1; A–1 = –13–25. AA–1 = A–1A = 1001 = I.
