Quant Boosters by Shubham Ranka, IIM Calcutta

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) 1

Q2) 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 determined

let 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 Kg

Q3) 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 these

concept: 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 satisfies

Q5) How many times in a day, the hands of a clock are straight?
a. 22
b. 24
c. 44
d. 48

Straight == > 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. Friday

Calendar 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 = 16

Q8) 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 = 19832539
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 = 19832539
Using hit and trial:
== > n = 9, 2001
Hence n = 6,9,2001

Q9) 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) 12

k (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 = 10

Q11) 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?