CTET Solved Paper - UTET April 2015 Paper2
- Option : A
- Explanation : (A) Given condition is,
h2 = 2 (6 × p)
⇒ h2 = 2 (12 × p) (Here, p = Perpendicular)
We know that,
h2 = b2 + p2
⇒ h2 = 144 + p2
⇒ 24p = 144 + p2
⇒ p2 -24p + 144 = 0
⇒ p2 - 12p - 12p + 144 = 0
⇒ p(p - 12)-12(p - 12) = 0
Solving above equation, we get, p = 12
So, the perpendicular of triangle will be 12 cm
- Option : D
- Explanation : (D) Let five consecutive integers are (n - 4), (n - 3), (n - 2), (n - 1), n
According to question,
S = n - 4 + n -3 + n - 2 + n - 1 + n [n = Largest integer]
S = 5n - 10
S + 10 = 5n
n = (S + 10) / 5
- Option : A
- Explanation : (A) Total number of players = 5 and it is compulsory for each player to play match with every other player.
So, number of matches = 3C2
104. In the given figure, both ABCD and PQRS are the parallelograms. The value of θ will be
- Option : A
- Explanation : (A) From the figure, AD ∥ BC
So, ∠ADC = ∠BCR = 1200 [corresponding angles]
∴ ∠BCS = 1800 - 1200 = 600
Similarly, SP ∥ RQ
∠RQP = ∠SPB = 700 [corresponding angles]
Now, DC ∥ AP and SP is a transversal
∴ ∠RSP = 700 [alternate angles]
In △OSC,
∠OSC + ∠SCO + ∠COS = 1800
⇒ 700 + 600 + ∠COS = 1800
⇒ ∠COS = θ = 500
- Option : C
- Explanation : (C) Solving a routine type of problem using a formula does not demonstrate the achievement of skill objectives of teaching school mathematics in upper primary stages
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