TS Polycet (Polytechnic) 2020 Previous Question Paper with Answers And Model Papers With Complete Analysis

TS Polycet (Polytechnic) 2020 Previous Question Paper with Answers And Model Papers With Complete Analysis
TS Polycet (Polytechnic) Previous Year Question Papers And Model Papers:
While preparing for TS Polycet (Polytechnic), candidates must also refer to the previous year question papers of the same. Scoring well in TS Polycet (Polytechnic) and understanding weaknesses and strengths in the respective sections.

TS Polycet (Polytechnic) Previous Year Question Papers can be found on this page in PDF format. Students taking the exam to get into some of the best Polytechnic colleges/institutes in the state of Andhra Pradesh may practice these papers to get a clear idea of the structure of the exam, marking scheme, important topics, etc.


Section — I
MATHEMATICS





Q) 10th term of A.P.: 13, 8, 3, -2,…………… is ___________
A) -32
B) -23
C) 30
D) -30

View Answer
A) -32
Explanation: 13, 8, 3,-2…….. A.P.
Here a = 13, d = 8 — 13 = —5
Nth term an = a + (n – 1) d
∴ 10th term a10 = 13 + (10 – 1) (-5)
= 13 + 9 (-5)
= 13-45 = -32

Q) If the slope of the line through (2, -7) and (x, 5) is 3, then x =__________
A) 4
B) 5
C) 6
D) 7

View Answer
C) 6
Explanation: Let A (2, -7), B (x, 5)
Slope of the line passing through
A (x1,y1), B (x2,y2) is
AB= \frac{Y2-Y1}{X2-X1}
Given slope m = 3
3 = \frac{5-(-7)}{X-2}
3 = 12/x-2
=> 3x-6 = 12
=> 3x = 18
=> x = 6

Q) The centroid of the triangle whose vertices are (3, -5), (-7, 4), (10, -2) is ____________
A) (1, 1)
B) (1,-2)
C) (-2,1)
D) (2,-1)

View Answer
D) (2,-1)
Explanation: The centroid of the Ale formed by the points A (x1,y1), B (x2,y2), C (x3,y3) is
G = (\frac{x1+x2+x3}3,\frac{y1+y2+y3}3)
G = (\frac{3-7+10}3,\frac{-5+4-2}3)
G = (\frac63,\frac{-3}3)
= (2,-1)

Q) If the roots of 2 x2+ kx + 3 = 0 are real and equal, then k =
A) ±6√2
B) ±4
C) ±2√6
D) ±5

View Answer
C) ±2√6
Explanation: Given quadratic equation is
2 x2+ kx + 3 = 0
Comparing with
a x2 + bx + c = 0
a = 2, b = k, c = 3
Roots are real and equal, then
b2 – 4ac = 0
=> (k)2 – 4(2)(3) = 0
=> k2 – 24 = 0
k2 = 24
=> k = ± √24 = ±2 √6

Q) Product of the roots of √3 x2 – 2x -√3 =0 is
A) 1
B) -1
C) 2/√3
D) -2/√3

View Answer
B) -1
Explanation: Given quadratic equation is
√3 x2-2x-√3=0
Comparing with a x2 + bx + c = 0
a = √3, b = —2, c = — √3
Product of the roots = c/a
= -√3/√3
= -1

Q) 4, 7, 10,………….. are in A.P., then stun of 15 terms is _________
A) 385
B) 475
C) 375
D) 325

View Answer
C) 375
Explanation: 4, 7, 10……….A.P.
Here a = 4, d = 7 -4 = 3, n = 15
Sn = n/2 [2a + (n-1)d] ∴ S15 = 15/2 [2(4) + (15- 1) 3] = 15/2 [8 + 14 x 3] = 15/2 [8 + 42] = 15/2 x 50
= 15 x 25 = 375

Q) The point (2,-3) divides the line segment joining the points (-1, 3), (4; -7) in the ratio_______
A) 3 : 2
B) 2 : 3
C) 8 : 1
D) 1 : 4

View Answer
A) 3 : 2
Explanation: The ratio in which (x, y) divides the line segment joining (x1,y1), (x2,y2) is
X1 – X : X -X2
(x, y) = (2,-3), (x1,y1) = (-1, 3), (x2,y2) = (4, -7)
= -1 – 2 : 2 – 4
= -3 : -2 = 3 : 2

Q) If (8, 1), (k, —4), (2, -5) are collinear, then k= __________
A) 4
B) 3
C) 2
D) 1

View Answer
B) 3
Explanation: (8, 1), (k, -4), (2, -5) are collinear, then area of △ABC = 0
X1 (y2-y3)+ X2 (y3-y1)+ X3 (y1-y2) = 0
=> 8(—4 + 5) + k (-5-1) + 2 (1 + 4) = 0
=> 8(1) + k(-6) +2 (5) = 0
=> 8-6k+10 = 0
=>18-6k = 0
=> -6k = -18
=>k = 3

Q) 8 x2-6x-9 = ________
A) (2x-3)(x-3)
B) (2x-3)(x + 1)
C) (2x+1) (x- 1)
D) (2x – 3) (4x + 3)

View Answer
D) (2x – 3) (4x + 3)
Explanation: 8 x2-6x-9
= 8 x2– 12x + 6x-9
= 4x(2x-3) + 3 (2x-3)
= (2x – 3) (4x + 3)

Q) 1,-1/2,1/4,…………………. are in G.P., then find 8th term.
A) 1/128
B) 1/64
C) -1/128
D) -1/64

View Answer
C
Explanation: 1,-1/2,1/4,………………G.P.
Here s = 1, r=-1/2/1 = -1/2, n = 8
An = a rn-1
∴ A8 =1(-1/2)8-1 = (-1/2)7 = -1/128
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