Test Level 3: Quadratic Equations & Linear Equations - CAT MCQ

Given: u + v + w + x + y = 15
We need the maximum value of uvx + uvy + uwx + uwy.
(uv + uw)(x + y) = u(v + w)(x + y)
Let's use AM ≥ GM in u, (v + w), (x + y):

Hence, the maximum value of uvx + uvy + uwx + uwy is 125.

Put |x| = y.
Given equation: y² + y - 6 = 0
On solving, we get y = 2 or -3
i.e. |x| = 2 or -3
Modulus cannot be negative.
Therefore, |x| = 2
⇒ The roots (α and β) of the equation are 2 and -2.
Hence, α + β = 0

For real x, f(x) = x² + 2bx + 2c²
= x² + 2bx + b² + 2c² - b²
= (x + b)² + 2c² - b² ≥ 2c² - b²
and g(x) = -x² - 2cx + b²
= -(x² + 2cx + c²) + b² + c²
= b² + c² - (x + c)² ≤ b² + c²
Hence, min. f(x) > max. g(x)
if 2c² - b² > b² + c²
i.e. if c² > 2b²

If the roots of the given equation are α and β, then α > 2 and β < 2.
⇒ α - 2 > 0 and β - 2 < 0
⇒(α - 2)(β - 2) < 0
⇒ αβ - 2(α + β) + 4 < 0
⇒ a² + a - 8 - 2(a + 1) + 4 < 0
(∵ α + β = a + 1 and αβ = a² + a - 8)
⇒ a² - a - 6 < 0
⇒ (a - 3)(a + 2) < 0
⇒ -2 < a < 3

Let c, w and n be the number of questions answered correctly, wrongly and not attempted, respectively.
∴ c + w + n = 50  … (i)

6c – 2w – n = 32 × 6 = 192 … (ii)
Adding equations (i) and (ii), we have
7c – w = 242
w = 7c – 242
Since w and c are positive numbers, the minimum value of c for which w is positive is 35.
Hence, w = 3

Multiply the first equation by 5, second by -2 and third by -1.
5x + 10y - 15z = 5p ... (i)
-4x - 12y + 22z = - 2q ... (ii)
-x + 2y - 7z = - r ... (iii)
Adding all three equations, we get
5p - 2q - r = 0

Let us suppose:
W_G : Age of the wife of Ghosh Babu 
D_B : Age of the daughter of Banke Lal
G_G : Age of the great grandfather of Ghosh Babu
E^_B : Age of the eldest son of Banke Lal
W_B : Age of the second wife of Banke Lal
According to the question:
W_G = DB + 6 … (1)

From equations (3) and (5): 

From equations (4) and (6):

From equations (1), (2) and (7):

At the time of conversation, age of Ghosh Babu's great-grandfather would have been (G_G + 6) = 90 + 6 = 96 years

a + 3s + 7p = 1400 … (1)
a + 4s + 10p = 1700 … (2)
10a + 15s + 25p = ?
And a + s + p = ?
We have 2 equations and 3 unknown variables.
Subtracting equation (1) from (2), we get
s + 3p = 300
And 4 × equation (1) - 3 × equation (2) gives a - 2p = 500.
a + s + p = 800
Also, 10a - 20p = 5000
And 15s + 45p = 4500
10a + 15s + 25p = 9500

Let the contributions be C + 2, C + 1 and C.
Total vodka = 3C + 3 = 3 (C + 1)
The total vodka is divided into 4 equal parts.
Thus, each of them gets

Contribution of first drunkard towards fourth's portion 

Contribution of second drunkard towards fourth's portion  

According to the question: 

C + 5 = 2C + 2
C = 3
Contribution of second drunkard towards fourth's portion = (3 + 1)/4 = 1
Now, the price equals 5 roubles a litre.
Hence, the second drunkard received 5 roubles from the fourth drunkard.

Let the correct equation be X² + aX + b = 0 … (1)
Roots found by Rocky are 8 and 2. As such, the equation should be X² - (8 + 2)X + 8  2 or X² - 10X + 16 = 0.
Since he has committed a mistake only in the constant term, coefficient of X in both would be the same, or a = -10. 
Now, roots found by Anthony are -9 and -1. As such, the equation should be X² - (-9 - 1)X + (-9)(-1) = 0 or X² + 10X + 9 = 0.
As he has committed mistake only in the coefficient of X, therefore constant term in equation (1) and his solution would be the same, i.e. b = 9. Hence, the correct equation is X² - 10X + 9 = 0 or (X - 9)(X - 1) = 0.
So, the roots are 9 and 1.

Given,


x = aK, y = bK, z = cK

r², s² are the roots of the equation.
Sum of roots,
r² + s² = (r + s)² - 2rs

Product of roots,

So, required equation is: 

or m²x² - n²x + 2mcx + c² = 0

Since (y - k) is a factor of the quadratic equation, so substitute y = k in the equation.
So the equation y² - 11y + 4k = 0 transforms to k² - 11k + 4k = 0
Hence, k² = 7k and k = 7.
Now, substitute k = 7 in the equation y² - 11y + 4k = 0
The equation becomes y² - 11y + 28 = 0, the roots of which are 7 and 4.
Sum of the roots = 11;
Product of roots = 28
So, required difference = 28 - 11 = 17

Suppose, according to the initial plan, 'a' hectares of land were to be harvested each day and 'd' days were to be taken for the complete job.
So, ad = 170 ... (i)
Now, the equation formed for the new situation is as follows.
4a + (d - 4 - 8)2a = 170
Or, 4a + 2ad - 24a = 170 ... (ii)
Plugging in the value of ad as 170 from equation (i) into equation (ii), we get 4a + 2 × 170 - 24a = 170
Or, a = 8.5
Thus, 8.5 hectares had to be harvested by the harvesting team each day according to the initial plan.
Hence, answer option 2 is correct.

Let the numbers of coins with A, B, C and D be a, b, c and d, respectively.
Now, according to the question:
a + b + c + d = 36 … (1)
And d > c > b > a (Given)
b - a = 1 … (2)
⇒ b = a + 1
c - b = 9 … (3)
Add (2) and (3).
c - a = 10 … (4)
⇒ c = a + 10
d - c = 7 … (5)
Add (4) and (5).
d - a = 17
d = a + 17
The total numbers of coins is 36.
a + a + 1 + a + 10 + a + 17 = 36
⇒ 4a + 28 = 36
⇒ 4a = 8
⇒ a = 2