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SEBI Grade A : Quantitative Aptitude Sample Questions

Hello everyone! SEBI Grade A notification 2024 may be released soon. Keep watch for the upcoming notification, initiating your preparation early to enhance your chances of success. This is a golden opportunity for the candidates who wish to work with SEBI. Embrace this opportunity with enthusiasm, by gearing up for early preparation to pave the way for your achievements in this upcoming exam! Adequate preparation across all SEBI Grade A exam subjects is imperative for optimal performance. Therefore, if you want to do well in this exam, it is essential that aspiring candidates are prepared with all the subjects that are assessed in SEBI Grade A recruitment exam.

This article emphasizes the significance of the Quantitative Aptitude section in banking exams by spotlighting sample questions. Quantitative Aptitude holds paramount importance in banking assessments, testing candidates’ numerical skills and problem-solving abilities. By exploring these sample questions, candidates can comprehend the section’s nature, grasp various problem-solving methods, and prepare effectively for the challenging quantitative segment of SEBI Grade A exam.

SEBI Grade A : Exam Pattern

Phase 1 Exam Pattern

It is essential to have a thorough understanding of the SEBI Grade A exam syllabus and exam pattern in order to qualify for each stage of the recruitment process. 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-

Phase 2 Exam Pattern

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

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 2024 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?

  1.  2000 
  2.  1600 
  3. 1500 
  4.  2500 
  5. 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?

  1. 4 hours 48 minutes
  2. 5 hours 10 minutes
  3. 4 hours
  4. 5 hours 20 minutes
  5. 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? 

  1. 91/200
  2. 51/200
  3. 81/200
  4. 93/200
  5. 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? 

  1. 30
  2. 33 ⅓ 
  3. 35
  4. 44
  5. 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?

  1. 900
  2. 500
  3. 1000
  4. 400
  5. 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 ViewersPercentage by which number of subscribers is more than the number of non-subscribers
40050%
550300%
480200%
D360800%

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? 

  1. 168
  2. 148
  3. 154
  4. 164
  5. 172

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? 

  1. 312
  2. 298
  3. 294
  4. 308
  5. 310

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? 

  1. 40
  2. 20
  3. 25
  4. 60
  5. 50

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? 

  1. 20 ⅓
  2. 22 2/9
  3. 23 1/9
  4. 24 2/9
  5. 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% 

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