# Quant Boosters by Shubham Ranka, IIM Calcutta

• (a+b)(a-b)=3k {k=1 to 100}
Sum of factors and difference of factors =even . So, both even or both odd.
For odd values of k, a-b=1, and take a=3k+1/2, b=3k-1/2. 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 = 1-6c5/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' = 7-5 = 2
7c5
==> 1 - 6/ (12c5-6c1-7c5) = 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 + (n-1) * -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

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