Quant Boosters by Shubham Ranka, IIM Calcutta

Taylor series expansion for e^x = 1 + x + x^2/2! + x^3/3! +...
e1 = 1+1/2!+1/3!+1/4!+....
3,7,15,27,43,
Numerator:
Tn = Tn1 + 4(n1), n > =2
T1=3
Tn = 3+ 4 (n(n1)/2)
== > Generic Tn = [3+2(n(n1))]/n!
= 3/n! + 2/(n2)! (n > =2)
= 3(e1) + 2e
= 5e3

Q 22) In the year 1980, the age of a person is 1/89th of the year of his birth. What was the age of person in 2012?

Let x be the man's year of birth,
Then x/89 = 1980  x
== > x = 1958
== > In 2012 Age was: 54 years

Q23) Let a, b, c, d belongs to [1/2, 2]. Suppose abcd=1. Find the maximum and minimum value of (a + 1/b)(b + 1/c)(c + 1/d)(d + 1/a).

(ab+1)(bc+1)(cd+1)(ad+1)
(ab+1)(cd+1) = (ab+1)(1/ab + 1)
Now: (x+1)(1/x +1) = 1+x+1/x+1
We know, that: x+1/x > =2 (2 @ x=1)
== > (ab+1)(1/ab + 1) > 2+2 =4
Similarly: (bc+1)(ad+1) > =4
Since, in the given domain: a=b=c=d=1 satisfies:
== > Min. values= 4*4 = 16
Put a=b=c=d=2 to get the max. value

Q24) In the series 3,6,9,12.. how many of the first 100 terms can be expressed as a difference of 2 perfect squares ?

(a+b)(ab)=3k {k=1 to 100}
Sum of factors and difference of factors =even . So, both even or both odd.
For odd values of k, ab=1, and take a=3k+1/2, b=3k1/2. So, 50 values of k
For even values of k, should be a multiple of 4. So, 25 values of k.
Hence, a total of 75 terms can be expressed.

Q25) In the game of cricket , Runs scored in one ball can be 0(dot ball) , 1, 2, 3 , 4 or 6. What is the probability that there is at least one dot ball in an over in which 7 runs was scored. No extra deliveries were bowled.

a+b+c+d+e+f = 7  (A)
a,b,c,d,e,f > 0
==> 6c5
==> p = 16c5/N
N = no. of solns of (A)
= 12c5  x
x = Ways in which he scores 5 or 7
For x = 7 > 6c1 = 6
When any of the value is 5
a'+b'+c'+d'+e'+f' = 75 = 2
7c5
==> 1  6/ (12c56c17c5) = 769/775

Q 26) P and Q are the two opposite ends of a pond and the distance between them is 420 metres. Messi and Ronaldo start swimming towards each other at the same time from P and Q, with speeds in the ratio 5 : 9 respectively. As soon as any of them reaches an end, he turns back and starts swimming towards the other end. At what distance (in metres) from P will they meet when Ronaldo is in his 13th round?

For the same distance, Ratio of time will be inverse of speeds = 9 : 5
When Ronaldo covers 9 rounds, Messi covers 5.
==> When Ronaldo covers 13 , Messi will cover 13 x 5 / 9 = 65/9 rounds.
==> Distance = 420 x 65/9 = 9100/3 m
FOr P, nearest multiple of the track length = 420 x 7 (closest to 65/9 ) = 2960m.
==> Distance travelled from P = 9100/3  2960 = 43.33m

Q27) No of negative integral solution of 7a + 17b + 1000 = 0 is ?

a0
7x + 17y = 1000
mod 3y/7 = 6
x y
138 2
Tn = 138 + (n1) * 17 > 0
n < 9. xx
===> 9 solns

Q 28) 20% of notes are fake.Probability of accepting a fake for an original is .1 & original never rejected.The probability of accepting original is

p(F) = 0.2  p(O) = 0.8
p(Acceptance) = 0.8 + 0.2*.1 = .82
p(Acceptance of Original) = .8/.82 = .97

Q 29) There are three classes X, Y and Z. The average age of the students of classes X and Y together is 20 years. The average age of the students of classes Y and Z together is 35 years. The average age of the students of classes X and Z together is 15 years. Which of the following can be the average age of all the three classes together?
a) 16
b) 38
c) 30
d) 25

X+Y = 20(a+b) (1)
Y+Z=35(b+c) (2)
X+Z = 15(a+c) (3)
Let, X+Y+Z/a+b+c = M
Adding (1), (2) and (3) :
X+Y+Z = 5(7a+11b+10c)
= 35(a+b+c) + 5 (4b+3c)
M = 35 + 5 (4b+3c)/(a+b+c)
which is greater than 35 ===> (B)

Q 30) A hospital nursery has 3 boys and some girls. Overnight a child is born and is added to the group. The next morning, one child is chosen from the group at random, with uniform probability, and turns out to be a boy.Determine the probability p that the overnight addition to the group was also a boy

p = prob (overnight born = boy Next day chosen = boy)
= prob (overnight born = boy AND Next day chosen = boy) / prob (Next day chosen = boy)
= .5 * 4/T / (.5*4/T + .5 * 3/T) = 2/3.5 = 4/7 ~ .57