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Quant Marathon by Gaurav Sharma - Set 9

Gaurav Sharma, Director of Genius Classes Of Mathematics (GCM), Karnal ( Haryana ), completed his graduation from Kurukshetra University in Electronics. Being inclined towards Mathematics, he pursued his Masters in Mathematics and also completed his L.L.B. He believes that, when a man quits learning he will be more senile onwards. He is a mentor right from his college days and loves teaching and inspiring students. Team Genius Classes, along with the regular classes, organizes workshops, events and seminars time to time. You can find lot of good stuffs in his Facebook group - CAT Mausi

How many values can natural number x take so that (4x + 7)^2/(x + 5) is an integer?

  1. 1
  2. 2
  3. 3
  4. 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

  1. b^2 – 4c > 0 and 0 < k < (4c – b^2)/4
  2. b^2 – 4c < 0 and 0 < k < (4c – b^2)/4
  3. b^2 – 4c > 0 and k < (4c – b^2)/4
  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. 1/6
  2. 1
  3. 6
  4. 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

  1. 0
  2. 1
  3. 2
  4. 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 ?

  1. Sister in law
  2. Brother in law
  3. Brother
  4. 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 + …

  1. 375/128
  2. 125/64
  3. 625/128
  4. 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?

  1. 24
  2. 20
  3. 18
  4. 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.

  1. 11
  2. 00
  3. 1
  4. 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. 1/2
  2. 7/18
  3. 5/12
  4. 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