Placement Aptitude Readiness Test | Check Your Percentile

This mock test is designed to mimic placement aptitude tests for engineering students. Attempt the test, check your ranking, and invite your friends to join as well.

Section 1: Quant

Section 2: Reasoning

Section 3: Verbal

Leaderboard

Total Participants on LeaderBoard: 586

Rank	Name	College	Quantitative Aptitude	Logical Reasoning	Verbal Ability	Final Score
🥇	abc	abc	10	7	10	27
🥈	MS	NM	10	7	10	27
🥉	Rajasi P	PTVA's Institute of Management	10	7	10	27
4	Vidish	IIT Roorkee	10	6	10	26
5	Ijika	Bhu	9	7	10	26
6	Franklin	ACGCET	10	8	8	26
7	Kamakshi sayam	PTVA	8	8	10	26
8	Savi Band	PTVA’s Institute of Management	10	7	9	26
9	Satyajeet Kumar	NIT Rourkela	10	5	10	25
10	Yash Padmakar	Ptva’s institute of management	8	8	9	25

How much percentile is safe for Campus Placement Tests?

Generally, 20–50% of students qualify placement aptitude tests, depending on the company, college type, and available vacancies.

Percentile	Verdict
80+ (top 20 percentiel students)	You are likely to easily qualify most aptitude rounds.
50 – 80+ (top 20-50 percentile students)	You may qualify easier aptitude tests, but will face strong competition in tougher rounds like TCS, ZS, Mu Sigma, etc.
0-50	You will need to improve your aptitude skills to compete effectively.

Solutions

Please note that order might be different as questiosn ae randomized in the test

Quant Solutions

Sol : 1

Original Price = 200

Price after a 20% increase = 200+(20% of 200)=200+40=240

Price after a 10% discount on 240 = 240−(10% of 240)=240−24=216

Sol : 2

A=$5000 * (1+8/100)^2 = 5000*(1.08)^2$

= 5000×1.1664 = 5832$

Sol : 3

Based on given ration, let number of boys, girls and teachers be 33x, 53x and 2x.

Given, 2x=10, so x=5

Number of students = (33x + 53x)=86×5=430

Sol : 4

Loss = CP – SP = 5400 – 4500 = 900

Loss % = 900/ 5400*100 = 16.67%

Sol: 5

Subtracting the two equations:

$x+y – (x−y)=12 – 4$

$2y = 8$

$y = 4$

Sol : 6

Let the five consecutive numbers be x−2, x−1, x, x+1 and x+2

Their average = x=15

Smallest number = x−2=13; and largest = x+2=17

Sum of smallest and largest = 13+17 = 30

Sol: 7

Convert the speed into km/hr first

Speed = 30m/s = 30*3600/1000 = 108 km/hr

Time = 240/108 = 2.22 hour

Sol : 8

Total work = 6 men×8 days=48 man-days

For 4 men, days required = 48 /4 =12

Sol : 9

Use direct formula of area of circle = $\pi*r^2$

Sol: 10

Number of arrangements = 4!

Reasoning Solutions

Sol : 1

Z (26), Y (25), W (23), T (20), P (16)

The pattern is: -1, -2, -3, -4

Following the pattern, the next term will be P−5=M

Sol : 2

Every letter is increased by 1

Sol : 3

The man is the son of Sarah’s mother or father, which makes him Sarah’s brother.

Sol : 4

The problem states a clear requirement of 5 years of experience. Although Jane has the required qualification, she lacks the experience, so based on the strict criteria, she should not be selected.

Sol : 5

The statement implies that planning aids in success, but it doesn’t imply exclusivity. Hence, assumption B is correct.

Sol : 6

U is second from the right, so position 5 is U.Since Q is between P and R, and S is to the left of T, the order is:

P – Q – R and S – T – U

With U in the fifth position, R must be on the right end (position 6).

Sol : 7

2024 is a leap year, so it has 366 days.

From February 15, 2024, to February 15, 2025, there are 366 days.

366 days = 52 weeks + 2 extra day. Thus, the day will shift by 2 days

Sol : 8

Starting at south, turns go as follows: