Quant Boosters by Shubham Ranka, IIM Calcutta

(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?

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 = 240

Q 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?
a) 40
b) 42
c) 44
d) 48

2 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 44

Q 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
a) 9 cm
b) 10 cm
c) 11 cm
d) 12 cm

Let 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 and 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
a:b:c:d=72:168:378:441
TRICK:
[a:b:c:d = 3 * 4 * 6: 7 * 4 * 6 : 7 * 9 * 6 : 7 * 9 * 7]
a+b+c+d=52950
d = 52950 * 441/1059 = 22050

Q 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)

Q18 ) Find the minimum value of x^2  2x + 5 + 2x^2  4x + 9 + 3x^2  6x + 7 .