Quant Boosters  Rajesh Balasubramanian  Set 5

Let us split this into two cases. Case 1, when x is greater than 0 and Case 2, when x is lesser than 0.
Case 1
x > 0. Now, x = x
x^2 – 7x – 18 = 0
(x – 9) (x + 2) = 0
x is either –2 or +9.Case 2
x < 0. Now, x = –x
x^2 + 7x – 18 = 0
(x + 9) (x – 2) = 0
x is either –9 or +2.However, in accordance with the initial assumption that x < 0, x can only be –9 (cannot be +2).
Hence, this equation has two roots: –9 and +9.
Alternatively, we can treat this as a quadratic in x, the equation can be written as x2 – 7 x – 18 = 0.
Or, (x – 9) (x + 2) = 0
x = 9 or –2. x cannot be –2.
x = 9, x = 9 or –9.Correct Answer: There are 2 solutions.

Q5) x^2  9x + k = 0 has real roots. How many integer values can 'k' take?
a) 40
b) 21
c) 20
d) 41

Discriminant, D = 81 – 4k
If roots are real, D > 0
81 – 4k > 0
4k < 81
k < 20.25Hence, –20.25 < k < 20.25
The integer values that k can take are –20, –19, –18 … 0 … 18, 19 and 20.41 different values (Remember to include 0. Answer choice 'A' will definitely be 40

Q6) x^2  11x + p = 0 has integer roots. How many integer values can 'p' take?
a) 6
b) 4
c) 8
d) More than 8

To start with the discriminant should be a perfect square. Let the dicriminant be 'D'.
From the quadratic formula: (−11 ± √D)/2, we see that the numerator has to be an even number for the roots to be integers.
This implies that the discriminant should be a perfect square and be square of an odd number. (Only then we will have odd + odd = even in the numerator)
D = 121 – 4p = 121 – 4p
4p cannot be negative => D can take values 121, 81, 49, 25, 9, 1
p can be 0, 10, 18, 24, 28, 30p can take 0, ±10, ± 18, ± 24, ± 28, ± 30
Correct Answer: More than 8 values are possible.

Q7) 2x + 5y = 103. Find the number of pairs of positive integers x and y that satisfy this equation.
a) 9
b) 10
c) 12
d) 20

Rearranging the equation, we get:
2x = 103 – 5yThis says that when you subtract a multiple of 5 from 103, you get an even number. You have to subtract an odd multiple of 5 from 103 in order to get an even number. There are 20 multiples of 5 till 100, ten of which are odd. (Note that you cannot subtract 105, or higher multiples as they result in a negative value for x.)
So, y can have ten integer values. x also has 10 integer values, each corresponding to a particular value of y.
y = 1, gives us a potential value for x, so do y = 3, 5, 7….19. y can take 10 values totally.

Q8) Consider three numbers a, b and c. Max (a,b,c) + Min (a,b,c) = 13. Median (a,b,c)  Mean (a,b,c) = 2. Find the median of a, b, and c.
a) 11.5
b) 9
c) 9.5
d) 12

The numbers can be arranged in ascending order thus: Min < Median < Max
According to the question, Min + Max = 13
Median  (Min + Med + Max)/3 = Median  (12 + Med)/3 = 2.
Solving, we get Median = 9.5.

Q9) a1x + b1y + c1z = d1, a2x + b2y + c2z = d2, a3x + b3y + c3z = d3.
Which of the following statements if true would imply that the above system of equations does not have a unique solution?
i. a1/a2 = b1/b2 = c1/c2 ≠ d1/d2
ii. a1/a2 = a2/a3 ; b1/b2 = b2/b3
iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers

If we have three independent equations, we will have a unique solution. In other words, we will not have unique solutions if:
 The equations are inconsistent or
 Two equations can be combined to give the third
Now, let us move to the statements.
i. a1/a2 = b1/a2 = c1/c2 ≠ d1/d2
This tells us that the first two equations cannot hold good at the same time.
Think about this:
x + y + z = 3;
2x + 2y + 2z = 5.
Either the first or the second can hold good. Both cannot hold good at the same time.
So, this will definitely not have any solution.ii. a1/a2 = a2/a3 and b1/b2 = b2/b3
a1, a2, a3 are in GP, b1, b2, b3. This does not prevent the system from having a unique solution.
For instance, if we have
x + 9y + 5z = 11
2x + 3y – 6z = 17
4x + y – 3z = 15This could very well have a unique solution.
iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers
This gives us practically nothing. This system of equations can definitely have a unique solution.So, only Statement i tells us that a unique solution is impossible.

Q10) Equation x^2 + 5x – 7 = 0 has roots a and b. Equation 2x^2 + px + q = 0 has roots a + 1 and b + 1. Find p + q.
a) 6
b) 0
c) 16
d) 2

Given, x^2 + 5x – 7 = 0 has roots a and b. We know that, Sum of roots in a quadratic equation = a + b = (−5)/1 = 5.
Product of the roots = ab = (−7)/1 = 7.
Now, The second equation 2x^2 + px + q = 0 has roots a + 1 and b + 1.
Sum of the roots = a + 1 + b + 1 = a + b + 2 = (−p)/2 = 5 + 2 = 3 = (−p)/2 => p = 6 => p = 6.
Product of the roots = (a + 1)(b + 1) = ab + a + b + 1 = q/2. We know the values of ab and a+b. Substituting this, we get, 7 + (5) + 1 = q/2 => 11 = q/2 => q =  22.
Hence, p = 6 and q = 22. => p + q = 6 + (22) = 16.

Q11) Sum of the roots of a quadratic equation is 5 less than the product of the roots. If one root is 1 more than the other root, find the product of the roots?
a) 6 or 3
b) 12 or 2
c) 8 or 4
d) 12 or 4

Sum of the roots = (product of the roots) – 5
Let r1 and r2 be the two roots
r1 = r2 + 1Therefore,
r1 + 1 + r2 = (r1r2) ∗ r2 − 5
r2 + 1 + r2 = (r2 + 1) ∗ r2 − 5
2r2 + 6 = (r2)^2 + r2
(r2)^2 − r2 − 6 = 0
Solving, we get r2 = 3 or 2
Therefore, r1 = 4 or 1
Hence product of the two roots is either 12 or 2.

Q12) How many real solutions are there for the equation x^2 – 7x  30 = 0?
a) 3
b) 1
c) 2
d) none

Case 1: x > 0
x^2 – 7x  30 = 0
(x  10 ) (x  3 ) = 0
x = 10, x = 3
But x = 3 is not possible as we have considered x > 0, thus 1 solution for this case.Case 2: x < 0
x^2 + 7x  30 = 0
(x + 10 )(x  3)=0
x = 10 and x = 3
Only x = 10 is permissible.
Thus this equation has 2 real solutionsAlternatively, we can think of the above as a quadratic in x
x^2 – 7x  30 = 0 can be factorized as
(x 10) ( x + 3) = 0
x cannot be 3, x can only be 10. x can take 2 real values.

Q13) If (3x + 2y  22)^2 + (4x  5y + 9)^2 = 0 and 5x  4y = 0. Find the value of x + y.
a) 7
b) 9
c) 11
d) 13

Do not jump and start expanding the squares.
In equation (1), both the squares will yield either positive number or 0. Since sum of two positive number cannot be 0, each of the term must be 0
∴ 3x + 2y  22 = 0
Multiplying by 2, and adding both equation
6x + 4y  44 = 0
5x  4y = 0
=) x = 4
Putting the value of x in the equation
5 * 4 – 4y = 0
y = 5
∴ x + y = 9

Q14) Let x^3  x^2 + bx + c = 0 has 3 real roots which are in A.P. which of the following could be true
a) b = 2, c = 2
b) b = 1, c = 1
c) b = 1, c = 1
d) b = 1, c = 1