Campus Placement Aptitude Test (2026) | Check All India Rank

Evaluate your placement aptitude readiness and performance at all India level

Instructions

Enter your details correctly to receive accurate results.
Ensure a stable internet connection. Do not refresh the page during the test.
Download result and certificates once you have attempted all three sections.

Section 1: Quant

Section 2: Reasoning

Section 3: Verbal

Leaderboard

Total Participants on LeaderBoard: 601

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: