SSC CGL 2019 EXAM PAPER : Held on 03-March-2020 Shift-3 (Quantitative Aptitude)

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

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Q.1 If the difference between 62% and 80% of a number is 198, then the difference between 92% and 56% of the number will be:

1. 3564
2. 360
3. 396
4. 1100
Correct answer is Option 3 i.e. 396
Let the number be x, then
According to the question
x × (80/100) – (62/100) = 198
⇒ 18x/100 = 198
⇒ x = 198 × (100/18)
⇒ x = 1100
⇒ 1100 × (92/100) – 1100 × (56/100)
⇒ 11 (92 – 56)
⇒ 11 × 36
⇒ 396
Short Trick:
⇒ 80% – 62% = 18%
⇒ 92% – 56% = 36%
⇒ 18% = 198
⇒ 36% = 198 × 2 = 396
Q.2 Two racers run at a speed of 100 m/min and 120 m/min, respectively. If the second racer takes 10 minutes less than the first to complete the run, then how long is the race?
1. 2 km
2. 4 km
3. 6 km
4. 1 km
The correct answer is option 36 Km
Speed ratio of two racers = 100 : 120 = 5x : 6x
As we know,
Time is inversely proportional to speed.
Time ratio of two racers = 6x : 5x
⇒ 6x – 5x = 10 min
⇒ x = 10 min
Time taken by first racer to complete race = 6 × 10 = 60 min
Distance = Speed × Time
∴ Distance = 60 × 100 = 6 km
Q.3 In the figure, what is the value of cot R?

1. 17/18
2. 8/15
3. 15/8
4. 15/17
Correct answer is option 3 i.e. 15/8
As we know,
PR2 = PQ2 + QR2
⇒ 172 = 82 + QR2
⇒ QR = 289 – 64
⇒ QR = √225 = 15
⇒ Cot R = QR/PQ = 15/8
Q.4 If 2 sin θ – 8 cos2 θ + 5 = 0, 0° < θ < 90°, then what is the value of (tan 2θ + cosec 2θ)?
1. \frac { 5\sqrt { 3 } }{ 3 }
2. 2√3
3. 3√3
4. \frac { 4\sqrt { 3 } }{ 3 }
Correct answer is Option  i.e.  \frac { 5\sqrt { 3 } }{ 3 }
Detailed Method:
2 sin θ – 8 cos2 θ + 5 = 0
⇒ 2 sin θ – 8 (1 – sin2 θ) + 5 = 0
⇒ 2 sin θ – 8 + 8 sin2 θ + 5 = 0
⇒ 8 sin2 θ + 2 sin θ – 3 = 0
⇒ 8 sin2 θ + 6 sin θ – 4 sin θ – 3 = 0
⇒ 2 sin θ (4 sin θ + 3) – 1 (4 sin θ + 3) = 0
⇒ (2 sin θ – 1) (4 sin θ + 3) = 0
Taking,
⇒ (2 sin θ – 1) = 0
⇒ sin θ = 1/2
⇒ sin θ = sin 30
⇒ θ = 30
Now,
tan 2θ + cosec 2θ
⇒ tan 60 + cosec 60
⇒ √3 + 2/√3
⇒ 5/√3
⇒ 5√3/3
Short Trick:
2 sin θ – 8 cos2 θ + 5 = 0
Put θ = 30 and satisfied, then
⇒ tan 60 + cosec 60
⇒ √3 + 2/√3
⇒ 5/√3
\frac { 5\sqrt { 3 } }{ 3 }
Q.5 A dealer sold 6 sewing machines for Rs. 63,000 with a profit of 5%. For how much should be sell 8 machines if he intends to earn 15% profit?
1. Rs. 88,200
2. Rs. 92,400
3. Rs. 69,300
4. Rs. 92,000
Correct answer is option 4 i.e.Rs 92,000
SP of 6 sewing machines = Rs. 63,000
SP of 1 sewing machines = 63,000/6 = Rs. 10,500
CP of 1 sewing machine = 10,500 × (100/105) = 10,000
New SP of 1 sewing machine = 10,000 × (115/100) = 11,500
∴ New SP of 8 sewing machines = 11,500 × 8 = 92,000 
Q.6 Two members are in the ratio 5 : 7. If the first number is 20, then the second number will be:

1. 18
2. 22
3. 8
4. 28
Correct answer is option 4 i.e. 28
Ratio of two numbers = 5x : 7x
⇒ 5x = 20
⇒ x = 20/5
⇒ x = 4
∴ Second number = 7 × 4 = 28
Q.7 A can finish a work in 20 days and B can finish the same work in 25 days. They began together, but B left the work after 5 days. How many more days will A take to finish the remaining work?
1. 11
2. 16
3. 8
4. 21
Correct answer is option 1 i.e. 11
A can complete the whole work in = 20 days
A’s 1 day’s work = 1/20
B can complete the whole work in = 25 days
B’s 1 day’s work = 1/25
Let A complete the whole work in x days.
According to the question
⇒ 5/20 + 5/25 + x/20 = 1
⇒ 1/4 + 1/5 + x/20 = 1
⇒ (5 + 4 + x)/20 = 1
⇒ 9 + x = 20
⇒ x = 20 – 9
⇒ x = 11 days
Short Trick :
Total work = 100
Work done by A and B in 5 days = (5 + 4) × 5 = 45
Remaining work = 100 – 45 = 55
∴ Remaining work, A complete in = 55/5 = 11 days
Q.8 The number of students enrolled in different faculties in a school is as follows:
Science
Arts
Commerce
Vocational
Boys
Girls
Boys
Girls
Boys
Girls
Boys
Girls
35
18
25
47
45
40
10
30
The percentage of students studying in Science and Vocational Subjects is:
1. 25%
2. 37.2%
3. 50%
4. 93%
Correct answer is option 2 i.e. 37.2%
Total number of students in Science and Vocational = 35 + 18 + 10 + 40 = 93
Total number of students in all four subjects = 35 + 18 + 25 + 47 + 45 + 40 + 10 + 30 = 250
∴ Percentage of students of Science and Vocational subjects = (93/250) × 100 = 37.2%
Q.9 What should replace * in the number 94*2357, so that number is divisible by 11?

1. 7
2. 3
3. 1 
4. 8
 
Correct answer is option 2 i.e. 3
Divisibility law of 11: If the difference of the alternating sum of digits of the number is a multiple of 11 (e.g. 2343 is divisible by 11 because 2 – 3 + 4 – 3 = 0, which is a multiple of 11).
The number 94*2357 is divisible by 11, then
⇒ 9 – 4 + * – 2 + 3 – 5 + 7 = 0
⇒ 19 + * – 11 = 0
⇒ (*) = – 8 
⇒ or * = 11 – 8 = 3
Q. In the given figure, AP bisects ∠BAC. If AB = 4 cm, AC = 6 cm and BP = 3 cm, then the length of CP is:

1. 5 cm
2. 7 cm
3. 3 cm
4. 4.5 cm
Correct answer is Option 4.  4.5
If AP bisects ∠BAC, then
AB/AC = BP/PC
⇒ 4/6 = 3/PC
⇒ PC = 18/4
⇒ PC = 4.5
Q.11 In ΔABC, if AB = AC and ∠BAC = 40°, then the measure of ∠B is: 

1. 50°
2. 60°
3. 40°
4. 70°
correct answer is option 4 i.e. 70°
Given, ∠A = 40°
In ΔABC,
If AB = AC, then
∠B = ∠C
∠A + ∠B + ∠C = 180°
⇒ 40° + ∠B + ∠B = 180°
⇒ 2∠B = 180° – 40 = 140°
⇒ ∠B = 140°/2 = 70°
Q.12  (a + b – c + d)2 – (a – b + c – d)2 = ?
1. 4a (b + d – c)
2. 2a (b + c – d)
3. 4a (b – d + c)
4. 2a (a + b – c)
Correct answer is Option 1 i.e.  4a (b + d – c)
(a + b – c + d)2 – (a – b + c – d)2
Put a = 1, b = 2, c = 3 and d = 4, then
⇒ (1 + 2 – 3 + 4)2 – (1 – 2 + 3 – 4)2
⇒ 16 – 4
⇒ 12
From option 2.
4a (b + d – c)
⇒ 4 × 1 (2 + 4 – 3)
⇒ 4 × 3
⇒ 12 (satisfied)
Q.13 The simple interest on a certain sum at the end of three years at 5% p.a. is Rs. 1,200. The compound interest on the same sum for the same period at the same rate is (interest compounded yearly).
1. Rs. 1,261
2. Rs. 820
3. Rs. 1,260
4. Rs. 1,800
Correct Answer is option 1 i.e. Rs 1,261
Tree Method:
CI for three years = 1200 + 60 + 1 = 1261
Ratio method:
Rate = 5%
SI for 3 years = 1200
SI for 1 year = 1200/3 = 400
5% of 400 = 20
⇒ 5% of 20 = 1
CI for 3 years = 3 : 3 : 1 = 3 × 400 + 3 × 20 + 1 × 1 = 1200 + 60 + 1 = 1261
Detailed Method:
R = 5%, t = 3 years SI = 1200
As we know,
SI = Prt/100
⇒ 1200 = (P × 5 × 3)/100
⇒ P = 8000
Again, As we know,
A = P (1 + r/100)t
⇒ A = 8000 (1 + 5/100)3
⇒ A = 8000 × (21/20) × (21/20) × (21/20)
⇒ A = 9261
⇒ CI = 9261 – 8000
CI = 1261
Q.14 In a circle with radius 5 cm, a chord is at a distance of 3 cm from the centre. The length of the chord is:

1. 8 cm
2. 3 cm 
3. 7 cm
4. 4 cm
Correct answer is Option 1 i.e. 8 cm
Radius of circle OA = 5 cm
Let the length of the chord be AB
OP = 3 cm
As we know,
OA2 = OP2 + AP2
⇒ 52 = 32 + AP2
⇒ AP2 = 25 – 9 = 16
⇒ AP = √16 = 4 cm
⇒ AB = 2AP = 2 × 4 = 8 cm
Q.15 The value of \frac { 36÷42of6×7+24×6÷18+3÷(2−6)−(4+3×2)÷8 }{ 21÷3of7 }   is:
1. 7
2. \frac { 1 }{ 7 } 
3.  7\frac { 1 }{ 2 } 
4.  8\frac { 1 }{ 2 } 

Correct answer is Option 1 i.e.  7

=\frac{36÷42of6×7+24×6÷18+3÷(2−6)−(4+3×2)÷8}{21÷3of7}

=\frac { 36÷42×6×7+24×\frac { 6 }{ 18 } +\frac { 3 }{ -4 } −(4+6)÷8 }{ 21÷3×7 }

=\frac { \frac { 36 }{ 42\times 6 } ×7+8-\frac { 3 }{ 4 } −10÷8 }{ 21÷21 }

=\frac { 1+8-\frac { 3 }{ 4 } -\frac { 10 }{ 8 } }{ 1 }

=1+8–\frac { 3 }{ 4 } -\frac { 5 }{ 4 }

=\frac { (4+32–3–5) }{ 7 }

=7

Q.16 In the given figure, AP and BP are tangents to a circle with centre O. If ∠APB = 62° then the measure of ∠AQB is:


1. 28°
2. 118°
3. 31°
4. 59°
Correct answer is Option 3 i.e. 7 cm
If AP and BP are tangents to a circle with centre O, then
∠OAP = ∠OBP = 90°
In quadrilateral APBO
∠OAP + ∠OBP + ∠APB + ∠BOA = 360°
⇒ 90° + 90° + 62° + ∠BOA = 360°
⇒ ∠BOA = 360° – 242° = 118°
As we know,
∠AQB = ∠BOA/2
⇒ ∠AQB = 118°/2 = 59°
Short Trick:
∴ ∠AQB = 90° – 62°/2 = 90° – 31° = 59°
Q.17 From the following table, how many patients were in the age group 40 – 60?
Age (years)
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
Less than 70
No. of patients
11
15
22
29
35
45
50

1.  16
2.  6
3.  45
4.  29
Correct answer is Option 2 i.e.  16
The patient in the age group 40 – 60 are = 45 – 29 = 16
Q18. The table below shows income (in rupees) for a particular month, together with their sources in respect of 5 employees A, B, C, D and E
Employee
A
B
C
D
E
Salary
52,000
48,500
42,000
31,000
25,000
Overtime
0
0
1,500
2,500
3,200
Arrears
5,500
4.500
4,000
3.000
1,5000
Bonus
3,500
3,000
2,500
2,000
2,000
Miscellaneous income
5,000
3,000
2,000
1,500
0
Total
66,000
59,000
52,000
40,000
31,700

How many employees have their salary more than four times their other incomes?

1. 4
2. 3
3. 2
4. 1
Correct answer is Option 3 i.e. 2
Salary of B = 48,500
Total income of B = 59,000
Other income of B excluding salary = 59,000 – 48,500 = 10,500
Four times of other income of B = 4 × 10,500 = 42,000
Salary of C = 42,000
Total income of C = 52,000
Other income of C excluding salary = 52,000 – 42,000 = 10,000
Four times of other income of C = 4 × 10,000 = 40,000
Now, we can say only two employees B and C whose salary more than four times their other incomes.
Q.19 The value of -\frac { 5 }{ 2 } +\frac { 3 }{ 2 } \div 6\times \frac { 1 }{ 2 }   is equal to:
1. – 19/8
2. – 9/8
3. – 1/12
4. – 1/3
Correct answer is Option 1 i.e. -19/8
-\frac { 5 }{ 2 } +\frac { 3 }{ 2 } \div 6\times \frac { 1 }{ 2 }
= (– 5/2) + 3/2 × 1/6 × 1/2
= (– 5/2) + 1/8
= (– 20 + 1)/8
= (– 19)/8
Q.20 The average of five consecutive even numbers is M. If the next five even numbers are also included, the average of ten numbers will be:

1. M + 5
2. M + 10
3. 10
4. 11
Correct answer is Option 1 i.e. M+5
Let the five consecutive even numbers are 2, 4, 6, 8, 10.
Average of consecutive even numbers = (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6
Hint: As we know, Average of any AP is the median so, the average = 6
So, M = 6
If the next five even numbers are also included, then the numbers are
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Average of 10 consecutive numbers = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)/10 = 110/10 = 11
Hint: As we know, Average is the median so, the average = 11
We can write 11 = 6 + 5 or M + 5
Q.21 The value of 27a3 – 2√2b3 is equal to:
1. (3a – √2b)(9a2 – 2b2 – 3√2 ab)
2. (3a – √2b)(9a2 – 2b2 + 6√2 ab)
3. (3a – √2b)(9a2 + 2b2 + 3√2 ab)
4. (3a – √2b)(9a2 + 2b2 + 6√2 ab)
Correct answer is Option 3  i.e. (3a – √2b)(9a2 + 2b2 + 3√2 ab)
As we know,
a3 – b3 = (a – b) (a2 + b2 + ab)
⇒ 27a3 – 2√2 b3
⇒ (3a – √2 b) (9a2 + 2b2 + 3√2 ab)
Q.22 The curved surface area of a hemisphere with radius 7 cm is: (Take π = 22/7)
1. 616 cm2
2. 308 cm2
3. 462 cm2
4. 385 cm2
 
Correct answer is Option 2 i.e. 308 cm2
Radius of hemisphere r = 7 cm
As we know
∴ Curved surface area of hemisphere = 2πr2 = 2 × (22/7) × 7× 7 = 308 cm2
Q.23 A person marked his goods at a price that would give him 40% profit. But he declared a sale and allowed 20% discount on the marked price. What is the profit percentage of the person in the whole transaction?
1. 12%
2. 20%
3. 32%
4. 30%
Corrrect answer is Option 2 i.e. 12%
Let the cost price of the article be Rs. 100
MP of the article = 100 × (140/100) = Rs. 140
SP of the article = 140 × (80/100) = Rs. 112
Profit = 112 – 100 = 12 
∴ Profit percentage = 12/100 × 100 = 12%
Q.24 As per data in the table, what is the percentage of students who got 20 or more marks?
Scores
0 – 5
5 – 10
10 – 15
15 – 20
20 – 25
25 – 30
30 – 35
35 – 40
No. of students
13
15
18
12
14
19
6
3

1. 58%
2. 14%
3. 54%
4. 42%
Correct answer is Option 4 i.e. 42%
Total number of students = 13 + 15 + 18 + 12 + 14 + 19 + 6 + 3 = 100
Number of students who got 20 or more marks = 14 + 19 + 6 + 3 = 42
∴ Required percentage = 42/100 × 100 = 42%
Q.25  The Value of \frac { tan30∘+tan60∘}{ cos30∘ } is

1. 8/√3
2. 8/3
3. √3 + 3
4. 1 + √3 
Correct answer is Option2  i.e. \frac{8}{3}
\frac { tan30∘+tan60∘}{ cos30∘ }
\frac { \frac { 1}{ \sqrt { 3 } } +\sqrt { 3 } }{ \frac { \sqrt { 3 } }{ 2 } }
=  (4/√3) × (2/√3)
8/3

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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