Test: SSC CGL Previous Year Questions: Algebra (2023-18) - 2 - SSC CGL MCQ

A number is the factor of its square. 
Squaring the identity (mx + n)
(mx + n)² = (m² x² + 2mnx + n²)

(x + y + z)²
= x² + y² + z² + 2(xy + yz + zx)
(a + b + c)² = 9 + 2 × 8
a + b + c = 5
Now, (a + b + c) - 3 = 5 - 3 = 2

Formula :-
(a − b)³ = a³ − b³ − 3ab (a − b)
a − b = 2 and a³ - b³ = 80…(given)
Now, (2)³ = 80 - 3ab(2) ⇒ ab = 12

x = 1 …..(given)
x² + y² + z² = xy + yz + zx ... (given)
Put , y = 1 and z = 1 . which satisfies the given condition.

4x² + y² = 40
Adding (4xy) both sides, we get
(2x)² + y² + 4xy = 40 + 4xy (xy = 6 given)
(2x + y)² = 64 ⇒ 2x + y = 8
Short trick :
4x + y² = 40 and xy = 6 ….(given)
Put x = 3 and y = 2 which satisfy the both given equations.
Now , the value of 2x + y = 2 × 3 + 2 = 8.

mx^m - nxⁿ = 0


{using eq . (i) and (ii)}

(a + 1)² + (a + 2)² = 16
⇒ a² + 2a + 1 + a² + 4a + 4 = 16
⇒ 2a² + 6a + 5 = 16
⇒ 2 (2a² + 6a + 5 ) = 16 × 2
⇒ 4a² + 12a + 10 = 32
So, 4a² + 12a + 10 + 30 = 32 + 30
⇒ 4a² + 12a + 40 = 62

2x - 1 = 0 ⇒ x = 1/2,
Put x = 1/2 in the given equation.
3x⁴ - 2x² + 4x - 1

So, the remainder = 11/16.

putting a = b = 1 and c = 5 in the given expression

4(z + 7) (2z – 1) = (4z + 28)(2z - 1)
= 8z² - 4z + 56z - 28 = 8z² + 52z - 28
Now, 8z² + 52z - 28 = Az² + Bz + C
On comparing both side, we have :
A = 8, B = 52 , C = - 28
So, A + B + C = 8 + 52 - 28 = 32

Formula used :
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
a² + b² + c² = 9² - 2 x 23 = 81 - 46 = 35

x - (1/x) = 

x³ - 1/(x³) = 4³ + 3 x 4 = 64 + 12 = 76

Formula : x³ + y³ + z³ = 3xyz
(when x + y + z = 0)

x + 2/x = √129.

I. x³ + y³ + z³ = 12³ - 2³ - 10³
= 1728 - 8 - 1000 = 720
II. 4 (x² + y²) = 4 (x + y)² [−2xy]
= 4[48² - 64] = 4[2304 - 64]
= 4 x 2240 = 8960
Clearly, we can see that neither statement I nor II is correct.

On cubing above equation, we get

So, x⁶ + x^-6 = -2

2X + 3Y = 12 ……(1)
3X - 2Y = 5 ………(2)
Multiplying eqn (1) by 3 and eqn (2) by 2, we get
6X + 9Y = 36 …...(3)
6X - 4Y = 10 ……(4)
Solving equation (3) and (4), we get
X = 3 and Y = 2
So, option (a) is the correct answer.

(4x + 2y)³ + (4x − 2y)³
= 16(Ax³ + Bxy²)
⇒ [64x³ + 8y³ + 24xy(4x + 2y)] + [64x³ - 8y³ - 24xy(4x - 2y)]
= 16(Ax³ + Bxy²)



On comparing, we get A = 8 and B = 6

Given,
a³ + b³ = 218 and a + b = 2,
(a + b)³ = a³ + b³ + 3ab(a + b)
2³ = 218 + 3ab × 2
⇒ – 210 = 6ab ⇒ ab = –35

Short trick :
by hit and trial method a = 7 and b = – 5

 … squaring both sides
x + 1/x = (√5)² + 2 = 7
dividing numerator and denominator by x
 = 46

x⁴ - 79x² + 1 = 0
⇒ x⁴ + 1 = 79x² …..divide eq by x²
  … add 2 both sides
= 81 … take square root both sides ⇒ x + 1/x = 9

4x - 3y = 12
Squaring both sides
16x² + 9y² - 24xy = 144
16x² + 9y² = 144 + 24×(5)
16x² + 9y² = 144 + 120 = 264
 264/8 = 33

(No option matches)
For 
= (−7)² - 2 = 49 - 2 = 47

2x - y = 2
Both side divided by 2, then

x² − √7x + 1 = 0

x³ + y³ = (x + y) (x² + y² - xy)
= (x + y) [(x + y)² - 3xy]
⇒ x³ + y³ = 4(4² - 3 × (15/4)) = 19

Given that, 
 So, xy = - 24
x³ - y³ + x²y² = (x - y) {(x - y)² + 3xy} + x²y²
x³ - y³ + x²y² = 11 (121 - 72) + (-24)²
x³ - y³ + x²y² = 11 × 49 + 576
= 539 + 576 = 1115
Short-trick:
Put x = 8 and y = -3
x - y = 11 ⇒ 8 - (-3) = 11 ⇒ xy = - 24
x³ - y³ + x²y² = 512 - (-27) + 64 × 9
= 512 + 27 + 576 = 1115

If x - 1/x = 5 then

 dividing the numerator and denominator by x³

x - 1/x = 1
(On squaring both sides)
⇒ x² + 1/x² = 1² + 2 = 3
Similarly,