Algebra Questions for SSC Stenographer

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Algebra Questions for SSC Stenographer
Algebra Questions for SSC Stenographer

Algebra Questions for SSC Stenographer

SSC Stenographer Algebra Questions and Answers download PDF based on previous year question papers of SSC exam. Top – 15 Very important Algebra Questions for Stenographer.

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Question 1: 8+57+38+108+169

a) 4

b) 6

c) 8

d) 10

Question 2: What is the value of (941+149)2+(941149)2(941×941+149×149)?

a) 10

b) 2

c) 1

d) 100

Question 3: If a (2+ √3) = b (2 – √3) then the value of 1/(a2+1)+1/(b2+1) is

a) -5

b) 1

c) 4

d) 9

Question 4: If a2+1=a, then the value of a12+a6+1 is :

a) -3

b) 1

c) 2

d) 3

Question 5: If a=2+323 and b=232+3, then the value of a2+b2+a×b is

a) 185

b) 195

c) 200

d) 175

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Question 6: If x+1x=99, find the value of 100x2x2+102x+2

a) 1/6

b) 1/2

c) 1/3

d) 1/4

Question 7: If 2x – 2(4 – x) < 2x – 3 < 3x + 3; then x can take which of the following values?

a) 2

b) 3

c) 4

d) 5

Question 8: If (x + y):(x – y) = 5:2, find value of (4x + 5y) / (x – 4y)

a) 43/5

b) -5/43

c) -43/5

d) 5/43

Question 9: Which of the following equations has the sum of its roots as 5?

a) x25x+6=0

b) x26x5=0

c) x2+5x+6=0

d) x2+6x5=0

Question 10: (2x+5)2×(4x1)(3x316x2+25x21)

a) 13x3+92x255x+4

b) 13x392x255x4

c) 13x3+92x2+55x4

d) 13x392x2+55x+4

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Question 11: What is the value of 299996 x 300004?

a) 89999999984

b) 89999699984

c) 89999999884

d) 89999999974

Question 12: If a+b = 8 and ab =15, then a3+b3 is

a) 224

b) 244

c) 152

d) 128

Question 13: If 3x+5(43x)>24x<3xx3; then the value of x is

a) 3

b) 0

c) 2

d) -1

Question 14: Simplify (b5x2a3z4)(b3x2a4z5)/(a2b3z2)

a) b5x4a5z5

b) b5x4a5z7

c) b5x4a4z7

d) b4x4a5z7

Question 15: If xy = 56 and x2+y2=113, then what will be the value of (x + y)?

a) 29

b) 21

c) 36

d) 15

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Answers & Solutions:

1) Answer (A)

Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4

2) Answer (B)

Expression : (941+149)2+(941149)2(941×941+149×149)

= (9412+1492+2.941.149)+(9412+14922.941.149)9412+1492

= 2(9412+1492)9412+1492

= 2

3) Answer (B)

Let a=(23) and b=(2+3)
a2=743, and b2=7+43

1a2+1+1b2+1=(a)2+(b)2+2((a)2+1)((b)2+1)

(a2+1)(b2+1) = (843)(8+43) = 16(23)(2+3)=16

a2+b2+2=16
Thus 1a2+1+1b2+1=1

4) Answer (D)

Expression : a2+1=a

=> a2a+1=0

Multiplying by (a+1) on both sides

=> (a+1)(a2a+1)=0

=> a3+13=0

=> a3=1

To find : a12+a6+1

= (a3)4+(a3)2+1

= (1)4+(1)2+1

= 1+1+1=3

5) Answer (B)

a=2+323 on rationalising we will get a = (2+3)2

b=232+3 on rationalizing we will get b = (23)2

now putting values of a and b in , a2+b2+a×b

a2+b2+a×b = 195

6) Answer (C)

Given : x+1x=99

To find : 100x2x2+102x+2

= 50xx2+1+51x

Dividing numerator and denominator by x

= 50x+1x+51

Substituting value of (x+1x), we get :

= 5099+51=50150

= 13

7) Answer (A)

Expression 1 : 2x3<3x+3

=> 3x2x > 33

=> x > 6 ———-(i)

Expression 2 : 2x2(4x)<2x3

=> 4x8 < 2x3

=> 4x2x < 83

=> x < 52 ——(ii)

Combining inequalities (i) and (ii), we get : 6 < x < 52

Thus, only value that x can take among the options = 2

=> Ans – (A)

8) Answer (C)

Given : x+yxy=52

=> 2x+2y=5x5y

=> 2y+5y=5x2x => 7y=3x

=> y=3x7

To find : 4x+5yx4y

= [4x+5(3x7)]÷[x4(3x7)]

= (4x+15x7)÷(x12x7)

= (43x7)÷(5x7)

43x7×75x=435

=> Ans – (C)

9) Answer (A)

Sum of roots in an equation : ax2+bx+c=0 is ba

(A) : x25x+6=0

=> Sum of roots = 51=5

(B) : x26x5=0

=> Sum of roots = 61=6

(C) : x2+5x+6=0

=> Sum of roots = 51=5

(D) : x2+6x5=0

=> Sum of roots = 61=6

=> Ans – (A)

10) Answer (C)

Expression : (2x+5)2×(4x1)(3x316x2+25x21)

= [(4x2+20x+25)×(4x1)](3x316x2+25x21)

= [(16x34x2)+(80x220x)+(100x25)]+(3x3+16x225x+21)

= (16x33x3)+(80x24x2+16x2)+(100x25x20x)+(25+21)

= 13x3+92x2+55x4

11) Answer (A)

Expression :  299996 x 300004

= (300000 – 4) x (300000 + 4)

= (300000)2(4)2

= 90000000000 – 16 = 89999999984

=> Ans – (A)

12) Answer (C)

It is given that : (a+b)=8 and ab=15

Using the expression, (a3+b3)=(a+b)(a2+b2ab)

= (8)(a2+b215)

Also, (a2+b2)=(a+b)22ab

= 8[((a+b)22ab)15]

= 8[(82(2×15))15]

= 8(643015)=8×19

= 152

=> Ans – (C)

13) Answer (C)

Expression 1 : 3x + 5(4 – 3x) > 2 – 4x

=> 3x+2015x > 24x

=> 12x4x < 202

=> 8x < 18

=> x < 94 ———–(i)

Expression 2 : 2 – 4x < 3x – x/3

=> 4x+3xx3 > 2

=> 20x3 > 2

=> x > 310 ———-(ii)

Combining inequalities (i) and (ii), we get : 310 < x < 94

The only value that x can take among the options = 2

=> Ans – (C)

14) Answer (B)

Expression : (b5x2a3z4)(b3x2a4z5)/(a2b3z2)

= (a)3+4(b)5+3(x)2+2(z)4+5÷a2b3z2

= a7b8x4z9÷a2b3z2

= (a)72(b)83(x)4(z)92

= a5b5x4z7

=> Ans – (B)

15) Answer (D)

Given : (x2+y2)=113 and xy=56

Using (x+y)2=x2+y2+2xy

=> (x+y)2=113+(2×56)

=> (x+y)2=113+112=225

=> (x+y)=225=15

=> Ans – (D)

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