Quant Marathon by Gaurav Sharma  Set 8

Two sides of a triangle are given by the roots of the equation x^2 – 2 root(3)x + 2 = 0. The angle between the sides is Pi/3. The perimeter of the triangle is
 6 + root(3)
 2 root(3) + root(6)
 2 root(3) + root(10)
 None of these
Let ABC be the triangle such that it’s sides a = BC and b = CA are the roots of the equation x^2 – 2 root(3)x + 2 = 0
a + b = 2 root(3) and ab = 2
it is also given that angle C = Pi/3 so Cos C = ½
(a^2 + b^2 – c^2)/2ab = ½
a^2 + b^2 – c^2 = ab
(a + b)^2 – c^2 = 3ab
(2 root(3))^2 – c^2 = 3 x 2 = c^2 = 6
c = root(6)
perimeter of triangle ABC = a + b + c = 2 root(3) + root(6)
Find the range of x for  2x – 4 + 3 < 7
 (0, 8 )
 ( 4, 0)
 [4, 0]
 (0, 4)
We have  2x – 4 + 3 < 7
7 < 2x – 4 + 3 < 7
10 < 2x – 4 < 4
2x – 4 is always greater than – 10, so we need to look only one part of the inequality
2x – 4 < 4
4 < 2x – 4 < 4
0 < 2x < 8
0 < x < 4
x e (0, 4)
If a, b, c, d are four consecutive terms of an increasing AP then the roots of the equation (x – a) (x – c) + 2( x – b) (x – d) = 0 are
 Non real complex
 Real and equal
 Integers
 Real and distinct
Given than a < b < c < d
f(x) = (x – a) (x – c) + 2 (x – b) (x – d)
f(b) = (b – a) (b – c) < 0
And f(d) = (d – a) (d – c) > 0
f(x) = 0 has one root in (b, d). Also, f(a) f(c) < 0
So the other root lies in (a, c)
Hence roots of the equation are real and distinct
If four squares are chosen at random on a chess board, find the probability that they lie on a diagonal line
 364/64C4
 182/64C4
 364/64C2
 91/64C4
Total number of ways to select 4 squares out of 64 squares 64C4
Clearly only diagonals 4, 5, 6 … 12 contain 4 or more squares
Number of ways in which we can select 4 squares from the diagonals in one direction =
4C4 + 5C4 + 6C4 + 7C4 + 8C4 + 7C4 + 6C4 + 5C4 + 4C4
= 1 + 5 + 15 + 35 + 70 + 35 + 15 + 5 + 1
= 182
Similarly we can select 182 squares from the diagonals in the other direction
Total number of ways = 182 + 182 = 364
So, required probability = 364/64C4
If ABCD is a parallelogram and Q, R are the circumcenters of triangles ABC and ADC respectively. Then AQCR is always
 Rectangle
 Trapezium
 Rhombus
 None of these
Since the triangles ABC and ADC are congruent their circumcircles will be congruent.
Therefore, AQ, QC, AR and CR will be equal as they are all radii of the circumcircles.
So AQCR is a rhombus.
An Infinite GP has its first term a and sum = 8, what is the range of values a can take
 (0, 16)
 [0, 16]
 (1, 15)
 (0, 8 )
Sum of infinite GP = a/ ( 1 – r) = 8
1 < r < 1 , r e (1, 1)
So, 1 – r e (0, 2)
a = 8 ( 1 – r)
Now if 1 < r < 1
So, 0 < 1 – r < 2
8 x 0 < 8 ( 1 – r) < 8 x 2
0 < 8 ( 1 – r) < 16
0 < a < 16
N! is expressed in base 12 ends in exactly 7 trailing zeroes. How many values can N take
 7
 5
 0
 11
N! is a multiple of 12^7 but not 12^8
N! is a multiple of (2^2 x 3)^7 but not ( 2^2 x 3)^8
2^14 x 3^7 divide N! but 2^16 x 3^8 does not.
Now by trial and error method, Let us start with N = 15
Highest power of 2 that divides N! = 7 + 3 + 1 = 11
Highest power of 3 that divide N! = 5 + 1 = 6
15! = 2^11 x 3^6 x p
16! = 2^15 x 3^6 x p
17! = 2^15 x 3^6 x q
18! = 2^16 x 3^8 x q
18! Is a multiple of 12^8 and 17! Is not a multiple of 12^7
So, there are no number N such that N! is a multiple of 3^7 but not 3^8
There is no number N such that N! is a multiple of 12^7 but not 12^8
0 values of N satisfy the above condition.
Triangle ABC has a perimeter 18. What is the highest value that the inradius can take?
 3
 Root(2)
 Root(3)
 3 root(3)
Perimeter = 18
Area = A
rs = A
semi perimeter = 9
r = A/s = A/9
the higher the area, the higher the radius
For a given perimeter, the maximum area will for an equilateral triangle
In radius = a / 2root(3) = 6/2root(3) = 3/root(3) = root(3)
The number of values of k for which [x^2 – (k – 2)x + k^2] * [x^2 + kx + (2k – 1)] is a perfect square is
 2
 1
 0
 None of these
For the given equation [x^2 – (k – 2)x + k^2] * [x^2 + kx + (2k – 1)] should have roots common or each should have equal roots. If both roots are common, then
1/1 =  ( k – 2)/k = k^2/(2k – 1)
k =  k + 2 and 2k – 1 = k^2 = > k = 1
If both the equations have equal roots, then
( k – 2)^2 – 4k^2 = 0 and k^2 – 4( 2k – 1) = 0
(3k – 2) (k – 2) = 0 and k^2 – 8k + 4 = 0 ( no common value )
k = 1 is the only possible value.
If a is an integer lying in [5, 30], then the probability that the graph of y = x^2 + 2(a + 4)x – 5a + 64 is strictly above the xaxis.
 1/6
 7/36
 2/9
 3/5
x^2 + 2(a + 4)x – 5a + 64 ≥ 0
if D ≤ 0 then
(a + 4)^2 – ( 5a + 64) < 0
Or a^2 + 13a – 48 < 0
(a + 16) (a – 3) < 0
16 < a < 3
5 ≤ a ≤ 2
Then the favorable cases is equal to the number of integers in the interval [5, 2] i.e 8
Total number of cases is equal to the number of integers in the interval [ 5, 30] i.e 36
Required probability = 8/36 = 2/9