SSC CGL Previous Year Questions: Algebra- 5 - SSC CGL MCQ

y = 4x,
When, x = 1, y = 4

x + y + z = a – b + b – c + c – a = 0
∴ x³ + y³ + z³ –3xyz = 0

xy + yz + zn = 0

Given 
We know, a³ +b³ = (a + b)(a² +b² - ab)
∴ (a + b) (ab – ab) ⇒ 0 (using given)

a² +b² + c² = 2(a -b - c) - 3
⇒ a² + b² + c² - 2(a -b - c) + 3 = 0
⇒ a² + b² + c² - 2a + 2b + 2c + 3 = 0
⇒ (a² +1- 2a) + (b² +1+ 2b) + (c² +1+ 2c) = 0
⇒ (a -1)² + (b +1)² + (c +1)² = 0
This is possible when (a -1)² = 0,(b +1)² = 0 and (c +1)² = 0.
⇒ a = 1, b = –1, c = –1
Thus, 2a – 3b + 4c = 2(1) – 3(–1) + 4(–1)
= 2 + 3 – 4 = 1.

= a³ + b³ + 3a²b + 3b² – (a³ – b³ – 3a²b + 3b²a)
= a³ + b³ + 3a²b + 3b² – a³ + b³ + 3a²b – 3b²a

Let the numbers are x, y.
x – y = 3 … (1)
x² – y² = 39
⇒ (x – y) (x + y) = 39
⇒ x + y = 13 … (2)
Adding eqn (1) and (2)
x + y + x – y = 16
⇒ x = 8
∴ y = 5
Hence, 8 is the larger number.

⇒ 9x + 15 = 10x – 4
⇒ 15 + 4 = 10x – 9x
⇒ x = 19

So, x = k (b + c)
⇒ x – y = k (b + c) – k (c + a)
= k (b – a)
y = k (c + a)
⇒ y – z = k (c + a) – k (a + b)= k (c – b)
z = k (a + b) = z – x = k (a + b) – k (b + c) = k (a – c)
So,check option (a)

k = k = k
option (a) is true.

Given equation is y=3x
When x=1,⇒y=3×1=3
When x=2,⇒y=3×2=6
When x=3,⇒y=3×3=9

Plot the points A(1,3), B(2,6), C(3,9) on the graph paper.Join these points by a straight line AC

y = x + |x|
case 1, x>0
y = x + x = 2x
case 2, x<0
y = x - x = 0
so, the graph of y = x + |x| is lies in 

As indicated in the graph, the line passing through the points cuts Y-axis only.

Þ 16x = 32y
 ...(1)
Now, to calculate value of  Divide numerator & denominator by y.

Putting value of x/y from equation (1)

a + b = 5
Squaring on both sides
(a + b)² = (5)²
a² + b² + 2ab = 25
13 + 2ab = 25
2ab = 25 – 13 = 12 ... (1)
Again, a² + b² = 13
Subtracting (– 2ab) from both sides
a² + b² – 2ab = 13 – 2ab
(a – b)² = 13 – 12 from equation (1)
(a – b)² = 1
TRICK Þ a = 3
b = 2 (a > b)
a – b = 1

is minimum when x = 0

fartrise the given expression

Therefore,

Multiply by 2 on both the sides

Seprate individual square

So, for Add equal to 0, individual becomes 0

from equation (1) & (2)
a=b=c 

If a + b + c = 0
then a³ + b³ + c³ = 3abc
Dividing both sides by abc

Multiply by 3/5 on both sides

Squaring on both sides

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

We have x³ + y³ + z³ – 3xyz = (x + y + z)
(x² + y² + z²–xy – yz - zx)
Here x = a – 4, y = b – 3, z = c –1
So, given expression is (x + y + z)
(x² + y² + z² – xy – yz – zx)
= (a – 4 + b – 3 + c – 1) (x² + y² + z² – xy – yz – zx)
= (a + b + c – 8) (x² + y² + z² – xy – yz – zx)
= (8 – 8) (x² + y² + z² – xy – yz – zx)
= 0

x = 4 ...(1)
y = 3 ...(2)
3x + 4y = 12 ...(3)
Putting x = 0 in 3rd equation we get y = 3
Putting y = 0 in 3rd equation we get x = 4
The triangle will be formed by joining the points (3, 0) and (0, 4).
So, base = 3 and altitude = 4
Area = 

x² + y² + z² – xy – yz – zx

Applying compodendo and Dividendo

5x + 7y = 35 ...(i)
4x + 3y = 12 ...(ii)
By equation (i) × 4 – (ii) × 5

⇒ y = 80/13 = Height of triangle
Point of intersection on x-axis of equation
5x + 7y = 35
⇒ 5x + 7 × 0 = 35
⇒ 5x = 35
⇒ x = 7
∴ (7, 0)
Similarly, point of intersection of
4x + 3y = 12 = (3, 0)
∴ Base = 7 – 3 = 4

On multiplying by 3/5,

On squaring,

 [on dividing by 2]
 [On Squaring]

a² + b² + c² = 2(a – b –c) –3
⇒ a² + b² + c² – 2a + 2b + 2c + 3 = 0
⇒ a² – 2a + 1 + b² + 2b + 1 + c² + 2c + 1 = 0
⇒ (a –1)² + (b + 1)² + (c + 1)² = 0
[If x² + y² + z² = 0 ⇒ x = 0; y = 0; z =0]
∴ a – 1 = 0 ⇒ a = 1
b + 1 = 0 ⇒ b = – 1
c + 1 = 0 ⇒ c = – 1
∴ 2a – 3b + 4c = 2 + 3 – 4 = 1