SNAP 2022 Question Paper

H - 8
D - 8/2=4
R - 18
I - 18/2=9
Using the same logic,
Z - 26
M - 26/2 = 13
Hence, option A is the right answer.

The logic followed here is,

S + 2 = U, U + 2 = W, W + 2 = Y. All the letters are added by 2.

B - 2 = Z, Z -2 = X, X - 2 = V. All the letters are subtracted by 2.

Hence, the correct option is C.

The place values of B and D are $$2$$ and $$2^2=4$$
The place values of E and Y are $$5$$ and $$5^2=25$$
The place values of C and I are $$3$$ and $$3^2=9$$
But the place values of D and U are $$4$$ and $$21$$. Hence, Option B is correct answer.

All Triangles are Rectangle is not always true. All Triangle being Rectangle is a possibility. Hence, only Conclusion I follows. Option A is correct.

To determine the day of the week when Sofu and Kitto finally met, we need to calculate the total number of days between their initial meeting and the day they were supposed to meet.

Sofu met Kitto 25 days ago. Let's denote this day as Day 1.

Kitto informed Sofu that they would meet after 10 days from today, which is a Wednesday. Let's denote this day as Day 2.

To find the day of the week when they finally met, we need to add the number of days between Day 1 and Day 2 and determine the resulting day of the week.

The number of days between Day 1 and Day 2 is calculated as follows:

25 days + 10 days = 35 days.

Since there are 7 days in a week, we divide 35 by 7 to determine the remainder.

35 days ÷ 7 = 5 weeks with no remainder.

This means that after 5 weeks, the day of the week will remain the same.

Since Day 1 was a Wednesday, and 5 weeks have passed, the day of the week when Sofu and Kitto finally met is also a Wednesday.

Therefore, they met on a Wednesday.

Let's assume A is a knight, which implies A always tells the truth, implying B is a knight, which is not possible ( SInce only one knight is there).

Now if A is a knave, which implies B is surely not Knight, which implies B is a joker => C is the knight ( which is impossible since C's statement is definitely false.)

Therefore, A must be a joker, which implies C is a knave and B is a knight.

The correct option is D.

When the time of the clock is 3 o'clock, the hour hand is at 3, and the minute hand is at 12, making an angle of 90 degrees. We know that the relative velocity of the minute hand w.r.t the hour hand is 5.5 degrees per minute.

Hence, it will take (90/5.5) minutes to make the angle between the hands 0 degrees.(the hands will be together)

(90/5.5)= (180/11) minutes past 3. => $$16\frac{4}{11}$$ minutes past 3.

The correct option is C.

The hour hand covers 30 degrees in 1 hour or 0.5 degrees / min

The min hand covers 360 degrees in 1 hour of 6 degrees/ min

Hence, we can say that the the minute hand gains 6 - 0.5 = 5.5 degrees on hour hand, every minute.

At 8 pm the the hour hand will be ahead of minute hand by 30*8 = 240 degrees.

Hence, the minute hand will have to cover 60 degrees more w.r.t the hour hand for them to point in opposite direction/ for the angle between them to be 180 degrees. 

let x be the number of minutes taken by the minute hand to do the same. 

5.5x = 60 

x = 60/5.5 = 120/11 = 10 10/11 

Given, R sits second to the right of S.

The possible cases are (S, R) are (4, 6), (3, 5), (2, 4) and (1, 3)
If (S, R) sit at (4, 6), U has to sit at 5 as U and R are neighbors. Since the number of people between S and U is same as the number of people between R and T, T has to sit at 3. But T sits at one of the ends, which is not possible.

If (S,R) sit at (3, 5), U can sit at 6 or 4. If U sits at 6, T should sit at 2 as the number of people between S and U is same as the number of people between R and T. This is not possible as T should sit at one of the ends. If U sits at 4, T has to sit at 6. This is not possible as P and Q should not be neighbors.

If (S, R) sit at (2, 4), U can sit 3 or 5. If U sits at 3, T has to sit at 5, which is not possible as T sits at one end. If U sits at 5, T has to sit at 1, but T and S are not neighbors, hence this case is not possible.

If (S, R) sit at (1, 3), U can sit at 2 or 4. U cannot sit at 2 because then T has to sit at 4, which is not possible. If U sits at 4, T has to sit at 6. P, Q can sit at either 2 or 5 in any order.
Thus, 2 cases are possible:

Case 1:

Case 2:

Option c is step 6 by given description of the machine.