Test Level 2: Quadratic Equations & Linear - 2 CAT MCQs solutions - Free

⇒ (xy - x - y + 1)(z - 1) = (xy + x + y + 1)(z + 1)
xyz - xy - xz + x - yz + y + z - 1 = xyz + xy + xz + x + yz + y + z + 1
⇒ -2(xy + yz + zx ) = 2
∴ xy + yz + zx = -1

If there are t teachers, d doctors and l lawyers, then
t + d + l = 2160/36 = 60 … (1)
If the average ages of teachers, doctors and lawyers respectively are x, y and z years, then
xt + yd + lz = 2160 ... (2)
Given: (x + 1)t + (y + 6)d + (z + 7) l = 41  60 = 2460
∴ t + 6d + 7l = 300, use the equation 2
i.e. 5d + 6l = 240, use the equation 1

Since d and l are always positive integer values, minimum value of d is possible when l is maximum.
l(maximum) = 35
Hence, from equation (3),

Let a - b = x,
b - c = y,
c - a = z.
Then, x + y + z = 0 and the expression becomes 

Now, if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

|x|² is always positive. 3|x| is also positive.
As L.H.S. is always greater than zero, no real solution can exist.
Thus, total number of real solutions = 0
Thus, answer option 1 is correct.

lx² + mx+ n = 0
Let the roots be α and β.

12m² = 49nl

m² + 12m + 20 ≤ 0
(m + 2)(m + 10) ≤ 0
Hence, -10 ≤ m ≤ -2
And the set of possible values are {-10, -9, -8, -7, -6, -5, -4, -3, -2}
So, the total possible values for m are 9.

Let the number of stamps that A had be x.
And, let the number of stamps that B had be y.
Then, x + 1 = y - 1
x - y = - 2  … (i)
And, y + 1 = 2(x - 1)
-2x + y = - 3  … (ii)
Equations (i) + (ii) gives,
x = 5
Substitute the value of x in equation (i), we get
y = 7
Therefore, A had 5 stamps and B had 7 stamps. Thus, the total number of stamps with A and B is 12.

px + qy + rz = x³ - xyz + y³ - xyz + z³ - xyz
= x³ + y³ + z³ - 3xyz
= (x + y + z)(x² + y² + z² - xy - yz - zx)
= (x + y + z)(p + q + r)

a + b = 2c
∴ a - c = c - b

= 2

The given equation is:
x² - (λ + 1)x + λ + 4 = 0
Both roots are negative, if
(i) discriminant ≥ 0
(ii) sum of the roots is negative, and
(iii) product of the roots is positive
i.e. if (λ + 1)² - 4(λ + 4) ≥ 0
and λ + 1 < 0 because (-(b/a) = -(-(λ + 1)) => (λ + 1)
and λ + 4 > 0
i.e λ² -2λ -15 ≥ 0 and λ < -1 and λ > -4
i.e. (λ - 5) (λ + 3) ≥ 0 and -4 < λ < -1
i.e. (λ ≤ -3 or λ ≥ 5) and -4 < λ < -1
i.e. -4 < λ ≤ -3 ( -4 is not included and -3 is included in the range)

x² = y + z
i.e. x + x² = x + y + z ……… (1)
y² = z + x
y + y² = z + x + y ……… (2)
z² = x + y
z + z² = x + y + z ……… (3)
Eqn. (1) = Eqn. (2) = Eqn. (3)
⇒ x(1 + x) = y(1 + y) = z(1 + z) = x + y + z = k (let)

Adding L.H.S. and R.H.S., we get

Putting the value of y in  we get

Let the cost of each tennis ball be t cents. So, cost of each leather ball will be (t + 70) cents.
Let us suppose Michael bought k tennis balls and (30 – k) leather balls. So, total cost = kt + (30 – k)(t + 70)
Or 3200 = kt + 30t + 2100 – kt – 70k
1100 = 30t – 70k

When t = 60, we get feasible value of k.
So, cost of each tennis ball = 60 cents.

p + q = -p
And, pq = q
⇒ p = 1 ( q ≠ 0)
Hence, 
1 + q = -1
⇒ q = -2
x² + px + q = x² + x - 2

Given expression:
x² + (2y)² + (3z)² - (2y)(3z) -(3z)x - x(2y)
= a² + b² + c² - bc - ca - ab, where a = x, b = 2y, c = 3z

5x⁴ + 3x³ - 7x² - 3x + 5
= x²(5x² + 3x - 7 - 3/x + 5/x²)
= x²[5(x² + 1/x2 - 2) + 3(x - 1/x) + 3]
= x²[5(x - 1/x)2 + 3(x - 1/x) + 3]
= x²(5y2 + 3y + 3)

⇒ x² + 1 = 3x  ........(1)
Multiply by x , we get
⇒ x³ + x = 3x²  ........(2)
Multiply by x, we get
⇒ x⁴ + x² = 3x³  ........(3)
Again, Multiply by x, we get
⇒ x⁵ + x³ = 3x⁴  ........(5)
Now , Add the eq. (1), (2), (3) and (4) , we get

F + M + E + Y = 60 (F = Father's age, M = Mother's age, E = Elder son's age and Y = Younger son's age)
E = Y + 3,
M = E + Y + 17 = 2Y + 20,
F - M = E
F = M + E = (2Y + 20) + (Y + 3) = 3Y + 23
Now, F + M + E + Y = (3Y + 23) + (2Y + 20) + (Y + 3) + Y = 60
⇒ Y = 2 and F = 3Y + 23 = 29 years

3x³ - ax² - 74x + b is a multiple of x² + 2x - 24.
x² + 2x - 24 = 0
Or, x² + 6x - 4x - 24 = 0
Or, x(x + 6) - 4(x + 6) = 0
Or, (x - 4)(x + 6) = 0
Or, x = 4, x = -6 satisfies 3x3 - ax2 - 74x + b.
When x = 4:
192 - 16a - 296 + b = 0 … (1)
When x = -6:
-648 - 36a + 444 + b = 0 … (2)
Solving the above two equations, we get a = -5, b = 24.