# Quant Boosters - Gaurav Sharma - Set 3

• Number of Questions - 30
Solved ? - Yes
Topic - Quant Mixed Bag
Source - Genius Tutorial Preparation Forum

• Q1) Find the sum of all rational terms in the expansion of (3^1/5 + 2^1/3)^15

• T(r + 1) = 15Cr (3^1/5)^(15 – r) (2^1/3)^r
T (r + 1) = 15Cr (3)^( 3 – r/5) 2^(r/3)
For the rational terms r = 0, 15
Then T(0 + 1) = 15C0 3^3 2^0 = 27
T(15 + 1) = 15C15 3^0 2^5 = 32
Sum of rational terms = T1 + T16
= 27 + 32 = 59

• Q2) If P is real root of 2x^3 – 3x^2 + 6x + 6 = 0 then [ P ] where [ . ] denotes the greatest integer function is equal to
a) 0
b) -1
c) 1
d) -2

• f(x) = 2x^3 – 3x^2 + 6x + 6
Then d/dx [f(x)] = 6x^2 – 6x + 6 = 6 (x^2 – x + 1) > 0 for all Real number x
f(x) is increasing on R
f(x) has only one real root
Now f(0) = 6 and f(- 1) = -5
Real root of f(x) lies between 0 and -1
[ P ] = - 1

• Q3) The number of solutions of the equation log(3 + root(x)) + log(1 + x^2) = 0 is (base of log is 3)
a) 0
b) 1
c) 2
d) More than 2

• log(3 + root(x)) + log(1 + x^2) = 0
x ≥ 0
Also 3 + root(x) > 3 and 1 + x^2 > 1 for all x > 0
log(3 + root(x)) > 1 and log(1 + x^2 ) > 0 for all x > 0
log(3 + root(x)) > 1 and log(1 + x^2 ) > 0 for all x ≥ 0
Hence the given equation has no solution

• Q4) If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side of the perimeter is
a) root(3) : 2 + root(3)
b) 1 : 6
c) 1 : 2 + root(3)
d) 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)

• Q5) 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

• Q6) 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) =
a) x^1/4 + x^3/4
b) – ( x^1/4 + x^3/4)
c) x + x^3
d) 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

• Q7) The first two terms of a HP are 2/5 and 12/23 respectively. Then the largest term is
a) 5th term
b) 6th term
c) 4th term
d) 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

• Q8) 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
a) Periodic function
b) An even function
c) An odd function
d) 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.

• Q9) The sum of the numerical coefficients in the expansion of ( 1 + x/3 + 2y/3)^12 is
a) 1
b) 2
c) 2^1/2
d) 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

• Q10) 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
a) 4/25
b) 4/35
c) 4/33
d) 4/1155

61

61

71

64

58

61

1

61