SSC CGL Previous Year Questions: Algebra- 3 - SSC CGL MCQ

2x – 3y = 0 is passing through origion because its satisfy x = 0 and y = 0.

x = 0 ...(1)
2x + 3y = 6 ...(2)
x + y = 3 ...(3)

2x + 3y = 6 ...(ii)
x = 0, y = 2
x = 3, y = 0
From eqn. (iii) x+ y = 3
x = 0, y = 3
x = 3, y = 0
Area made by these three lines
= Area of triangle OBC – Area of OAC

x = – 1

Using the formula,
 
 

 
 = 0
m- a² - b² - c² = 0
m = a² + b² + c²

For no solution, a = b

m³ – 3m² + 3m + 3n + 3n² + n³
⇒ (– 4)³ – 3 (– 4)² + 3 (– 4) + 3 (– 2) + 3 (– 2)² + (– 2)³
⇒ – 64 – 48 – 12 – 6 + 12 – 8
⇒ – 126

Intercept can represent in the form of 
To get x and y intercept, we have
3x + 4y = 12

So, triplets of 3, 4 and 5.
Hence, 5 is the length of portion of straight line.

= c– (c – 1)
= c – c + 1
= 1

(2a – 1)² + (4b – 3)² + (4c + 5)² = 0

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b²+ c² – ab – bc – ca)
But a + b + c = 0
So, a³ + b³ +c³ – 3abc = 0
So, 

Let the two numbers are x and y
So, x + 2y = 8 … (i)
x – y = 2 … (ii)
Solving both equations
x = 4; y = 2
So, numbers are 4, 2.

Contribution of each boy = Number of boys
Total contribution raised = ₹ 12544
So, number of boys = 

x² – 4x – 1 = 0 can be written as x² – 4x + 4 – 1 = 4
So (x - 2)² – 1 = 4





= 1458 – 1440 = 18.

= 125 + 125 + 360 + 360
= 250 + 720
= 970

p = – 0.12 ; q = – 0.01; r = – 0.015
So, p < r ; p < q
= p < r < q

x + (1/x) = 2
Take cube on both sides

Squaring on both sides

Now multiplying both sides by x


x² – 2x + 1 = 0
x² – x – x +1 = 0
x(x – 1) –1(x – 1)
(x –1)² = 0
x = 1
So 

4x² + 8x = (2x)² + 2(2x) (2) + (2)²
= (2x + 2)² ∴ x² + y² + 2xy = (x + y)²
So, 4 should be added to make if perfect square.

999x + 888y = 1332
111(9x + 8y) = 1332
9x + 8y = 1332/111 = 12
9x + 8y = 12 ...(i)
888x + 999y = 555
8x + 9y = 5 ...(ii)
Solving (i) and (ii)
x = 4, y = – 3
x + y = 4 – 3 = 1

3^{2x – y} = 3^{x + y} = √27 = 3^3/2

4x – 2y = 3 ...(i)
2x + 2y = 3 ...(ii)
Solving equation (i) and (ii)
x = 1 y = 1/2

Mean of x and (1/x) = M

Mean of x² and 

Given,

We know that property of ratio,

So,

= (x + y + z)/(xa + xb + xc + ya + yb + yc + za + zb + zc)
⇒ x / (xa + yb + zc) = y / (ya + zb + xc) = z / (za + xb + yc)
= x / (xa + yb + zc) = y (ya + zb + xc) = z / (za + xb + yc)
= (x + y + z) / {x(a + b + c) + y(a + b + c) + z(a + b + c)}
⇒ x(xa + yb + zc) = y(ya + zb + xc) = z(za + xb + yc)
= (x + y + z) / {(a + b + c) (x + y + z)
⇒ x / (xa + yb + zc) = y / (ya + zb + xc) = z / (za + xb + yc)
= 

We have, x = 3 + 2√2


= 6[6² – 3] + 6 = 198 + 6 = 204

x⁴ – 2x²y² + y⁴ = (x² – y²)² = [(x + y)(x – y)]²

Here, a + b + c = 0
∴ a³ + b³ + c³ – 3abc = 0

x² –2x + 1 = 0; (x – 1)² = 0 ; x = 1

= 1 + 1 = 2

2a² – 5ab + 2b²
2(a² – 2ab + b²) – ab
2 (a – b)² – ab

2 × 4 × 5 – 1 = 39

x = 4

x + y + z = 0
y + z = –x
Squaring both sides
y² + z² + 2yz = x²
⇒ y² + z² = x² – 2yz ...(1)