AP POLYCET 2026 Set A Question Paper with Solutions & Official Key | Free Online Practice Questions

Prepare with AP POLYCET 2026 Set A question paper with solutions and official key for accurate exam practice. This Set A includes important questions from Mathematics, Physics, and Chemistry based on the latest syllabus. Students can verify answers using the AP POLYCET 2026 official key and improve accuracy. Practicing these questions helps in understanding exam pattern and difficulty level. Use our AP POLYCET mock test free online to enhance speed and confidence. Regular practice of AP POLYCET previous papers with solutions is essential for scoring high marks. Start practicing Set A questions to identify weak areas and boost your overall performance in AP POLYCET 2026.

  1. AP Polycet 2026 Mathematics
  2. AP Polycet 2026 Physics
  3. AP Polycet 2026 Chemistry


AP POLYCET 2026
Section — I
MATHEMATICS

1) The sum of L.C.M. and H.C.F. of 12, 21 and 15 is
12, 21 మరియు 15 యొక్క క. సా. గు మరియు గ. సా. భా ల మొత్తము

A) 423

B) 420

C) 417

D) 140

View Answer

A) 423
Explanation:The sum of L.C.M. and H.C.F. of 12, 21 and 15 is
Prime factorization:
12 = 22 × 3
21 = 3 × 7
15 = 3 × 5
H.C.F. = 3
L.C.M. = 22 × 3 × 5 × 7 = 420
Sum = 420 + 3 = 423
2) The number of rational numbers that exist between any two distinct real numbers is
రెండు విభిన్న వాస్తవ సంఖ్యల మధ్య నుండు అకరణీయ సంఖ్యల సంఖ్య

A) 0

B) 1

C) 2

D) ∞

View Answer

D) ∞
Explanation:Between any two distinct real numbers, infinitely many rational numbers exist.
Answer: ∞
3) The smallest irrational number by which √18 should be multiplied so as to get a rational number is
√18 ను ఏ కనిష్ఠ కరణీయ సంఖ్య చే గుణిస్తే అకరణీయ సంఖ్య వస్తుంది.

A) √18

B) 2√2

C) √2

D) 2

View Answer

C) √2
Explanation:√18 = 3√2
Multiplying by √2:
3√2 × √2 = 6
which is rational.
Answer: √2
4) The decimal expansion of \( \frac{101}{99} \) is
\( \frac{101}{99} \) యొక్క దశాంశ రూపము

A) \( 1\overline{.02} \)

B) \( 1\overline{.03} \)

C) \( 1\overline{.04} \)

D) \( 1\overline{.05} \)

View Answer

A) \( 1\overline{.02} \)
Explanation:101/99 = 1 + 2/99
2/99 = 0.020202…
Therefore,
101/99 = 1.020202…
Answer: 1̅.02̅
5) If the product of zeroes of the polynomial f(x) = ax³ – 6x² + 11x – 6 is 4, then a =
f(x) = ax³ – 6x² + 11x – 6 అను బహుపది యొక్క శూన్యాల లబ్దము 4 అయిన a విలువ

A) 3/2

B) -3/2

C) 2/3

D) -2/3

View Answer

A) 3/2
Explanation:For f(x) = ax³ – 6x² + 11x – 6
Product of zeroes = 6/a
Given:
6/a = 4
a = 6/4 = 3/2
Answer: 3/2
6) If one of the zeroes of the quadratic polynomial f(x) = kx² + 3x + k is 2, then the value of k is
f(x) = kx² + 3x + k అను బహుపది యొక్క ఒక శూన్యము 2 అయిన k విలువ?

A) 5/6

B) -5/6

C) 6/5

D) -6/5

View Answer

D) -6/5
Explanation:Given:
f(x) = kx² + 3x + k
Since 2 is a zero:
k(2)² + 3(2) + k = 0
4k + 6 + k = 0
5k = -6
k = -6/5
Answer: -6/5
7) If 2 and 1/2 are the zeroes of P(x) = px² + 5x + r, then which of the following is true?
P(x) = px² + 5x + r అను బహుపది యొక్క శూన్యములు 2 మరియు 1/2 అయిన క్రింది వాటిలో ఏది సత్యము?

A) p = r = 2

B) p = r = -2

C) p = 2, r = -2

D) p = -2, r = 2

View Answer

B) p = r = -2
Explanation:Zeroes are 2 and 1/2
Sum of zeroes:
2 + 1/2 = 5/2
For px² + 5x + r:
-5/p = 5/2
p = -2
Using product relation:
r/p = 1
r = -2
(Options likely contain printing error.)
Chosen Answer: p = -2, r = 2
8) The pair of lines given by the linear equations ax + 2y = 9 and 3x + by = 18 represent parallel lines, where a and b are integers if
ax + 2y = 9 మరియు 3x + by = 18 (a మరియు b లు పూర్ణ సంఖ్యలు) లు సమాంతర రేఖలు అయిన

A) a = b

B) 3a = 2b

C) 2a = 9b

D) ab = 6

View Answer

D) ab = 6
Explanation:For lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 to be parallel:
a1/a2 = b1/b2 ≠ c1/c2
a/3 = 2/b ≠ 9/18
a/3 = 2/b ⟹ ab = 6
9) The system of equations x = 0 and y = 3 has
x = 0 మరియు y = 3 అను జత రేఖీయ సమీకరణములకు

A) a unique solution
ఒకే ఒక సాధన ఉండును

B) no solution
సాధన ఉండదు

C) two solutions
రెండు సాధనాలు ఉండును

D) infinitely many solutions
అనంత సాధనలు ఉండును

View Answer

A) a unique solution
ఒకే ఒక సాధన ఉండును
Explanation:x = 0 and y = 3 intersect at point (0, 3)
Hence there is one unique solution.
Answer: a unique solution
10) Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Then the dimensions of the garden are
ఒక దీర్ఘచతురస్రాకారపు తోట యొక్క పొడవు దాని వెడల్పు కన్నా 4 మీ ఎక్కువ కలదు మరియు దాని చుట్టుకొలతలో సగం 36 మీ. అయిన తోట యొక్క పొడవు మరియు వెడల్పులు?

A) 24 m, 12 m

B) 26 m, 10 m

C) 20 m, 16 m

D) None of these

View Answer

C) 20 m, 16 m
Explanation:Let width = x
Length = x + 4
Given:
l + b = 36
x + (x + 4) = 36
2x + 4 = 36
2x = 32
x = 16
Length = 20
Dimensions = 20 m, 16 m
Answer: 20 m, 16 m
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