SEBI Grade A 2022: Quantitative Aptitude Sample Questions

January 10, 2022

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SEBI Grade A Phase 1 (Prelims) Exam will be held on February 20, 2022. This is a golden opportunity for the candidates who wish to work with SEBI. Therefore, if you want to do well in this exam, it is essential that you are prepared with all the subjects that are judged in SEBI Grade A Phase 1 exam.

In this article, we will focus on some sample questions of the Quantitative Aptitude section. So, if you are appearing for the SEBI Grade A 2022 exam, we encourage you to go through the sample questions provided below carefully.

SEBI Grade A: Quantitative Aptitude Sample Questions

SEBI Grade A 2022: Exam Pattern

Phase 1 Exam Pattern

Before going through the sample questions, it will be beneficial to understand the SEBI Grade A Phase 1 exam pattern.

In Phase 1 (Prelims), the candidates are tested based on the following four subjects-

English Language
General Awareness
Reasoning Ability
Quantitative Aptitude

Phase 2 Exam Pattern

The SEBI Grade A Phase 2 (Mains) exam will consist of two stages-

English Paper (Descriptive)
Professional Knowledge (Stream specific)

You can find the detailed SEBI Grade A 2022 exam pattern here.

SEBI Grade A: Quantitative Aptitude

Quantitative Aptitude (QA) of the candidates is tested in Phase 1 of the SEBI Grade A recruitment process. Quantitative Aptitude is a scoring but tricky section since you have to solve many questions in a limited time.Therefore, it becomes all the more important to prepare this section well. Given below are some sample questions from the Quantitative Aptitude section to help you understand the level of questions that can be asked in the SEBI Grade A 2022 exam.

Quantitative Aptitude: Sample Questions

Q1) Ram sells only two kinds of bags—pure leather and faux leather, 75% and 80% of the total number of bags sold, in February and March respectively, were of pure leather. Average number of pure leather bags sold in both the months is 5850 and the total number of bags sold in March is 50% more than that sold in February. How many faux leather bags did Ram Sell in February?

2000
1600
1500
2500
1400

Answer: (3) 1500

Explanation:

Total no. of bags sold in Feb. = x

Total no. of bags sold in March = y

Given, y = x + 50/100 × x = 150/100 x

No. of pure leather bags sold in Feb. = 75/100 x And no. of pure leather bags in March 80/100 y

Given,

Avg. no. of pure leather bags = 5850

So total no. of pure leather bags = 5850 × 2 According to Question = 11700

75/100x + 80/100 y = 11700

=> 75/100 x + 80/100 × 150/100 x = 11700

=> x = 6000 bags

So, no. of pure leather bags sold in

Feb. = 75/100×6000 = 4500

No. of faux leather bags sold in Feb. = 6000 – 4500 = 1500

Q2) The respective ratio between the time taken by a bus to travel a certain distance at x kmph and the time taken by the bus to travel the same distance at (x + 20) kmph is 4 : 3. How much time will the bus take to travel 480 km at a speed of (x + 30) kmph?

4 hours 48 minutes
5 hours 10 minutes
4 hours
5 hours 20 minutes
6 hours

Answer: (4) 5 hours 20 minutes

Explanation:

Given, d = 4x

And d = 3(x + 20)

=> 4x = 3x + 60

=> x = 60 km

Now, t = 480 / (x + 30) = 480/90 = 5.33 hr = 5 hr 20 min.

Q3) A box contains 200 balls and each ball is marked with a number from 1 to 200. One ball was picked at random from the box, what is the probability that the ball picked has a number which is divisible by either 3 or 5?

91/200
51/200
81/200
93/200
53/100

Answer: (4) 93/100

Explanation:

Total no. divisible by 3 from 1 to 200 = 66

Total no. divisible by 5 from 1 to 200 = 40

And total no. divisible by both 3 and 5 from 1 to 200 = 13

Required probability = (66 + 40 – 13)/200 = 93/200

Q4) Some articles were bought at 6 for Rs. 5 and sold at 5 for Rs. 6. Gain percentage is?

30
33 ⅓
35
44
50

Answer: (4) 44

Explanation: Gain %

= [(m×) – (n×n)/(n×n)] × 100 =(6)² –(5)² /(5)²×100

= 36 – 25 /25 × 100 = 44%

Q5) Rani had a certain sum of money (Rs. x), out of which she pays Rs. 800 to the grocer. She paid 1/3 rd of Rs. x to her maid and gardener in the respective ratio of 2 : 1. If after all the mentioned expenses she is left with Rs. 2200, how much did she pay to her maid?

900
500
1000
400
200

Answer: (3) 1000

Explanation:

Rs. 800 = given to grocer

1/3x × 2/3 = 2/9x

Given to maid

1/3 x × 1/3 = 1/9x

Given to gardener

x – 100 – 2/9x – 1/9x = 2200 => 9x – 7200 – 3x = 19800

=> 6x = 27000

=> x = 4500

Amount she gave to maid = 2/9x = 2/9 × 4500 = Rs. 1000

Directions (Q6 to Q9) Study the following table and answer the given questions.

Channels	No. of Viewers	Percentage by which number of subscribers is more than the number of non-subscribers
A	400	50%
B	550	300%
C	480	200%
D	360	800%

Note: Number of viewers = Number of scribers + Number of non- subscribers.

Q6) The number of viewers of Channel E is five-ninth of that of Channel D. If the number of non-subscribers for Channel E is equal to that of Channel D, What is the number of subscribers for Channel E?

Answer: (4) 164

Explanation:

No. of viewers of channel E = 5/9 × 360 = 200 No. of non-subscribers for E

No. of non-subscribers for D

Let, x = no. of subscribers,

And y = no. of non-subscribers

=> x = y + 800/100 y (given)

=> x + y = 360 …(i)

=> x = 9y

From (i) y = 36

So, no. of non-subscribers for E = 36

No. of subscribers = 200 – 36 = 164

Q7) What is the average number of subscribers for Channels A, C and D?

Answer: (4) 308

Explanation:

For channel A :

No. of subscribers = x1

And no. of non-subscribers = y1

x1 + y1 = 400

x1 = y1 + 50/100 y1

= 150/100 y1

150/y1 + y1 = 400

=> y1 = 400×100/250 = 160

x1 = 400 – 160 = 240

For Channel C:

No. of subscribers = y2

y2 + y2 = 480

-> x2 = y2 + 200/100 y2

which equals to 3y2

= 4y2 = 480

=> y2 = 120

X2 = 480 – 120 = 360

For channel D:

No. of subscribers = x3

And no. of non-subscribers = y3

x3 + y3 = 360

-> x3 = y3 + 800 / 100 y3

= 9y3

=> 10y3 = 360

y3 = 36

x3 = 360 – 36 = 324

Avg. = x1 + x2 + x3 / 3

= 240 + 360 + 324 / 3

= 924 / 3 = 308

Q8) What percent of viewers of channel A have not subscribed for the channel?

Answer: (1) 40

Explanation:

Let subscribers of channel A = x

And non-subscribers of channel A = y

x + y = 400 …(i)

x = 50/100 y + y

= 150/100 y

From (i),

150/100 y + y 400

=> y = 160 (non-subscribers)

% = 160/400 × 100 = 40%

Q9) The number of subscribers for Channel B is what percent more than that for Channel C?

20 ⅓
22 2/9
23 1/9
24 2/9
21 1/9

Answer: (2) 22 2/9

Explanation:

For channel B:

Subscribers= x

non-subscribers= y

x + y = 550 …(i)

x = 300/100 y + y = 4y

4y + y = 550

=> y = 110 (non-subscribers)

x = 550 – 110

= 440 (subscribers)

For channel C:

subscribers= x1

non-subscribers= y1

x1 + y1 = 480 …(ii)

x1 = y1 + 200 y1 = 3y1

4y1 = 480

=> y1 = 120 (non-subscribers)

x1 = 480 – 120 = 360 (subscribers)

% increase = 440 – 360 / 360 × 100

=22 2/9%

Study with ixamBee for SEBI Grade A 2022

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