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

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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/m2?(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 4Rs. 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 (R2 – r2) = (22/7) × (80.52 – 702) = 22/7 × 150.5 × 10.5 = 4966.5 m2
Cost of levelling 1 m2 = Rs. 6
∴ Cost of levelling 4966.5 m2 = 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 cos2 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 cos2 θ = 7

⇒ 2 sin θ + 15 (1 – sin2 θ) = 7

⇒ 2 sin θ + 15 – 15 sin2 θ = 7

⇒ 15 sin2 θ – 2 sin θ – 8 = 0

⇒ 15 sin2 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,

H2 = P2 + B2

⇒ 52 = 42 + B2

⇒ B2 = 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
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  sec⁡θ−tan⁡θ/sec⁡θ+tan⁡θ=3/5,
 then the value of  cosecθ+cot⁡θ/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,
H2 = P2 + B2
⇒ 42 = 12 + B2
⇒ B2 = 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
⇒ x2 = 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 – 8a3 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 – 8a3
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 x4 + x2y2 + y4 = 21 and x2 + xy + y= 7, then the value of  (1/x2 +1/y) is: 

1. 5/2
2. 7/4
3. 5/4
4. 7/3
Correct answer is Option 3 i.e. 5/4
As we know,
x4 + x2y2 + y4 = (x2 – xy + y2) (x2 + xy + y2)
x4 + x2y2 + y4 = 21 and x2 + xy + y2 = 7      — (1)
21 = (x2 – xy + y2) × 7
(x2 – xy + y2) = 21/7
(x2 – xy + y2) = 3      — (2)
Subtract equation (2) from equation (1)
2xy = 4
⇒ xy = 2
⇒ (xy)2 = 4
From equation (1)
x2 + y2 = 7 – 2
⇒ x2 + y2 = 5
⇒ (1/x2 + 1/y2)
⇒ (x2 + y2)/(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: 
We can’t consider x = 1, 2, 3, 4, 6, and 8 as remainder is 8. So, x > 8.
Q.22
1. 324
2. 322
3. 318
4. 327
 
Correct answer is Option 2 i.e. 322
x2 + 3x + 1 = 0,
Dividing by x
x + 3 + 1/x = 0
⇒ x + 1/x = (-3)
As we know,
(x + 1/x)2 = (-3)2
⇒ x2 + 1/x2 + 2 = 9
⇒ x2 + 1/x2 = 9 – 2
⇒ x2 + 1/x2 = 7
Again, as we know,
(x2 + 1/x2)3 = 73
⇒ x6 + 1/x6 + 3(x2 + 1/x2) = 343
⇒ x6 + 1/x6 + 3 × 7 = 343
⇒ x6 + 1/x6 = 343 – 21
⇒ x6 + 1/x6 = 322
Q.23  Anu fixes the selling price of an article at 25% above its cost of production. If the cost of production goes up by 20% and she raises the selling price by 10% then her percentage profit is (correct to one decimal place):
1. 16.4%
2. 14.6%
3. 13.8%
4. 15.2%
Corrrect answer is Option 2 i.e. 14.6%
Let cost price of the article be Rs. 100
Selling price of the article = 100 × 125/100 = Rs. 125
Profit = 125 – 100 = Rs. 25
New cost price of the article = 100 × (120/100) = Rs. 120
New selling price of the article = 125 × (110/100) = Rs. 137.5
New profit = 137.5 – 120 = 17.5
∴ Profit percentage = (17.5/120) × 100 = 14.6%
Q.24  A and B start moving towards each other from places X and Y respectively, at the same time. The speed of A is 20% more than that of B. After meeting on the way. A and B take  hours and x hours now to reach Y and X respectively. What is the value of x?

1.
2.
3.
4.
Correct answer is Option 2 i.e.
Let the speed of B be 5 unit
Speed of A = 5 × (6/5) = 6 unit
Time taken by A after meeting = 5/2 hrs
Time taken by B after meeting = x hrs
As we know,
(Speed of A/Speed of B)2 = Time taken by B/Time taken by A
⇒ (6/5)2 = x/(5/2)
⇒ 36/25 = 2x/5
⇒ x = 18/5
⇒ x =
Q.25  A dealer marks an article 40% above the cost price and sells it to a customer, allowing two successive discounts of 20% and 25% on the marked price. If he suffers a loss of Rs. 140, then the cost price (in Rs) of the article is: 

1. 900
2. 840
3. 872
4. 875
Correct answer is Option 4 i.e. 875
Let CP of the article = Rs. 100x
Let the MP of the article = 100x × (140/100) = Rs. 140x
SP of the article after getting two successive discount of 20% and 25% = 140x × (80/100) × (75/100) = Rs. 84x
Loss = 100x – 84x = 16x
16x = 140
x = 140/16
CP of the article = 140/16 × 100 = Rs. 875
 

Download Complete SSC CGL 2019 Paper Held on 03 March 2020 Shift 1 paper here Click Download below.

manish aggarwal
Founder of edumo, SSC CGL Aspirant, Educator, BCA Graduate
http://edumo.in

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