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


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20 Questions MCQ Test SSC CGL (Tier - 1) - Previous Year Papers (Topic Wise) - SSC CGL Previous Year Questions: Algebra- 1

SSC CGL Previous Year Questions: Algebra- 1 for SSC CGL 2024 is part of SSC CGL (Tier - 1) - Previous Year Papers (Topic Wise) preparation. The SSC CGL Previous Year Questions: Algebra- 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL Previous Year Questions: Algebra- 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SSC CGL Previous Year Questions: Algebra- 1 below.
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SSC CGL Previous Year Questions: Algebra- 1 - Question 1

If (5a – 3b) : (4a – 2b) =2 : 3, then a : b is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 1



SSC CGL Previous Year Questions: Algebra- 1 - Question 2

If a – b = 5 an d ab = 6, then (a3 – b3) is equal to:        (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 2

(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

SSC CGL Previous Year Questions: Algebra- 1 - Question 3

If a + b + c = 10 and ab + bc + ca = 32 then a3 + b3 + c3 – 3abc is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 3

(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

SSC CGL Previous Year Questions: Algebra- 1 - Question 4

If  then  is equal to:        (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 4

On squaring both sides, we get


Again, squaring on both sides,

SSC CGL Previous Year Questions: Algebra- 1 - Question 5

If  then x2 + (1/x2) is equal to:        (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 5

Given that: 
Squaring both sides, we get

Again, squaring both sides, we get

SSC CGL Previous Year Questions: Algebra- 1 - Question 6

If a : b = 3 : 2, then (5a + 2b) : (3a + 4b) is equal to:        (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 6


SSC CGL Previous Year Questions: Algebra- 1 - Question 7

If a + b + c = 13 and ab + bc + ca = 54, then a3 + b3 + c3 – 3abc is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 7

(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

SSC CGL Previous Year Questions: Algebra- 1 - Question 8

If x + (1/x)= 3, then x3 + (1/(x3)) is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 8




= 3 (7 – 1) = 3 x 6 = 18

SSC CGL Previous Year Questions: Algebra- 1 - Question 9

If a2 + b2 = 169, ab = 60, (a > b), then (a2 – b2) is equal to :       (SSC CHSL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 9

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

SSC CGL Previous Year Questions: Algebra- 1 - Question 10

If (2x + 3)3 + (x – 8)3 + (x + 13)3 = (2x + 3) (3x – 24) (x +13), then what is the value of x?       (SSC Sub. Ins. 2018 )

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 10

(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

SSC CGL Previous Year Questions: Algebra- 1 - Question 11

If a3 + b3 = 5824 and a + b = 28, then (a – b)2 + ab is equal to:         (SSC Sub. Ins. 2018 )

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 11

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

SSC CGL Previous Year Questions: Algebra- 1 - Question 12

If x – (1/x) = 6, then x3 – (1/x3) is equal to:       (SSC Sub. Ins. 2018 )

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 12




Now. 
= 6(38 + 1) = 6 × 39 = 234.

SSC CGL Previous Year Questions: Algebra- 1 - Question 13

If  then what is the value of           (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 13

Here,

Then,

Now,

SSC CGL Previous Year Questions: Algebra- 1 - Question 14

If x2 – 3x + 1 = 0, then what is the value of      (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 14

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

= (3)3 – 3 × 3
= 27 – 9 = 18

SSC CGL Previous Year Questions: Algebra- 1 - Question 15

If a = 2017, b = 2016 and c = 2015, then what is the value of a2 + b2 + c2 – ab – bc – ca?    (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 15

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

SSC CGL Previous Year Questions: Algebra- 1 - Question 16

If the expression (px3 – 8x2 – qx + 1) is completely divisible by the expression (3x2 – 4x + 1), then what will be the value of p and q respectively?          (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 16

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

SSC CGL Previous Year Questions: Algebra- 1 - Question 17

If (p2 + q2) / (r2 + s2) = (pq) / (rs), then what is the value of (p – q) (p + q) in terms of r and s?       (SSC Sub. Ins. 2017)

SSC CGL Previous Year Questions: Algebra- 1 - Question 18

If 'a' and 'b' are positive integers such that a2 – b2 = 19, then the value of 'a' is :       (SSC MTS 2017)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 18

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

SSC CGL Previous Year Questions: Algebra- 1 - Question 19

The line passing through (–2, 5) and (6, b) is perpendicular to the line 20x + 5y = 3. Find b?       (SSC CHSL 2017)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 19

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

SSC CGL Previous Year Questions: Algebra- 1 - Question 20

If x – y = 6 and xy = 40, then find x2 + y2?       (SSC CHSL 2017)

Detailed Solution for SSC CGL Previous Year Questions: Algebra- 1 - Question 20

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

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