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Algebra SSC CGL Tier 2 Questions With Explanations | Maths Preparation

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Algebra SSC CGL Tier 2 Questions With Explanations | Maths Preparation

Algebra SSC CGL Tier 2 Questions With Explanations | Maths Preparation

Dear Readers,

Numerical ability section is considered to be one of the toughest subjects of Competition Exams but it can be the most scoring area if prepared well. Scoring in this section can be easy or tough depending on the complexity of the questions and your understanding of the questions. Today, in this article we are sharing some important Algebra Questions. Practice these questions well and take test to keep yourself at par with your Competition.

You can access Important questions for Advance Maths - Mensuration, Profit and Loss and Number System on our website.

We wish you good luck for the upcoming SSC-Exams.

 

Q1.If 3(x-y)= 27 and  3(x+y)= 243 , then the value of x is

(a) 0

(b) 2

(c) 4

(d) 6

Option:(c)

Explanation :

3(x+y) = 27;

or , 3(x – y) = 33;

X- y= 3 ...............(i)

∴ 3(x+y)= 243;  

3(x+y)= 35 ;

or , x+y = 5

By adding the equations (i) and (ii), X+y = 5;

2x = 8 ; X = 8/2 = 4

Q2.If 8a – b2 = 24, 8b + b2 = 56 then a + b = ?

(a) 3

(b) 7

(c) 10

(d) 80

Option:(c)

Explanation :

On adding both equations;8a-b2= 24

1

⇒x(a +b)= 80 ;

⇒a+b = 10

Q3. If 2 then  3 = ?

(a) 1

(b) 0

(c) 2

(d) None of these

Option:(b)

Explanation :

2

⇒By putting x = 1 the equations will be satistied.

3 =  4

2nd method: let x47  =y;

⇒y+ 1/y = 2 or , y2-2y + 1 = 0

or , (y-1)2 = 0

⇒y= 1

⇒x47 = 1 ⇒1 ,x= 1

Q4.If point A is in third quadrant and 3 tan A – 4 = 0 then the value of 5 sin 2A + 3 sin A + 4 cos A = ?

(a) 0

(b) - 24/5

(c) 24/5

(d) 48/5

Option:(a)

Explanation :In third quadrant sin A < 0 and cos A < 0; 

Sec A = 6= -5/3

Cos S = - 3/5;

Sin A = 7

5 sin 2A + 3 Sin A + 4 Cos A;

= 10 sin A cos A + 3 Sin A = 4 Cos A ;

=10×(-4/5) ×(-3/5)+ 3(-4/5) + 4 ×(-3/5) = 24/5-12/5-12/5=0

Q5. if α and γ are the root of equations x2 +px +q = 0 then the value of α / β + β / α is 

(a)8

(b)9

(c)10

(d)11

Option:(a)

Explanation :

Since α and γ are the roots of equations ;

x2 +px +q = 0

then sum of the roots ( α +γ ) = -p; And product of the roots (α γ) = q

15

Q6.If y + 1/z=1 and  x+ 1/y=1 then the value of  xyz is

(a) 1

(b) – 1

(c) 0

(d) 1/2

Option:(b)

Explanation :

Given that; y + 1/z = 1 ⇒yz + 1 = z ....................(i);

And x +1/y = 1  ⇒xy+1 = y ............... (ii)

From equations (i)y = z-1/z; putting this value in eq. (ii)xy + 1 = z-1/z;

⇒xyz + z = z -1;xyz= -1

Q7.Which of the following statement is true?

(a) 960 < 2735

(b) 960 35                

(c) 960 > 2735

(d) 960 2735

Option:(c)

Explanation :

960 = (32)60 = 3120 ;  2735 = (33)35 = 3105 ;  960 < 2735 is wrong.;

960 > 2735 is right.

Q8.If ab + bc + ca = 0 then the value of 16 is 

???  ab + bc + ca = 0 ?? ?? 16 ?? ??? ???? ???

 (a) 0

(b) 1

(c) 3

(d) a + b + c

Option:(a)

Explanation :Here given that ; 

ab + bc + ca = 0; 

⇒ ab + ca = - bc;  ⇒ a2 + ab + ca = a2 – bc; 

⇒ a(a + b + c) = a2 – bc

17.......................(i)similarly 18 and  so on ...............

By adding eq. (i), (ii) and (ii) we find out the value of the given equations is 0

Q9. If α and β  are the roots of equation  x2+ px + q = 0 then what will be the equations whose  roots are α2+ αβ and β2+ αβ?

(a) x2 + p2x + p2q = 0

(b) x2 – q2x + p2q = 0

(c) x2 + q2x + p2q = 0

(d) x2 – p2x + p2q = 0

Option:(d)

Explanation :If α and β  are the roots of equation x2+ px + q = 0

α +β = -p .........(i);

And  α β = q ............(ii)  That equation whose roots are a2 +α β andβ 2+ α β ;

Sum of roots =a2+ α β+β 2+ α β= (α+ β )2 = p2

Product of roots =  (a2+ α β)+(β 2+ α β)=α β (α+ β )2 =α p2;

(Hence required equation X2 – (sum of roots)x + Product of roots = 0;

Or, x2 – p2x + qp2 = 0

Q10.The graph of the equation 4x – 5y = 20 intersects the x – axis at the point

(a) (2,0)

(b) (5,0)

(c) (4,5)

(d) (0,5)

Option:(b)

Explanation :

 When a straight line cuts x-axis the coordinates of point of intersection = (x, 0), i.e., y = 0.

Putting y = 0 in 4x – 5y = 20

4x = 20 ⇒ x = 5

point of intersection = (5, 0)

[Note: Putting y = 0 in 4x – 5y = 20, point of intersection on x-axis = (5, 0)

Putting x = 0 in 4x – 5y = 20,

Point of intersection on y-axis = (0, -4).

Look at the graph of the equation:

19

Q11.If (2x)(2y) = 8 and (9x)(3y) = 81, then (x, y) is :

(a) (1,2)

(b) (2,1)

(c) (1,1)

(d) (2,2)

Option:(a)

Explanation :

2x.2y = 8   ⇒ 2x+y = 2⇒ x + y = 3 ……..(i) ;

9x. 3y = 34   ⇒ 32x. 3y = 34   ⇒ 2x + y = 4 ……(ii)

By equation (ii) – (i),  x = 1

From equation (i),  1 + y = 3   ⇒ y = 3 – 1 = 2

Method 2: You can check through options also. ⇒ y = 2  ⇒ (1, 2)

Q12.If 2m – 2m-1 – 4 = 0 then the value of mm is

(a) 4

(b) 3

(c) 6

(d) 27

Option:(d)

Explanation : 2m – 2m-1 = 4; 

⇒ 2m-1(2 – 1) = 4;  ⇒ 2m-1 = 22

⇒ m – 1 = 2; 

⇒ m = 3;  mm = 33 = 27;

Q13.If a2 + b2 + c2 = ab + bc + ca, then the value of a3 + b3 + c3 is

(a) 3abc

(b) 3a2b2c2

(c) 3(abc)3

(d) None of these 

Option:(a)

Explanation :

a3 + b3 + c2 – 3abc;

= (a2 + b2 + c2 – ab – bc – ca)(a + b + c);

= [(a2 + b2 + c2) – (ab + bc + ca)] (a + b + c)

But a2 + b2 + c2 = ab + bc + ca (given);

a3 + b3 + c3 – 3abc = (a + b + c) ;

a3 + b3 + c3 = 3abc

Q14.The expression x4 - 2x2 + k will be a perfect square when the value of k is

(a) 2

(b) 1

(c) -1

(d) -2

Option:(b)

Explanation :

(a – b)2 = a2 – 2ab + b2

x4 – 2x2 + k = (x2)2 – 2.x2.1 + k

K= (1)2= 1

Q15.If x + x + 2/x = 3 then the value of x3-x2 -4x+4 is :

(a) 2

(b) 3

(c) 4

(d) 0

Option:(d)

Explanation :

x +2/x= 3 ;  X2 – 3x + 2 = 0; X2 – 3x + 2 = 0;  X2 – 2x – x + 2 = 0;  (x – 2)(x – 1) = 0;  X = 1 or x = 2

By putting x = 1 = (1)3 – 12 – 4×1 + 4

Q16. If a, b, c are real numbers and a + b + c = 0 then the value of a3 + b3 + c3 is:

(a) 1

(b) ab2 + bc2 + ca2

(c) 0

(d) 3abc

Option:(d)

Explanation :

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca);

a3 + b3 + c3 – 3abc = 0 (a+b +c= 0 ) a3 + b3 + c3 = 3abc

Q17.If a + b = 12, ab = 22, then (a2 + b2) is equal to

(a) 188

(b) 144

(c) 34

(d) 100

Option:(d)

Explanation :

a + b = 12, ab = 22

a2 + b2 = (a + b)2 – 2ab

= (12)2 – 2×22=144-44=100

Q18.If x2 + ax + b is a perfect square, then which one of the following relations between a and b is true?

(a) a2 = b

(b) a2 = 4b

(c) b2 = 4a

(d) b2 = a

Option:(b)

Explanation :

ax2 + bx + c will be a perfect square, if b2 = 4 ac

x2 + ax + b will be a perfect square if a2 = 4b

Look : x2 + 2 √bx+b

= x2+2.x√b +(√b)2

= (x+√b)2

Q19.If a = x + y, b = x – y, c = x + 2y, then a2 + b2 + c2 – ab – bc – ca is:

(a) 4y2

(b) 5y2                

(c) 6y2            

(d) 7y2

Option:(d)

Explanation :

a – b = x + y – x + y = 2y

B – c = x – y – x – 2y = -3y

C – a = x + 2y – x – y = y

Now,

A2 + b2 + c2 – ab – bc – ca

=1/2(2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca)

=1/2[(a – b)2 + (b – c)2 + (c – a)2]

=1/2((2y)2 + (-3y)2 + y2)= 1/2×14y2=7y2

Q20.For real a, b, c if a2 + b2 + c2 = ab + bc + ca, then value of a+c/b is

(a) 1

(b) 2

(c) 3

(d) 0

Option:(b)

Explanation :

a2 + b2 + c2 = ab + bc + ca

⇒ 2a2 + 2b2 + 2c2

= 2ab + 2bc + 2ca

⇒ a2 – 2ab + b2 + b2 – 2bc + c2 + c2 – 2ac + a2 = 0

⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0

⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0

⇒ a – b = 0 ⇒ a = b

B – c = 0 ⇒ b = c

C – a = 0 ⇒ c = a

⇒ a = b = c

a+c/b =a+a/a=2

Q21For a > b, if a + b = 5 and ab = 6, then the value of (a2 – b2) is:

(a) 1

(b) 3

(c) 5

(d) 7

Option:(c)

Explanation :

 (a – b)2 = (a + b)2 – 4ab

 ⇒ 52 – 4×6=1

 ⇒ a – b = 1

(a2-b2)=(a+b)(a-b)=5

Q22.If a2 + b2 = 2 and c2 + d2 = 1, then the value of (ab – bc)2 + (ac + bd)2 is

(a) 4/9

(b) 1/2

(c) 1

(d) 2

Option:(c)

Explanation :

 (ad – bc)2 + (ac + bd)2

= a2d2 + b2c2 – 2abcd + a2v2 + b2d2 + 2abcd

= a2d2 + b2c2 + a2c2 + b2d2

= a2d2 + b2d2 + b2c2 + a2c2

= d2 (a2 + b2) + c2 (b2 + a2)

= (a2 + b2)(c2 + d2)

= 2×1= 2

 Q23.If x and y are positive real numbers and xy= 8, then the minimum value of 2x + y is

(a) 9

(b) 17

(c) 10

(d) 8

Option:(d)

Explanation :

xy=8=1×8=2×4=1/2×16

= 1/3×24

Minimum value of 2x +y

=2×2+4= 8

Q24.The value of (xb+c)b-c (xc+a)c-a(xa+b)a-b(x≠0) is     

(a) 1

(b) 2

(c) –1

(d) 0

Option:(a)

Explanation :(xb+c)b-c. (xc+a)c-a. (xa+b)a-b

1

Q25.If a and b be positive integers such that a2-b2=19, then the value of a is

(a) 19

(b) 20

(c) 9

(d) 10

Option:(d)

Explanation :

Tricky approach

a2 - b2 = 19

⇒ 102 – 92 = 10

⇒ a = 10