Research Aptitude Test - 28

AP 20,15,10

Here a = 20

d = - 5

15^th term will be a + 14d which is equal to 20 + (-5) x 14 = -50

Let numbers are

a - d, a and a + d

now it is given that sum = 15 & sum of there square is 83 i.e,

3a = 15 …(i)

(a - d)² + a² + (a + d)² = 83 …(ii)

3a² + 2d² = 83

75 +2d² = 83

2d² = 8

d² = 4, d = 2

So, least term is a - d = 3

Out of letters in the word ‘BHARAT' two letters, that is, A's are alike.

∴ Number of permutations = 6!/21 = 360.

Number of words in which B and H are never together = Total number of words - number of words in which B and H are together

= 360 - (5!/2!).2! = 360 - 120 = 240.

The first event of going from Delhi to Mumbai can be performed in 15 ways as he can go by any of the 15 buses. But the event of coming back from Mumbai can be performed in 14 ways ( a different bus is to be taken).

Hence, both the events can be performed in 15 x 14 = 210 ways.

Since the first exercise contains 15 questions, the number of ways of choosing the first question In 15.

Since the second exercise contains 12 questions, the number of ways of choosing the second question is 12. Hence, by the fundamental principle, two questions can be selected in 15 x 12 = 180 ways.

If D occupies the first place n(A) = 7!, Total words = n(S) = 8!

∴ Required probability

Out of theTetters in the word ‘UNIVERSITY’ two letters ‘I’ are alike.

∴ Number of permutations = 10!/2 …(i)

Number of words in which two ‘I’ are never together = Total number of words - Number of words in which two ‘I’ are together

∴ Required probability =

= 8/10 = ⅘.

We know P(A) = 1-P(A)

= 1 - 0.65 = 0.35 and

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ 0.65 = 0.35 + P(B) - 0

[∵ A and B are mutually exclusive events]

⇒ P(B) = 0.65 - 0.35 = 0.30.