# Quant Boosters by Shubham Ranka, IIM Calcutta

• 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 work--Kanti --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)/(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

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