Quant Boosters by Shubham Ranka, IIM Calcutta

Shubham Ranka is currently pursuing his PGDM from IIM Calcutta and holds an electrical engineering degree from IIT Gandhinagar. Shubham aced CAT with 99.65 percentile and considers himself as an extreme mathlover, something you would expect from an IITian but his craze for the subject is relatively higher. In his free time, he mentors students for competitive exams like IITJEE, CAT, SSC, Aptitude tests and he served as a TIME faculty for 1.5 years. He likes table tennis, carrom and is also an ardent Poker fan. With an innovative mind to learn and try something new, Shubham’s dream is to be an entrepreneur.
Q1) N^2= 1+ 4018 × 4019 × 4020 × 4021, What is the value of N?
Method1:
N^2 = (4019.5^2.5^2) (4019.5^21.5^2) + 1. This is slightly lesser than 4019.5^2
Now, observe that N^2 must end with either 1 or 9.
And, N ~ 4019.5^2 = 16156380.25
so, the integer ending with either 1/9 just before 16156380.25 = 16156379. Cross check !!
Method2:
N^2 = 1+ a(a+1)(a+2)(a+3) == > N= (a+1)(a+2) 1Q2) A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kg.The clerk weights the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117,118, 120 and 121 kg. What is the weight of the heaviest box?
a. 60 kg
b. 62 kg
c. 64 kg
d. Cannot be determinedlet the weight of the 5 boxes be: a < b < c < d < e:
GIven, a+b = 110 and d+e = 121
Adding all the 10 eqns will give: a+b+c+d+e = 289
== > c= 58
Now, next min combination is 112 (a+c)== > b = 56, a = 54
Now, max = 121  > 59 + 62 or 60 + 61 (does not satisfy other combinations)
== > 62 KgQ3) If the points (k + 2, k +1), (k + 3, k) and (k + 4, k – 1) are collinear, then the value of k is
a) –3 only
b) an integer only
c) any real number
d) None of theseconcept: slope of line joining any 2 points should be same
1/1 = 1/1 ==== > any real number (C)Q4) if f(2x^2 5x + 3) = 3x+2, then Find f(x^2)
2x^2 5x + 3 = t
2x^2 5x + 3  t = 0
x = 5 + rt(1+8t))/4
== > f(t) = 3[5 + rt(1+8t))/4] + 2 = [23 + 3rt(1+8t)]/4
x^2 = 1/16[26+8t+10rt(1+8t)]
== > f(x^2) = f(1/8[13+4t+5rt(1+8t)]) = [23 + 3rt(p)]/4
where, p = 1+8*1/8[13+4t+5rt(1+8t)] = 14+4t+5rt(1+8t)
Converting to x :
1+8t = 16x^2 40x + 25 = (4x5)^2
== > p = 14+4(2x^2 5x + 3) + 5(4x5) = 8x^2 20x +26 + (20x25) = 8x^2 +1 or 8x^2 40x +51
f(x^2) = 23+3rt(8x^2 +1 or 8x^2 40x +51) / 4
Since, f(0) = 5
f(x^2) = [23  3rt(8x^2+1)]/4 satisfiesQ5) How many times in a day, the hands of a clock are straight?
A. 22
B. 24
C. 44
D. 48Straight == > 0 degree or 180 degree.
It makes an angle of 180 degrees 11 times in one circle. 00:30:xx, 1:35:xx, 4:50:xx, 6:00, 7:05:xx, ... 11:25:xx == > 22 times.
0 degrees  > 00:00: 1:05, ..10:50:xx. == > 2*11 = 22 times
== > 22+22 = 44 times (C)Q6) The last day of a century cannot be
A. Monday
B. Wednesday
C. Tuesday
D. FridayCalendar cycle repeats in 400 years.
100 years  > 5 odd days  > Friday
200 years  > 3 odd days  > Wed
300 odd days  > 1 odd day  > Mon
400 odd days  > 0 odd day  > Sunday
== > (C)Q7) The shopkeeper charged 12 rupees for a bunch of chocolate. but I bargained to shopkeeper and got two extra ones, and that made them cost one rupee for dozen less than first asking price. How many chocolates I received in 12 rupees?
n chocolates = Rs 12 à price per dozen = 144/n
I got n+2 in rs 12 = > Price per dozen = 144/n+2
144/n+2 = 144/n  1
== > n = 16Q8 ) Determine the positive integer values for n for which n^2+2 divides 2+2001n
(n^2 + 2) < = 2 + 2001n == > n < = 2001
^2 + 2 divides 2 + 2001n == > It will also divide 2+2001n  (n^2+2) = 2001nn^2
n^2 + 2 divides 2001nn^2 (1)
Case1:
If n is even = 2k:
4k^2 + 2 divides (4002k  4k^2)
2k^2 + 1 divides 2001k2k^2 = k(20012k)
== > 2k^2 + 1 divided 20012k (A)
Again, Since 2k^2 + 1 divides 2001k2k^2, it will also divide 2001k2k^2 + (2k^2 +1)= 2001k+1
=== > 2k^1 + 1 divides (20012k) and (2001k+1)
Trying to remove k:
2k^1 + 1 will also divides 2001(20012k) + 2(2001k+1) = 2001^2 + 2 = 4004003 = 19*83*2539
By hit and trial, k = 0,3 == > n = 6 (positive integer)
Case2:
If n is odd:
n^2 + 2 dvides 2001nn^2 = n(2001n)
== > n^2 + 2 divides 2001n (1)
Also, since n^2 + 2 dvides 2001nn^2, it also divides: 2001nn^2 + (n^2+2) = 2001n+2 (2)
Trying to remove n:
2001*(1) + (2):
n^2 + 2 divides 2001^2 + 2 = 19*83*2539
Using hit and trial:
== > n = 9, 2001
Hence n = 6,9,2001Q9) Ram and Kanti working together can complete a job in 11(13/16) days. Ram started working alone and quit the job after completing 1/3rd of it, and then Kanti completed the remaining work. They took 25 days to complete the work in this manner. How many days could Kanti have taken to complete the job working alone?
Let ram takes 3x no. of days to complete the task
And Kanti takes 3y no. of days to complete the task
== > 3xy/x+y = 11(13/16) = 189/16
== > 16xy = 63x+63y (1)
== > xy should be a multiple of 63 = 7*9 == > either of x or y should be a multiple of 7
1/3rd work  Ram x days
2/3rd workKanti 2y days
== > x+2y = 25 (2)
Now, either of x or y should be 7/21 == > x = odd == > 7
x = 7 == > y =9 == > satisfies 1
Kranti can complete the work in 3y = 27 days, working alone.Q10) If the sum of two distinct natural numbers is 50, then what can be the maximum possible HCF of these 2 numbers ?
(a) 5
(b) 10
(c) 11
(d) 12k (a+b) = 50
a and b are distinct and a,b > = 1 and a+b > =3 == > k < =17
k will be a factor of 50 < =27 == > 10,5,1
Max. of k = 10Q 11) In how many ways can 6 letters A, B, C, D, E and F be arranged in a row such that D is always somewhere between A and B?
6!  2 * x
x = D _ _ _ _ _ = 5! = 120
= _ D _ _ _ _ = 3c1 (Sice A/B can not come in the first place now) * 4! = 72
= _ _ D _ _ _ = 3c2 * 3! * 2! = 36
= _ _ _ D _ _ = 3! * 2! = 12
== > x = 120 + 72 + 36 + 12 = 240
== > Total no. of ways = 720 = 480 = 240Q 12) 2 years ago, fathers age was 6 times of his son's age. 6 years hence the ratio between the ages of father and son is 10:3. What is fathers present age?
40
42
44
482 years ago, let ages be: F S
F = 6s
F+8/S+8 = 10/3
3(6S+8 ) = 10(S+8 )
S = 7
F = 42
== > Current age is 44Q 13) A square is divided into 108 identical rectangles. The ratio of length to breadth of the rectangle is 3:4. If the diagonal of the rectangle measures 1 cm, then the length of the diagonal of the square is closest to
1) 9 cm
2) 10 cm
3) 11 cm
4) 12 cmLet the side of square be a, and rectangle be 3y, 4y
== > a^2 = 108*12y^2  (1)
Length of diagonal, 5y=1 == > y=1/5
== > length of diagonal of square = rt(2)a = rt(2*108*12) * 1/5 = 10.18  > (B )Q 14) If 3,2,2 are three altitudes of a triangle then find the area & perimeter of the triangle.
Corresponding sides will be: 2x,3x,3x
s = 4x
== > A = rt(4x*2x*x*x) = 2rt(2)x^2 = 1/2 * 3*2x
== > x = 3/2rt(2)
A = 3x = 9rt(2)/4
p = 2s = 8x = 6rt(2)Q 15) The earnings of a and b are in ratio 3:7 and that of b and c is 4:9 and that of d and c is ,7:6 if the sum of the earnings of a b c and d is 52950 then what are the earnings of d ?
a:b = 3:7
b:c = 4:9
c:d = 6:7
ac:d=72:168:378:441
TRICK:
[ac:d = 3*4*6:7*4*6:7*9*6:7*9*7]
a+b+c+d=52950
d = 52950* 441/1059 = 22050Q 16) f(x)=ax^2+bx+c (b > a) & f(x) ≥ 0 for all x. Find the minimum value of (a+b+c)/(ba)
a > 0, b^2 < = 4ac
By symmetry: a=c== > b=2a (D=0)
4/1== > 4Explanation:
We know, a > 0, b^2 < =4ac. So, c > 0. Now sice b > a == > a,,b,c > 0
Now, Considering the boundary case: b^2 = 4ac. And, in this eqn: interchanging a and c does not change my final outcome. So, I safely assumed a=c to check for the boundary case. Now, it could either be minima or maxima. Check for other values to figure out that.Q 17) a1 = p, b1=q.
a(n)=pb(n1) and b(n) = qb(n1) for all even n > 1
a(n)=pa(n1) and b(n) = qa(n1) for all odd n > 1
Find sigma[a(n)+b(n)] for even n.a1=p, b1=q
a2=pq, b2 = q^2
a3 = p^2q, b3 = pq^2
a4 = p^2q^2, b4 = pq^3
a5 = p^3q^2, b5 = p^2q^3
a6 = p^3q^3, b6 = p^2q^4
an = p^n/2.q^n/2,
bn = sigma[a(n)+b(n)] = pq(1+pq+p^2q^2+....(pq)^n/2 1) + q^2 (1+pq+p^2q^2 + ...(pq)^n/2 1)
= (pq+q^2)* pq^(n/21)/(pq1)
= q(p+q)[pq^(n/2)1]/ (pq1)Q 18 ) Find the minimum value of x^2  2x + 5 + 2x^2  4x + 9 + 3x^2  6x + 7 .
(x1)^2 + 4 
+ 2(x1)^2 + 7
+ 3(x1)^2 + 4
min. @ x=1  > 15
PS: Assuming x to be a real no.Q 19) If a + b + c = 6 where a,b,c are positive then maximum value of (a+1)(b+2)(c+3) is
AM > = GM
(a+1 + b+2 + c+3)/3 > = (a+1)(b+2)(c+3)^1/3
4 > = (a+1)(b+2)(c+3)^1/3
(a+1)(b+2)(c+3) < = 4^3
(a+1)(b+2)(c+3) < = 64
Maximum value is 64Alternate approach:
a + 1 =x, b + 2 = y and c + 3 = z
x + y + z = 12
For maximum value equate the terms x = y = z = 4
So product max is 4 x 4 x 4 = 64Q 20) If 2x^2 + 3y^2 = 1. If M is the maximum value and N is the minimum value of x^2 + 3xy + 2y^2. What is M+N  MN
Let x^2 + 3xy + 2y^2 = k
Given that 1 = 2x^2 + 3y^2
So, x^2 + 3xy + 2y^2 = k = k*1 = k(2x^2 + 3y^2)
== > x^2 (12k) + 3xy + y^2 (23k) = 0
Dividing by y^2:
(x/y)^2 (12k) + (x/y)*3 + (23k) = 0
For the above quadratic eqn in (x/y) to have a real solution:
D > = 0 thus
9  4(12k)(23k) > = 0'
so 24k^2  28k  1 < = 0. Thus the interval is [N,M] so N and M are roots of the equation, thus M + N = 7/6 and MN = 1/24
so M + N  MN= 7/6  1/24 = 27/24 = 9/8Q 21) Find the value of the series 3/1!+7/2!+15/3!+27/4!+43/5!+...
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
= 5e3Q 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 yearsQ 23) 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. valueQ 24) 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.Q 25) 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) = 769/775Q 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 = 43.33mQ 27) No of negative integral solution of 7 * a + 17 * b + 1000=0 is ?
a0
7x+17y=1000
mod 3y/7 = 6
x y
138 2
Tn = 138 + (n1)*17 > 0
n< 9. xx
===> 9 solnsQ 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 = .97Q 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?
1) 16
2) 38
3) 30
4) 25X+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