Quant Boosters  Vikas Saini  Set 5

Suppose x^2 / y = t.
t = 4x – 3.
x = 1, t = 1, y = 1.
x = 3, t = 9, y = 1.
Only two integer solutions.

Q23) If a, b are integers then how many ordered pairs (a, b) satisfy the equation a^2 + ab + b^2 = 1 ?

a^2 + ab + ab + b^2 – ab = 1
(a+b)^2 = 1 + ab.
a = 0, b = 1.
a = 1, b = 0.
a = 0, b = 1.
a = 1, b = 0.
a = 1, b = 1.
a = 1, b=1.
6 ordered pairs.

Q24) x^2 * y^3 = 8, where x, y > 0. What is the minimum value of 4x + 3y ?

4x + 3y = P.
4x = 2k
x = k/2.
3y = 3k.
y = k.
(k/2)^2 k^3 = 8.
K^5/4 = 8.
K = 2.
4x + 3y = 2k + 3k = 5k = 10.

Q25) [ log (1) + log (1+3) + log (1+3+5) ... log(1+3+5…….19)  2 (log 1 + log 2 ... log 7) = m + nx + ay. If log 2 = x and log 3 = y, then find the value of m, n & a ?

(log 1 + log 4 + log 9……..log 100) – 2(log 1 + log 2…….log 7)
= 2(log 1 + log 2 + log 3……..log 10) – 2( log 1 + log 2…………log 7)
= 2 (log 8 +log 9 + log 10)
= 2(3log2 + 2log3 + 1)
= 2 + 6log 2 + 4log 3
= 2 + 6x + 4y
m = 2, x =6, y =4.

Q26) All three roots of the cubic equation x^3  10x^2 + 31x  k = 0 are prime numbers. What is the value of K ?

Let’s three roots l,m and n.
l + m + n = 10.
lm + mn + nl = 31.
lmn = k.
all roots are prime.
Hence l = 2, m = 3, n = 5.
lmn = 30.

{x} = x – [x]
(a) 0
(b) 1
(c) 2
(d) 3

Hence 0.

Q28) When ‘2’ is added to each of the three roots of x^3  Ax^2 + Bx  C = 0, we get the roots of x^3 + Px^2 + Qx  18 = 0. If A, B, C, P and Q are all non zero real numbers then what is the value of (4A + 2B + C).

Let three roots of first equation are l, m, n respectively.
18 = (l+2)(m+2)(n+2)
18 = (lm+2l+2m+4)(n+2)
18 = lmn + 2lm + 2ln + 4l +2mn + 4m + 4n + 8.
10 = lmn + 2(lm+ln+mn) + 4(l+n+m)
10 = C+2B+4A.

Q29) If a, b and c are root of the equation 3x^3 + 42x + 93 = 0, then what is value of a^3 + b^3 + c^3 ?

Here coefficient of x^2 = 0.
means a+b+c = 0.
Then a^3 + b^3 + c^3 = 3abc = 93

Q30) ax^2 + bx + c = 0 is a quadratic equation with rational coefficients such that a + b + c = 0, then which of the following is necessarily true ?
(a) Both the roots of this equation are less than 1.
(b) One of the roots of the equation is c/a.
(c) Exactly one of the root is 1.
(d) b & c both.

let f(x) = ax^2 + bx + c
f(1) = a + b + c = 0.
Product of roots = c/a.
Hence option d.

@vikas_saini Thanks bro...very neat and nice explanation....keep doing this...perfect questions with perfect solutions...Thanks alot...:)

shouldn't we also consider the values less than 8? 1, 2, 3, 4, 5,6, 7?
as these would give negative results too.. please clarify.