**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 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? **

- 168
- 148
- 154
- 164
- 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? **

- 312
- 298
- 294
- 308
- 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? **

- 40
- 20
- 25
- 60
- 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? **

- 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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If the need arises, you can also seek **expert guidance regarding SEBI Grade A 2022 preparation**.

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**Also Read**

Eligibility & Syllabus: SEBI Grade A Legal 2022

Notification, Exam Pattern & Eligibility: SEBI Grade A 2022

General Awareness Preparation: SEBI Grade 2022 (Phase 1)