Arun has certain number of books. It is not less than 15 and not more than 30. If he wants to give these books equally to 2 of his friends, then 1 book will remain. If he wants to give these books equally to 3 of his friends, then 2 books will remain. If he wants to give these books equally to 4 of his friends, then 3 books will remain. What is the number of books he had ?

Here, n = 3, m = 6, R₁ = 4%
? R₂ = [(m - 1)/(n - 1)] x R₁
= [(6 - 1)/(3 - 1)] x 4
=(5/2) x 4
= 10%

Since, the two simple interest are equal.
Then, (4000 x 3 x R)/100 = (5000 x 12 x 2)/100
? R = 10%

Given Exp. = 6/7 + [(y - x)/(y + x)]
= 6/7 + [1 - (x/y) / 1 + (x/y)]
= 6/7 + [1 - (3/4)] / [1 + (3/4)]
= 6/7 + 1/7 =1.

As Sum = [(100 x SI)/(Time x Rate)]
here, let R =x%, T=x yr, and, SI=Rs x
? Sum=[(100 × x)/(x × x)]
=(100/x)

When three dice are rolled, the number of possible outcomes = 6³ = 216
Number of possible outcomes in which 2 does not appear on any dice = 5³ = 125
? Number of possible outcomes in which atleast one dice shows 2 = 216 - 125 = 91

Let sum = P
Then, SI = (P x R x T)/100
= {P x (27/2) x 4} / 100
= {P x 54} / 100 = 27P / 50
? Amount = P + 27P / 50 = 77P/50
According to the question 77P/50 = 3080
? P = (3080 x 50)/77 = ? 2000

Let us draw a figure below as per given question.
Let AB = CD = h meter be the heights of the towers. E is a point such that DE = 100 meter;
?CED = 60° and ?AEB = 30°
Now, BE = x meter (say)
From right triangle CDE.
h = 100 tan 60°
? h = 100?3 meter
From right triangle ABE,
x = h cot 30° put the value of h, we will get
x = 100?3 X ?3
x = 100 X 3 = 300 meters
Distance between the tower = DE + EB = 100 + 300 = 400 meters
Height of the tower = h = 100?3 meter

Taking p = 11
2^p - 1 = 2¹¹ - 1
= 2047

Since 2047 is divisible by 23 it is not prime. Thus, required least positive prime number is 11.