Quant Marathon by Gaurav Sharma  Set 9

How many values can natural number x take so that (4x + 7)^2/(x + 5) is an integer?
 1
 2
 3
 4
(4x + 7)^2/(x + 5) = [ (x + 5) (16x – 24) + 169) / (x + 5) ]
For the equation to have integer values, 169 should be divisible by (x + 5)
So can take 2 values, 8 and 164
Option (b) is correct.
If the equation x^2 + bx + c = k has four real roots, then
 b^2 – 4c > 0 and 0 < k < (4c – b^2)/4
 b^2 – 4c < 0 and 0 < k < (4c – b^2)/4
 b^2 – 4c > 0 and k < (4c – b^2)/4
 None of these
For the equation to have four real roots, the line y = k must intersect y = x^2 + bx + c at four points.
D > 0 and k e (0, D/4)
Find the missing term: 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ?
4 ( 3 and 5 are prime )
6 ( 5 and 7 are prime )
12 ( 11 and 13 are prime )
18 ( 17 and 19 are prime )
30 ( 29 and 31 are prime )
42 ( 41 and 43 are prime )
…
108 ( 107 and 109 are prime )
138 ( 137 and 139 are prime )
If a, b, c, d, e, f are non negative real numbers such that a + b + c + d + e + f = 1, then the maximum value of ab + bc + cd + de + ef is
 1/6
 1
 6
 1/4
If we take a = b = 12 and c = d = e = f = 0 then the given equation holds so the maximum value is ¼
Option b and c do not hold as they are integers and option a i.e 1/6 is less than ¼ so option (a) is also rejected. Hence option d is correct.
The number of integral values of x satisfying root(x^2 + 10x – 16) < x – 2 is
 0
 1
 2
 3
root(x^2 + 10x – 16) < x – 2
x^2 + 10x – 16 ≥ 0
x^2  10x + 16 ≤ 0
2 ≤ x ≤ 8 … (1)
Also, x^2 + 10x – 16 < x^2 – 4x + 4
2x^2 – 14x + 20 > 0
x^2 – 7x + 10 > 0
x > 5 or x < 2 … (2)
From (1) and (2), 5 < x ≤ 8
x = 6, 7, 8
A is the father of C and D is son of B and E is brother of A. if C is sister of D, how is B related to E ?
 Sister in law
 Brother in law
 Brother
 Can’t say
A is the father of C and C is sister of D. So A is father of D but D is son of B
So, B is the mother of D and wife of A
E is the brother of A so B is the sister in law of E.
The sum of the following series is
(1 x 2 x 3)/5^1 + (2 x 3 x 5)/5^2 + (3 x 4 x 5)/5^3 + (4 x 5 x 6)/5^4 + …
 375/128
 125/64
 625/128
 875/64
S = (1 x 2 x 3)/5^1 + (2 x 3 x 5)/5^2 + (3 x 4 x 5)/5^3 + (4 x 5 x 6)/5^4 + …
S = 6/5 + 24/5^2 + 60/5^3 + 120/5^4 + …
S/5 = 6/5^2 + 24/5^3 + 60/5^4 + 120/5^5 + …
S – S/5 = 4S/5 = 6/5 + 18/5^2 + 36/5^3 + 60/5^4 + 90/5^5 + …
4S/25 = 6/5^2 + 18/5^3 + 36/5^4 + 60/5^5 + …
4S/5 – 4S/25 = 16S/25 = 6/5 + 12/5^2 + 18/5^3 + 24/5^4 + 30/5^5 + …
16S/125 = 6/5^2 + 12/5^3 + 18/5^4 + 24/5^5 + 30/5^6 + …
64S/125 = 6/5 + 6/5^2 + 6/5^3 + …
S = (6/5 / ( 1 – 1/5) ) X 125/64 = 375/128
Using the vertices of a regular hexagon as vertices, how many triangles can be formed with distinct areas?
 24
 20
 18
 28
Let us number the vertices from 1 – 6, there are three different types of triangles that can be formed.
Type 1 : (1, 2, 3) [ this is same as (2, 3, 4), (3, 4, 5) … etc)
Type 2 : (1, 2, 4) [ this is same as (1, 2, 5) (2, 3, 5) (3, 4, 6)]
Type 3 : (1, 3, 5) and (2, 4, 6) have same shape and area
So there will be three triangles of different areas that can be formed from a regular hexagon.
There are 6 triangles of Type 1, 12 of type 2 and 2 of type 3.
Adding these we get 6 + 12 + = 20 triangles.
Total number of triangles possible = 6C3 = 20
There are 11 numbers of the blackboard – six zeroes and five ones. You have to perform the following operation 10 times : cross out any two numbers. If they were equal write a zero on the blackboard and if they were not equal, write a one. This operation can be performed in any order as you wish. What is the number you have at the end.
 11
 00
 1
 Cant say
The operation could be performed in a number of different ways, but 1 thing is always the same : After each operation the sum of the numbers on the blackboard is always odd.
Also, we are losing 1 digit in each operation. So, finally we must have a single digit – either 0 or 1
But it should be odd so our result must be 1.
A natural number is chosen at random from the first 100 natural numbers. The probability that x + 100/x > 50 is
 1/2
 7/18
 5/12
 11/20
We have x + 100/x > 50
x^2 + 100 > 50x
(x – 25)^2 > 525
x – 25 < root(525) or x – 25 > root(525)
x < 25 – root(525) or 25 + root(525)
As x is positive and root(525) = 22.9
We must have x ≤ 2 or x ≥ 48
Thus 2 + 53 = 55 favorable cases.
Probability = 55/100 = 11/20