Quant Boosters by Shubham Ranka, IIM Calcutta

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 .

(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 = 64

Q 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 = 9/8

Q 21) Find the value of the series 3/1! + 7/2! + 15/3! + 27/4! + 43/5! + ...