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
 root(3) : 2 + root(3)
 1 : 6
 1 : 2 + root(3)
 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) =
 x^1/4 + x^3/4
 – ( x^1/4 + x^3/4)
 x + x^3
 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
 5^{th} term
 6^{th} term
 4^{th} term
 7^{th} 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
 Periodic function
 An even function
 An odd function
 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
 2
 2^1/2
 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
 4/25
 4/35
 4/33
 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
 Equilateral
 Right angled and isosceles
 Right angled and not isosceles
 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) = log_{10}25 [ x e R ]
 0
 1
 2
 Infinitely many
5^x + 5^(x) = 5^x + 1/5^x
Which is always greater than or equal to 2
Also log_{10}100 = 2
And log_{10}25 < log_{10}100
Log_{10}25 < 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, root(3)/2)
 (2/3, 1/root(3)
 (2/3, root(3)/2)
 (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))