CTET Solved Paper MCQ - UTET April 2015 Paper2

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

64¹² ÷ 4¹⁵ = 64^?
⇒ (2⁶)¹² ÷ (2²)¹⁵ = (2⁶)^?
⇒
⇒ (2)⁴² = (2)^{6 × ?}
Comparing the power of both side,
42 = 6 × ? ⇒ ? = 7

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

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.