TS Polycet (Polytechnic) 2018 Previous Question Paper with Answers And Model Papers With Complete Analysis

Q). If \frac x7+\frac y3 = 5 and \frac x2-\frac y9=6 , then x, y is …………………
A) (9, 14)
B) (14,9)
C) (7, 9)
D) (14,18)

View Answer
B) (14,9)
Explanation: \frac x7+\frac y3= 5
\frac{3x+7y}{21} = 5
⇒ 3x+7y = 105
\frac x2-\frac y9=6
\frac{9x-2y}{18}=6
9x – 2y = 108 …………….(1)
3(3x + 7y = 105) = 9x + 21y = 315 …………….(2)
by solving (1) and (2)
9x — 2y= 108
9x + 21y = 315
(-) (-) (-)
____________
-23y = -207
Y = 9
Now substitute y = 9 in 3x + 7y = 105
3x + 7 x 9 = 105
⇒ 3x + 63 = 105
⇒ 3x = 105 – 63 = 42
⇒ x = 14
∴ x = 14 and y = 9

Q). The graph of y + x2 = 0 lies in the quadrants
A) Q1,Q2
B) Q2,Q3
C) Q3,Q4
D) Q4,Q1

View Answer
C) Q3,Q4
Explanation: y = m x2, m < 0, the graph lies in Q3 and Q4
∴ y = – x2 lies in Q3 and Q4

Q). Solution of the equations √2x + √3y = 0 and √3x – √8y = 0 is ………………………
A) x = 1, y = 0
B) x = 0;y=l
C) x=1,y=1
D) x = 0, y = 0

View Answer
D) x = 0, y = 0

Q). If 2X + 3y = 17 and 2X + 2 – 3y + 1 = 5, then y = …………………….
A) 2
B) 3
C) 1
D) 7

View Answer
A) 2
Explanation: Let 2X = a and 3y = b
2X + 3y = 17 ⇒ a + b = 17 ⇒ b = 17 – a
2X + 2 – 3y x 31 = 5
⇒ 2X . 2 2 – 3y . 31 = 5
⇒ 4a -3b = 5
=> 4a-3(17-a) = 5=>4a-51+3a = 5
=> 7a = 56 => a = 56/7 = 8
∴a = 8 => 2x = 8 => 2x = 23 => x = 3
=> 4(8) -3b = 5 => 32 -3b = 5
=> 27 = 3b => b = 9
∴ b = 9=>3y = 9=>3y = 32=>y = 2

Q). x + y = 2015 has ….. number of solutions.
A) 10
B) 2014
C) 20
D) infinite

View Answer
D) infinite

Q). The condition for a x2 + bx + c = 0 to be a quadratic equation is
A) a, b, c ∈R
B) a=0,c ≠0
C) a ≠0
D) a=0,b ≠0

View Answer
C) a ≠0

Q). The difference between two consecutive positive integers is
A) 2
B) 3
C) 1
D) 4

View Answer
C) 1

Q). If α,β are the roots of ax2 + bx + c = 0, then 1/α + 1/β =
A) –b/c
B) b/c
C) c/b
D) –c/b

View Answer
A) –b/c
Explanation: 1/α + 1/β = \frac{\alpha\;+\;\beta}{\alpha\beta}
= \frac{-b/a}{c/a}
= -b/c

Q). If 65 diagonals are drawn to a polygon of ‘n’ sides, then what is the value of ‘n’?
A) 10
B) 20
C) 13
D) None

View Answer
C) 13
Explanation: No. of sides of polygon = n
No. of diagonal of a polygon having ‘n’ sides = \frac{n(n-3)}2
\frac{n(n-3)}2 = 65 ⇒ n2 – 3n = 130
⇒ n2-3n-130 = 0 ⇒ n2 + 10n-13n-130=0
⇒ n (n + 10) – 13 (n + 10) =0
⇒ (n + 10) (n – 13) = 0 ⇒ n = 13
∴ No. of sides are n = 13

Q). Find the 10th term in A.P. 5, 1, -3, -7,…………….
A) -31
B) 31
C) 41
D) 71

View Answer
A) -31
Explanation: a = 5, d=1-5 = 4
10th term = a10 = a + 9d,= 5 + 9(-4)
= 5 -36
= -31

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