We will start with power plays now. It shall consist of 10 Problems and You are expected to solve them in 25-30 Mins. Each correct answer carries 5 marks. The first two wrong answers carry (-1) marks each, the next two (-2) marks each and so on. Maximum time is 30 mins, but you are expected to solve in 25 mins!
Good Luck !
Please hit the comment button and post your keys, I will post the official keys and solution in 7 days time!
And We will have a better mechanism hopefully, by the next powerplay!
Power Play I
1) Two different positive numbers a and b differ from their reciprocals by 1. Find a+b
A) 1 B) √6 c) √5 D) 3 E) None of these.
2) How many positive integer multiples of 1001 can be expressed in the form 10^j-10^i where i and j are integers such that 0<=i<j<99?
A) 15 B) 90 C) 64 D) 720 E) 784
3) How many subsets of the set {1,2,3...,30} have the property that the sum of all elements is more than 232?
A) 2^30 B) 2^29 C) 2^29 -1 D) 2^29+1 E ) None of these
4) Let the point P be (4,3). Choose a point Q on y=x line and another point R on the line y=0 such that sum of lengths PQ+QR+PR is minimum. Find this minimum length.
A) 5√2 B) 3+3√2 C) 5 D) 4 E) 10
5) Let f be a function defined on odd natural numbers which return natural numbers such that f(n+2)=f(n)+n
and f(1)=0 . Then f(201)?
A) 10000 B)20000 C) 40000 D) 2500 E) None of these
6) A semicircle is inscribed in a right triangle so that its diameter lies on the hypotenuse and the centre divides the hypotenuse into segments 15 cm and 20 cm long. FInd the length of the arc of the semicricle included between its points of tangency with the legs.
A) 2π B) 3π C) 4π D) 6π E) none of these
7) The product of n numbers is n and their sum is 0. Then n is always divisible by
A) 3 B) 4 C) 5 D) 2 E) None of the foregoing
8 ) Let P(x) be a polynomial with integer coefficients such that P(17)=10 and P(24)=17 . It is further know that P(n)=n+3 has two distinct integer solutions a and b. Find a.b
A) 10 B) 220 C) 190 D) 48 E) 418
9) Let A and B be two points on the plane. Let S be the set of points P such that PA^2+PB^2 is at most 10. Find the area of S
A) 2π B) π C) 4π D) 6π E) none of these
10) Let T={9^k: k is an ineteger 0<=k<=4000}. Given that 9^4000 has its 3817 digits and its leftmost digit is 9. Find the number of elements in T which also have leftmost digit as 9.
A) 92 B) 93 C) 183 D) 184 E) 185
Good Luck !
1- option(3) root(5) a^2-a+1 = 0
ReplyDelete4) B) 3+3√2
ReplyDeleteQ lies on the circle with PR=3 as the diameter.
1) Two different positive numbers a and b differ from their reciprocals by 1. Find a+b
ReplyDeleteA) 1 B) √6 c) √5 D) 3 E) None of these.
A) 1
3) How many subsets of the set {1,2,3…,30} have the property that the sum of all elements is more than 232?
A) 2^30 B) 2^29 C) 2^29 -1 D) 2^29+1 E ) None of these
E ) None of these
4) Let the point P be (4,3). Choose a point Q on y=x line and another point R on the line y=0 such that sum of lengths PQ+QR+PR is minimum. Find this minimum length.
A) 5√2 B) 3+3√2 C) 5 D) 4 E) 10
c) 5
5) Let f be a function defined on odd natural numbers which return natural numbers such that f(n+2)=f(n)+n
and f(1)=0 . Then f(201)?
A) 10000 B)20000 C) 40000 D) 2500 E) None of these
A) 10000
7) The product of n numbers is n and their sum is 0. Then n is always divisible by
A) 3 B) 4 C) 5 D) 2 E) None of the foregoing
B) 4
[...] Originally Posted by dare2 which test are you talking about?? can i have the link of that? Power Play I ( 30 August 200 Ultimate Quant Marathon Blog For IIM CAT [...]
ReplyDeleteMy take..
ReplyDelete1)A
4)B
5)A
6)D
7)B
8)E
Most of the problems I got, I got them in 1-2 minutes. But at least 5 of them here are of QQAD standard. Great job Implex..Expect more of these from you:)
1. was very easy a-1/a = 1 and ten sum of roots. 5 second problem
ReplyDelete4. Prade did it better. I did it by a bit of trial and error and narrowed it.
5. Easy (n-1/2)^2
6. You get the sides as 35,28 and 21 and R=12..Circumference/4
7. Did it by trial and error
8. We can easily say that its a quadratic equation. Bit of intensive calculation involved. You can avoid it by making use of the answer choices.
6) A semicircle is inscribed in a right triangle so that its diameter lies on the hypotenuse and the centre divides the hypotenuse into segments 15 cm and 20 cm long. FInd the length of the arc of the semicricle included between its points of tangency with the legs.
ReplyDeleteA) 2π B) 3π C) 4π D) 6π E) none of these
Option D - 6pi
let the radius be r
and the point of tangency be P and Q and triangle be ABC. P lies on AB and Q on BC
let AP = m and BQ = n
m^2 = 15^2-x^2
n^2 = 20^2-x^2
m = 9 and n = 16 x =12 arcPQ = 6pi
[...] Power Play I ( 30 August 2008) [...]
ReplyDeleteMy attempts
ReplyDelete1. C
2. B
3. E
4. B
5. A
For Q 10 I think it's one of the last three options 183, 184 or 185.
ReplyDeleteFor Q 2 I got 93 values but as 90 was the closest, opted for that.
ReplyDelete