**SSC CGL 2019 EXAM PAPER : Held on 03-March-2020 Shift-2 (Quantative Aptitude)**

Table of Contents

**Q.1 A can complete a certain piece of work in 40 days. B is 25% more efficient than A and C is 28% more efficient than B. They work together for 5 days. The remaining work will be complete by B alone, in: **

1. 20 1/2 days

2. 16 1/5 days

3. 16 3/5 days

4. 20 3/4 days

##### Correct answer is **Option 3 i.e. 16 3/5**

##### A can complete a certain work in = 40 days

##### Efficiency of A (A’s 1 day’s work) = 1/40

##### Efficiency of B (B’s 1 day’s work) = 1/40 × (125/100) = 1/32

##### Efficiency of C (C’s 1 day’s work) = 1/32 × (128/100) = 1/25

##### Let remaining work B complete in x days.

##### According to the question

##### 5 (1/40 + 1/32 + 1/25) + x/32 = 1

##### ⇒ 5 [(20 + 25 + 32)/800] + x/32 = 1

##### ⇒ 77/160 + x/32 = 1

##### ⇒ x/32 = 1 – 77/160

##### ⇒ x/32 = 83/160

##### ⇒ x = (83/160) × 32

##### ⇒ x = 83/5

##### ⇒ x = 16 3/5 days

**Short Trick:**

##### Let Efficiency of A = 4

##### Efficiency of B = 100 × 5/4 = 5

##### Efficiency of C = 5 × 32/25 = 32/5

##### Efficiency ratio of A, B and C = 4 : 5 : 32/5 = 20 : 25 : 32

##### Total work = 20 × 40 = 800

##### Work done by A, B and C in 5 days = (20 + 25 + 32) × 5 = 385

##### Remaining work = 800 – 385 = 415

##### ∴ Remaining work B complete in = 415/25 = **16 3/5 ****days**

**Q.2 A race track is in the shape of a ring whose inner and outer circumferences are 440 m and 506 m, respectively. What is the cost of levelling the track at Rs 6/m**^{2}?(take π = 22/7)

^{2}?(take π = 22/7)

##### 1. Rs. 18,966

2. Rs. 24,832

3. Rs. 19,866

4. Rs. 29,799

##### The correct answer is__ option 4__: **Rs. 29,799**

##### Let internal and external radius of the track be r m and R m respectively.

##### Internal circumference of the track = 2πr

##### 2 πr = 440

##### ⇒ 2 × (22/7) × r = 440

##### ⇒ r = 70 m

##### External circumference of the track = 2πR

##### 2 πR = 506

##### ⇒ 2 × (22/7) × R = 506

##### ⇒ R = 506 × (7/44)

##### ⇒ R = 80.5 m

##### Area of track = π (R^{2} – r^{2}) = (22/7) × (80.5^{2} – 70^{2}) = 22/7 × 150.5 × 10.5 = 4966.5 m^{2}

##### Cost of levelling 1 m^{2} = Rs. 6

##### ∴ Cost of levelling 4966.5 m^{2} = 4966.5 × 6 = Rs. 29,799

**Q.3 The compound interest on a certain sum at 10% p.a. for 2 1/3 years is Rs. 1,201.60, interest compounded yearly. The sum is**

1. Rs. 4,200

2. Rs. 4,800

3. Rs. 5,400

4. Rs. 4,500

##### Correct answer is **option 2 i.e. 4800**

##### As we know,

##### A = P (1 + r/100)^{t}

##### Rate for first 2 yeas = 10%

##### Rate for 1/3 year = 10/3 %

##### CI = Rs. 1201.60

##### 1201.60 + P = P × (11/10) × (11/10) × (31/30)

##### ⇒ 1201.60 = 3751P/3000 – P

##### ⇒ (3751 – 3000P)/3000 = 1201.60

##### ⇒ 751P/3000 = 1201.60

##### ⇒ P = 1201.60 × 3000/751

##### ∴ P = 4800

**Short Trick:**

##### 3751 – 3000 = 751 unit

##### ⇒ 751 unit = 1201.60

##### ⇒ 1 unit = 1201.6/751

##### ⇒ 3000 unit = (1201.6/751) × 3000 = 4,800

**Q.4 **The given table represents the number of engineers recruited by four companies A, B, C and D over the years. Study the table carefully and answer the question that follows.

##### The ratio of the total number of engineers recruited by companies A and B in 2015 and 2018 to the total number of engineers recruited by C and D in 2014 and 2018, is

##### 1. 28 : 19

2. 13 : 21

3. 17 : 14

4. 9 : 14

##### Correct answer is **option 3 i.e. 17:14**

##### Number of engineers recruited by companies A and B in 2015 and 2018 = 132 + 118 + 148 + 112 = 510

##### Number of engineers recruited by companies C and D in 2014 and 2018 = 85 + 105 + 105 + 125 = 420

##### Required ratio = 510 : 420 = **17 : 14**

**Q.5 Two bottles of the same capacity are 35% and 33 1/3% full of orange juice, respectively they are filled up completely with apple juice and then the contents of both bottles are emptied into another vessel. The percentage if apple juice in the mixture is: **

1. 34 4/6

2. 64 1/3

3. 60 2/3

4. 65 5/6

##### Correct answer is option 4 i.e. 65 5/6

##### 33 1/3%= 33+1/3

##### Average percentage of orange juice, in new vessel = (35 + 33 + 1/3)/2 = 34 + 1/6 = 34 1/6%

##### Percentage of apple juice, in new vessel = 100% – 34 1/6% = 65 5/6%

**Q.6 In ΔABC, AB = AC and AL is perpendicular to BC at L. In ΔDEF, DE = DF and DM is perpendicular to EF at M. If (area of ΔABC) : (are of ΔDEF) = 9 : 25 and ∠BAC = ∠EDF, then **

1. 6

2. 4

3. 3

4. 5

##### Correct answer is option 2 i.e. 4

##### Hence, (area of ΔABC) : (are of ΔDEF) = (AL)^{2} : (DM)^{2}

##### (AL)^{2} : (DM)^{2} = 9 : 25

##### AL/DM = 3/5

##### Let AL = 3 and DM = 5

##### ∴ (DM + AL)/(DM – AL) = (5 + 3)/(5 – 3) = 8/2 = **4**.

**Q.7 The given table represents the number of engineers recruited by four companies A, B, C and D over the years. Study the table carefully and answer the question that follows.**

##### The number of the years in which the number of engineers recruited by company D is less than the average number of engineers recruited by B in the given six years is:

1. 3

2. 1

3. 4

4. 2

##### Correct answer is __option 1 i.e. 3__

##### Total number of engineers recruited by company B = 90 + 118 + 98 + 106 + 112 + 118 = 642

##### Average number of engineers recruited by company B = 642/6 = 107

##### There are three years 2014, 2015 and 2016 the number of engineers recruited by company D is less than the average number of engineers recruited by B in the given six years.

**Q.8 The value of (18÷2of1/4)×(2/3÷3/4×5/8)÷(2/3÷3/4of3/4)is : **

1. 10 2/3

2. 8 5/8

3. 16 7/8

4. 2 7/64

#### Correct answer is option 3 i.e. 16 7/8

**Q.9 If 2 sin θ + 15 cos**^{2} θ = 7, 0° < θ < 90°, then tan θ + cos θ + sec θ = ?

^{2}θ = 7, 0° < θ < 90°, then tan θ + cos θ + sec θ = ?

1. 3 3/5

2. 3

3. 3 4/5

4. 4

##### Correct answer is option 2 i.e. 604

2 sin θ + 15 cos^{2} θ = 7

⇒ 2 sin θ + 15 (1 – sin^{2} θ) = 7

⇒ 2 sin θ + 15 – 15 sin^{2} θ = 7

⇒ 15 sin^{2} θ – 2 sin θ – 8 = 0

⇒ 15 sin^{2} θ – 12 sin θ + 10 sin θ – 8 = 0

⇒ 3 sin θ (5 sin θ – 4) + 2 (5 sin θ – 4) = 0

⇒ (3 sin θ + 2) (5 sin θ – 4) = 0

Taking,

(5 sin θ – 4) = 0 (as in 0° < θ < 90° sinθ will be + ve)

⇒ sin θ = 4/5

Perpendicular = P, Hypotenuse = H and Base = B.

sin θ = P/H = 4/5

P = 4 and H = 5

As we know,

H^{2} = P^{2} + B^{2}

⇒ 5^{2} = 4^{2} + B^{2}

⇒ B^{2} = 25 – 16 = 9

⇒ B = √9 = 3

Now,

tan θ + cos θ + sec θ

⇒ 4/3 + 3/5 + 5/3

⇒ 9/3 + 3/5

⇒ 3 + 3/5

⇒ 3 3/5

**Q. 10 ΔABC is an equilateral triangle and AD ⊥ BC, where D lies on BC. If AD = 4√3 cm. then what is the perimeter (in cm) of ΔABC?**

##### 1. 27

2. 24

##### 3. 30

4. 21

##### Correct answer is __Option 2. 24__

__Option 2. 24__

##### If ABC is an equilateral triangle and AD ⊥ BC, then AD will be perpendicular bisector of BC.

##### As we know,

##### AD = √3/2 × a

##### ⇒ 4√3 = √3/2 × a

##### ⇒ a = 4√3 × 2/√3

##### ⇒ a = 8 cm

##### So perimeter of triangle = 8 × 3 = 24 cm

**Q.11 PQRS is a cyclic quadrilateral in which PQ = x cm, QR = 16.8cm, RS = 14 cm, PS = 25.2 cm, and PR bisects QS. What is the value of x?**

1. 18

2. 21

3. 28

4. 24

##### correct answer is **option 2 i.e. 21**

##### From the following figure

##### As we know,

##### ΔPOQ∼ΔSOR

##### PQ/SR = OQ/OR

##### OQ/OR = x/14

##### Given, OQ = OS

##### OS/OR = x/14 — (1)

##### As we know,

##### ΔPOS∼ΔQOR

##### OS/OR = PS/QR

##### OS/OR = 25.2/16.8 — (2)

##### From equation (1) and equation (2)

##### x/14 = 25.2/16.8

##### x = (25.2 × 14)/16.8

##### ∴ x = 21 cm

**Q.12 If secθ−tanθ/secθ+tanθ=3/5,**

** then the value of cosecθ+cotθ/cosecθ−cotθ**** is:**

1. 24 + √15

2. 31 + 8√15

3. 27 + √15

4. 33 + 4√15

##### Correct answer is **Option 2 i.e. 31 + 8√15**

##### (sec θ – tan θ)/(sec θ + tan θ) = 3/5

##### Or

##### (sec θ + tan θ)/(sec θ – tan θ) = 5/3

##### Using Componendo or Dividendo

##### sec θ/tan θ = (5 + 3)/(5 – 3)

##### ⇒ (1/cos θ)/(sin θ/cos θ) = 8/2

##### ⇒ 1/sin θ = 4

##### ⇒ sin θ = 1/4

##### Perpendicular = P, Hypotenuse = H and Base = B.

##### sin θ = P/H = 1/4

##### P = 1 and H = 4

##### As we know,

##### H^{2} = P^{2} + B^{2}

##### ⇒ 4^{2} = 1^{2} + B^{2}

##### ⇒ B^{2} = 16 – 1 = 15

##### ⇒ B = √15

##### cosec θ = H/P = 4

##### cot θ = B/P = √15

##### Now,

##### (cosec θ + cot θ)/(cosec θ – cot θ)

##### (4 + √15)/(4 – √15)

##### ⇒ [(4 + √15)/(4 – √15)] × [(4 + √15)/(4 – √15)]

##### ⇒ (4 + √15)^{2}

##### ⇒ 16 + 15 + 2 × 4 √15

##### ⇒ 31 + 8√15

**Q.13 The given table represents the number of engineers recruited by four companies A, B, C and D over the years. Study the table carefully and answer the question that follows.**

##### The total number of engineers recruited by company A in 2014 to 2017 is what percentage more than the total number of engineers recruited by all four companies in 2019?

##### 1. 3

2. 3.5

3. 2.5

4. 4

##### Correct Answer is **option 4 i.e. 4**

##### Number of engineers recruited by company A in 2014 to 2017 = 120 + 132 + 128 + 140 = 520

##### Number of engineers recruited by all four companies in 2019 = 150 + 118 + 110 + 122 = 500

##### Required percentage = [(520 – 500)/500] × 100 = 4%

**Q.14 If x is the mean proportional between 12.8 and 64.8 and y is the third proportional to 38.4 and 57.6, then 2x : y is equal to:**

1. 2:3

2. 4:5

3. 3:4

4. 1:2

##### Correct answer is **Option 1 i.e. 2:3**

##### If x is the mean proportional between 12.8 and 64.8, then

##### 12.8 : x : : x : 64.8

##### ⇒ 12.8/x = x/64.8

##### ⇒ x^{2} = 12.8 × 64.8

##### ⇒ x = √[16 × 0.8 × 0.8 × 81]

##### ⇒ x = 4 × 0.8 × 9

##### If y is the third proportional to 38.4 and 57.6, then

##### 38.4 : 57.6 : : 57.6 : y

##### ⇒ 38.4/57.6 = 57.6/y

##### ⇒ y = (57.6 × 57.6)/38.4

##### ⇒ y = 86.4

##### Now,

##### 2x : y = 2 × 4 × 0.8 × 9 : 86.4 = 2 : 3

**Short Trick :**

##### Mean Proportional x = √12.8 × 64.8 = 28.8

##### Third proportional y = (57.6 × 57.6)/38.4 = 86.4

##### Now,

##### 2x : y = 2 × 28.8 : 86.4 = 2 : 3

**Q.15 The average of the first four numbers is three times the fifth number. If the average of all the five numbers is 85.8, then the fifth number is;**

1. 33

2. 34

3. 39

4. 29

##### Correct answer is **Option 1 i.e. 33**

##### Let the fifth number be x, then

##### Average of first four number is = 3x

##### According to the question

##### 4 × 3x + x = 85.8 × 5

##### ⇒ 12x + x = 429

##### ⇒ 13x = 429

##### ⇒ x = 429/13

##### ∴ x = 33

**Q.16 Quadrilateral ABCD circumscribes a circle. If AB = 8 cm, BC = 7 cm and CD = 6 cm, then the length of AD is:**

1. 6 cm

2. 7.5 cm

3. 7 cm

4. 6.8 cm

##### Correct answer is **Option 3 i.e. **7 cm

##### AB = 8 cm, BC = 7 cm and CD = 6 cm.

##### As we know,

##### AD + BC = CD + AB

##### ⇒ AD + 7 = 6 + 8

##### ⇒ AD = 14 – 7

##### ⇒ AD = 7 cm

**Q.17 The expression (a + b – c)**^{3} + (a – b + c)^{3} – 8a^{3} is equal to:

^{3}+ (a – b + c)

^{3}– 8a

^{3}is equal to:

1. 6a(a – b + c) (c – a – b)

2. 3a(a + b – c)(a – b + c)

3. 6a(a + b – c) (a – b + c)

4. 3a(a – b + c) (c – a – b)

##### Correct answer is **Option 1 i.e. ****6a(a – b + c) (c – a – b)**

##### (a + b – c)^{3} + (a – b + c)^{3} – 8a^{3}

##### Let a = b = c = 1

##### 1 + 1 – 8

##### ⇒ (-6)

##### Checking option 1.

##### 6a(a – b + c) (c – a – b)

##### ⇒ 6 × 1 × (-1)

##### ⇒ (-6)

**Q18. If x**^{4} + x^{2}y^{2} + y^{4} = 21 and x^{2} + xy + y^{2 }= 7, then the value of (1/x^{2 }**+1/**y^{2 }**)**** is: **

^{4}+ x

^{2}y

^{2}+ y

^{4}= 21 and x

^{2}+ xy + y

^{2 }= 7, then the value of (1/

1. 5/2

2. 7/4

3. 5/4

4. 7/3

##### Correct answer is **Option 3 i.e. 5/4**

##### As we know,

##### x^{4} + x^{2}y^{2} + y^{4} = (x^{2} – xy + y^{2}) (x^{2} + xy + y^{2})

##### x^{4} + x^{2}y^{2} + y^{4} = 21 and x^{2} + xy + y^{2} = 7 — (1)

##### 21 = (x^{2} – xy + y^{2}) × 7

##### (x^{2} – xy + y^{2}) = 21/7

##### (x^{2} – xy + y^{2}) = 3 — (2)

##### Subtract equation (2) from equation (1)

##### 2xy = 4

##### ⇒ xy = 2

##### ⇒ (xy)^{2} = 4

##### From equation (1)

##### x^{2} + y^{2} = 7 – 2

##### ⇒ x^{2} + y^{2} = 5

##### ⇒ (1/x^{2} + 1/y^{2})

##### ⇒ (x^{2} + y^{2})/(xy)^{2}

##### ⇒ 5/4

**Q.19 ****The given table represents the number of engineers recruited by four companies A, B, C and D over the years. Study the table carefully and answer the question that follows.**

##### The total number of engineers recruited by company B in 2014 and 2017 is what percentage of the total number of engineers recruited by C during 2015 to 2019?

1. 38.4

2. 38.2

3. 39.2

4. 37.8

##### Correct answer is **Option 3 i.e. 39.2**

##### Number of engineers recruited by company B in 2014 and 2017 = 90 + 106 = 196

##### Number of engineers recruited by company C during 2015 to 2019 = 93 + 94 + 98 + 105 + 110 = 500

##### Required percentage = 196/500 × 100 = 39.2%

**Q.20 The value of the expression cosec (85° + θ) – sec(5° – θ) – tan (55° + θ) + cot(35° – θ)is:**

1. 3/2

2. 0

3. -1

4. 1

##### Correct answer is **Option 2 i.e. 0**

##### cosec (85° + θ) – sec(5° – θ) – tan (55° + θ) + cot(35° – θ)

##### ⇒ cosec (85 + θ) – cosec [90 – (5 – θ)] – tan (55 + θ) + tan [90 – (35 – θ)]

##### ⇒ cosec (85 + θ) – cosec (85 + θ) – tan (55 + θ) + tan (55 + θ)

##### ⇒ 0

**Q.21 When 200 is divided by a positive integer x, the remainder is 8. How many values of x are there?**

##### 1. 7

2. 5

3. 8

4. 6

##### Correct answer is **Option i.e. 8**

##### When 200 is divided by a positive integer x, the remainder is 8. So,

##### 200 – 8 = 192

##### As we know,

##### 192 is divisible by (more than 8) 12, 16, 24, 32, 48, 64, 96 and 192.

##### So, we can say x = 8.

**Mistake point: **

**Mistake point:**