Quant Boosters by Shubham Ranka, IIM Calcutta

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! + ...

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
= 5e3

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?

Let x be the man's year of birth,
Then x/89 = 1980  x
== > x = 1958
== > In 2012 Age was: 54 years

Q23) 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. value

Q24) 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.

Q25) 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.