SSC CGL Prelims Quantitative Aptitude Practice Paper

This is the Practice Paper for the Aptitude section in SSC CGL prelims exam. It comprises 25 questions as per the latest SSC CGL pattern with shortcut solutions. 25 questions are asked in each section in SSC CGL pre-examination. Aspirants can use this as a sample paper for have an idea of questions asked in SSC CGL Prelims exam.

Aspirants usually face problems in solving the Quant section with the right speed & accuracy and hence lose on some sure shot marks. This sample paper will make you ready for the real exam by improving your speed & accuracy.

Que 1. If 3a+5b/11a-2b =5/6, Find a : b?
A) 20/27        
B) 31/37
C) 40/37
D)31/27

Answer: C

Explanation:
Given, (3a+5b)/(11a-2b) = 5/6
By cross multiplication we get,
6(3a + 5b) = 5(11a- 2b)
18a + 30b = 55a – 10b
40b = 37a
It gives us, a/b =40/37.

Hence the correct answer is option C.

Que 2. The average of 23 numbers is 12. The average of the first 20 numbers is 11 and the 21st number is 3 more than the 23rd and the 22nd number is 2 more than the 23rd number. Find the 22nd number.
A) 17  
B) 18
C) 19
D) 20

Answer: C

Explanation :

Given, Average of 23 numbers =12 
So, Sum of the 23 numbers = 23 × 12 = 276
Also, the average of 20 numbers = 11
Then, the sum of 20 numbers= 20 × 11 = 220.
The sum of the remaining 3 numbers = total of 23 numbers – total of 20 numbers

= 276  –  220 = 56.              
Let the 23rd number be x,
Then A.T.Q’s 22nd number is (x + 2) and the 21st number is (x + 3)
3x + 5 = 56
3x =51
x = 17

22nd number = x + 2 = 17 + 2 = 19.

Hence the correct answer is option C.

Que 3. Tap A can fill the tank in 15 minutes and Tap B can fill the tank in 30 minutes and another tap C can empty the filled tank in 20 minutes. For the first 2 minutes, Tap A and Tap B are opened then tap C is also opened. Then how much total time does it takes to fill the tank completely.

A) 16 minutes
B) 18 minutes
C) 20 minutes  
D) 22 minutes

Answer: B

Explanation:

Let the total work be 60 units (LCM of 15, 30, and  20)
So, the efficiency of A, B & C is 4 units, 2 units & -3 units respectively (As tap C is working to empty the tank)
Now, Work done by A & B in minutes = 6 × 2   =  12 units
Remaining work = 60 – 12 = 48 units
Time to do remaining work by A , B & C  = Remaining work / Total efficiency of A , B & C
= 48/ (4 + 2 – 3)
= 48/3 = 16 minutes                                  
Total time is taken to do the work = 16 + 2 = 18 minutes  
Hence the correct answer is option B.

Que 4. If a person travels at a speed of 15 km/hr then it reaches 5 minutes late and if a person travels at a speed of 20 km/hr then it reaches 2 minutes early. Then what is the actual time a person takes?

A) 17 minutes  
B) 19 minutes
C) 21 minutes
D) 23 minutes

Answer: D

Explanation :
As the distance is the same, so we can write
Speed × (actual time + extra time) = Speed × (actual time – less time)
15(t+5) = 20(t-2)
 3t+15 = 4t-8
 t = 23 minutes
Hence the correct answer is option D.

Que 5. The difference between compound interest and simple interest at a rate of 10% p.a. for 3 years is Rs 3100. Find the principle.

A) 80000       
B) 90000
C) 100000      
D) 110000

Answer: C

Explanation: 

Difference = 3 x P(R)²/(100)² + P (R/100)³  (Formula for 3 years difference)
Putting the value of R = 10, upon solving we get P = 100000
Principle =  Rs 100000
Hence the correct answer is option C.

Que 6. A dealer buys some articles at a price of Rs 25 each. If he sells 20% of articles at Rs 20 each and 40% of the remaining at Rs 25 each then at what price he must sell the remaining articles in order to gain 20% on whole?

A) 30
B) 32.5
C) 35  
D) 37.5

Answer: D

Explanation: 

Let the number of articles the dealer buys be 100
Total CP = 100 × 25 = 2500
SP of 20 articles = 20 × 20 = 400                            
SP of 32 articles ( 40% of remaining 80 articles) = 32 × 25 = 800                                                                                   
Overall SP of the 100 article to get 20% profit = 2500 × 1.2 = 3000

Remaining SP of 48 article = 3000 – 400 – 800 = 1800

SP of the remaining article = 1800/48 = 37.5.
He should sell the remaining at 37.5 per article
Hence the correct answer is option D

Que 7. If the Cost Price of an item is Rs 200 & the profit percentage is 20. If the item was marked up at Rs 300, find the discount percentage?

A) 10%
B) 15%
C) 20%
D) 25%

Answer: C

Explanation: 
SP of the article at 20% profit = 200 × 1.2 = 240.

Discount offered = MP – SP = 300 – 240 = 60.

Discount% = 60/300 × 100 = 20%.

Hence the correct answer is option C

Que 8. ABC is a triangle with sides AB and BC of 4 cm and 5 cm respectively and angle ABC = 60°. Find the area of triangle ABC.

A) 3√3  
B) 5√3
C) 4√3
D) 7√3

Answer: B

Explanation:

Area of the triangle with two sides and including angles given = 1/2 × first side × second side × sin α (α is the angle between these 2 sides)
Area of triangle ABC =   ½  × 4 × 5 × sin 60° = 5√3.
Hence the correct answer is option B

Que 9. ABC is a triangle, OB and OC are the bisectors of angle ABC and angle ACB. If angle BAC =75, then find angle BOC.

 A)125
 B)127.5
 C) 130  
 D) 132.5

Answer: B
Explanation:
By using the angle bisector theorem,

Angle BOC   =   90 + 1/2 angle BAC  
                      =   90 + 1/2 × 75
                      =   90 + 37.5
                      =   127.5
Hence the correct answer is option B

Que 10. A sphere of radius 14 cm is melted and small hemispheres of 7 cm made from it. Then the maximum number of hemispheres which can be made from it?

A) 12
B) 14
C) 16
D) 18

Answer: C

Explanation: 
 Let the number of hemispheres that can be made be N
 The volume of sphere = Volume of hemisphere × N
 4/3 π × 14 × 14 × 14   = 2/3 π × 7 × 7 × 7 × N
 This gives, N = 16

Hence the correct answer is option C

Que 11. The curved Surface Area of the cylinder is 198 cm² of which radius is 7 cm. Then find the volume of the cylinder (in cm³)?

A) 672       
B) 689
C) 693           
D) 682

Answer: C

Explanation:

CSA of cylinder = 2πrh
198 = 2 × 22/7 × 7 × h 
h=   9/2
The volume of the cylinder =πr²h
Volume = 22/7 × 7 × 7 × 9/2
Volume  = 693 cm³.
Hence the correct answer is option C

Que 12. In triangle ABC bisectors of internal angle, ABC and external angle ACD meet at E.  If angle BAC = 52° then find angle BEC?

A) 24        
B) 28      
C) 26
D) 30

Answer: C

Explanation:

Angle BEC   =  1/2  × Angle BAC  
                    =  1/2 × 52  
                    =   26° 

Hence the correct answer is option C

Que 13. From the top of the tower of height 24 cm, the angle of depression of 2 persons is 30° and 45° standing on the same side of the tower. Find the distance between 2 persons.

A) 24                     
B) 24 (√3)  
C) 48
D) 24 (√3  – 1)  

Answer: D

Explanation:

Let AB be the tower,
In triangle ABC
tan 45° = 24/ BC
BC = 24 cm
Now, in triangle ABD 
Tan 30° = 24/ BD
1/√3  =  24/ BD
BD  =  24√3
CD = 24√3   –   24
CD = 24(√3 -1) cm.
Hence the correct answer is option D

Que 14. Tanα= (5/12) , (π/2 < α < π) Then, Sinα + cosα = ?
A) -8/13
B) 8/13
C) 17/13
D) 7/13

Answer: C

Explanation:

Tanα = (5/12) = P/B

By using Pythagoras theorem we get,

H = 13
Sinα= 5/13
Cosα= (12/13)
sinα + cosα = 5/13 + 12/13 = (5 + 12)/13 = 17/13.

Hence the correct answer is option C

Que 15. If x + y = 2 Then, x²⁴ + y²⁴ ?
A) 1
B) 2
C) 512
D) 1024

Answer: B
Explanation:

x + y = 2 is possible only when x = y = 1
Then x²⁴ + y²⁴ = 2
Hence the correct answer is option B

Que 16. The total number of votes cast in an election in which 2 candidates A & B are competing is 10000. Out of which 20% of votes are invalid. Candidate A won by 2000 votes then find out how many votes did B get?
A) 1000         
B) 3000        
C) 1500
D) 2500    
Answer: B

Explanation:
Total number of votes = 10000
Invalid votes = 2000
Valid Votes = 8000
A.T.Q

A + B = 8000

A – B = 2000

2A = 10000

A = 5000

Vote received by B = 3000.

Hence the correct answer is option B

Que 17. A person spends 30% of his income on food, 10% of the remaining he spends on clothes, and the rest he divided among his 3 sons. If each son gets Rs 21000, then find his income?
A) 80000     
B) 100000   
C) 90000
D) 110000

Answer: B

Explanation: 
Let his income be x 
x × (70/100) × (90/100) × (1/3) = 21000
x = Rs 100000
Hence the correct answer is option B

Que 18. If  x² – 9x + 2 = 0 Then find x²+ 4/x² = ?
A) 73          
B) 75      
C) 77     
D) 79

Answer: C
Explanation:
x²  –  9x + 2 = 0 can be written as x + 2/x = 9

on squaring both sides we get, 
⇒ x² + 4/x² + 4 = 81 
⇒ x² + 4 /x² = 77.
Hence the correct answer is option C

Que 19. x³ + 1/x³ = 110 then find x + 1/x.
A) 4                
B) 5            
C) 6     
D) 7
Answer: B
Explanation:

(x+ 1/x)³ = x³+ 1/ x³ + 3 (x + 1/x)

By options, 5 satisfies this equation

Hence the correct answer is option B

Que 20. 68x61y is divisible by 72. Find 3x + 4y.

A) 49
B) 53
C) 51
D) 55

Answer: C

Explanation:

For it to be divisible by 72, it should be divisible by 8 & 9. 

For the divisibility of 8, we check if the last 3 digits should be divisible by 8. 

61y should be divisible by 8, this gives us y = 6, as 616 is divisible by 8. 
Now, we have 68×616, for it to be divisible by 9, the sum of the digits should be divisible by 9.
That is, 27 + x should be divisible by 9, this gives us x = 9. 
Hence, x = 9 & y=6 
So 3x + 4y is 3 × 9 + 4 × 6
= 51

Hence the correct answer is option C

Que 21. 18 × 18/(18 of 18) + 18 /(6×3) + 3 × 18 – 5 × 10.

A) 2
B) 4
C) 6
D) 8

Answer: C

Explanation: 

18 × 18/324  + 18/18 + 54 – 50

1 + 1 + 4 = 6.

Hence the correct answer is option C

Directions (22-25): 

The following table shows the data of students in five schools and the ratio of absent and present students on a particular day.

Solution ( 22 – 25 ) :

In School A,

Number of Absent students = 156 × (5/12) = 65

Number of Present students = (156 – 65) = 91

In School B,

Number of Absent students = 147 × (4/7) = 84

Number of Present students = ( 147 – 84) = 63

In School C,

Number of Absent students = 132 × (5/11) = 60

Number of Present students = ( 132 – 60) = 72

In School D,

Number of Absent students = 238 × (8/17) =112

Number of Present students = (238 – 112) =126

In School E,

Number of Absent students = 207 × (11/23) = 99

Number of Present students = (207 – 99) = 108

School	Total Students	Absent Students	Present Students
A	156	65	91
B	147	84	63
C	132	60	72
D	238	112	126
E	207	99	108

Que 22. Maximum numbers of students present in which school?

A) A
B) E  
C) C
D) D

Answer: D

Explanation:  
Maximum numbers of present students in school D.
Hence the correct answer is option D

Que 23. What is the ratio of Absent students in B to present students in E?

A) 7 : 22
B) 14 :23
C) 7: 9
D) 15 :28

Answer: C

Explanation:

Absent students in B = 84
Present students in E = 108
The required ratio is = 84 : 108 = 7 : 9

Hence the correct answer is option C.

Que 24. Find the total number of present students in D?

A) 276
B) 126
C) 72
D) 63

Answer: B

Explanation: 

The total number of present students in school D = 126 

Hence the correct answer is option B.

Que 25.  Find the total number of absent students of A & B.

A) 153
B) 149
C) 152
D) 160

Answer: B

Explanation:

Total Number of absent student in A = 65

Total number of absent student in B = 84

Required sum = 65 + 84 = 149
Hence the correct answer is option B.

Hope these sample questions on Quantitative Aptitude for SSC CGL Prelims will help you to prepare well for the SSC CGL exam.