JIPMAT 2021 Question Paper

Let the speed be "v" kmph , and time taken be 't' hours to cover distance of D kms.

D = v(t)

1. If the train moved 4 kmph faster, it would take 30 minutes less :  D = (v + 4)(t - 0.5)
2. If it moved 2 kmph slower, it would have taken 20 minutes more : D = (v - 2)(t + 0.33)

     Now, D = vt + 4t -v/2 - 2
By substituting D = Vt in the above equation : V = 8t - 4

Similarly,  D = Vt -2t + 0.33(V) - 0.66 
By substituting D = Vt in the just above equation : 2t = 0.33(v) - 0.66 
                                                                        :  V = 2 + 6t

Therefore, (8t - 4) = (2 + 6t) 
Hence, t = 3 hours 
So, V = 2 + 6(3) = 20 kmph

Therefore, Distance (D) = V(t) = 20(3) = 60 kms.

Let the race course of length be "L" . So lets assume time taken by A to finish it be "t" seconds . 
                              $$t=\dfrac{\ L}{V_a}$$

$$V_b(t)=V_b[\frac{\ L}{V_a}]=L-15$$ ... equation 1
$$V_c(t)=V_c[\frac{\ L}{V_a}]=L-29$$ .....equation 2

In race between B & C for length of L, B wins by 15 m : 
                                     $$V_b(t_2)=L$$...........equation 3
                                    $$V_c(t_2)=L-15$$.........equation 4 

Dividing equation 1 by equation 2 : $$\dfrac{\ V_b}{V_c}=\dfrac{\ L-15}{L-29}$$ 

Dividing equation 3 by equation 4 : $$\dfrac{\ V_b}{V_c}=\dfrac{\ L}{L-15}$$ 

         So, $$\dfrac{\ L-15}{L-29}$$ = $$\dfrac{\ L}{L-15}$$

         it implies ,  $$(L-15)^2=L^2-29L$$
                           L = 225.

Hence, option C.

$$2^{x}=3^{y}=6^{-z}$$ = K

X = $$\log_2k$$
Y = $$\log_3k$$
Z = -($$\log_6k$$)

Hence, $$\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$ = $$\log_k2$$ + $$\log_k3$$ - $$\log_k6$$
                                          = $$\log_k6$$ - $$\log_k6$$
                                          = 0.

The given series is like:

$$19=17+2^1$$

$$15=19-2^2$$

$$23=15+2^3$$

$$7=23-2^4$$

$$39=7+2^5$$

So, the next number will be $$39-2^6=39-64=-25$$

So, option B is the correct answer.

The logic for this question is like:

M(13)+6 --> S(19)

O(15)+6 --> U(21)

Q(17)+6 --> W(23)

Again,

S(19)+6 --> Y(25)

U(21)+ 6 --> A(27 or 26+1 i.e. 1 after a complete cycle)

W(23)+6 --> C(29 or 26+3 i.e. 3 after a complete cycle)

So, the next word will be like:

Y(25)+6 --> E(31 or 26+5 i.e. 5 after a complete cycle)

A(1)+6 --> G(7)

C(3)+6 --> I(9)

So, the correct word is EGI, option B.

This question is solved by hit and trial method.

We can put the four letters present in the four options in four different blanks and check which option gives a logical answer.

The correct answer choice is option D.

This is because upon putting the letters present in option D, in the blank, the sequence is: a(b)c(c)ba(b)ca(a)cb

Each triplet has each of the three letters a,b and c.

So, option D is the correct answer.

210=14*15=14*(14+1)

380=19*20=19*(19+1)

And 19=14+5

Now, 182=13*14

So, the number which holds the same relation with 182 as 210 and 380 holds is = (13+5)*(13+5+1) = 18*19 = 342

So, option D is the correct answer.

The correct answer is Sift. A drill functions as a tool for the action of boring, while a sieve acts as a tool for the process of sifting.

Drill : Bore

A drill serves as the tool, and boring is the action it executes to form a hole.

Sieve : Sift

Likewise, a sieve is the tool, and sifting is the action it carries out to separate particles.

December = 12th month of year

February = 2nd month of year

March = 3rd month of year

November = 11th month of year

Now, 2,3 and 11 are all prime numbers.

12 is not a prime number.

So, December, 2021 is the odd one out.

Option A is the correct answer choice.

The big cube is cut into 36 small cubes in total. Out of these, 32 are unit cubes and 4 are bigger blocks (each made of 2×2×2=8 unit cubes).

Now, the two outer layers (top and bottom) remain as unit cubes. That’s why we have 32 single-unit cubes there.

Now, these two outer layers are painted red (opposite faces).

We know that a cube has exactly one side coloured if it lies on a face but not on any edge or corner (because edges and corners show 2 or 3 faces).  On each outer red face, the middle 2×2 portion has only the red face painted. That gives four cubes on the top face and four cubes on the bottom face.

So, total cubes with exactly one painted face = 4 + 4 = 8.