CTET Solved Paper - UTET April 2015 Paper2
- Option : B
- Explanation : (B) From the given information, we can form the figure as,
Now, in △ABE, AB = AE (As these are the sides of regular pentagon)
So, from base angle theorem, we know
∠BEA = ∠EBA
∠BAE = 1800 (As we know each angle of regular pentagon is 1800.)
Now, from angle sum property in △ABE,
∠BEA + ∠EBA + ∠BAE = 1800,
So, ∠BEA + ∠BEA + 1800 = 1800,
2∠BEA = 720
∠BEA = 360
∠BEA = ∠EBA 360 ....... (i)
Similarly, in △DEC,
∠CED = ∠ECD 360 ....... (ii)
And in △CDB,
∠BDC = ∠DBC = 360 ....... (iii)
Then,
∠DEA = ∠BEA + ∠CEB + ∠CED
We know, ∠DEA = 1800(Angle of regular pentagon)
Then, 360 + ∠CEB + 360 = 1080
∠CEB = 360
Similarly, ∠CEB = ∠CAD
= ∠ADB = ∠ACE
= ∠DBE = 360
And sum of all 5 vertical angles of star = 360 + 360 + 360 + 360 + 360 = 1800
92. If 6412 ÷ 415 = 64?, then what will come in place of question mark(?)?
- Option : C
- Explanation : (C) 6412 ÷ 415 = 64?
⇒ (26)12 ÷ (22)15 = (26)?
⇒
⇒ (2)42 = (2)6 × ?
Comparing the power of both side,
42 = 6 × ? ⇒ ? = 7
- Option : B
- Explanation : (B) If a2 = 13 + 23 + 33 = 1 + 8 + 27
⇒ a2 = 36 ⇒ a= 6
Now, b2 = 13 + 23 + 33 + 43
= 1 + 8 + 27 + 64 = 100
⇒ b = 10
And c2 = 13 + 23 + 33 + 43 + 53
= 1 + 8 + 27 + 64 + 125 = 225
⇒ c = 15
Then, a + b + c = 6 + 10 + 15 = 31
- Option : C
- Explanation : (C) Desired probability lies in the range of 1-7 only.
Then, again from 8, remainder starts with a count 1, 2 ....
Total favourable cases are 7 and in only 1 case remainder is 1. Thus, probability that the remainder is 1 = 1/7.
∴ The probability that the remainder is not 1
95. Which of the following is not a 'proposition'?
- Option : C
- Explanation : (C) A proposition is a statement which can be either true or false but not both. So, (A), (B) and (C) are propositions but (C) is a question.
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