TS POLYCET 2023 Solutions That Boost Confidence

TS POLYCET 2023 Solutions That Boost Confidence

Are you searching for accurate and easy-to-understand TS POLYCET 2023 Question Paper Solutions? This comprehensive guide is designed to help students master the exam with confidence.

Along with solutions, this post also includes a motivational storytelling approach that connects with students emotionally and encourages them to overcome exam fear. By practising these solutions, students can improve speed, accuracy, and problem-solving skills.

Whether you are revising concepts or preparing seriously for the next POLYCET exam, these TS POLYCET 2023 Question Paper Solutions will act as a powerful resource to guide your preparation and boost your performance.

Real Numbers

1. If ‘n’ is a prime number, then √n is

(1) Prime number          (2) Composite number

(3) Rational number      (4) irrational number

‘n’ అనేది ఓకే ప్రధాన సంక్య అయిన √n   అనేది

(1)  ప్రధాన సంఖ్య                             (2) సంయుక్త  సంఖ్య

(3) అకరణీయ సంఖ్య                   (4)  కరణీయ సంఖ్య

   Answer: (4)

  Solution:   If ‘n’ is a prime number, then √n  is an irrational number

2. Among 1/2, 1/3, 1/4, 1/5 the non-terminating decimal is

1/2, 1/3, 1/4, 1/5 అనే సంఖ్యలలో అంతం కాని దశాంశం

(1) 1/2          (2) 1/3      (3) 1/4       (4) 1/5

Answer: (2)

Solution:  A fraction in its simplest form has a terminating decimal if the prime factors of its denominator are only 2, 5, or both. If the denominator has any other prime factor, the result is a non-terminating repeating decimal.

3. The value of log₆₂₅ ⁽⁵⁾

log₆₂₅ ⁽⁵⁾యొక్క విలువ

(1) 1/2          (2) 1/4      (3) 1/3       (4) 1/5

Answer: (2)

Solution: let log₆₂₅(5) =  x

                      5 = 625^x

                      5 = (5⁴)^x

                      5 = 5^4x

                      1 = 4x

                       x = 1/4

4. √2 + √3 is

(1) Rational number           (2) Irrational number

(3) Prime number        (4) composite number

     √2 + √3 is అనునది ఒక

(1) అకరణీయ సంఖ్య              (2) కరణీయ సంఖ్య       (3) ప్రధాన సంఖ్య              (4) సంయుక్త  సంఖ్య

Answer: (2)

Solution:  √2 + √3 is an irrational number

5. HCF of 7, 8, 9 is

7, 8, 9  ల గా.సా.భా.

(1) 9          (2) 7        (3) 1       (4)  2

Answer: (3)

Solution:  factors of 7 = 1, 7

                    factors of 8 = 1, 2, 4, 8

                    factors of 9 = 1, 3, 0039

                    HCF 0f 7, 8, 9 = 1

TS POLYCET 2023 Question Paper Solutions

  Sets

1. If A={P, O, L, Y, T, E,C, H, N, I} and B = {E, X, A, M}then A∩B =

A={P, O, L, Y, T, E,C, H, N, I} మరియు  B = {E, X, A, M}అయితే, A∩B =

(1) {P}               (2) {E}                      (3) {X}                       (4) {T}

Answer: (2)

Solution:  Given A = {P, O, L, Y, T, E, C, H, N, I} and B = {E, X, A, M}

               A∩B = {P, O, L, Y, T, E, C, H, N, I}∩{E, X, A, M} = {E}

2. If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7} then A – B =

A = {1, 2, 3, 4, 5} మరియు    B = {4, 5, 6, 7} అయితే, A – B =

(1) {1, 2, 3}         (2) {3, 4, 5}

 (3) {5, 6, 7}       (4) {2, 3, 4}

Answer: (1)

Solution:   Given A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}

            A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7} = {1, 2, 3}

TS POLYCET 2023 Question Paper Solutions

TS POLYCET 2022 Solutions – Complete Questions Explained Simply

  Polynomials

1. Product of zeroes of polynomial 5x² – 1 is

5x² – 1 అనే వర్గ బహుపదీ శూన్యాల లబ్దము

(1) 1        (2) ½         (3) 1/5    (4) – 1/5

Answer: (4)

Solution:  Given polynomial is 5x² – 1

                     Product of zeroes = c/a = – 1/5

2. (x + a) is a factor of f(x), if

(x + a)  అనేది f(x) యొక్క కారణాంకం అయినచో

  (1) f(a) = 0            (2) f(–a) = 0

   (3) f(1/a) = 0        (4) f(– 1/a) = 0

  Answer: (2)

  Solution:    Given (x + a) is a factor of f(x),then f(–a) = 0

3. If α, β are the zeroes of the quadratic polynomial ax² + bx + c, a≠0, then α² + β² =

ax² + bx + c, a≠0 అనే వర్గ బహుపది యొక్క శూన్యాలు α, β అయిన α² + β² =

(1)   (b² + 2ac)           (2)  (c² + 2ab)

(3) (b² – 2ac)             (4) (c² – 2ab)

   Answer: (2)

   Solution: Given polynomial is ax² + bx + c, a≠0

α + β = – b/a

αβ = c/a

α²+ β² = (α + β)² – 2αβ

α²+ β² = (– b/a)² – 2(c/a)

α²+ β² =  b²/a² – 2c/a

α²+ β² =  (b² – 2ac)

  Linear equations in Two Variables

1. If 5x + py + 8 = 0 and 10x + 15y + 12 = 0 has no solution, then p =

5x + py + 8 = 0 మరియు 10x + 15y + 12 = 0 అను సమీకరణాలకు సాధన లేనిచో, p విలువ

(1) 15/2      (2) 13/2          (3) 7/2        (4) 5/2

Answer: (1)

Solution:   Given 5x + py + 8 = 0 and 10x + 15y + 12 = 0 have no solution

a₁ = 5; b₁ = p; c₁ = 8

a₂ = 10; b₂ = 15; c₂ = 12

p = 15/2

2. If ax + b = 0 then x =

ax + b = 0 అయిన, x విలువ

(1) – a       (2) a     (3) b/a    (4) – b/a

Answer: (4)

Solution:    Given ax + b = 0

⟹ ax = – b

⟹ x = – b/a

3. The solution of the system of equations and is

  మరియు సమీకరణాల సాధన

(1) (1/4, 1/3)     (2)  (1/3, 1/4)       (3) (1/2, 1/3)      (4) (1/3, 1/2)

Answer: (3)

Solution:  Given equations are  and

                    Let  = a and  = b

⟹2a + 3b= 13 and 5a – 4b = – 2

a = 46/23 = 2

2(2) + 3b= 13

4 + 3b= 13

4 + 3b = 13 – 4

3b= 9

b= 3

⟹ x = 1/2, y = 1/3

4. The line x = 7 is

 x = 7 అను  రేఖ

(1) Parallel to X – axis( X – అక్షమునకు సమాంతరము)

 (2) Parallel to Y-axis ( Y – అక్షమునకు సమాంతరము)

(3) Passes through origin ( మూల బిందువు గుండా పోతుంది)

 (4) Passes through (0, 7)  [ (0, 7) బిందువు గుండా పోతుంది]

    Answer: (2)

    Solution:  The line x = 7 is parallel to the Y–axis

5. The two lines 2x + 3y = 7, 8x + 12y = 1are ______ Lines.

2x + 3y = 7, 8x + 12y అను రేఖలు ____ రేఖలు

(1) Perpendicular (లంబ)    

(2) Parallel (సమాంతర)

 (3) Intersecting (ఖండన )

(4) None(ఏది కాదు)

 Answer: (2)

 Solution:  Given lines are 2x + 3y = 7, 8x + 12y = 1are have no solution

a₁ = 2; b₁ = 3; c₁ = 7

a₂ = 8; b₂ = 12; c₂ = 1

Given lines are parallel

Quadratic Equations

1. If the equation 3x² + 2x + k=0 has real roots then k is

3x² + 2x + k=0 సమీకరణము వాస్తవ మూలాలు కలిగి ఉన్నచో k విలువ

(1) k < 1/3      (2) k > 1/3         (3) k ≤ 1/3         (4) k ≥1/3

     Answer: (3)

    Solution:  Given Equation is 3x² + 2x + k=0

                   a = 3, b = 2,  c = k

The given equation has real roots

b² – 4ac ≥ 0

2² – 4(3)(k) ≥ 0

4 – 12k ≥ 0

4(1 – 3k) ≥ 0

1 – 3k≥ 0

1 ≥3k

k ≤ 1/3

2. The condition a² + bx + c = 0 to be a quadratic equation is

a² + bx + c = 0 ఒక వర్గ సమీకరణము కావలెను అనిన నియమము ఏది?

(1) a ≠ 0, a, b, c∈ R                   (2) a = 0, b = 0, c ≠ 0

(3) a = 0, b ≠ 0, c ≠ 0             (4) a = b= c = 0

    Answer: (3)        

3. If the roots of the quadratic equation px² + qx + r = 0, are equal then q² =

px² + qx + r = 0 వర్గ సమీకరణము యొక్క మూలాలు సమానమైన q² =

(1) 2pr      (2) 3pr           (3) 4pr      (4) 8pr

Answer: (2)

Solution:  Roots of the quadratic equation px² + qx + r = 0 are equal

q² – 4pr = 0 ⟹ q² = 4pr

4. The sum of roots of the quadratic equation 3x²– 6x + 1=0 is

3x²– 6x + 1=0 వర్గ సమీకరణము యొక్క మూలాల మొత్తము

(1)   2       (2) 3± √6  (3) – 3         (4)  1/3

Answer: (1)

Solution:   Equation is 3x²– 6x + 1=0

                   a = 3, b = – 6,  c = 1

sum of l roots = – b/a = – (6)/3 =2

   Progressions

1. Find the 10^th term of the AP 5, 1, 2, – 3, – 7,…… is

      5, 1, 2, – 3, – 7,…… అంక శ్రేడి యొక్క 10 వ పదం

(1) – 31                           (2)  31

(3)   – 27                           (4) – 35

Answer: (1)

Solution:  Given AP is 5, 1, 2, – 3, – 7,……

a = 5, d = 1 – 5 = – 4

10^th term = a + 9d

= 5 + 9(– 4)

= 5 – 36

= – 31

2. If a, b, c are in GP, then a/b =

a, b, c లు గుణ శ్రేడి లో ఉన్నచో  a/b =

(1) b/c        (2) c/b          (3) b/a            (4) c/a

Answer: (1)

Solution:     Given a, b, c are in GP

a₂/a₁ = a₃/a₁

b/a = c/b

⟹ a/b = b/c

3. If the 2^nd term and 5^th term of a GP are 24, 81 then the r =

          గుణ శ్రేడి లో 2 వ పదం మరియు 5 వ పదం 24, 81అయితే  r =

 (1) 16           (2) 3         (3) 20         (4) 3/2

    Answer: (4)

   Solution:    Given a₂ = 24, a₅ = 81

ar = 24

 ar⁴ = 81

ar⁴/ar = 81/24

r³ = 27/8

r = 3/2

4. Which term of P. √3, 3, 3√3,………. is 729

√3, 3, 3√3,………. గుణశ్రేఢిలో ఉంటే  729 ఎన్నవ పదము  

(1) 10     (2) 12       (3) 14      (4) 16

Answer: (2)

Solution:   Given G.P is √3, 3, 3√3,…..

a = √3, r = 3/√3 = √3

a_n = 729

ar^{n – 1}  = 3⁶

ar^{n – 1}  = 3⁶

(√3) (√3)^{n – 1} = 3⁶

(√3)ⁿ  = 3⁶

(3^1/2)ⁿ  = 3⁶

3^n/2  = 3⁶

n  = 12

12^th term of the given G.P. is 729

5. The sum of first 100 natural numbers is

మొదటి 100 సహజ సంఖ్యల మొత్తము

(1) 2250     (2) 5100       (3) 5000      (4) 5050

     Answer: (4)

Solution:

We know that sum of first n natural number = [n(n + 1)]/2

sum of first 100 natural number = [100(100 + 1)]/2

= 50(101)

= 5050

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