Problem Solving

Thanks for the help:)

Srinivasa, please post only one question per thread. Posting more than 1 question per thread can create too much confusion while discussing the Questions.

So, I would advise you to edit this thread and create new threads, and distribute these questions among them.

Solution for Question 1:

length 40 to be divided by two parts
lets for circle with radius r, it is 2*pi*r
then remaing length for square would be 40 -2*pi*r
lets length of each side of square is a
then 4a=40 -2*pi*r
a= (10-(pi*r)/2)

hence total area pi*r^2 + (10-(pi*r)/2)^2

hence E
...............................................................
solution question 2:

Let's begin with the triangle on the left.

We know the sides are 1 and (sqrt 3) from point P.
If you know your special right triangles, you will quickly see that this is a 30-60-90 right triangle.

The angle opposite '1' is 30 degrees.



Let's move on to the triangle on the right.

We know that a straight line has 180 degrees.

Since we know the lower angle of the triangle on the left is 30 degrees, and we also know the angle between the two line segments is 90 degrees, the lower angle of the triangle on the right must be 60 degrees in order to sum to 180 degrees. (30 + 90 + x = 180; x = 60)

This means the triangle on the right is also a 30-60-90 triangle. The hypotenuse of this triangle is the same as the other triangle's (which is '2' by the Pythagorean Theorem), since both are radii of the same circle.

Using the same properties of a 30-60-90 triangle, you can find the side lengths and finally the point (s,t) which gives the value for s = 1.

Hence B
..........................................................
Solution for 3:
90(1/(v-3)-1/(v+3))=1/2
after solving this above equation

v=33
hence time for downstream 90/36 =2.5
hence A
...............................................................
solution for 4

Initial it was 4.
lets it increase by constant length d

end of the first year 4+d
end of the 2nd year = 4+2d
.................
similarly end of the 6th year 4+6d
and similarly end of the 4th year 4+4d

hence according to the problem
4+6d=(4+4d)+1/5(4+4d)
>> 6d= 4
>> d=2/3
hence D

Thanks a lot sanjoy18 for the great explanations! It really helps.

Thanks Rich and Rahul - will post one question per post in future:)