SSC CGL 2019 EXAM PAPER : Held on 03-March-2020 Shift-1 (Quantitative Aptitude)

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SSC CGL 2019 EXAM PAPER : Held on 03-March-2020 Shift-1 (General Awareness)

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Q.1 If the number 1005×4 is completely divisible by 8, then the smallest integer in place of x will be:
1. 0
2. 1
3. 4
4. 2

 

the correct answer is option 1 i.e. 0
Divisibility rule of 8 = A number divisible by 8 if its last three digits are divisible by 8
If the number 1005×4 is divisible by 8, then its last digit 5×4 also divisible by 8
504 is divisible by 8 so the number 100504 is also divisible by 8.
Hence, 0 will be the correct answer.
Q.2 The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.
 If 2013 and 2014 are put together, which type of cars constitutes exactly 25% of the total number of cars produced in those 2 years?
1. E
2. B
3. D
4. C
The correct answer is option 3:   D.

Total production of cars in 2013 and 2014 = 130 + 170 = 300

Hence, 25% of 300 = 300 × 25/100 = 75

Production of car D in 2013 and 2014 are = 40 + 35 = 75

Q.3 If x = 4 cos A + 5 sin A and y = 4 sin A – 5 cos A, then the value of x2 + y2 is:

1. 25
2. 41
3. 0
4. 16

Correct answer is option 2 i.e. 41

Given, x = 4 cos A + 5 sin A and y = 5 sin A – 5 cos A
Hence, x2 + y2 = (4 cos A + 5 sin A)2 + (4 sin A – 5 cos A)2
⇒ 16 cos2 A + 25 sin2 A + 40 cos A sinA + 16 sin2 A + 25 cos2A – 40 sin A cos A
⇒ 16 (sin2 A + cos2 A) + 25 (sin2 A + cos2 A)
⇒ 16 + 25
⇒ 41
Q.4 The radius of a circular garden is 42 m. The distance (in m) covered by running 8 rounds around it, is: (Take π = 22/7)

1. 2112
2. 1124
3. 3248
4. 4262
Correct answer is option 1 i.e. 2112
Radius of the circular garden r = 42 m
Circumference of circular garden = 2 πr = 2 × (22/7) × 42 = 264 m
1 round cover the distance = circumference of the circular garden = 264 m
8 round cover the distance = 8 × 264 = 2112 m
Short trick:
Answer divisible by 11 and 8 also. Only option is divisible by 11 and 8 is 2112.
Q.5   The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.
The percentage decrease in the production of which type of car in 2017, with reference to 2016, was the maximum?

1. D
2. E
3. A
4. C
Correct answer is option 3 i.e. A

Decreasing percentage of type car D in 2017 with reference to 2016 = [(38 – 22)/38] × 100 = 16/38 × 100 = 42.1%

Decreasing percentage of type car C in 2017 with reference to 2016 = (44 – 40)/44 × 100 = 4/44 × 100 = 9.09%

Decreasing percentage of type car A in 2017 with reference to 2016 = (36 – 12)/36 × 100 = 24/36 × 100 = 66.67%

Decreasing percentage of type car E in 2017 with reference to 2016 = (50 – 28)/50 × 100 = 22/50 × 100 = 44%

The percentage decrease in the production of type A car in 2017, with reference to 2016, was the maximum

Q.6 If x, y, z are three integers such that x + y = 8, y + z = 13 and z + x = 17, then the value of x2/yz is:

1. 7/5
2. 18/11
3. 1
4. 0
Correct answer is option 2 i.e. 18/11
x + y = 8       —(1)
y + z = 13       —(2)
z + x = 17       —(3)
from equation (1)
y = 8 – x
Put y = 8 – x in equation (2)
8 – x + z = 13
z – x = 13 – 8
z – x = 5       —(4)
From equation (3) and equation (4)
z = 11 and x = 6
Put x = 6 in equation (2)
y + 11 = 13
y = 2
Now,
x2/yz
⇒ (6 × 6)/(2 × 11)
18/11
Q.7 In ΔABC, MN∥BC, the area of quadrilateral MBCN = 130 cm2. If AN : NC = 4 : 5, then the area of ΔMAN is:

1. 40 cm2
2. 45 cm2
3. 32 cm2
4. 65 cm2
Correct answer is option 3 i.e. 32 cm2
Given,
the area of quadrilateral MBCN = 130 cm2.
AN : NC = 4 : 5
Hence, AC = 4 + 5 = 9
In ΔABC, If MN∥BC, then
Area of ΔAMN/area of ΔABC = (AN/AC)2
Area of ΔAMN/area of ΔABC = (4/9)2 = 16/81
Area of ΔAMN = 16 unit and area of ΔABC = 81 unit
Now, Area of quadrilateral MBCN = area of ΔABC – area of ΔAMN
⇒ 81 – 16 unit = 130 cm2
⇒ 65 unit = 130 cm2
⇒ 1 unit = 130/65 = 2 cm2
∴ Area of ΔAMN = 16 × 2 = 32 cm2
Q.8 A, B and C can individually complete a piece of work in 24 days, 15 days and 12 days, respectively. B and C started the work and worked for 3 days and left. The number of days required by A alone to complete the remaining work is:

1. 11
2. 15 1/2
3. 18
4. 13 1/5

Correct answer is option 4 i.e. 13 1/5

Let, total work = LCM of (24, 15 and 12) = 120
Hence, efficiency of A = 5
Efficiency of B = 8
Efficiency of C = 10
Work done by B and C in 3 days = (8 + 10) × 3 = 54
Remaining work = 120 – 54 = 66
∴ Remaining work done by A in = 66/5 =  13 1/5  days.
Q.9 If x – y = 4 and xy = 45, then the value of x3 – y3 is:

1. 822
2. 604
3. 151
4. 82
 
Correct answer is option 2 i.e. 604
Short Trick:
x – y = 4 and xy = 45
Put x = 9 and y = 5
x3 – y3
⇒ 93 – 53
⇒ 729 – 125
⇒ 604
Detailed solution:
x – y = 4 and xy = 45
(x3 – y3) = (x – y) [(x – y)2 + 3xy]
x3 – y3 = 4 [42 + 3 × 45] [Hint: Answer will be divisible by 4, only 1 option 604 divisible by 4.]
x3 – y3 = 4 [16 + 135]
x3 – y3 = 4 × 151 = 604
 

Q.10 A person sells an article at 10% below its cost price. Had he sold it for Rs. 332 more, he would have made a profit of 20%. What is the original selling price (in Rs.) of the article?

1.  996
2. 1,328

3. 1,028
4.  896

Correct answer is Option 1.  996
Let the Cost price of the article be Rs. 100x, then according to the question
100x × (120/100) = 100x × (90/100) + 332
⇒ 120x – 90x = 332
⇒ 30x = 332
⇒ x = 332/30
Selling price of the article = 90x = 90 × (332/30) = 996 
Q.11  If the base radius of 2 cylinders are in the ratio 3 : 4 and their heights are in the ratio 4 : 9, then the ratio of their volumes is:

1. 2 : 1
2. 1 : 4
3. 1 : 2
4. 4 : 1
correct answer is option 2 i.e. 1:4
As we know,
Volume of cylinder = πr2h
Ratio of radius of 2 cylinder, r1 : r2 = 3 : 4
Ratio of heights of 2 cylinder, h1 : h2 = 4 : 9
Ratio of volume of two cylinder, V1 : V2 = π(r1)h1 : π(r2)2 h2 = 32 × 4 : 42 × 9 = 
1 : 4

 

Q.12 A train crosses a pole in 12 sec, and a bridge of length 170 m in 36 sec. Then the speed of the train is:

1. 10.8 km/h
2. 30.75 km/h
3. 25.5 km/h
4.32.45 km/h
Correct answer is Option 3 i.e. 25.5 km/hr
Short Trick:
If train cross its length in 12 seconds and 170 m bridge in (36 – 12 = 24) seconds.
Speed of train = [170/24] × [18/5] = 25.5 km/hr
Detailed method:
Let the length of the train be x m.
As we know,
Speed = Distance/time
Speed = x/12       —(1)
Speed = (x + 170)/36       —(2)
Frome equation (1) and equation (2)
x/12 = (x + 170)/36
⇒ 3x = x + 170
⇒ 2x = 170
⇒ x = 170/2
⇒ x = 85 m
From equation (1)
Speed = 85/12 × (18/5) km/hr
∴ Speed = 25.5 km/hr

Q.13 If ‘+’ means ‘-‘, ‘-‘ means ‘+’, ‘×’ means ‘÷’ and ‘÷’ means ‘×’, then the value of is: 
1. -15/19
2. 5/3
3. -5/3
4. 15/19

Correct Answer is option 1 i.e. -15/19

Q.14 A shopkeeper marks the price of the article in such a way that after allowing 28% discount, he wants a gain of 12%. If the marked price is Rs. 224, then the cost price of the article is:

1. Rs. 168
2. Rs. 144
3. Rs. 120
4. Rs. 196
Correct answer is Option 2 i.e. Rs 144
Short Trick:
Ratio of MP to CP = (100% + Profit%) : (100% – Discount%) = (100 + 12) : (100 – 28) = 112 : 72
112 unit = 224
⇒ 1 unit = 224/112 = 2
Now, 72 unit = 2 × 72 = 144
∴ Cost price of the article is Rs. 144.
Detailed Solution:
MP of the article = Rs. 224
SP of the article = Rs. 224 × (72/100)
∴ CP of the article = 224 × (72/100) × (100/112) = Rs. 144
Q.15 Rs. 4,300 becomes Rs. 4,644 in 2 years at simple interest. Find the principle amount that will become Rs. 10,104 in 5 years at the same rate of interest

1. Rs. 7,200
2. Rs. 8420
3. Rs. 9,260
4. Rs. 5,710
Correct answer is Option 2 i.e. Rs 8420
P = 4300, A = 4644 and t = 2 years
SI = 4644 – 4300 = 344
As we know,
SI = Prt/100
⇒ 344 = (4300 × r × 2)/100
⇒ r = 344/86
⇒ r = 4%
Again,
A = 10,104, r = 4% and t = 5 years
As we know
A = P + Prt/100
⇒ 10104 = P + (P × 4 × 5)/100
⇒ 10104 = P + P/5
⇒ 6P/5 = 10104
⇒ P = 10104 × (5/6)
∴ P = 8420
Q.16 Out of 6 numbers, the sum of the first 5 numbers is 7 times the 6th number. If their average is 136, then the 6th number is:

1.116
2. 102
3. 96
4. 84
Correct answer is Option 2 i.e. 102
Let the sum of 6th number be x
The sum of first 5 numbers = 7x
According to the question
(x + 7x)/6 = 136
⇒ 8x = 136 × 6
⇒ x = (136 × 6)/8
⇒ x = 102
∴ The 6th number is 102.
Q.17 If A + B = 45°, then the value of 2(1 + tan A) (1 + tan B) is: 

1. 4
2. 0
3. 2
4. 1
Correct answer is Option 1 i.e. 4
2(1 + tan A) (1 + tan B)
If A + B = 45°
Put A = 45° and B = 0°
2 (1 + tan 45°) (1 + tan 0°)
⇒ 2 (1 + 1) (1 + 0)
⇒ 2 × 2 × 1
⇒ 4
Detailed solution:
Taking tan both sides:
tan(A + B) = tan 45°  
(tan A + tan B)/(1 – tan A tan B) = 1
tan A + tan B = 1 – tan A tan B
tan A + tan B + tan A tan B = 1
adding “1” on both sides
1+ tan A + tan B + tan A tan B = 1 + 1
(1 + tan A) + tan B(1 + tan A) = 2
(1 + tan A)(1 + tan B) = 2 
∴ 2 × (1 + tan A)(1 + tan B) = 2 × 2 = 4
Q18. If the length of a rectangle is increased by 40%, and the breadth is decreased by 20%, then the area of the rectangle increases by x%. Then the value of x is:

1. 16
2. 20
3. 8
4. 12
Correct answer is Option 4 i.e. 12
Short trick:
Area increased by = 40 – 20 – [(40 × 20)/100] = 20 – 8 = 12%
Detailed method:
Let the length and breadth of the rectangle be 10 unit each.
Area of the rectangle = lb = 10 × 10 = 100 sq. unit
New increased length = 10 × (40/100) = 14 unit
New decreased length = 10 × (80/100) = 8 unit
New area of the rectangle = 14 × 8 = 112
Area increased by = 112 – 100 = 12
∴ Area increased by (in %) = (12/100) × 100 = 12%
Q.19 The area of ΔABC is 44 cm2. If D is the midpoint of BC and E is the midpoint of AB, then the area (in cm2) of ΔBDE is:

1. 5.5
2. 44
3. 11
4. 22
Correct answer is Option 3 i.e. 11
If D is the midpoint of BC and E is the midpoint of AB, then
DE ∥ AC
Let BC = 2 unit and BD = 1 unit
As we know,
area of ΔBDE/area of ΔBCA = (BD/BC)2
⇒ area of ΔBDE/44 = (1/2)2
∴ area of ΔBDE = (1/4) × 44 = 11 cm2
Q.20 The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.
In the year 2015, which type of car constitutes exactly 20% of the total number of cars produced that year?

1. E
2. A
3. B
4. D
Correct answer is Option 1 i.e. E
Total production of total car in 2015 = 100
Hence, 20% of 100 = 20
In the year 2015, E type of car constitutes exactly 20% of the total number of cars produced that year.
Q.21 If A lies in the first quadrant and 6 tan A = 5, then the value of
1. 16
2. 1
3. 4
4. -2
Correct answer is Option 2 i.e. 1
Perpendicular = P, Base = B and Hypotenuse = H
Given, 6 tan A = 5
⇒ tan A = 5/6
⇒ tan A = P/B = 5/6
⇒ P = 5 and B = 6
Using Pythagoras theorem,
H2 = P2 + B2
⇒ H2 = 52 + 62
⇒ H2 = 25 + 36
⇒ H2 = 61
⇒ H = √61
sin A = P/H = 5/√61 and cos A = B/H = 6/√61
8sinA4cosAcosA+2sinA8sin⁡A−4cos⁡Acos⁡A+2sin⁡A
8×5614×661661+2×561⇒8×561−4×661661+2×561
⇒ (40 – 24)/(6 + 10)
⇒ 16/16
⇒ 1
Short Trick:
6 tan A = 5
tan A = 5/6
sinA/cosA = 5/6
Let sin A = 5 and cos A = 6
8sinA4cosAcosA+2sinA⇒8sin⁡A−4cos⁡Acos⁡A+2sin⁡A
8×54×66+2×5⇒8×5−4×66+2×5
⇒ (40 – 24)/(6 + 10)
⇒ 16/16
⇒ 1

Q.22 The ratio of the number of boys to the number of girls in a school of 640 students, is 5 : 3. If 30 more girls are admitted in the school, then how many more boys should be admitted so that the ratio of boys to that of the girls, becomes 14 : 9.

1. 30
2. 25
3. 15
4. 20
 
Correct answer is Option 4 i.e. 20
Total students = 640
Ratio of boys to girls = 5 : 3
Number of boys = [5/8] × 640 = 400
Number of girls = 640 – 400 = 240
Let x boys admitted in the school.
According to the question
(400 + x)/(240 + 30) = 14/9
⇒ 400 + x = [14/9] × 270
⇒ x = 420 – 400
⇒ x = 20
20 students admitted in the school.
Q.23 A, B and C are three points on a circle such that the angles subtended by the chord AB and AC at the centre O are 110° and 130°, respectively. The value of ∠BAC is:

1. 70°
2. 75°
3. 60°
4. 65°
Corrrect answer is Option 3 i.e. 60°

from the following figure.
∠AOB = 110° and ∠AOC = 130°
As we know,
∠AOB + ∠AOC + ∠BOC = 360°
∠BOC = 360° – 110° – 130° = 120°
∠BAC = ∠BOC/2
∴ ∠BAC = 120°/2 = 60°
Q.24 The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.
The percentage increase in the total car in 2016 over 2012, is:

1. 50%
2. 45%
3. 33.33%
4. 62.33%
Correct answer is Option 1 i.e. 50%
Total production of car in 2012 = 120
Total production of car in 2016 = 180
∴ The percentage increase of total car in 2012 over 2016 = [(180 – 120)/120] × 100 = 50% 
Q.25 If x2a = y2b = z2c ≠ 0 and x2 = yz, then the value of
1. 3
2. 3ac
3. 3bc
4. 3ab
Correct answer is Option 1 i.e. 3
x2a = y2b = z2c ≠ 0
Let x2a = y2b = z2c = k, then
x = k1/2a, y = k1/2b, z = k1/2c
Now,
x2 = yz
x × x = y × z
k1/2a × k1/2a = k1/2b × k1/2c
k(1/2a + 1/2a) = k(1/2b + 1/2c)
Comparing on power
1/2a + 1/2a = 1/2b + 1/2c
⇒ 2/2a = (c + b)/2bc
⇒ 1/a = (c + b)/2bc
⇒ 2bc = ac + ab
Now,
⇒ (2bc + bc)/bc
⇒ 3bc/bc
⇒ 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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