SSC CGL Previous Year Questions: Algebra- 1 - SSC CGL MCQ

(a - b)² = a² + b² - 2 ab ⇒ (5)² = a² + b² -2 x 6
∴ a² + b² = 25 + 12 = 37
(a³ - b³) = (a - b) (a² + ab + b²)
= 5 (37 + 6) = 215

(a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)
a² + b² + c² = (a + b + c)² - 2 (ab + bc + ca)
= (10)² - 2 (32) = 100 - 64
a² + b² + c² = 36
Now, a³ + b³ + c³ - 3abc
= (a + b + c) (a² + b² + c² - ab - bc - ca)
= 10 (36 - 32) = 40

On squaring both sides, we get


Again, squaring on both sides,

Given that: 
Squaring both sides, we get

Again, squaring both sides, we get

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(13)² = a² + b² + c² + 2(54)
a² + b² + c² = 169 – 108 = 61
Now, a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)
= 13(61 – 54) = 13 x 7 = 91

= 3 (7 – 1) = 3 x 6 = 18

From question,
(a + b)² = a² + b² + 2ab
= 169 + 2 × 60
(a + b)² = 289 ⇒ (a + b) = 17 ...(i)
and
(a – b)² = a² + b² – 2ab
= 169 – 120 = 49
(a – b) = √49 = 7 ...(ii)
from (i) and (ii), we get
(a + b) (a – b) = 17 × 7
a² – b² = 119

(2x + 3)³ + (x – 8)³ + (x + 13)³ = (2x + 3) (3x – 24) (x + 13)
(2x + 3)³ + (x – 8)³ + (x + 13)³ = 3(2x + 3)(x – 8)(x +13)
We know that if a³ + b³ + c³ = 3 abc
then, a + b + c = 0
(2x + 3) + (x – 8) + (x + 13) = 0
4x + 8 = 0
x = (-8)/4
∴ x = – 2

a³ + b³ = 5824
(a + b) (a² + b² – ab) = 5824
28(a² + b² – ab) = 5824
(a² + b² – ab) = 5824/28 = 208
Now, (a – b)² + ab = a² + b² – 2ab + ab
= a² + b² – ab = 208

Now. 
= 6(38 + 1) = 6 × 39 = 234.

Here,

Then,

Now,

If x² – 3x + 1 = 0 then 
Dividing equation by x

= (3)³ – 3 × 3
= 27 – 9 = 18

Here, a = 2017, b = 2016, and c = 2015
∴ a² + b² + c² – ab – bc – ca
⇒ (2017)² + (2016)² + (2015)² – 2017× 2016 – 2016 × 2015 – 2015 × 2017 = 3

Let p(x) = px³ – 8x² – qx + 1
Since, (3x² – 4x + 1) is factor of p (x), so p (a) = 0
∴ 3x² – 4x + 1 = 0
3x² – 3x – x + 1 = 0
3x (x – 1) – 1 (x – 1) = 0
(3x – 1) (x – 1) = 0
∴ x = 1/3, 1
∴ p(x) = 1/3, 1
i.e.,


⇒ p – 24 – 9q + 27
⇒ p – 9q = –3 ...(i)
p (1) = px³ – 8x² – qx + 1
⇒ p (1)³ – 8 (1)² – q × 1 + 1
⇒ p – 8 – q + 1
⇒ p – q = 7 ....(ii)
From Eq. (i) and (ii),
p = 33/4, q = 5/4

According to question,
a² – b² = 19
(a + b) (a – b) = 19
Since 19 is prime, one of (a + b) (a – b) is 19
Therefore, (10)² – (9)² = 19
∴ a = 10

Here,
20x + 5y = 3
⇒ 5y = – 20x + 3
∴ y = – 4x + (3/5)
Slope of 20x + 5y = 3 ⇒ –4
We know, product of slopes = –1 for perpendicular lines 
Hence, the slope of the line which passes through (–2, 5) and (6, b) = 
Now, 
⇒ b – 5 = 2
∴ b = 5 + 2 = 7

Here,
x – y = 6, xy = 40, x² + y² = ?
(x – y)² = (6)²
x² + y² – 2. x. y = 36
∴ x² + y² = 36 + 2xy
⇒ 36 + 2 × 40
= 116