Skip to main content

Quant Marathon by Gaurav Sharma - Set 7

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

 

If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side of the perimeter is

  1. root(3) : 2 + root(3)
  2. 1 : 6
  3. 1 : 2 + root(3)
  4. 2 : 3

Let the angle of triangle ABC be 4x, x and x then

4x + x + x = 180

x = 30

So, the angles are A = 120, B = 30 and C = 30

Now a/Sin A = b/Sin B = c/Sin C = k

a = [root(3)/2] x k

b = k/2

c = k/2

required ratio = a / (a + b + c) = root(3)/[root(3) + 2] = root(3) : 2 + root(3)

ABC x DEF = 123456. If A = 1, find A + B + C + D + E + F

ABC x DEF = 123456

123456 = 2^6 x 3 x 643

ABC = 64 x 3 x 643

We know A = 1 and 2^6 x 3 = 192

ABC = 192 and DEF = 643

If one of the root of the equation x^2 + x f(a) + a = 0 is the cube of the other for all real number x, then f(x) =

  1. x^1/4 + x^3/4
  2. – ( x^1/4 + x^3/4)
  3. x + x^3
  4. None of these

Let p and p^3 be the roots of the given equation

p + p^3 = -f(a) and p^4 = a

f(a) = - p – p^3 = - (a)^1/4 – a^3/4

f(a) = - (a^1/4 + a^3/4) where a = p^4 > 0

f(x) = - (x^1/4 + x^3/4), x > 0

The first two terms of a HP are 2/5 and 12/23 respectively. Then the largest term is

  1. 5th term
  2. 6th term
  3. 4th term
  4. 7th term

Let the HP be 1/a, 1/(a+d), 1/(a + 2d), 1/(a + 3d), …

Then 1/a = 2/5 and 1/(a + d) = 12/23

a = 5/2 and d = -7/12

nth term of HP = 1/ [a + (n – 1)d = 12/(37 – 7n)

So the nth term is largest when 37 – 7n has the least value

Hence, 12/(37 – 7n) is largest for n = 5

If a real valued function f(x) satisfies the equation f( x + y) = f(x) + f(y) for all x element of R, then f(x) is

  1. Periodic function
  2. An even function
  3. An odd function
  4. None of these

We know that the function f(x) satisfying the property f (x + y) = f(x) + f(y) for all x, y e R

Has the formula f(x) = x f(1) for all x e R

Clearly it is an odd function.

The sum of the numerical coefficients in the expansion of ( 1 + x/3 + 2y/3)^12 is

  1. 1
  2. 2
  3. 2^1/2
  4. None of these

We have (1 + x/3 + 2y/3) ^12

To find sum of numerical coefficients put x = 1 and y = 1

Required sum = (1 + 1/3 + 2/3) ^12

= 2^12

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 are

  1. 4/25
  2. 4/35
  3. 4/33
  4. 4/1155

A number will be divisible by both 2 and 3 if it is divisible by 6 (LCM). There are 16 numbers in first 100 natural numbers which are divisible by 6

Number of ways of selecting 3 numbers such that all of them are divisible by both 2 and 3 are 16C3

Required probability = 16C3/100C3 = (16 x 15 x 14) / ( 100 x 99 x 98 ) = 4/1155

If in a triangle ABC, a^2 + b^2 + c^2 = ac + root(3) ab

Then the triangle is

  1. Equilateral
  2. Right angled and isosceles
  3. Right angled and not isosceles
  4. None of these

We have a^2 + b^2 + c^2 = ac + root(3) ab

(a^2)/4 + c^2 – ac + 3(a^2)/4 + b^2 – root(3)ab = 0

(a/2 – c)^2 + (root(3)a/2 – b)^2 = 0

a/2 – c = 0 and root(3)a/2 – b = 0

a = 2c and root(3)a = 2b

a = 2b/root(3) = 2c = k (say)

a = k, b = root(3) k/2 and c = k/2

b^2 + c^2 = a^2

Hence, the triangle is right angled but not isosceles

The number of solutions of equation 5^x + 5^(-x) = log1025 [ x e R ]

  1. 0
  2. 1
  3. 2
  4. Infinitely many

5^x + 5^(-x) = 5^x + 1/5^x

Which is always greater than or equal to 2

Also log10100 = 2

And log1025 < log10100

Log1025 < 2

Hence LHS and RHS will never be equal

So, no solution of the given equation is possible.

The orthocenter of a triangle with vertices (1, root(3)), (0,0) and (2,0) is

  1. (1, root(3)/2)
  2. (2/3, 1/root(3)
  3. (2/3, root(3)/2)
  4. (1, 1/root(3)

Let the vertices of the triangle be A (1, root(3)), B (0,0) and C (2,0)

AB = BC = AC = 2

So the triangle is equilateral

Now it’s orthocenter coincides with the centroid whose coordinates are:

[(1 + 0 + 2) / 3, (root (3) + 0 + 0)/ 3)

= (1, 1/root (3))