Quant Marathon by Gaurav Sharma - Set 7


  • Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)


    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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