SSC CHSL 2022 Exam Memory Based Quantitative Aptitude Questions (Download PDF): Check memory based Quantitative Aptitude questions that are coming in SSC CHSL 2022 Exam that is being held from 24^{th} May to 10^{th} June 2022.
SSC CHSL 2022 Exam Memory Based Quantitative Aptitude Questions (Download PDF): SSC CHSL 2022 Tier1 Exam began on 24^{th} May 2022 and will be held till 10^{th} June 2022. In this article, we are going to share the important memorybased Questions as per the feedback received by the candidates who have appeared for SSC CHSL 2022 Exam. Candidates are advised to definitely cover these questions for scoring high marks in the Exam.
SSC CHSL 2022 Memory Based Quantitative Aptitude/Maths Question Paper
Questions asked in the Quantitative Aptitude (Basic Maths) section were of easy to moderate level. Attempt of above 18 Questions is a good attempt. Let’s have a look at the Important Questions that are being covered in the SSC CHSL 2022 Exam:
Memory Based General Intelligence & Quantitative Aptitude Language Questions (25 Questions of 2 Marks each) 

1 
Ratio & Proportion 
02 
2 
Average 
02 
3 
Number System 
03 
4 
Simplification 
04 
5 
Time & Work 
01 
6 
Speed & Distance [Train] 
01 
7 
S.I. & C.I. 
02 
8 
Profit & Loss 
01 
9 
Algebra 
02 
10 
Geometry 
01 
11 
Mensuration 
01 
12 
DI [Pie Chart] 
01 
13 
DI [Bar Graph] 
02 
14 
Misc. 
02 
Total Questions 
25 
Sample Questions to Solve SSC CHSL 2022 Memory Based Quantitative Aptitude/Maths Paper
1. A and B together can complete a piece of work in 20 days. B and C together can complete in 30 days. If A is twice as good a workman as C, then in how many days will B alone can complete the same work?
 30 Days
 60 Days
 40 Days
 10 Days
Answer: b)
Solution:
A+B’s 1 day’s work =1/20
B+C’s 1 day’s work = 1/30
Since, A is twice as good a workman as C.
∴ A = 2C
B+2C =1/20
C = 1/20 – 1/30 = 1/60
B can complete the same work = 1/30 – 1/60 = 1/60 = 60 Days
2. Find out the ratio of length and the area of a rectangle if the ratio of length & perimeter of rectangle is 1:6 and the area of rectangle is 800 sq.m?
 1:20
 2:10
 5:11
 2:13
Answer (a)
Solution:
Let the length is x & perimeter be 6x
Let the breadth be B
Then 2(x + B) = 6x
x + B = 3x
B = 2x
Now
Length × breadth = area
x × 2x = 800
2x^{2} = 800
x = 20
Hence length = 20 m
Ratio of length: area
20:400
1:20
3. The marked price of an article for sale is 20% of its cost price. How much percent does the dealer gain by allowing a discount of 15%?
 5%
 7%
 10%
 25%
Answer: d)
Solution:
Let the CP of the article be Rs.100
Marked Price = (100 ×100)/80=Rs.125
Then, SP after discount = (125 ×85)/100 = Rs. 106.25
Gain Percent = (106.25100)/100 ×100=6.25%
4. What will be the simple interest on an amount of Rs. 2000 in 3 years at interest 4% per annum?
 280
 240
 250
 220
Answer: (b)
Solution:
Simple Interest = P ×R × T/100
(2000 ×4 ×3)/100 = Rs.240
5. Two trains starting at the same time from two stations 400 km apart and going in opposite direction cross each other at a distance of 230 km from one of the stations. What is the ratio of their speeds?
 11:9
 23:17
 18:4
 None of these
Answer (b)
Solution:
In same time, they cover 230 km and 170 km respectively.
For the same time speed and distance is inversely proportional.
So ratio of their speed = 230:170 = 23 : 17
6. The value of 16^{2} X 8^{4}/2^{16} is
 41/5
 20/3
 23/7
 16
Answer: (d)
Explanation:
16^{2} X 8^{4}/2^{16} = (2^{4})^{2} X (2^{3})^{4}/2^{16}
= (2^{8}×2^{12})/2^{16} = 2^{20}/2^{16} = 2^{4}= 16
7. The average weight of 10 boys is increased by 1.5 kg when one of them whose weight is 16 kg is replaced by a new boy. The weight of the new boy is:
 5 kg
 16 kg
 20 kg
 31 kg
Answer: (d)
Solution:
The weight of new boy = weight of ex boy + 1.5 * 10
= 16 + 15 = 31 kg
8. A shopkeeper sold a book at Rs. 144 in such a way that his percentage profit is same as the cost price of the book. If he sells it at twice the percentage profit of its previous percentage profit then new selling price will be:
 208
 250
 192
 180
Answer (a)
Solution: From taking options,
i.e. CP = 80
i.e New SP = 80+2 ×64=208
9. From the below equation, find the value of :
(0.1 ÷ 0.01 × 0.5 ÷ 2.5) × ((10 × 0.004) ÷ 0.1 + (0.05 ÷ 0.1)^{2}) = 1.3^{(x2)}
 1
 1
 3
 3
Answer: d)
Explanation:
(0.1 ÷ 0.01 × 0.5 ÷ 2.5) × ((10 × 0.004) ÷ 0.1 + (0.05 ÷ 0.1)^{2}) = 1.3^{(x2)}
(10 X 0.2) X (0.4 + 0.25) = 1.3^{(x2)}
1.3^{1} = 1.3^{(x2)}
=> x 2 = 1
x = 3
10. If 58^{3}+ 19^{3} + x^{3} – 57 × 58x = 0, then find the value of .
 77
 39
 39
 77
Answer: d)
Solution:
If a^{3} + b^{3} + c^{3} – 3abc = 0,
Then a + b + c = 0
58 + 19 + x = 0
x = 77x
Directions (1115): The table shows the number of candidates appeared for the interview in different division of a Government Organisation in Delhi in various years:
Years 
Divisions (Number of candidates appeared for the interview) 

Marketing 
Administrative 
Human Resources 
Finance 
IT 

2010 
150 
25 
50 
45 
75 
2011 
225 
40 
45 
62 
70 
2012 
450 
65 
30 
90 
73 
2013 
470 
73 
32 
105 
70 
2014 
500 
80 
35 
132 
74 
2015 
505 
75 
36 
130 
76 
11. In which year did the total number of candidates appeared for the interview is approximately twice the total number of candidates appeared for the interview in the year 2010?
 2011
 2012
 2013
 2014
Answer: b)
Explanation:
Total number of candidates appeared for the interview in 2010 = 345
Total number of candidates appeared for the interview in 2012 = 708
Therefore, in 2012 twice the candidates appeared for the interview as compared with candidates in 2010.
12. In which division did the number of candidates appeared for the interview (approximately) remain the same during 2010 and 2015?
 Marketing
 Finance
 IT
 Human Resources
Answer: c)
Explanation:
Division 
2010 
2015 
Marketing 
150 
505 
Finance 
45 
130 
IT 
75 
76 
Human Resources 
50 
36 
13. In how many years was the number candidates appeared for the interview in the Marketing division was more than 70% of the total number of candidates appeared for the interview?
 2
 3
 4
 5
Answer: a)
Explanation:
Year 
Candidates for Marketing 
Total Number of Candidates 
% More 
2010 
150 
345 
130.00 
2011 
225 
442 
96.44 
2012 
450 
708 
57.33 
2013 
470 
750 
59.57 
2014 
500 
821 
64.20 
2015 
505 
822 
62.77 
14. In which year did each division had a larger number of candidates appeared than in the immediately preceding year?
 2015
 2012
 2014
 2011
Answer: c)
Explanation: In 2014, each division had a larger number of candidates appeared than in 2013.
15. Which division has showed the maximum percentage increase in the number of candidates appeared for the interview from 2010 till 2015?
 Marketing
 Administrative
 Finance
 IT
Answer: b)
Explanation:
Division 
% increase from 2010 till 2015 
Marketing 
70.23 
Administrative 
200 
Finance 
188.89 
IT 
1.33 