# Quant Boosters by Shubham Ranka, IIM Calcutta

• 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)/(b-a)

• a > 0, b^2 < = 4ac
By symmetry:- a=c== > b=2a (D=0)
4/1== > 4

Explanation:

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(n-1) and b(n) = qb(n-1) for all even n > 1
a(n)=pa(n-1) and b(n) = qa(n-1) 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/2-1)/(pq-1)
= q(p+q)[pq^(n/2)-1]/ (pq-1)

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

• |(x-1)^2 + 4 | + |2(x-1)^2 + 7| + |3(x-1)^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 64

Alternate 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 (1-2k) + 3xy + y^2 (2-3k) = 0
Dividing by y^2:-
(x/y)^2 (1-2k) + (x/y)*3 + (2-3k) = 0
For the above quadratic eqn in (x/y) to have a real solution:-
D > = 0 thus
9 - 4(1-2k)(2-3k) > = 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! + ...

• Taylor series expansion for e^x = 1 + x + x^2/2! + x^3/3! +...
e-1 = 1+1/2!+1/3!+1/4!+....
3,7,15,27,43,
Numerator:-
Tn = Tn-1 + 4(n-1), n > =2
T1=3
Tn = 3+ 4 (n(n-1)/2)
== > Generic Tn = [3+2(n(n-1))]/n!
= 3/n! + 2/(n-2)! (n > =2)
= 3(e-1) + 2e
= 5e-3

• Q 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?

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