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

B) (14,9)
Explanation:= 5
⇒ = 5
⇒ 3x+7y = 105
=6
=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

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

D) x = 0, y = .0

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

D) infinite

C) a ≠0

C) 1

A) –b/c
Explanation:1/α + 1/β =
=
= -b/c

C) 13
Explanation:No. of sides of polygon = n
No. of diagonal of a polygon having ‘n’ sides =
= 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

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