JIPMAT 2023 Question Paper

1. D ran faster than E but slower than B.

So, the order here is: B > D > E

2. C ran slower than E but was not the slowest

So, C is slower than E, but not last.

So, the order should be B > D > E > C, where there should be someone slower than C. The only person left is A. So, A should be the slowest person. 

B > D > E > C > A

Hence, we can say B is the fastest. 

The relation in BJMS : EMPV is as follows:

B(2) → E(5): 5 - 2 = +3

J(10) → M(13): 13 - 10 = +3

M(13) → P(16): 16 - 13 = +3

S(19) → V(22): 22 - 19 = +3

Applying the same pattern to AWPL:

A(1) + 3 = 4 → D

W(23) + 3 = 26 → Z

P(16) + 3 = 19 → S

L(12) + 3 = 15 → O

Hence, the answer is DZSO.

Week I: 28.3 (Buses) + 15.4 (Trucks) + 40.2 (Cars) = 83.9 (in hundreds)

Week II: 24.5 (Buses) + 12.3 (Trucks) + 36.1 (Cars) = 72.9 (in hundreds)

Week III: 20.4 (Buses) + 14.2 (Trucks) + 42.9 (Cars) = 77.5 (in hundreds)

Week IV: 24.4 (Buses) + 24.4 (Trucks) + 39.8 (Cars) = 88.6 (in hundreds)

Ranking:

1st Highest: Week IV (88.6)

2nd Highest: Week I (83.9)

3rd Highest: Week III (77.5)

4th Highest: Week II (72.9)

Sum of Buses per Week: 

Week I: 28.3 (in hundreds)

Week II: 24.5 (in hundreds)

Week III: 20.4 (in hundreds)

Week IV: 24.4 (in hundreds)

Total: 28.3+24.5+20.4+24.4=97.6 (in hundreds)

Average = Total Buses / Number of Weeks

Average = 97.6/4 = 24.4 (in hundreds)

24.4 × 100 = 2440

In the code, each letter’s position in the alphabet is increased by 3. 

For example, C is the 3rd letter, and 3 + 3 = 6; A is 1st, so 1 + 3 = 4; T is 20th, so 20 + 3 = 23. So the code becomes 6423.

Applying the same rule to LION:

L (12) + 3 = 15

I (9) + 3 = 12

O (15) + 3 = 18

N (14) + 3 = 17

So, the code will be 15121817

For a year’s calendar to be the same, it must start on the same day and have the same leap year status.

2008 is a leap year with 366 days. Since 366 days = 52 weeks + 2 days, the next year (2009) will start 2 days later than 2008.

2008 starts on Monday

So, 2009 starts on Monday + 2 days = Wednesday

2009 (non-leap): starts Wednesday

2010 (non-leap): starts Wednesday + 1 day = Thursday

2011 (non-leap): starts Thursday + 1 day = Friday

2012 (leap): starts Friday + 1 day = Saturday

2013 (non-leap): starts Saturday + 2 days (because 2012 is leap) = Monday

2014 (non-leap): starts Monday + 1 day = Tuesday

2015 (non-leap): starts Tuesday + 1 day = Wednesday

Since 2015 starts on Wednesday like 2009 and both are non-leap years, their calendars will be the same.

Highway length = 120 km

Car A: Moves 30 km on the highway, then 15 km right (off-road), 30 km left (parallel), then returns to the highway.

So, a total of 60 km from point X

Car B: Moves 35 km from point Y, which means 120 - 35 = 85 km from point X.

So, Distance between them = 85 - 60 = 25 km

We are told everyone is facing inside. And let's fix the spot of U. U was sitting opposite of P , so we can use this to determine the relative positions of the others. 

P was in the immediate left of R, which means R is to the immediate right of P. Also, S was sitting to the immediate right of U. So putting these in , we have,

 Now, T was sitting exactly between P and S, which means the only remaining seat is for Q. Completing the diagram, we have the following: 

Hence, we can say Q was sitting between U and R, which makes option C the correct choice. 

We can look at the difference:

11−10 = 1

13−11 = 2

17−13 = 4

25−17 = 8

31−25 = 6

33−31 = 2

Those differences follow the pattern of powers of 2 reduced modulo 10:

$$1(2⁰),2(2^1),4(2^2),8(2^3),16→6(2⁴mod10),32→2(2⁵mod10)$$

Next is 64 →4 (2⁶ mod 10)

So the next difference should be 4. Hence the next term = 33 + 4 = 37.

EIWNTR → WINTER

LCUOD → CLOUD

UMRSSME → SUMMER

PIGRSN → SPRING

Three of these — WINTER, SUMMER, and SPRING — are seasons. CLOUD is not a season — it's a weather phenomenon.

Hence, the odd one out is CLOUD.