Test Level 1: Quadratic Equations & Linear Equations - 1 (September 17) - CAT MCQ

The equation is x + y = xy. The only two integer pairs satisfying this are (0, 0) and (2, 2).
Hence, there are 2 pairs.

Since it is known that A + B = Z, it follows that A + B + P = T.
We also know that T + A = F.
So, in the equation B + P + F = 24, we can replace F with T + A.
The equation then becomes B + P + T + A = 24 or B + P + T = 22, since A = 2.
Then, we have
B + P + 2 = T ... (i)
Since B + P + T = 22
-B - P + 22 = T ... (ii)
So, from equations (i) and (ii),
24 = 2T
So, T = 12
Now, 12 + A = 14 = F and Q - 12 = 7
So, Q = 19
Therefore, S = 5

Step 1:
Multiplying both sides by 3, we get:

x + 368 = 93x
Step 2:
Subtracting 93x from both sides, we get:
x + 368 - 93x = 93x - 93x
x + 368 - 93x = 0
-92x + 368 = 0
Step 3:
Subtracting 368 from both sides, we get:
-92x + 368 - 368 = 0 - 368
-92x = -368
Step 4:
Dividing both sides by -92, we get:

x = 4

From the given information,
3 = (x² - y²)(x² - 2xy + y²)
3 = (x - y)(x + y)(x - y)² = x + y (Because, it is given that x - y = 1)
Hence, 2x = (x + y) + (x - y) = 3 + 1 = 4 and 2y = (x + y) - (x - y) = 3 - 1 = 2
x = 2 and y = 1, so xy = 2.

If p, q and r are the roots of the cubic equation x³ - 3x² + 5x + k - 2 = 0, then the product of the roots are as follows.
pqr = 2 - k (Using the formula of product of roots of a cubic equation)
k = 2 - 1
k = 1

(x³ + y³ + z³ - 3xyz) = (x + y + z)(x² + y² + z² - xy - yz - zx)
= (x + y + z)[(x + y + z)² - 3(xy + yz + zx)]
= 9(9² - 3 x 11) = 9(81 - 33) = 432

1/2 (a + b + c){(a - b)² + (b - c)² + (c - a)²}
1/2 (a + b + c){a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca}
= (a + b + c)(a² + b² + c² - ab - bc - ca)
= a³ + b³ + c³ - 3abc

Solving, we get:

x = ay ... (i)
y = bx ... (ii)
Put the value of x in (ii).
y = b(ay)
⇒ a = 1/b  ...(iii)
According to the question,

Put the value of 'a' from (iii).