# Quant Boosters - Vikas Saini - Set 2

• Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source -

• Q1) If equation Px^2 + Qx + R = 0, where P, Q and R are integers and P + Q + R is equal to the smallest possible non negative integer. If the roots are equal then find the sum of roots.

• Let's roots are m and m.
m + m = - Q / P .
2 m = - Q / P.
Q = -2Pm
m x m = R / P
m^2 = R / P .
R = Pm^2.
P + Q + R = 1.
P - 2Pm + Pm^2 = 1.
P( 1 - m)^2 = 1.
P = 1, m =2.
then Q = -4.
Sum of roots = 4.

• Q2) Let p and q be the roots of quadratic equation x^2 - (a - 2) x - (a+1) = 0. Find minimum possible value of p^2 + q^2.

• p + q = a - 2.
pq = - (a + 1).
p^2 + q^2 = (p + q)^2 - 2pq
= (a - 2)^2 + 2(a + 1)
= a^2 - 4a + 4 + 2a + 2
= a^2 - 2a + 6.
= (a-1)^2 + 5.

if a = 1,
then p^2 + q^2 = 5.
Hence minimum value = 5

• Q3) (a+b)x^2 + (c+d)x + e = 0 is a quadratic equation such that a:b = 1:3, b:c = 3:5, c:d = 5:11 and d:e = 11:7. Find the absolute difference between the two roots of the equation.

• a = k, b = 3k, c = 5k, d = 11k, e = 7k.
(k+3k)x^2 + (5k+11k)x + 7k = 0.
4k x^2 + 16k x +7k = 0.
4x^2 + 16x + 7 = 0.
4x^2 + 14x + 2x + 7 = 0.
2x(2x+7)+(2x+7) = 0.
(2x+7) (2x+1) = 0.
x = -7/2 , -1/2.
absolute difference = -1/2 + 7/2 = 3

• Q4) f(x) = px^2 + qx + r. If f(x) attains its maximum value at x = 2, then what is the sum of the roots of f(x) = 0.

• f(x) = px^2 + qx + r.
f '(x) = 2px + q.
put f '(x) = 0
x = - q / 2p.
function f(x) attains maximum value at x = -q / 2p.
2 = -q / 2p ... (i)
by putting f(x) = 0.
px^2 + qx + r = 0.
sum of roots = -q / p.
from equation (i)
-q / 2p = 2.
-q / p = 4.

• Q5) The roots of the equation x^2 + bx - 14 = 0 are m and n. If 4m + 3n = 13, then which of the following must be true ?
(a) b = -31 / 12
(b) m = -2
(c) n = 14
(d) either a or b.

• m + n = -b.
m x n = -14.
m = -14 / n.
4m + 3n = 13
4(-14 / n) + 3n = 13.
= > -56 + 3n^2 = 13n.
= > 3n^2 - 13n - 56 = 0.
= > 3n^2 - 21n + 8n - 56 = 0.
= > 3n(n - 7) + 8(n - 7) = 0.
= > n = 7 , -8/3.
if n = 7, m = -2 and b = -5.
n = -8/3, m = 42 / 8 and b = -31/12.
Hence option d.

• Q6) How many values of k exist such that both the roots of the equation x^2 - 35x + k = 0 are prime numbers ?

• suppose roots are m & n.
m + n = 35.
35 is an odd number and it is sum of one odd and one even number.
2 is only even number which is prime.
if m = 2, n =33.
but 33 is not prime here.
Hence there is no value of k for which both the roots of equation are prime numbers

• Q7) Suppose k and 4 are two distinct roots of the quadratic equation ax^2 + bx + c = 0 with a > 0 and c > 0. What you say about b ?
(a) b.
(b) b - 4; b > 0 for other value of k
(c) b > 0 for all values of k
(d) b > 0 for k.

• two roots given 4 & k.
[(x - 4 ) (x - k )] = ax^2 + bx + c.
= > [x^2 - (4+k)x + 4k] = ax^2 + bx + c.
a > 0, c > 0.
suppose a = 1.
= > x^2 - (k+4)x + 4k = x^2 + bx + c.
c = 4k.
b = -(k+4)
(k+4) > 0, any value of k.
Hence option a.

• Q8) Suppose m & n are roots of equation x^2 - 10x +30 = 0 then what is the value of (1+m+m^2) (1+n+n^2) ?

• m + n = 10.
mn = 30.
(1+m+m^2) (1+n+n^2)
= (1+n+n^2)+(m+mn+mn^2)+(m^2+m^2n+(mn)^2)
= 1+(m+n)+(m^2 + n^2)+mn + (mn)^2 + mn (m+n)
= 1 + (m+n) + (m+n)^2 - 2mn + mn +(mn)^2 + mn(m+n)
= 1 + 10 + 10^2 - 2 x 30 + 30 + 30^2 + 30(10)
= 1281.

• Q9) If a & b are the roots of x^2 - kx + 72 = 0, and c & d are the roots of x^2 - 8x + k = 0. If a, b, c and d are natural numbers then how many values can a + b + c + d take ?

• a + b = k.
ab = 72.
c + d = 8.
cd = k.
a+b = cd.
(a,b)= (1,72),(2,36),(3,24),(4,18 ),(6,12),(8,9).
(c,d )= (1,7),(2,6),(3,5),(4,4).
a + b = 73,38,27,22,18,17.
cd = 7,12,15,8.
no real value of a+b+c+d.

• Q10) If x = 1 / ( 4 - 2√3) then find the value of x^4 - 4x^3 + 7x^2 - 6x + 7/4.

61

51

58

61

61

61

61