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# sir,can u tell me those chapters in maths of XIth andXIIth which can get me a cut off marks

10 years ago

there are none of them in individual.every year they give equal or partially equal questions in all chapters.they used to give a bit more questions in the calculus chapter but as we got used to solve them they are giving twisted questions from even easy chapters like complex numbers. iits hav on their mind to introduce aptitude questions in jee. they are really tough check out these.

1-You are a cook in a remote area with no clocks or other way of keeping time other than a four minute sandglass timer and a seven minute sandglass timer. (The kind you turn over - hourglass shaped) You do have a stove, however, with water in a pot already boiling. Somebody asks you for a nine-minute egg, and you know this person is a perfectionist and will be able to tell if you undercook or overcook the eggs by even a few seconds. What is the least amount of time it will take to prepare the egg? And how will you prepare it so that it is neither undercooked or overcooked?

2-A princess is as old as the prince will be when the princess is twice the age that the prince was when the princess's age was half the sum of their present ages.

What are their ages?

10 years ago

check out the first question's solution

The answer is 9 minutes. First, flip both hourglasses over and drop the egg into the water. When the four minute timer runs out, flip it again. When the seven minute timer runs out, flip it over. The egg has been cooking seven minutes. Now when the four minute timer runs out again (after eight minutes) flip the seven minute timer back over. Since the seven minute timer has been running only a minute between flips, there's a minute worth of sand left. And when that minute runs out, the egg will have been cooking for exactly nine minutes.

10 years ago

here's the 2nd question's solution

It’s a bit messy… but using simple algebra… I’m sure there are much easier ways of accomplishing the solution.

Princess: a
Prince: b

0.5*(a+b) a – (0.5*(a+b)) = 0.5*a-0.5*b b – (0.5*a-0.5*b) = 1.5*b-0.5*a 3*b-a
(3*b-a) – a = 3*b-2*a b + (3*b-2*a) = 4*b-2*a a = 4*b-2*a or 3*a = 4*b b+10
0.5*b+5 b – (0.5*b+5) = 0.5*b-5 a – (0.5*b-5) = a-0.5*b+5 b = a-0.5*b+5 or a = 1.5*b-5

Now we have two equations:
3*a = 4*b and a = 1.5*b-5.

Substitute “a” in the first equation for the value of “a” from the second equation:
3*(1.5*b-5) = 4*b

Solving this equation, we get:
b=30

Since a = 1.5*b-5:
a=40

10 years ago

heres the altenative solution for the second question

create the following table from the riddle:

 Current Future Past Princess x 2z (x+y)/2 Prince y x z

then create three equations, since the difference in their age will always be the same.

d = the difference in ages

x – y = d
2z – x = d
x/2 + y/2 – z = d

then create a matrix and solved it using row reduction

 x y z 1 -1 0 d -1 0 2 d .5 .5 -1 d

It reduces to:

 x y z 1 0 0 4d 0 1 0 3d 0 0 1 5d/2

This means that you can pick any difference you want (an even one presumably because you want integer ages).

Princess age: 4d
Prince age: 3d

Ages that work

 Princess Prince 4 3 8 6 16 12 24 18 32 24 40 30 48 36 56 42 64 48 72 54 80 60