AP 10th Class Question Paper 2025-26 FA1 Maths

FOR AP 10TH CLASS FA1 EXAM PREPARATION

AP 10th Class 2025-26
Question Paper · FA1
Maths
EXAM YEAR TYPE SUBJECT

AP 10th Class FA1 2025-26 Question Paper Maths

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AP FA1 2025-26 Question Paper & Answer Key
Class 10 · Subject: Mathematics (PL-30)

Name of the Student: ______________________ Roll No: __________
Class: X Subject: Mathematics
Max. Marks: 35 Time: 1 hr 15 min

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I. Answer the following questions. 7 × 1 M = 7M
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c. o of 12, 15, and 21 is ________ [B]
1. The highest common factor
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A) 5
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B) 3
C) 420
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D) 7 a
Answer: B) 3

2. The exponent of 2 in the prime factorization of 144 is ________ [A]
A) 4
B) 5
C) 3
D) 6
Answer: A) 4

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3. Assertion (A): Sum of the zeroes of a quadratic polynomial 2x² + 3x − 4 is −3/2. Reason (R):
Sum of zeroes of a quadratic polynomial ax² + bx + c is −b/a. Now choose the correct answer from
the following: [C]
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A) Both assertion (A) and Reason (R) are true, Reason is correct explanation of the assertion.

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B) Both assertion (A) and Reason (R) are true, Reason is not correct explanation of the assertion.
C) Assertion (A) is true, but Reason (R) is false.
D) Assertion (A) is false, but Reason (R) is true.
Answer: C) Assertion (A) is true, but Reason (R) is false.

4. Form a pair of linear equations in two variables for the given information. “5 pencils and 7 pens
together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46.”
Answer: 5x + 7y = 50, 7x + 5y = 46

5. Find the discriminant of the quadratic equation 2x² − 4x + 3 = 0.

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a = 2, b = −4, c = 3.

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Discriminant D = b² − 4ac
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D = (−4)² − 4 × 2 × 3
D = 16 − 24 = −8
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6. Match the following. [C]
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a i) Common Difference of [ ] p) −4
the A.P: 3, 1, −1, −3, …
ii) Common Difference of [ ] q) 4
the A.P: 5, 1, −3, −7, …
iii) Common Difference of [ ] r) −2
the A.P: −10, −6, −2, 2, …
A) i)→p, ii)→r, iii)→q
B) i)→r, ii)→q, iii)→p

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C) i)→r, ii)→p, iii)→q
D) i)→q, ii)→r, iii)→p
Answer: C) i)→r, ii)→p, iii)→q

7. State the Fundamental Theorem of Arithmetic.
Answer: Every composite number can be expressed (factorised) as a product of primes, and this
factorization is unique, apart from the order in which the prime factors occur.

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II. Solve the following problems. 4 × 2 M = 8M

8. Find a quadratic polynomial, the sum and product of whose zeroes are −3 and 2, respectively.
α + β = −3 and αβ = 2.
Required quadratic polynomial is x² − (α + β)x + αβ = 0.
Required quadratic polynomial is x² − (−3)x + 2 = 0.
Required quadratic polynomial is x² + 3x + 2 = 0.

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9. Check whether (x − 2)² = x² + 3x + 1 is a quadratic equation or not.

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(x − 2)² = x² + 3x + 1

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x² − 4x + 4 = x² + 3x + 1
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x² − x² + 4x + 3x − 1 + 4 = 0
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7x + 3 = 0
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10. Find the a
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It is not of the form ax² + bx + c = 0. Hence it is not a quadratic equation.

zeroes of the quadratic polynomial x² − 2x − 8
x² − 2x − 8 = 0
x² + 2x − 4x − 8 = 0
(x² + 2x) + (−4x − 8) = 0
x(x + 2) − 4(x + 2) = 0
(x + 2)(x − 4) = 0
(x + 2) = 0 or (x − 4) = 0
x = −2 or x = 4

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11. “The product of two consecutive positive integers is 306”. We need to find the integers.
Represent this situation in the form of a quadratic equation.
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Let first consecutive positive integer = x
Second consecutive positive integer = x + 1
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“The product of two consecutive positive integers is 306” means
x(x + 1) = 306
x² + x = 306
x² + x − 306 = 0

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III. Solve the following problems. 3 × 4 M = 12M

12. Find two consecutive positive integers, the sum of whose squares is 365.
First consecutive positive integer = x
Second consecutive positive integer = x + 1
“Sum of squares is 365” means
x² + (x + 1)² = 365
x² + x² + 2x + 1 = 365
2x² + 2x + 1 − 365 = 0
2x² + 2x − 364 = 0
x² + x − 182 = 0
x² − 13x + 14x − 182 = 0
(x² − 13x) + (14x − 182) = 0
x(x − 13) + 14(x − 13) = 0
(x − 13)(x + 14) = 0
(x − 13) = 0 or (x + 14) = 0
x = 13 or x = −14
We must find positive integers only. So x = 13 only.
First consecutive positive integer = x = 13
Second consecutive positive integer = x + 1 = 14

13. Determine the AP whose 3rd term is 5 and 7th term is 9.
a₃ = 5 ⟹ a + 2d = 5 …… (1)
a₇ = 9 ⟹ a + 6d = 9 …… (2)
Subtracting (1) from (2): 4d = 4 ⟹ d = 1
Substituting d in (1): a + 2(1) = 5 ⟹ a = 3
AP = a, a+d, a+2d, … = 3, 4, 5, …

14. Answer the following questions by observing the graph.

a) Name the shape of the graph? — Parabola.
b) Write the number of zeros of the polynomial. — Two
c) Write the zeroes of the polynomial. — −2, 2
d) Write the product of zeros of the polynomial. — Product = −2 × 2 = −4

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IV. Solve the following problem. 1 × 8 M = 8M

15) a). Solve the following pair of linear equations graphically. x + 3y = 6 and 2x − 3y = 12.
Order pairs for x + 3y = 6 Order pairs for 2x − 3y =
12
x 0 3 x 0 6
y 2 1 y −4 0
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Answer: Solution: x = 6, y = 0
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mForm the pair of linear equations in the following problem and find theirmsolutions
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s e of girls is 4 more
s ethan the number of boys, find the number of boys and girls who took part
graphically.
ain quiz.
10 Students of Class X took part in a Mathematics quiz. If the number
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a Total number of students in class X = 10 ⟹ x + y = 10 …… (1)
“The number of girls is 4 more than the number of boys” means y − x = 4 ⟹ −x + y = 4 …… (2)

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Answer: No. of boys in the class x = 3, No. of Girls in the class y = 7

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