SSC CGL Prelims Quantitative Aptitude Question Paper 2020

SSC conducts the exam of SSC CGL every year through which graduates from any recognised university can get various Group B & Group C posts. Candidates preparing for SSC CGL exam know the importance of attempting previous year papers. To help you with that, we are here with a solved paper of SSC CGL 2020 Tier 1 Quant Section. Go through all the questions and check your conceptual clarity. 

Que 1. Radha purchased a computer table for Rs.10000 and a centre table for Rs.5000. She sold computer table with 8% profit. With what profit percent should she sell the centre table as to gain 10% on the whole transaction?  

A) 14%                                                                       

B) 18%  

C) 12%

D) 10%

Answer:A 

Explanation :

Selling price of computer table=10000×108/100 = Rs.10800    

Total Cost price = (10000 + 5000) = Rs.15000

Selling price of two articles= 15000×110/100 =Rs.16500

Now,Selling price of centre table=(16500-10800) = Rs 5700   

Profit on centre table= (5700-5000) = Rs.700

Profit percent on centre table=700/5000 ×100% = 14%

:. she should sell the centre table at 14% profit.

Que 2. cot³θ/cosec²θ + tan³θ/sec²θ + 2sinθcosθ = ? 

A) sin²θcosθ

B) sinθcosθ 

C) cosecθsec²θ 

D) cosecθsecθ 

Answer:D 

Explanation: 

cot³θ/cosec²θ + tan³θ/sec²θ + 2sinθcosθ 

= (cos³θ/sin³θ) × sin²θ + (sin³θ/cos³θ)×cos²θ + 2sinθcosθ 

= cos³θ/sin θ + sin³θ/cos θ + 2sinθcosθ 

= (cos⁴θ+sin⁴θ)/sin θ cos θ + 2sinθcosθ 

= (cos⁴θ+sin⁴θ+2sin²θ cos²θ )/sinθ cosθ 

= (sin²θ + cos²θ)²/sinθcosθ 

= 1/sinθcosθ 

= cosecθsecθ 

Que 3. A heap of wheat is in the form of a cone whose base diameter is 8.4 m and height is 1.75 m. The heap is to be covered by canvas. What is the area ( in m²) of the canvas required? (Use π=22/7)

A) 115.50 

B) 60.06

C) 60.60

D) 115.05

Answer: B

Explanation:

Radius=8.4/2 =4.1 m
r=radius, l=slant height , h=height
We know,

l² = r² + h² 

l² = (4.2)² + (1.75)² 

l²= 20.7025

Slant height(l)=√20.7025 

:.Slant height(l) =4.55 m

Curved surface area = πrl (where, r=radius, l=slant height).

Curved surface area=(22/7)×4.1×4.55        =60.06 m² 

:. The area of the canvas required = 60.06 m²

Que 4. Chamanlal, Arshad and Jagjit contested an election. All the votes polled we’re valid. Arshad got 35% of the total votes. For every 35 votes Chamanlal got 14 votes. The winner got 4950 more votes than the person who got least number of votes. Find the total number of votes polled.

A) 33000

B) 38000

C) 13378

D) 99000

Answer:A

Explanation:

Let, total number of votes polled = P

Arshad got votes= P × 35/100 = 7P/20

Chamanlal got votes= P/35 × 14 = 2P/5

Jagjit got votes= P – 7P/20 – 2P/5 = P/4

Here clearly shows that, Jagjit got minimum number of votes and Chamanlal got maximum number of votes.

According to question,

2P/5 – 4950 = P/4

=> 2P/5 – P/4 = 4950

=> (8P-5P)/20 = 4950

:. P = (4950×20)/3 = 33000

:. The total number of votes polled = 33000

Que 5. A train leaves station A at 8 am and reaches station B at 12 noon. A car leaves station B at 8:30 am and reaches station A at 12 noon. At what time do they meet?

A) 9:52 am

B) 9:38 am

C) 10:22 am

D) 10:08 am

Answer : D

Explanation:

Let total distance between two stations = D km

Time taken by train to cover D km = (12-8) = 4 h

Time taken by car to cover D km =(12 – 8:30) =7/2 h

Ratio of time of train and car = 4 : 7/2 = 8 : 7

Ratio of Speed = 7:8   ( Time & Speed inversely proportional)

Total distance = 7×4 = 28 km

Train leaves station 30 mins before the car , so in 30 mins train covered = 7×30/60 = 3.5 km

Remaining distance = (28-3.5) = 24.5 km

Train and car are running towards each-other, so the relative speed = (7+8) = 15 km/h

Time taken = 24.5/15 = 1 hour 38 mins.

Required time = (8:30 +1 hour 38 mins) =10:08 am

:.They meet at 10:08 am.

Que 6. If 4x⁴ – 37x² + 9 = 0, x>√(3/2), then what is the value of 8x³ – 27/x³.

A) -215

B) 215

C) 35

D) -35

Answer: B

Explanation:

Let, x² = p

4p² – 37p + 9 = 0

=> 4p² – 36p – p + 9 = 0

=> (4p – 1)(p – 9) = 0

:. p = 1/4 , 9

Here p=x²=1/4 is not acceptable because in question given x²=3/2 .

So, x² = 9  =>x = 3

Now, 8x³ – 27/x³ = 8×3³ – 27/3³ = 215

:. The required value is = 215

Que 7. A and B can complete a work in 15 days and 10 days respectively.They started doing the work together but after 4 days B had to leave. Then A working with a new worker C completed the remaining work in 3 days. If C works alone, in how many days he can do 40% of the same work?

A) 10 days

B) 9 days

C) 12 days 

D) 8 days

Answer:B

Explanation:

Total work(L.C.M of 15,10) = 30

A’s 1 day work = 30/15 = 2

B’s 1 day work = 30/10 = 3

(A+B)’s 1 day work = 2+3 = 5

(A+B)’s 4 days work = 4×5 = 20

Remaining work = (30-20) = 10

(A+C)’s 3 days work = 10

(A+C)’s 1 day work = 10/3

C’s 1 day work = (10/3 – 2) = 4/3

40% of 30 = 30 × 40/100 = 12

C’s required time = 3/4 × 12 = 9 days 

:. C alone completed 40% of the total work in 9 days.

Que 8. Three shopkeeper A,B and C marked an identical article at Rs.4820. A,B and C sold their article on successive discounts of 20% & 20% ; 25% & 15% ; 30% & 10% respectively. Which shopkeeper gives maximum discount and how much ( in Rs)?

A) C, 1783.40

B) C, 1760.50

C) A, 1735.20

D) B, 1800.50 

Answer: A 

Explanation:

Total discount % of A =20+20-(20×20)/100 =36%

Total discount % of B =25+15-(25×15)/100 =36.25%

Total discount % of C =30+10-(30×10)/100 =37%

Discount by A = 4820×(36/100) = Rs.1735.20 

Discount by B = 4820×(36.25/100) = Rs.1747.20 

Discount by C = 4820×(37/100) = Rs.1783.40 

:. Shopkeeper C gives the maximum discount of Rs.1783.40.

Que 9. If a – 12/a = 1, where a>0, then find the value of  a² + 16/a² .

A) 15

B) 19

C) 11

D) 17

Answer:D 

Explanation:

a – 12/a = 1

=> a² – a – 12 = 0

=> a² – 4a + 3a – 12 = 0

=> (a + 3)(a – 4) = 0

=> a = 4 , -3

Since a>0 , so a= -3 not is acceptable.

a² + 16/a² = 4² + 16/4² = 17

:. The required value = 17

Que 10. AB is a diameter of a circle. C and D are points on the opposite sides of the diameter AB,such that ∠ACD = 25°, E is a point on the minor arc BD. Find the value of ∠BED. 

A) 105°

B) 115°

C) 125°

D) 130°

Answer:B 

Explanation:

∆ABD is a right angle triangle where ∠ADB = 90°

Angle ACD and ABD are equal because they lie in same arc.

∠ACD =∠ABD = 25°

In ∆ABD,

∠ABD +∠ADB +∠BAD = 180°

=>∠BAD = 180° – 90° – 25° = 65°

Here, ABED is a cyclic quadrilateral whose opposite angles sum is 180° always.

∠BAD +∠BED = 180°

=> ∠BED = 180° – 65° = 115°

:. The value of ∠BED = 115°

Que 11. If one of the angles of a triangle is 74°, then the angle between the bisectors of the other two interior angle is –

A) 53°

B) 16°

C) 127°

D) 106°

Answer:C 

Explanation:

AD and BE are the two angle bisectors of angle A and B respectively. 

In ∆ABC,

∠A + ∠B + ∠C = 180°

=> ∠A + ∠B = 180° – 74° = 106°

In ∆APB,

∠PAB +∠ABP +∠BPA = 180°

=> ∠BPA = 180° – ∠PAB – ∠ABP 

=> ∠BPA = 180° – a/2 – b/2    ( from the fig.)

=> ∠BPA = 180° – 106°/2 = 127°

:. The angle between the two bisectors of the other two interior angle is = 127°

Que.12 If (16√2x³+81√3y³)÷(2√2x+3√3y)=Ax²+By²+Cxy ,then find the value of 2A-3B-2√6C.

A) 7

B) 37

C) 25

D) 79

Answer:A 

Explanation:

(16√2x³+81√3y³)÷(2√2x+3√3y) = Ax²+By²+Cxy 

=>{(2√2x)³+(3√3y)³}÷(2√2x+3√3y)   =Ax²+By²+Cxy 

=> {(2√2x+3√3y)(8x²+27y²-6√6xy)}÷(2√2x+3√3y)  = Ax²+By²+Cxy 

=> 8x² + 27y² – 6√6xy = Ax² + By² + Cxy 

=> A = 8 ; B = 27 and C = -6√6 

Now, putting the values in 2A-3B-2√6C.

the required value = 7

Que 13. If 2sin(3x-15)°=1, 0°<(3x-15)°<90°, then find the value of cos²(2x+15)°+cot²(x+15)°.

A) 1

B) -7/2

C) 7/2

D) 5/2

Answer:C 

Explanation:

2sin(3x-15)° = 1 

=> Sin(3x-15)° = 1/2 = sin30°

=> 3x – 15 = 30

=> x = 45/3 = 15°

Now,  Cos²(2x+15)° + cot²(x+15)°

=cos²(30+15)° + cot²(15+15)°

= cos²45° + cot²30°

= (1/√2)² + (√3)² 

= 1/2 + 3

= 7/2

:. The required value = 7/2

Que 14. The following pie chart represent the percentage wise distribution of 300 students of class X in a school in six different sections A,B,C,D,E and F.

The table given below shows the number of boys of class X in six different sections A, B, C, D, E and F.

Section	A	B	C	D	E	F
No of Boys	36	26	34	28	X	20

If in section E,the ratio of number of boys to the number of girls is 3:4,then the ratio of the number of boys in the section E to the number of girls in section C is –

A) 18 : 23

B) 23 : 24

C) 23 : 18

D) 24 : 23

Answer: A 

Explanation : 

Total students in section E =300×14/100 = 42

Number of boys in section E = 42×3/7 = 18

Number of girls in section C = (57 – 34) = 23

Required ratio = 18 : 23

:. The ratio of boys in section E to the girls in section C is = 18:23

Que 15. The bar graph shows the number of students enrolled for a science course in institutes A and B during 5 years from 2014 to 2018.

What is the ratio of the total number of students enrolled in institute B in 2015 and 2017 to that of students enrolled in institute A in 2014 and 2016? 

A) 111 : 91 

B) 137 : 92 

C) 92 : 137 

D) 91 : 111 

Answer: B 

Explanation : 

In institute B total students enrolled in 2016 and 2017 = (360+325) = 685 

In institute A total number of students enrolled in 2014 and 2016 = (160+300) = 460 

:. Required ratio = 685 : 460 = 137 : 92

Q.16

In a ∆ABC point D lies on AB and points E and F lie on BC such that DF is parallel to AC and DE is parallel to AF. If BE=4 cm, CF = 3 cm. Find the length ( in cm) of EF. 

A) 2 cm 

B) 5 cm 

C) 3 cm 

D) 1.5 cm 

Answer: A 

Explanation: 

In ∆ABC, DF||AC, so 

BD/AB = BF/BC         ———– (1) 

In ∆ABF , DE||AF , so 

BE/BF = BD/AB       ————– (2) 

Solving equations (1) and (2) , we get 

BE/BF = BF/BC 

=> BE × BC = BF² 

=> 4 × (4 + 3 + EF) = (4 + EF)² 

=> 28 + 4EF = 16 + 8EF + EF² 

=> EF² + 4EF – 12 = 0 

=> EF² + 6EF – 2EF – 12 = 0 

=> (EF – 2) ( EF +6) = 0 

=> EF = 2 , -6 

Length can not be negative so EF = 2 cm 

:. The length of EF = 2 cm

Que 17. Table shows income received by 4 employees of a company during the month of December 2020 and all their income sources. 

Source	Amit	Suresh	Nitin	Varun
Salary	35000	38500	29000	42000
Arrears	6000	6300	5000	7500
Bonus	1000	1100	1000	1240
Overtime	1800	1950	1400	1500

By what percent is the bonus of Varun less than all bonus of Amit and Nitin taken together?

A) 45 %

B) 48 %

C) 42 %

D) 38 %

Answer: D 

Explanation : 

Varun’s bonus = 1240 

(Amit & Nitin)’s bonus = (1000+1000) = 2000 

Required less percent =(2000-1240)/2000×100% = 38% 

Que 18. Hridaya opened her piggy bank and found coin of denomination Rs.1 ,Rs.2, Rs.5 and Rs.10 in the ratio 10:5:2:1. If there are 72 coins in all, then how much money was there in the piggy bank in the form of coins? 

A) 160 

B) 72 

C) 90 

D) 100 

Answer : A 

Explanation : 

Let, 10x, 5x, 2x and x be the numbers of Rs.1 , Rs.2 , Rs.5 and Rs.10 coins respectively. 

According to question, 

10x + 5x + 2x + x = 72 

=> x = 72/18 = 4 

Number of coins ofRs.1, Rs.2, Rs.5 and Rs.10 coins are 40,20,8 and 4 respectively.

 Total money = 40×1+20×2 + 8×5 + 4×10 = 160 

:. The total money in the piggy bank in form of coins is = 160

Que 19. What is the ratio of the average of first eight prime numbers to the average of first ten even natural numbers ?

A) 1 : 7 

B) 7 : 8 

C) 8 : 13 

D) 9 : 17 

Answer: B 

Explanation: 

First eight prime numbers= 2,3,5,7,11,13,17,19 

Average = (2+3+5+7+11+13+17+19)/8 = 77/8 

First ten even natural numbers= 2,4,6,8,10,12,14,16,18,20 

Average = (2+4+6+8+10+12+14+16+18+20)/10 = 110/10 = 11 

:. Required ratio = 77/8 : 11 = 7 : 8

Que 20.  If the nine digit number 7p5964q28 is completely divisible by 88, what is the value of (p² – q) for the largest value of q, where p & q are natural numbers ? 

A) 9

B) 81 

C) 5 

D) 72 

Answer : D 

Explanation : 

Number divisible by 88 means it should be divisible by 8 and 11 both. 

Largest possible value of q should be 9 [ number divisible by 8 when it’s last 3 digits divisible by 8] 

For q = 9, number should be divisible by 11 

( 7+5+6+9+8) – ( p+9+4+2) 

= (20 – p) 

(20 – p) should be divisible by 11 if p = 9. 

Now, p² – q = 9² – 9 = 72 

:. The required value is 72. 

Que 21. Table shows the number of trees planted in 4 cities from 2016 to 2020. 

Years	Delhi	Ahmedabad	Pune	Kolkata
2016	1800	2500	1800	2000
2017	2500	2300	1850	1800
2018	2300	2400	1840	1760
2019	2440	1950	1900	1600
2020	2250	2100	2000	1750

In which year were the maximum number of trees planted? 

A) 2016 

B) 2020 

C) 2017 

D) 2018 

Answer: C 

Explanation : 

Trees planted in 2016 =(1800+2500+1800+2000) = 8100 

Trees planted in 2017 =(2500+2300+1850+1800) = 8450 

Trees planted in 2018 =(2300+2400+1840+1760) = 8300 

Trees planted in 2020 =(2250+2100+2000+1750) = 8100 

:. In 2017 maximum number of trees were planted. 

Que 22. A sum at a certain rate of simple Interest becomes Rs.14880 after 3 years and Rs.16800 after 5 years . Find the simple Interest on the same sum at 10% p.a for 4 years.

A) 4860 

B) 4800 

C) 5100 

D) 5400 

Answer: B 

Explanation : 

Amount for 3 years, 

14880 = 3PR/100 + P     ——- (1) 

Amount for 5 years, 

16800 = 5PR/100 + P    ——– (2) 

Subtracting equation (1) from (2), we get,

2PR/100 = 16800 – 14880 

=> PR = 96000 

Putting the value of PR in eqn.(1) ,we get 

14880 = 96000 × 3/100 + P 

=> P = 14880 – 2880 

=> P = 12000 

Simple Interest = PRT/100 =(12000×10×4)/100 =4800 

:. The required Simple Interest = Rs.4800 

Que 23. 3(⅚) + [ 3(⅔) + { 15/4( 5(⅘) ÷ 14(½)}] = ? 

A) 9 

B) 7 

C) 8 

D) 6 

Answer : A 

Explanation : 

3(⅚) + [ 3(⅔) + { 15/4( 5(⅘) ÷ 14(½)}] 

= 23/6 + [ 11/3 + { 15/4 ( 29/5 × 2/29 )}] 

= 23/6 + [ 11/3 + { 15/4 × 2/5 }] 

= 23/6 + [ 11/3 + 3/2 ] 

= 23/6 + 31/6 

= 54/6 

= 9 

:. The required value is 9. 

[ Concept:- BODMAS 

B = BRACKETS (), {}, [ ] 

O = OF 

D = DIVISION 

M = MULTIPLICATION 

A = ADDITION 

S = SUBTRACTION ]

Que 24.  In ∆ABC , AB=20 cm , BC=21 cm and AC=29 cm. What is the value of cotC + cosecC – 2tanA = ? 

A) 3/5 

B) 9/20 

C) 2/5 

D) 7/22 

Answer: C 

Explanation: 

(AC)² = 29² = 841 

(AB)² + (BC)² = 20²+21² = 841 

Here, (AC)² = (AB)² + (BC)² 

:. ∆ABC is a right angle triangle. 

cotC = BC/AB = 21/20 

cosecC = AC/AB = 29/20 

tanA = BC/AB = 21/20 

Now, 

cotC + cosecC – 2tanA 

 = 21/20 + 29/20 – 2×21/20 

= 29/20 – 21/20 

= 8/20 

= 2/5 

:. The required value = 2/5 

Que 25. From an external point A,two tangents AB and AC have been drawn to a circle touching the circle at B and C respectively. P and Q are points on AB and AC respectively such that PQ touches the circle at R. If AB=11 cm , AP=7 cm and AQ=9 cm ,then what is the length of PQ? 

A) 5 cm 

B) 6 cm 

C) 7 cm 

D) 8 cm 

Answer: B 

Explanation : 

tangents are always be equal when they are drawn from an external point. 

AB = AC ; BP = PR ; CQ = RQ 

AB = AC = 11 cm 

BP = AB – AP = 11 – 7 = 4 cm 

CQ = AC – AQ = 11 – 9 = 2 cm 

BP + CQ = 4 + 2 = 6 cm 

PR + RQ = PQ = 6 cm 

:. The length of PQ = 6 cm.