Reasoning PYQs

Logic followed here is,

(−89 × 2) + 2 = −178 + 2 = −176

(−176 × 2) + 2 = −352 + 2 = −350

(−350 × 2) + 2 = −698

(−698 × 2) + 2 = −1394

(−1394 × 2) + 2 = −2786

(−2786 × 2) + 2 = −5570

Hence, the correct answer is “Option 1”.

Pattern followed here is,

Hence, the correct answer is “Option 2”.

The logic followed in this pattern is a combination of alphabetical shifts and the use of alphabetical opposites.

For 'AZ' related to 'FA':

For 'JS' related to 'OH':

Applying the same logic to 'KM':

Thus, KM is related to PN.

Hence, the correct answer is 'Option 3'.

Pattern followed here is,

Hence, the correct answer is “Option 1”.

The logic followed in this pattern is a combination of alphabetical shifts through subtraction.

For 'HJ' related to 'FD':

For 'AC' related to 'YW':

Applying the same logic to 'EX':

Thus, EX is related to CR.

Hence, the correct answer is 'Option 3'.

The logic followed in this pattern is a series of alphabetical shifts applied to the positional values of the letters in each cluster.

For the set '(MB, OF, RK)':

For the set '(DL, FP, IU)':

Applying the same logic to '(MF, OJ, RO)':

Thus, the group '(MF, OJ, RO)' follows the same Logic/Rule/Relation.

Hence, the correct answer is 'Option 3'.

To determine if the year 2100 is a leap year:

The year 2100 is a century year because it is divisible by 100.

According to the rule, a century year must be divisible by 400 to be a leap year.

2100 ÷ 400 = 5.25

Since 2100 is not exactly divisible by 400, it is a common year.

Now, let us list the number of days in the required months for a common year:

Number of days in January = 31

Number of days in February = 28 (as it is a common year)

Number of days in March = 31

Number of days in April = 30

Sum of the number of days = 31 + 28 + 31 + 30

Total sum = 120 days

Hence, the total number of days in the month of January, February, March and April in the year 2100 is 120.

Hence, the correct answer is “Option 3”.

We need to find the day on 1^st January 1906. We first calculate the odd days up to 31^st December 1905.

Breakdown of 1905 years: 1600 years + 300 years + 5 years.

1. Odd days in 1600 years = 0 (Multiple of 400)
2. Odd days in 300 years = 1
3. Odd days in 5 years (from 1901 to 1905):

Number of leap years in 5 years = 5/4 = 1 (which is the year 1904).

Number of ordinary years = 5 − 1 = 4.

Odd days = (1 × 2) + (4 × 1) = 2 + 4 = 6 odd days. 4. Total odd days until 31^st December 1905 = 0 + 1 + 6 = 7 ≡ 0 odd days. 5. Now, add the days for the year 1906 up to 1^st January:

Total odd days = 0 (from previous years) + 1 (for 1^st Jan) = 1 odd day.

Day code: 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday.

Since the total number of odd days is 1, the day was Monday.

Hence, the correct answer is Monday.

Hence, the correct answer is “Option 2”.

Given that there are 5 cars (J, K, L, M and N) with different speeds.

From the clues, we can deduce the following relationships:

1. Speed of J is more than K.

→ J > K

2. Speed of K is the second lowest.

→ _ > _ > _ > K > _

3. Speed of L is less than N but more than M.

→ N > L > M

4. Speed of J is less than only one car.

→ J is the second fastest.

Combining all these facts:

K is second lowest, so only one car is slower than K. Since M is slower than L, M must be the slowest.

Thus, the remaining order becomes:

→ N > J > L > K > M

Thus, the only car whose speed is more than J is N.

Hence, the correct answer is "Option 1".

7 boys B1, B2, B3, B4, B5, B6 and B7 are sitting around a circular table facing opposite to the centre (not necessarily in the same order).

1. B3 is not next to B7.
2. B4 is next to both B1 and B7.
3. B6 is second to the left of B7.
4. B5 is second to the left of B6. 

B5 is immediate left of B3.

Hence, the correct answer is “Option 4”.

Pattern followed here is,

Similarly,

Hence, the correct answer is “Option 4”.

Pattern followed here is,

Similarly,

Hence, the correct answer is “Option 2”.

First cluster pair: BG12 : EJ59

Second cluster pair: PQ48 : ST95

Third cluster pair: LM33 : ?

Hence, the correct answer is 'Option 3'.

 

For, AK40 : EE100

For, DL80 : HF200

Similarly,

For, BM60 : ?  

Hence, the correct answer is "Option 2".

The logic followed here is:

Logic: Second Number = (First Number ÷ 3) − 2

For the second pair 777 : 257

⇒ (777 ÷ 3) − 2

⇒ 259 − 2 = 257.

For the third pair 1812 : 602

⇒ (1812 ÷ 3) − 2

⇒ 604 − 2 = 602.

Applying the same logic to the first pair

2436 : ?

⇒ (2436 ÷ 3) − 2

⇒ 812 − 2 = 810.

Hence, the correct answer is 'Option 4'.

WEES is coded as ZJCO →

EACH is coded as HFAD →

Similarly, let us find the code for PREY:

Hence, the correct answer is “Option 3”.

The time duration from 6 AM to 4 PM on the same day is calculated as follows:

From 6 AM to 12 Noon = 6 hours

From 12 Noon to 4 PM = 4 hours

Total time elapsed = 6 + 4 = 10 hours

The clock gains 2 seconds every 5 minutes.

Gain in 1 minute = 2/5 seconds

Gain in 60 minutes (1 hour) = (2/5) × 60 = 24 seconds

Total gain in 10 hours = 10 × 24 = 240 seconds

To convert the total gain from seconds into minutes:

Total gain = 240 ÷ 60 = 4 minutes

The clock shows the correct time at 6 AM. At 4 PM, the actual time is 4:00 PM, but the clock will have gained 4 minutes.

Time shown by the clock = 4:00 PM + 4 minutes = 4:04 PM

Hence, the correct answer is “Option 1”.

Pattern followed here is,

Hence, the correct answer is “Option 3”.

Pattern followed here is,

Hence, the correct answer is “Option 3”.

Given that the Son's age after 11 years from the present is 16 years.

Son's present age (S) + 11 = 16

S = 16 − 11 = 5 years

The problem states that the Father's age 15 years ago was 6 times the Son's age 3 years ago.

F − 15 = 6 × (S − 3)

Substitute the value of S = 5 into the equation:

F − 15 = 6 × (5 − 3)

F − 15 = 6 × 2

F − 15 = 12

F = 12 + 15 = 27 years

Ratio of the present ages of the Father and Son = 27 : 5

Hence, the correct answer is “Option 3”.