SSC CGL Previous Year Questions: Algebra- 4 - SSC CGL

When x = 6.

⇒ 8 + 2P = 12
⇒ 2P = 12 – 8 = 4 Þ P = 2

(x – 3)² + (y – 5)² + (z – 4²)= 0
⇒ x – 3 = 0 ⇒ x = 3
y – 5 = 0 ⇒ y = 5
z – 4 = 0 ⇒ z = 4

C.P. of 1 bucket = ₹ x
C.P. of 1 mug = ₹ y
∴ 8x + 5y = 92 ...(i)
5x + 8y = 77 ...(ii)
By equation (i) × 5 – equation (ii) × 8,
40x + 25y – 40x – 64y = 460 – 616
⇒ –39y = –156
⇒ y = 4
From equation (i),
8x + 20 = 92
⇒ 8x = 92 – 20 = 72
⇒ x = 9
∴ C.P. of 2 mugs and 3 buckets
= 2 × 4 + 3 × 9
= 8 + 27 = ₹ 35

Let Puneet’s age = x yr.
Let Puneet’s father age = y yr.
x + y = 45 ⇒ y = (45 – x)
According to question,
xy = 126
Putting the value of y.
(x) (45 – x) = 126
45x – x² = 126
x² – 45x + 126 = 0
x² – 42x – 3x + 126 = 0
x (x – 42) – 3 (x – 42) = 0
x = 3, x = 42
Hence, Puneet’s age is 3yr.

It a = 0, b ≠ 0, c ≠ 0, then equation ax + by + c = 0 represents a line parallel to x-axis.

The least value of a + (1/a) is 2 where a = 1.

x² + 4 = 4x ⇒ x² – 4x + 4 = 0 ⇒ (x – 2)² = 0
x = 2

a + b + c = 0
i.e. a = – (b + c); b = – (c + a); c = – (a + b)
Now, 

⇒ – 1 + 2 – 1 = 0

12 xy – 1 = 24y
Now,

x⁴ – 2x² + k
(x²)² – 2x² + k ⇒ (x²)² – 2.1. x² + k
For above expression to make a perfect square, the k value is equal to 1.

ax + by +c = 0
When c = 0,
ax + by = 0

When x = 0, y = 0 i,e. this line passes through the origin (0, 0).

a² + b² + 4c² = 2a +2b – 4c – 3
⇒ a² + b² + 4c² – 2a –2b + 4c + 3 = 0
⇒ a² – 2a + 1 + b² – 2b + 1 + 4c² + 4c + 1 = 0
⇒ (a –1)² + (b – 1)² + (2c + 1)² = 0
∴ a – 1 = 0 ⇒ a = 1;
b – 1 = 0 ⇒ b = 1;
2c + 1 = 0 ⇒ c = -(1/2)

= 2(1/4)

x² – 3x + 1 = 0
⇒ x² + 1 = 3x
Dividing both sides by x,



= 9 – 2 + 3 = 10

Third proportional of a and b
a : b : c

x + y + z = 6
On squaring,
x² + y² + z² + 2xy + 2zy + 2zx = 36
⇒ 20 + 2 (xy + yz + zx) = 36
⇒ xy + yz + zx = 8
∴ x³ + y³ + z³ – 3xyz
= (x + y + z) (x² + y² + z² – xy – yz – zx)
= 6 (20 – 8) = 72

Expression = (x – 2) (x – 9)
= x² – 11x + 18 = ax² + bx + c
Minimum value = 

If a + b + c = 0, then
a³ + b³ + c³ – 3abc = 0

xy (x + y) = 1
⇒ x + y = 1/xy
Cubing both sides,

On squaring,

p = q + 5
⇒ p – q = 5
p² + q² = 55
∴ (p – q)² + 2pq = 55
⇒ 25 + 2 pq = 55
⇒ 2 pq = 30
⇒ pq = 15

x + 7954 × 7956
x + 7954 (7954 + 2)
x + 7954² + 2 × 7954 × 1
Putting x = 1
(x + 7954)² or (1 + 7954)²

∴ Equation is : y = –3

On squaring both sides

On squaring again

For n^r – tn + (1/4)to be a perfect square, r = 2 and t = ± 1
Look :

⇒ x² = x – a ⇒ x – x² = a

a² + b² + c² + 3
= 2a – 2b – 2c
⇒ a² – 2a + 1 + b² + 2b + 1 + c² * + 2c + 1 = 0
⇒ (a – 1)² + (b + 1)² + (c + 1)² = 0
∴ a – 1 = 0 ⇒ a = 1
b + 1 = 0 ⇒ b = –1
c + 1 = 0 ⇒ c = –1
∴ 2a – b + c = 2 + 1 – 1 = 2

On cubing both sides

p – 21 = 4
cubing both sides,
( p – 2q)³ = 64
⇒ p³ – 8q³ + 3p. 4q² – 3p² . 2q = 64
⇒ p³ – 8q³ + 12pq² – 6p²q = 64
⇒ p³ – 8q³ – 6pq (p – 2q) = 64
⇒ p³ – 8q³ – 6pq × 4 = 64
⇒ p³ – 8q³ – 24pq – 64 = 0

C.P of 1 cow = ₹ x
C.P of a goat = ₹ y
3x + 8y = 47200 ...(i)
⇒ 8x + 3y = 100200 ...(ii)
By equation (i) × 3 – (ii) × 8, 9x + 24y – 64x – 24y
= 141600 – 801600
⇒ 55x = 660000

a² +b² + c² = (a + b + c)² - 2(ab +bc + ca)
= 9² - 2(ab + bc + ca)
a² + b² + c² will be minimum if ab + bc + ca is maximum.
ab + bc + ca is maximum when a = 3, b = 3, and c = 3.
[∵ a + b + c = 9]
∴ minimum value of a² + b² + c²
= 81 – 2(3 × 3 + 3 × 3 + 3 × 3)
= 81 – 54 = 27

sin²θ cannot be – ve
Both sin²θ and cos²θ will be +ve when x = y

Since two digit number = 10x + y
According to question → y = 2x – 1 ..(i)
When digits are interchanged then new number
= 10y + x
then original number – [new number – original number]
= 20
→ 10x + y – [10y + x – (10x + y)] = 20
→ 10x + y – 10y – x + 10x + y = 20
19x – 8y = 20
19x – 8(2x – 1) = 20 (Using eq. (i))
19x – 16x + 8 = 20
3x = 12 → x = 4
From (i) y = 2 × 4 – 1 = y = 7
∴ original number = 10x + y = 10×4 + 7 = 47

x² – 50x + 600 = 0
x² – 30x – 20x + 600 = 0
x(x – 30) – 20(x – 30) = 0
x = 30, 20

a² + b² = 5ab

On squaring both sides.

Let the number be x and y.
According to question,
(x + y)² = 4xy
⇒ x² + y² + 2xy – 4xy = 0
⇒ (x – y)² = 0
⇒ x = y