Quant Boosters  Rajesh Balasubramanian  Set 5

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

Given the roots are in A.P. so let ad, a, a+d be the roots
From equation, sum of roots = 1
Sum of two roots taken at a time = +b
Product of two roots = c∴ (ad) + (a) + (a+d) = 1
=) 3a = 1
=) a = 1/3Also, (ad)a+ a(a+d)+ (ad)(a+d) = b
=) a^2 – ad + a^2 + ad + a^2 – d^2 = b
=) 3a^2 – d^2 = b
=) 3 * 1/9 – d^2 = b
=) d^2 = 1/3  bNow, since d is a real number,
1/3 − b > 0 =) b < 1/3
Also, (ad)(a)(a+d) = c
=) a(a^2 – d^2) = c
=) 1/3 ∗ (1/9 − d^2) = −c
=) c = d^2/3 − 1/27
=) c = (9d^2 − 1)/27
=) d^2 = 3c + 1/9Again, 3c + 1/9 > 0
c > −1/27 > 0
∴ Only option (c) b = 1 , c = 1 could be true .

Q15) If f(y) = x^2 + (2p + 1)x + p^2  1 and x is a real number, for what values of ‘p' the function becomes 0?
a) p > 0
b) p > 1
c) p ≥ 5/4
d) p ≤ 3/4

The function f(y) is a quadratic equation.
It is given that x is real.
So the discriminant of f(y) ≥ 0
i.e. D = b^2  4ac ≥ 0 or
(2p + 1)^2 – 4(p^2  1) ≥ 0
4p^2 + 4p + 1 – 4(p^2  1) ≥ 0
4p + 5 ≥ 0
Or p ≥  5/4

Q16) A merchant decides to sell off 100 articles a week at a selling price of Rs. 150 each. For each 4% rise in the selling price he sells 3 less articles a week. If the selling price of each article is Rs x, then which of the below expression represents the rupees a week the merchant will receive from the sales of the articles?
a) 175  x/2
b) x/3 + 156
c) 350  x^2/2
d) x^2  2x + 75

The selling price per article is Rs. x
The price of the article is Rs (x  150) more than Rs 1504% rise from Rs 150 ==> Rs 6
For each Rs 6 increase in price over Rs.150, he sells 3 articles lesser.So, the reduction in number of articles = (x – 150) / 6 * 3
So, in the new price of Rs.x he sells 100  (x – 150) / 6 * 3
Or he sells 100 – (x – 150)/2
= (200 – x + 150) / 2
= (350 – x)/ 2
= 175 – x/2

Q17) How many pairs of integer (a, b) are possible such that a^2 – b^2 = 288?
a) 6
b) 12
c) 24
d) 48

288 = 2^5 × 3^2. So it has 6 × 3 = 18 factors. Or, there are 9 ways of writing this number as a product of two positive integers.
Let us list these down.
1 × 288, 2 × 144, 3 × 96, 4 × 72, 6 × 48, 8 × 36, 9 × 32, 12 × 24 and 16 × 18Now, this is where the question gets interesting. If a, b are integers either a + b and a – b have to be both odd or a+ b and a –b have to be both even. So, within this set of possibilities
1 × 288, 3 × 96 and 9 × 32 will not result in integer values of a, b. So, there are 6 sets of numbers that work for us.
Moving on to these six sets; let us start with one example and see how many possibilities we can generate from this.Let us consider the set 2 × 144.
Let us solve for this for a, b being natural numbers first, then we will extend this to integers.
When a, b are natural numbers
a + b > a – b
So, a + b = 144, a – b = 2; a = 73 and b = 71Now, if a = 73, b = 71 holds good. We can see that a = 73, b = –71 also holds good. a = –73, b = 71 works and so does a = –73, b = –71.
There are 4 possibilities.
a = 73, b = 71
a = 73, b = –71
a = –73, b = 71
a = –73, b = –71
So, for each of the 6 products remaining, we will have 4 possibilities each. Total number of (a, b) that will satisfy this equation = 6 × 4 = 24.Answer choice (c)

Q18) If x^4 – 8x^3 + ax^2 – bx + 16 = 0 has positive real roots, find a – b.
a) 8
b) 6
c) 12
d) 14

Let p, q, r, s be the roots of the equation
p + q + r + s = 8
pqrs = 16
This happens only when p = q = r = s = 2(p + q + r + s)/4 = 2.
(pqrs)^1/4 = 2Arithemtic mean is equal to Geometric Mean. This is possible only when all the numbers are equal. (Note that the question says all roots are positive and real)
p = q = r = s = 2
pq + pr + ps + qr + qs + rs = a
24 = a
pqr + pqs + prs+ qrs = b
32 = b
So, a – b = 24 – 32 = – 8
Answer choice (a)

Q19) x^3 – 18x^2 + bx – c = 0 has positive real roots, p, q and z. If geometric mean of the roots is 6, find b.
a) 36
b) 216
c) 108
d) 72

p + q + z = 18
pq + qz + zp = b
pqz = c
According to the question, (pqz)^1/3 = 6. Hence, pqz = c = 216.
This is possible only when p = q= z = 6. {Arithmetic mean of p, q and z = 6, GM is also 6. This is possible only when all three numbers are equal.}
So, pq + qz + zp = 36 × 3 = 108 = b.
Hence, b = 108.

Q20) x^3 – 4x^2 + mx – 2 = 0 has 3 positive roots, two of which are p and 1/p. Find m.
a) 5
b) 11
c) 8
d) 2