# Quant Marathon by Gaurav Sharma - Set 7

• 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))

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