SBI PO Mains Memory Based Paper (Held On: 5th May 2025) (Previous Year Paper)

Given: Six different cars, P, Q, R, S, T, and U, were used in a logistics event. Each car travelled a different distance, i.e., 14 km, 18 km, 24 km, 28 km, 35 km, and 42 km, but not necessarily in the same order. Each car had a different number of helpers, ranging from 4 to 10.

1. Car P has a number of helpers that is a multiple of 5.  2. Car T travelled a distance which is a multiple of 7. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

 

3. Car U travelled more distance than Car R.

Car U > Car R

4. Car T travelled less distance than Car R. 

Car U > Car R > Car T

Now, we can eliminate Case 2 and Case 3.

 

5. Exactly one car travelled a distance that lies between the distances travelled by Car S and Car P. (i.e., Car S travelled 24 km.) 6. Car S has one helper more than Car T. 7. Car S travelled an even-numbered distance.  

 

8. Car S has more helpers than Car U. 9. Car Q travelled less distance than Car R. 

Car U > Car R > Car Q > Car T

i.e., Car U > Car P > Car R > Car S > Car Q > Car T

10. Car P has fewer helpers than Car U. 

This is our final arrangement,

 

Sum = 28 + 9 = 37

Hence, the correct answer is 37.

Given: Six different cars, P, Q, R, S, T, and U, were used in a logistics event. Each car travelled a different distance, i.e., 14 km, 18 km, 24 km, 28 km, 35 km, and 42 km, but not necessarily in the same order. Each car had a different number of helpers, ranging from 4 to 10.

1. Car P has a number of helpers that is a multiple of 5.  2. Car T travelled a distance which is a multiple of 7. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

 

3. Car U travelled more distance than Car R.

Car U > Car R

4. Car T travelled less distance than Car R. 

Car U > Car R > Car T

Now, we can eliminate Case 2 and Case 3.

 

5. Exactly one car travelled a distance that lies between the distances travelled by Car S and Car P. (i.e., Car S travelled 24 km.) 6. Car S has one helper more than Car T. 7. Car S travelled an even-numbered distance.  

 

8. Car S has more helpers than Car U. 9. Car Q travelled less distance than Car R. 

Car U > Car R > Car Q > Car T

i.e., Car U > Car P > Car R > Car S > Car Q > Car T

10. Car P has fewer helpers than Car U. 

This is our final arrangement,

 

Hence, the correct answer is Car T travelled the least distance.

Given: Six different cars, P, Q, R, S, T, and U, were used in a logistics event. Each car travelled a different distance, i.e., 14 km, 18 km, 24 km, 28 km, 35 km, and 42 km, but not necessarily in the same order. Each car had a different number of helpers, ranging from 4 to 10.

1. Car P has a number of helpers that is a multiple of 5.  2. Car T travelled a distance which is a multiple of 7. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

 

3. Car U travelled more distance than Car R.

Car U > Car R

4. Car T travelled less distance than Car R. 

Car U > Car R > Car T

Now, we can eliminate Case 2 and Case 3.

 

5. Exactly one car travelled a distance that lies between the distances travelled by Car S and Car P. (i.e., Car S travelled 24 km.) 6. Car S has one helper more than Car T. 7. Car S travelled an even-numbered distance.  

 

8. Car S has more helpers than Car U. 9. Car Q travelled less distance than Car R. 

Car U > Car R > Car Q > Car T

i.e., Car U > Car P > Car R > Car S > Car Q > Car T

10. Car P has fewer helpers than Car U. 

This is our final arrangement,

 

Hence, the correct answer is U, Q.

Given: Six different cars, P, Q, R, S, T, and U, were used in a logistics event. Each car travelled a different distance, i.e., 14 km, 18 km, 24 km, 28 km, 35 km, and 42 km, but not necessarily in the same order. Each car had a different number of helpers, ranging from 4 to 10.

1. Car P has a number of helpers that is a multiple of 5.  2. Car T travelled a distance which is a multiple of 7. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

 

3. Car U travelled more distance than Car R.

Car U > Car R

4. Car T travelled less distance than Car R. 

Car U > Car R > Car T

Now, we can eliminate Case 2 and Case 3.

 

5. Exactly one car travelled a distance that lies between the distances travelled by Car S and Car P. (i.e., Car S travelled 24 km.) 6. Car S has one helper more than Car T. 7. Car S travelled an even-numbered distance.  

 

8. Car S has more helpers than Car U. 9. Car Q travelled less distance than Car R. 

Car U > Car R > Car Q > Car T

i.e., Car U > Car P > Car R > Car S > Car Q > Car T

10. Car P has fewer helpers than Car U. 

This is our final arrangement,

 

Hence, the correct answer is U - 42 km.

Given: Six different cars, P, Q, R, S, T, and U, were used in a logistics event. Each car travelled a different distance, i.e., 14 km, 18 km, 24 km, 28 km, 35 km, and 42 km, but not necessarily in the same order. Each car had a different number of helpers, ranging from 4 to 10.

1. Car P has a number of helpers that is a multiple of 5.  2. Car T travelled a distance which is a multiple of 7. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

 

3. Car U travelled more distance than Car R.

Car U > Car R

4. Car T travelled less distance than Car R. 

Car U > Car R > Car T

Now, we can eliminate Case 2 and Case 3.

 

5. Exactly one car travelled a distance that lies between the distances travelled by Car S and Car P. (i.e., Car S travelled 24 km.) 6. Car S has one helper more than Car T. 7. Car S travelled an even-numbered distance.  

 

8. Car S has more helpers than Car U. 9. Car Q travelled less distance than Car R. 

Car U > Car R > Car Q > Car T

i.e., Car U > Car P > Car R > Car S > Car Q > Car T

10. Car P has fewer helpers than Car U. 

This is our final arrangement,

 

Hence, the correct answer is P.

Given: Eight people A, C, F, H, J, L, N, and R are sitting around a circular table facing the centre. According to the English alphabetical order, there are at least two letters between the names of any two persons who sit adjacent to each other.

1. F sits third to the left of L. 2. One person sits between N and C, and C is an immediate neighbour of L. 

3. A and F are immediate neighbours.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

4. Two persons sit between J and H.
5. F and H are not immediate neighbours of each other. 

6. R sits second to the right of J. 

Now, we can eliminate Case 1.

This is our final arrangement,

Hence, the correct answer is F.

A*5B → A is 8 m north of B

A#2B → B is 4 m south of A

A%3B → A is 6 m east of B

A!4B → B is 6 m west of A

The decoded form is,

H!13R → R is 16 m west of H

C%6Q → C is 8 m east of Q

L*1P → L is 4 m north of P

P!4E → E is 6 m west of P

T%5P → T is 8 m east of P

W*8J → W is 10 m north of J

Q!12D → D is 14 m west of Q

D#7L → L is 10 m south of D

R#4T → T is 6 m south of R

H*2W → H is 4 m north of W

The possible diagram is,

Hence, the correct answer is South-east.

A*5B → A is 8 m north of B

A#2B → B is 4 m south of A

A%3B → A is 6 m east of B

A!4B → B is 6 m west of A

The decoded form is,

H!13R → R is 16 m west of H

C%6Q → C is 8 m east of Q

L*1P → L is 4 m north of P

P!4E → E is 6 m west of P

T%5P → T is 8 m east of P

W*8J → W is 10 m north of J

Q!12D → D is 14 m west of Q

D#7L → L is 10 m south of D

R#4T → T is 6 m south of R

H*2W → H is 4 m north of W

The possible diagram is,

The distance is = √(8² + 6²) = √(64 + 36) = √100 = 10 m

Hence, the correct answer is 10 m.

A*5B → A is 8 m north of B

A#2B → B is 4 m south of A

A%3B → A is 6 m east of B

A!4B → B is 6 m west of A

The decoded form is,

H!13R → R is 16 m west of H

C%6Q → C is 8 m east of Q

L*1P → L is 4 m north of P

P!4E → E is 6 m west of P

T%5P → T is 8 m east of P

W*8J → W is 10 m north of J

Q!12D → D is 14 m west of Q

D#7L → L is 10 m south of D

R#4T → T is 6 m south of R

H*2W → H is 4 m north of W

The possible diagram is,

Except for H - T, in all other pairs, the first point is in the south-west of the second point.

Hence, the correct answer is H - T.

A*5B → A is 8 m north of B

A#2B → B is 4 m south of A

A%3B → A is 6 m east of B

A!4B → B is 6 m west of A

The decoded form is,

H!13R → R is 16 m west of H

C%6Q → C is 8 m east of Q

L*1P → L is 4 m north of P

P!4E → E is 6 m west of P

T%5P → T is 8 m east of P

W*8J → W is 10 m north of J

Q!12D → D is 14 m west of Q

D#7L → L is 10 m south of D

R#4T → T is 6 m south of R

H*2W → H is 4 m north of W

The possible diagram is,

Hence, the correct answer is D.

Given: A certain number of persons sit in a row and face north. Each of them has a different item.

1. Eight persons sit between S and T. 2. S sits third to the right of Q. 3. The one who has Slippers sits second to the right of T.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

4. The one who has Slippers sits fourth to the right of the one who has Alarm Clock and sits at one of the extreme ends of the row. 

Now, we can eliminate Case 2.

 

5. Nine persons sit between the ones who have Tape and Alarm Clock.  6. Three persons sit between the ones who have Tape and Stapler.   7. P sits fifth from one of the extreme ends and has Stapler. 

Here, we have a subcase of Case 1, i.e., Case 1a.    

8. P sits immediate right of the one who has Glue.  9. V sits second to the left of P, who doesn’t have Envelope.  10. V has Wallet. 

 

11. Two persons sit between the ones who has Doormat and Glue.  12. One person sits between L and the person who has Doormat. 

Now, we can eliminate Case 2.

13.  As many persons sit between V and the one who has Doormat as between P and the one who has Shoes.  14. The one who has Shoes sits to the right of the one who has Glue and second to the left of O.  15. Three persons sit between L and the one who has Envelope. 16. M sits second to the right of the one who has Envelope. 17. The number of persons sitting between the one who has Slippers and L is two more than the number of persons sitting between V and N.  

This is our final arrangement,

Hence, the correct answer is M - Slippers.

Given: A certain number of persons sit in a row and face north. Each of them has a different item.

1. Eight persons sit between S and T. 2. S sits third to the right of Q. 3. The one who has Slippers sits second to the right of T.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

4. The one who has Slippers sits fourth to the right of the one who has Alarm Clock and sits at one of the extreme ends of the row. 

Now, we can eliminate Case 2.

5. Nine persons sit between the ones who have Tape and Alarm Clock.  6. Three persons sit between the ones who have Tape and Stapler.   7. P sits fifth from one of the extreme ends and has Stapler. 

Here, we have a subcase of Case 1, i.e., Case 1a.    

8. P sits immediate right of the one who has Glue.  9. V sits second to the left of P, who doesn’t have Envelope.  10. V has Wallet. 

 

11. Two persons sit between the ones who has Doormat and Glue.  12. One person sits between L and the person who has Doormat. 

Now, we can eliminate Case 2.

13.  As many persons sit between V and the one who has Doormat as between P and the one who has Shoes.  14. The one who has Shoes sits to the right of the one who has Glue and second to the left of O.  15. Three persons sit between L and the one who has Envelope. 16. M sits second to the right of the one who has Envelope. 17. The number of persons sitting between the one who has Slippers and L is two more than the number of persons sitting between V and N.  

This is our final arrangement,

Hence, the correct answer is Eighth to the right.

Given: A certain number of persons sit in a row and face north. Each of them has a different item.

1. Eight persons sit between S and T. 2. S sits third to the right of Q. 3. The one who has Slippers sits second to the right of T.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

4. The one who has Slippers sits fourth to the right of the one who has Alarm Clock and sits at one of the extreme ends of the row. 

Now, we can eliminate Case 2.

5. Nine persons sit between the ones who have Tape and Alarm Clock.  6. Three persons sit between the ones who have Tape and Stapler.   7. P sits fifth from one of the extreme ends and has Stapler. 

Here, we have a subcase of Case 1, i.e., Case 1a.    

8. P sits immediate right of the one who has Glue.  9. V sits second to the left of P, who doesn’t have Envelope.  10. V has Wallet. 

 

11. Two persons sit between the ones who has Doormat and Glue.  12. One person sits between L and the person who has Doormat. 

Now, we can eliminate Case 2.

13.  As many persons sit between V and the one who has Doormat as between P and the one who has Shoes.  14. The one who has Shoes sits to the right of the one who has Glue and second to the left of O.  15. Three persons sit between L and the one who has Envelope. 16. M sits second to the right of the one who has Envelope. 17. The number of persons sitting between the one who has Slippers and L is two more than the number of persons sitting between V and N.  

This is our final arrangement,

Sum = 3 + 9 = 12

Hence, the correct answer is 12.

Given: A certain number of persons sit in a row and face north. Each of them has a different item.

1. Eight persons sit between S and T. 2. S sits third to the right of Q. 3. The one who has Slippers sits second to the right of T.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

4. The one who has Slippers sits fourth to the right of the one who has Alarm Clock and sits at one of the extreme ends of the row. 

Now, we can eliminate Case 2.

5. Nine persons sit between the ones who have Tape and Alarm Clock.  6. Three persons sit between the ones who have Tape and Stapler.   7. P sits fifth from one of the extreme ends and has Stapler. 

Here, we have a subcase of Case 1, i.e., Case 1a.    

8. P sits immediate right of the one who has Glue.  9. V sits second to the left of P, who doesn’t have Envelope.  10. V has Wallet. 

11. Two persons sit between the ones who has Doormat and Glue.  12. One person sits between L and the person who has Doormat. 

Now, we can eliminate Case 2.

13.  As many persons sit between V and the one who has Doormat as between P and the one who has Shoes.  14. The one who has Shoes sits to the right of the one who has Glue and second to the left of O.  15. Three persons sit between L and the one who has Envelope. 16. M sits second to the right of the one who has Envelope. 17. The number of persons sitting between the one who has Slippers and L is two more than the number of persons sitting between V and N.  

This is our final arrangement,

Hence, the correct answer is N.

Given: A certain number of persons sit in a row and face north. Each of them has a different item.

1. Eight persons sit between S and T. 2. S sits third to the right of Q. 3. The one who has Slippers sits second to the right of T.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

4. The one who has Slippers sits fourth to the right of the one who has Alarm Clock and sits at one of the extreme ends of the row. 

Now, we can eliminate Case 2.

5. Nine persons sit between the ones who have Tape and Alarm Clock.  6. Three persons sit between the ones who have Tape and Stapler.   7. P sits fifth from one of the extreme ends and has Stapler. 

Here, we have a subcase of Case 1, i.e., Case 1a.    

8. P sits immediate right of the one who has Glue.  9. V sits second to the left of P, who doesn’t have Envelope.  10. V has Wallet. 

11. Two persons sit between the ones who has Doormat and Glue.  12. One person sits between L and the person who has Doormat. 

Now, we can eliminate Case 2.

13.  As many persons sit between V and the one who has Doormat as between P and the one who has Shoes.  14. The one who has Shoes sits to the right of the one who has Glue and second to the left of O.  15. Three persons sit between L and the one who has Envelope. 16. M sits second to the right of the one who has Envelope. 17. The number of persons sitting between the one who has Slippers and L is two more than the number of persons sitting between V and N.  

This is our final arrangement,

Hence, the correct answer is 15.

Option 1: Incorrect

The passage does not indicate lack of smartphone access as a core issue, since initial downloads were high.

Option 2: Correct

High default rates and regulatory warnings strongly suggest weak credit profiling and risk assessment, which is critical in micro-lending.

Option 3: Incorrect

There is no mention of competitors affecting the company’s performance.

Option 4: Incorrect

While marketing was aggressive, there is no evidence that expenses alone caused failure.

Option 5: Incorrect

Although multiple issues existed, the primary and most direct cause highlighted is loan default and regulatory pressure.

Hence, the correct answer is Weak risk assessment and credit profiling led to high loan defaults.

Option 1: Incorrect

The platform complemented coaching institutes rather than replacing them.

Option 2: Incorrect

While personalization helped, this option alone does not fully explain partnerships and scalability.

Option 3: Incorrect

Celebrity endorsements are not mentioned in the passage.

Option 4: Incorrect

There is no reference to regulatory compulsion.

Option 5: Correct

Improved outcomes, personalization, retention, and partnerships together explain rapid adoption and investor interest.

Hence, the correct answer is A combination of personalization, performance improvement, and institutional partnerships.

The government has introduced a policy to make rainwater harvesting compulsory in new buildings to tackle declining groundwater levels and urban water scarcity.

I. Groundwater depletion has become a serious concern for urban planners.: Correct (It is valid because introducing such a policy shows that groundwater depletion is a serious issue for urban planning and governance.)

II. The government believes policy intervention can influence sustainable construction practices.: Correct (It is also valid since the government is using a policy mandate, which implies it believes regulations can encourage sustainable construction practices.)

III. Existing buildings will immediately face penalties for non-compliance.: Incorrect (It is invalid because the passage talks only about new buildings. There is no mention of penalties or requirements for existing buildings.)

Only Statements I and II are valid assumptions.

Hence, the correct answer is Only I and II.

The passage states that doctors in metro hospitals adapted quickly, but staff in smaller towns faced difficulties, which led to delays and workload complaints. This clearly suggests that the problem is not with the system itself, but with how it was adopted and supported in different locations.

Option 1: Incorrect

If the software were faulty, metro hospitals would also have faced problems. Since they adapted quickly, this option is unlikely.

Option 2: Incorrect

This could be partially true, but the passage does not explicitly mention training or literacy. On its own, it is incomplete.

Option 3: Incorrect

There is no mention of patient resistance in the passage.

Option 4: Incorrect

“Completely unavailable” is an extreme statement and not supported by the passage.

Option 5: Correct

This is the most plausible and balanced explanation: Staff in smaller towns may be less familiar or comfortable with new digital systems. They may not have received enough guidance, training, or technical support. Such resistance and lack of support can logically lead to delays and increased complaints.

Hence, the correct answer is Resistance to change combined with inadequate implementation support.

The passage highlights two key points: Sales of packaged foods declined in urban markets, and Sales of organic and locally sourced alternatives increased, driven by concerns about health, ingredient transparency, and sustainability. This clearly indicates a change in consumer preference toward more mindful and value-based purchasing.

Option 1: Correct

This is a logical inference. Packaged foods are usually associated with convenience, while organic and locally sourced products align with health and sustainability. The survey results directly support this shift in mindset.

Option 2: Incorrect

The passage mentions a decline, not a complete disappearance. This is an extreme conclusion and cannot be inferred.

Option 3: Incorrect

There is no information about pricing in the passage. Consumers may still consider price along with health and sustainability.

Option 4: Incorrect

The passage does not compare prices. In reality, organic products are often more expensive, so this cannot be inferred.

Option 5: Incorrect

The change is attributed to consumer concerns, not government intervention. No policy action is mentioned.

Hence, the correct answer is Consumers are shifting from convenience to conscious consumption.

Logic:

Step I: Arrange all the words in reverse dictionary order from left to right.

Step II: Break the 6-letter word in two words, of 3-letter each following the given condition-

• If the word starts with a vowel, then the first 3-letter word will form with the ‘even positioned letters from left’ of the given word. And, the second 3-letter word will form with the ‘rest/ odd-positioned letters from left’.

• If the word starts with a consonant, then the first 3-letter word will form with the ‘odd positioned digits from left end’ of the given word. And, the second 3-letter word will form with the ‘rest/ even-positioned letters from left’.

Step III: Arrange the words in dictionary order from left to right and change the middle letter of each 3-letter word to its immediately succeeding letter.

Step IV: If the words start with vowel, then arrange the letters within the word in alphabetical order (from left) and if the word starts with consonant, then arrange the letters within the word in reverse alphabetical order (from left).

Step V: Arrange the words in dictionary order from the left end and change the last letter of the word with the immediately succeeding letter.

Input: ANCHOR FROZEN TABLET LANTER ORBITS

Step I: TABLET ORBITS LANTER FROZEN ANCHOR  

Step II: TBE ALT RIS OBT LNE ATR FOE RZN NHR ACO

Step III: ADO AMT AUR FPE LOE NIR OCT RJS RAN TCE

Step IV: ADO AMT ARU PFE OLE RNI COT SRJ RNA TEC

Step V: ADP AMU ARV COU OLF PFF RNB RNJ SRK TED

This is our final step.

Hence, the correct answer is COU.

Logic:

Step I: Arrange all the words in reverse dictionary order from left to right.

Step II: Break the 6-letter word in two words, of 3-letter each following the given condition-

• If the word starts with a vowel, then the first 3-letter word will form with the ‘even positioned letters from left’ of the given word. And, the second 3-letter word will form with the ‘rest/ odd-positioned letters from left’.

• If the word starts with a consonant, then the first 3-letter word will form with the ‘odd positioned digits from left end’ of the given word. And, the second 3-letter word will form with the ‘rest/ even-positioned letters from left’.

Step III: Arrange the words in dictionary order from left to right and change the middle letter of each 3-letter word to its immediately succeeding letter.

Step IV: If the words start with vowel, then arrange the letters within the word in alphabetical order (from left) and if the word starts with consonant, then arrange the letters within the word in reverse alphabetical order (from left).

Step V: Arrange the words in dictionary order from the left end and change the last letter of the word with the immediately succeeding letter.

Input: ANCHOR FROZEN TABLET LANTER ORBITS

Step I: TABLET ORBITS LANTER FROZEN ANCHOR  

Step II: TBE ALT RIS OBT LNE ATR FOE RZN NHR ACO

Step III: ADO AMT AUR FPE LOE NIR OCT RJS RAN TCE

Step IV: ADO AMT ARU PFE OLE RNI COT SRJ RNA TEC

Step V: ADP AMU ARV COU OLF PFF RNB RNJ SRK TED

This is our final step.

Hence, the correct answer is PFE OLE RNI.

Logic:

Step I: Arrange all the words in reverse dictionary order from left to right.

Step II: Break the 6-letter word in two words, of 3-letter each following the given condition-

• If the word starts with a vowel, then the first 3-letter word will form with the ‘even positioned letters from left’ of the given word. And, the second 3-letter word will form with the ‘rest/ odd-positioned letters from left’.

• If the word starts with a consonant, then the first 3-letter word will form with the ‘odd positioned digits from left end’ of the given word. And, the second 3-letter word will form with the ‘rest/ even-positioned letters from left’.

Step III: Arrange the words in dictionary order from left to right and change the middle letter of each 3-letter word to its immediately succeeding letter.

Step IV: If the words start with vowel, then arrange the letters within the word in alphabetical order (from left) and if the word starts with consonant, then arrange the letters within the word in reverse alphabetical order (from left).

Step V: Arrange the words in dictionary order from the left end and change the last letter of the word with the immediately succeeding letter.

Input: ANCHOR FROZEN TABLET LANTER ORBITS

Step I: TABLET ORBITS LANTER FROZEN ANCHOR  

Step II: TBE ALT RIS OBT LNE ATR FOE RZN NHR ACO

Step III: ADO AMT AUR FPE LOE NIR OCT RJS RAN TCE

Step IV: ADO AMT ARU PFE OLE RNI COT SRJ RNA TEC

Step V: ADP AMU ARV COU OLF PFF RNB RNJ SRK TED

This is our final step.

Hence, the correct answer is Fifth to the left.

Logic:

Step I: Arrange all the words in reverse dictionary order from left to right.

Step II: Break the 6-letter word in two words, of 3-letter each following the given condition-

• If the word starts with a vowel, then the first 3-letter word will form with the ‘even positioned letters from left’ of the given word. And, the second 3-letter word will form with the ‘rest/ odd-positioned letters from left’.

• If the word starts with a consonant, then the first 3-letter word will form with the ‘odd positioned digits from left end’ of the given word. And, the second 3-letter word will form with the ‘rest/ even-positioned letters from left’.

Step III: Arrange the words in dictionary order from left to right and change the middle letter of each 3-letter word to its immediately succeeding letter.

Step IV: If the words start with vowel, then arrange the letters within the word in alphabetical order (from left) and if the word starts with consonant, then arrange the letters within the word in reverse alphabetical order (from left).

Step V: Arrange the words in dictionary order from the left end and change the last letter of the word with the immediately succeeding letter.

Input: ANCHOR FROZEN TABLET LANTER ORBITS

Step I: TABLET ORBITS LANTER FROZEN ANCHOR  

Step II: TBE ALT RIS OBT LNE ATR FOE RZN NHR ACO

Step III: ADO AMT AUR FPE LOE NIR OCT RJS RAN TCE

Step IV: ADO AMT ARU PFE OLE RNI COT SRJ RNA TEC

Step V: ADP AMU ARV COU OLF PFF RNB RNJ SRK TED

This is our final step.

Hence, the correct answer is OCT.

Logic:

Step I: Arrange all the words in reverse dictionary order from left to right.

Step II: Break the 6-letter word in two words, of 3-letter each following the given condition-

• If the word starts with a vowel, then the first 3-letter word will form with the ‘even positioned letters from left’ of the given word. And, the second 3-letter word will form with the ‘rest/ odd-positioned letters from left’.

• If the word starts with a consonant, then the first 3-letter word will form with the ‘odd positioned digits from left end’ of the given word. And, the second 3-letter word will form with the ‘rest/ even-positioned letters from left’.

Step III: Arrange the words in dictionary order from left to right and change the middle letter of each 3-letter word to its immediately succeeding letter.

Step IV: If the words start with vowel, then arrange the letters within the word in alphabetical order (from left) and if the word starts with consonant, then arrange the letters within the word in reverse alphabetical order (from left).

Step V: Arrange the words in dictionary order from the left end and change the last letter of the word with the immediately succeeding letter.

Input: ANCHOR FROZEN TABLET LANTER ORBITS

Step I: TABLET ORBITS LANTER FROZEN ANCHOR  

Step II: TBE ALT RIS OBT LNE ATR FOE RZN NHR ACO

Step III: ADO AMT AUR FPE LOE NIR OCT RJS RAN TCE

Step IV: ADO AMT ARU PFE OLE RNI COT SRJ RNA TEC

Step V: ADP AMU ARV COU OLF PFF RNB RNJ SRK TED

This is our final step.

Hence, the correct answer is Three.

Statement I: Two persons visited between L and O. N visited two persons after O. P visited before J.

Statement II: K visited before M, and M visited two persons after O. L visited on Monday. N visited after M. P visited before N. J visited before O.

From I and II, 

 

It's not sufficient because the lines contradict each other.

Statement I: Two persons visited between L and O. N visited two persons after O. P visited before J.

Statement III: M visited either on Monday or on Sunday. K visited before P. As many persons visited before K as after N. P visited after L but before O.

From I and III,

 

Hence, the correct answer is Statements I and III together are sufficient to answer the question.

Statement I: In a certain language, "mountain bright sky cloud" is coded as “za tp lm rk”, and "forest cloud dark sky" is coded as “rk lm se tp”.

Sky - lm

Statement II: In a certain language, "river is clean path" is coded as “mu ke jo fh”, and "clean garden path stone" is coded as “ke va jo nf”.

clean/path - ke/jo

river/is - mu/fh

garden/stone - va/nf

Statement III: In a certain language, "mountain blue forest river" is coded as “za wi rk mu”, and "bright river night" is coded as “tp mu gx”.

river - mu

From I, II, and III,

Forest is bright = rk fh tp 

Hence, the correct answer is Statements I, II and III together are necessary to answer the question.

Statement I: The Novel carton is placed two positions below the carton packed in the Green box. The Novel carton is not in the bottommost position. The number of cartons placed above the Green box is equal to the number of cartons placed below the White box.

 

Statement II: The Dictionary carton is placed immediately below the carton that is three positions above the White box. The Orange box is placed immediately above the Pink box. The Novel is in neither the Orange nor the Pink box.

 

Statement III: The Atlas carton is placed two positions above the carton packed in the Blue box. Atlas is not in the Green box. The Magazine carton is in neither the Green nor the Blue box. The Magazine carton is placed two positions above the Comic carton.

From I, and II,

 

From I, II, and III,

 

Hence, the correct answer is Data in All I, II, and III statements together are sufficient.

Statement I: Q sits to the right of P. Q sits at one of the ends. R is a Photographer. The person who faces the Dancer sits immediately left of the Photographer.

Statement II: L faces the person who sits to the immediate left of Q. Q faces the person whose hobby is Painter. L is not a Singer.

Statement III: R sits to the immediate left of the person facing M. Only one person sits between the Painter and the Dancer. The person whose hobby is Singer sits immediately left of the person who faces the Writer.

From II and III,

From I and III,

Hence, the correct answer is Data in either Statements II and III or Statements I and III together are sufficient.

Given: Ten flights are scheduled to take off from an airport between 4:00 pm to 5:00 pm. Each flight takes off at 5-minute intervals and is scheduled to land in different countries, Bhutan, Nepal, Malaysia, India, USA, Cambodia, Vietnam, Italy, Maldives, and Mauritius, but not necessarily in the same order.

1. Flight AX takes off five slots before the flight that is scheduled to land in Malaysia.
2. Flight AX takes off two slots before the flight, which is scheduled to land in India, and immediately after Flight CP’s slot.  3. Flight CP lands in Vietnam.  4. There are two slots between the flights that are scheduled to land in Malaysia and Cambodia.
3. Flight BR takes off two slots after the flight that is scheduled to land in Malaysia.  6. There is a vacant slot adjacent to the slot of the flight that is scheduled to land in Cambodia and India. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

Case 1:

 

Case 2:

 

Case 3:

 

7. Flight FN takes off before Flight BR and is scheduled to land in Bhutan.  8. Flight GT takes off before Flight FN, but not immediately before.  9. Flight GT takes off six slots before the flight that is scheduled to land in the Maldives.  10. The flight, which is scheduled to land in the Maldives, has a slot that is adjacent to Flight HR’s slot, and the flight, which is scheduled to land in the USA.  11. Flight JK and Flight HR don’t land in Malaysia. 

Now, we can eliminate Case 2 and Case 3.

Case 1:

 

12. Flight DV takes off immediately after Flight EM’s slot and two slots after the slot in which the flight that is scheduled to land in India.

Case 1:

 

13. Flight GT, Flight LM, and Flight AX don’t land in Nepal. 14. Flight GT doesn’t land in Italy. (From point 11)

This is our final arrangement,

 

Hence, the correct answer is Seven.

Given: Ten flights are scheduled to take off from an airport between 4:00 pm to 5:00 pm. Each flight takes off at 5-minute intervals and is scheduled to land in different countries, Bhutan, Nepal, Malaysia, India, USA, Cambodia, Vietnam, Italy, Maldives, and Mauritius, but not necessarily in the same order.

1. Flight AX takes off five slots before the flight that is scheduled to land in Malaysia.
2. Flight AX takes off two slots before the flight, which is scheduled to land in India, and immediately after Flight CP’s slot.  3. Flight CP lands in Vietnam.  4. There are two slots between the flights that are scheduled to land in Malaysia and Cambodia.
3. Flight BR takes off two slots after the flight that is scheduled to land in Malaysia.  6. There is a vacant slot adjacent to the slot of the flight that is scheduled to land in Cambodia and India. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

Case 1:

 

Case 2:

 

Case 3:

 

7. Flight FN takes off before Flight BR and is scheduled to land in Bhutan.  8. Flight GT takes off before Flight FN, but not immediately before.  9. Flight GT takes off six slots before the flight that is scheduled to land in the Maldives.  10. The flight, which is scheduled to land in the Maldives, has a slot that is adjacent to Flight HR’s slot, and the flight, which is scheduled to land in the USA.  11. Flight JK and Flight HR don’t land in Malaysia. 

Now, we can eliminate Case 2 and Case 3.

Case 1:

 

12. Flight DV takes off immediately after Flight EM’s slot and two slots after the slot in which the flight that is scheduled to land in India.

Case 1:

 

13. Flight GT, Flight LM, and Flight AX don’t land in Nepal. 14. Flight GT doesn’t land in Italy. (From point 11)

This is our final arrangement,

 

Hence, the correct answer is Flight JK - 4:55 pm - Cambodia.

Given: Ten flights are scheduled to take off from an airport between 4:00 pm to 5:00 pm. Each flight takes off at 5-minute intervals and is scheduled to land in different countries, Bhutan, Nepal, Malaysia, India, USA, Cambodia, Vietnam, Italy, Maldives, and Mauritius, but not necessarily in the same order.

1. Flight AX takes off five slots before the flight that is scheduled to land in Malaysia.
2. Flight AX takes off two slots before the flight, which is scheduled to land in India, and immediately after Flight CP’s slot.  3. Flight CP lands in Vietnam.  4. There are two slots between the flights that are scheduled to land in Malaysia and Cambodia.
3. Flight BR takes off two slots after the flight that is scheduled to land in Malaysia.  6. There is a vacant slot adjacent to the slot of the flight that is scheduled to land in Cambodia and India. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

Case 1:

 

Case 2:

 

Case 3:

 

7. Flight FN takes off before Flight BR and is scheduled to land in Bhutan.  8. Flight GT takes off before Flight FN, but not immediately before.  9. Flight GT takes off six slots before the flight that is scheduled to land in the Maldives.  10. The flight, which is scheduled to land in the Maldives, has a slot that is adjacent to Flight HR’s slot, and the flight, which is scheduled to land in the USA.  11. Flight JK and Flight HR don’t land in Malaysia. 

Now, we can eliminate Case 2 and Case 3.

Case 1:

 

12. Flight DV takes off immediately after Flight EM’s slot and two slots after the slot in which the flight that is scheduled to land in India.

Case 1:

 

13. Flight GT, Flight LM, and Flight AX don’t land in Nepal. 14. Flight GT doesn’t land in Italy. (From point 11)

This is our final arrangement,

 

Hence, the correct answer is Flight LM.

Given: Ten flights are scheduled to take off from an airport between 4:00 pm to 5:00 pm. Each flight takes off at 5-minute intervals and is scheduled to land in different countries, Bhutan, Nepal, Malaysia, India, USA, Cambodia, Vietnam, Italy, Maldives, and Mauritius, but not necessarily in the same order.

1. Flight AX takes off five slots before the flight that is scheduled to land in Malaysia.
2. Flight AX takes off two slots before the flight, which is scheduled to land in India, and immediately after Flight CP’s slot.  3. Flight CP lands in Vietnam.  4. There are two slots between the flights that are scheduled to land in Malaysia and Cambodia.
3. Flight BR takes off two slots after the flight that is scheduled to land in Malaysia.  6. There is a vacant slot adjacent to the slot of the flight that is scheduled to land in Cambodia and India. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

Case 1:

 

Case 2:

 

Case 3:

 

7. Flight FN takes off before Flight BR and is scheduled to land in Bhutan.  8. Flight GT takes off before Flight FN, but not immediately before.  9. Flight GT takes off six slots before the flight that is scheduled to land in the Maldives.  10. The flight, which is scheduled to land in the Maldives, has a slot that is adjacent to Flight HR’s slot, and the flight, which is scheduled to land in the USA.  11. Flight JK and Flight HR don’t land in Malaysia. 

Now, we can eliminate Case 2 and Case 3.

Case 1:

 

12. Flight DV takes off immediately after Flight EM’s slot and two slots after the slot in which the flight that is scheduled to land in India.

Case 1:

 

13. Flight GT, Flight LM, and Flight AX don’t land in Nepal. 14. Flight GT doesn’t land in Italy. (From point 11)

This is our final arrangement,

 

Hence, the correct answer is Flight GT.

Given: Ten flights are scheduled to take off from an airport between 4:00 pm to 5:00 pm. Each flight takes off at 5-minute intervals and is scheduled to land in different countries, Bhutan, Nepal, Malaysia, India, USA, Cambodia, Vietnam, Italy, Maldives, and Mauritius, but not necessarily in the same order.

1. Flight AX takes off five slots before the flight that is scheduled to land in Malaysia.
2. Flight AX takes off two slots before the flight, which is scheduled to land in India, and immediately after Flight CP’s slot.  3. Flight CP lands in Vietnam.  4. There are two slots between the flights that are scheduled to land in Malaysia and Cambodia.
3. Flight BR takes off two slots after the flight that is scheduled to land in Malaysia.  6. There is a vacant slot adjacent to the slot of the flight that is scheduled to land in Cambodia and India. 

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3.

Case 1:

 

Case 2:

 

Case 3:

 

7. Flight FN takes off before Flight BR and is scheduled to land in Bhutan.  8. Flight GT takes off before Flight FN, but not immediately before.  9. Flight GT takes off six slots before the flight that is scheduled to land in the Maldives.  10. The flight, which is scheduled to land in the Maldives, has a slot that is adjacent to Flight HR’s slot, and the flight, which is scheduled to land in the USA.  11. Flight JK and Flight HR don’t land in Malaysia. 

Now, we can eliminate Case 2 and Case 3.

Case 1:

 

12. Flight DV takes off immediately after Flight EM’s slot and two slots after the slot in which the flight that is scheduled to land in India.

Case 1:

 

13. Flight GT, Flight LM, and Flight AX don’t land in Nepal. 14. Flight GT doesn’t land in Italy. (From point 11)

This is our final arrangement,

 

Hence, the correct answer is There is a time difference of 25 minutes between the flights that are scheduled to land in Italy and Malaysia.

A major pharmaceutical company decided on its own to withdraw one of its top-selling medicines after internal testing indicated possible long-term health risks. This action was taken even though regulators had not instructed the company to do so. Although the recall caused immediate financial losses, it helped enhance the company’s reputation by showing responsibility toward public safety.

Option 1: Incorrect

There is no indication that the recall was aimed at reducing competition. This motive is not supported by the passage.

Option 2: Correct

This is the best conclusion. The voluntary recall, despite financial loss and no regulatory pressure, shows a strong ethical commitment to consumer safety.

Option 3: Incorrect

The passage explicitly states that no regulatory directive was issued, ruling this out.

Option 4: Incorrect

There is no mention of a ban. The recall was initiated by the company itself.

Option 5: Incorrect

The passage says the company’s public image improved, which implies consumer trust increased, not remained unchanged.

Hence, the correct answer is Ethical considerations influenced corporate decision-making.

Option (a): Incorrect

The passage does not assume that waste generation is equal across all areas of the city.

Option (b): Correct

The introduction of GPS-enabled trucks and sensor-based bins clearly assumes that technology-driven monitoring can improve efficiency in waste collection and cleanliness.

Option (c): Incorrect

There is no indication that peripheral areas generate less waste.

Option (d): Incorrect

The passage does not suggest that manual waste collection is entirely ineffective.

Option (e): Incorrect

Residents are not expected to monitor garbage trucks themselves.

Hence, the correct answer is Technology-driven monitoring can improve efficiency in waste collection.

Option (a): Incorrect

The passage does not suggest a lack of interest in insurance among rural policyholders.

Option (b): Incorrect

Digital platforms are shown to be effective for urban users, so this inference is invalid.

Option (c): Correct

Lower adoption in rural areas, despite success in cities, logically indicates limitations related to digital access or familiarity.

Option (d): Incorrect

The passage provides no comparison of claim volumes between urban and rural customers.

Option (e): Incorrect

There is no suggestion that physical branches should be shut down.

Hence, the correct answer is Lack of digital familiarity and access may limit rural adoption.

Let us first decode the given symbols and then draw a family tree.

 

Now draw the family tree using following notations:

‘K % L ^ M © N * O # P’ means K is the daughter of L, L is the son of M, M is grandfather of N, N is the wife of O, O is the brother of P.

Hence, K is the sister in law of O.

Let us first decode the given symbols and then draw a family tree.

 

Now draw the family tree using following notations:

Using statement 1:

‘K # X ¥ Z # L $ Y’ means K is the brother of X, X is the mother of Z, Z is the brother of L, L is the father of Y.

This statement is true.

So, “K # X ¥ Z # L $ Y” is the correct answer.

Let us first decode the given symbols and then draw a family tree.

 

Now draw the family tree using following notations:

‘P ¥ Q & R € S $ T % V’ means P is the mother of Q, Q is the husband of R, R is the sister of S, S is the father of T, T is the daughter of V.                                       

Hence, no of male members are 2.

Common Calculation

Let

Rs= Riya Saturday,

Ru= Riya Sunday,

Rm = Riya Monday.

Given: Ru:Rm=2:1

So, Ru=2k,Rm=k

Tina’s Monday: Tm = 20

Condition 1:

Tina Saturday = Average of Riya’s 3 days

Ts=[Rs + Ru + Rm]/3 = [Rs + 3k] / 3

Condition 2:

Difference on Monday = Difference on Saturday

Tm−Rm=Ts−Rs =

So,20 − k = [Rs+3k]/3− Rs

Or, Rs​ = 3k − 30

Also,

Ts=[Rs + 3k] / 3 =2k − 10

For Rs>0

So,3k−30>0

⇒ k > 10.

For Tm>Rm  

So, 20>k

⇒k<20.

So, 11≤k≤19

Calculation

1. If k=18 → R _ mon=18, R_sun=36, R _sat=24.

Then T _sat=26, T _mon=20. For total=81, T _sun=35, but this is not greater than R_sun=36. Hence, False.

II. R _sat=90 ⇒ 3k-30=90 ⇒ k=40, which is not allowed (k<20). Hence, False.

III. R _sat=21 ⇒ 3k-30=21 ⇒ k=17. Then R_sun=34, so Tina on Sunday >34 ⇒ minimum = 35. Hence, True.

Final Answer for Q1: Only III can be true.

Common Calculation

Let

Rs= Riya Saturday,

Ru= Riya Sunday,

Rm = Riya Monday.

Given: Ru:Rm=2:1

So, Ru=2k,Rm=k

Tina’s Monday: Tm = 20

Condition 1:

Tina Saturday = Average of Riya’s 3 days

Ts=[Rs + Ru + Rm]/3 = [Rs + 3k] / 3

Condition 2:

Difference on Monday = Difference on Saturday

Tm−Rm=Ts−Rs =

So,20 − k = [Rs+3k]/3− Rs

Or, Rs​ = 3k − 30

Also,

Ts=[Rs + 3k] / 3 =2k − 10

For Rs>0

So,3k−30>0

⇒ k > 10.

For Tm>Rm  

So, 20 > k

⇒ k < 20.

So, 11 ≤ k ≤ 19

Calculation

Tina total = T_sat + T_sun + T_mon = (2k - 10) + T_sun + 20 = 2k + T_sun + 10.

Given total=94 ⇒ T_sun=84 - 2k.

Possible values (10 ≤ k ≤ 19): {64, 62, 60, 58, 56, 54, 52, 50, 48, 46}.

So Tina’s Sunday sales are not unique. They depend on k.

Common Calculation

Vendor P

Basic pay = 3200 → Sales target = 8000

Units sold = 250 → Sales achieved = 250 × 20 = 5000

% target = 5000/8000 = 62.5% → Incentive = Rs. 7,500

Vendor Q

Sales target = 7000 → Basic pay = 7000/2.5 = 2800

Units sold = 300 → Sales achieved = 300 × 20 = 6000

% target = 6000/7000 = 85.71% → Incentive = Rs. 10,500

Vendor R

Basic pay = 3600 → Sales target = 9000

Sales achieved = 4500 → Units sold = 4500/20 = 225

% target = 4500/9000 = 50% → Incentive = Rs. 6,000

Vendor S

Basic pay = 4000 → Sales target = 10,000

Units sold = 330 → Sales achieved = 330 × 20 = 6600

% target = 6600/10000 = 66% → Incentive = Rs. 7,500

Calculation

I. Incentives of P and Q = 7500 + 10500 = 18,000 ≠ 15,000 → False

II. Units sold: P=250, Q=300, R=225, S=330 → R lowest → True

III. Incentive: S=7,500 > R=6,000 → True

<br>

Correct: II and III are true.

Common Calculation

Vendor P

Basic pay = 3200 → Sales target = 8000

Units sold = 250 → Sales achieved = 250 × 20 = 5000

% target = 5000/8000 = 62.5% → Incentive = Rs. 7,500

Vendor Q

Sales target = 7000 → Basic pay = 7000/2.5 = 2800

Units sold = 300 → Sales achieved = 300 × 20 = 6000

% target = 6000/7000 = 85.71% → Incentive = Rs. 10,500

Vendor R

Basic pay = 3600 → Sales target = 9000

Sales achieved = 4500 → Units sold = 4500/20 = 225

% target = 4500/9000 = 50% → Incentive = Rs. 6,000

Vendor S

Basic pay = 4000 → Sales target = 10,000

Units sold = 330 → Sales achieved = 330 × 20 = 6600

% target = 6600/10000 = 66% → Incentive = Rs. 7,500

Calculation

Q basic pay = 2800 → K = 1.2 × 2800 = 3360

Sales target = 2.5 × 3360 = 8400

Achieved 60% = 0.6 × 8400 = 5040

Units sold = 5040 / 20 = 252

Common Calculation

Vendor P

Basic pay = 3200 → Sales target = 8000

Units sold = 250 → Sales achieved = 250 × 20 = 5000

% target = 5000/8000 = 62.5% → Incentive = Rs. 7,500

Vendor Q

Sales target = 7000 → Basic pay = 7000/2.5 = 2800

Units sold = 300 → Sales achieved = 300 × 20 = 6000

% target = 6000/7000 = 85.71% → Incentive = Rs. 10,500

Vendor R

Basic pay = 3600 → Sales target = 9000

Sales achieved = 4500 → Units sold = 4500/20 = 225

% target = 4500/9000 = 50% → Incentive = Rs. 6,000

Vendor S

Basic pay = 4000 → Sales target = 10,000

Units sold = 330 → Sales achieved = 330 × 20 = 6600

% target = 6600/10000 = 66% → Incentive = Rs. 7,500

Calculation

Units = average of P and Q = (250 + 300)/2 = 275

→ Sales = 275 × 20 = 5500

Basic pay = average of R and S = (3600 + 4000)/2 = 3800

→ Target = 9500

% target = 5500/9500 ≈ 57.89%

→ Incentive = Rs. 6,000

Common Calculation

Vendor P

Basic pay = 3200 → Sales target = 8000

Units sold = 250 → Sales achieved = 250 × 20 = 5000

% target = 5000/8000 = 62.5% → Incentive = Rs. 7,500

Vendor Q

Sales target = 7000 → Basic pay = 7000/2.5 = 2800

Units sold = 300 → Sales achieved = 300 × 20 = 6000

% target = 6000/7000 = 85.71% → Incentive = Rs. 10,500

Vendor R

Basic pay = 3600 → Sales target = 9000

Sales achieved = 4500 → Units sold = 4500/20 = 225

% target = 4500/9000 = 50% → Incentive = Rs. 6,000

Vendor S

Basic pay = 4000 → Sales target = 10,000

Units sold = 330 → Sales achieved = 330 × 20 = 6600

% target = 6600/10000 = 66% → Incentive = Rs. 7,500

Calculation

Sales target = 7000 → Basic pay = 7000/2.5 = 2800

Units sold = 300 → Sales achieved = 300 × 20 = 6000

% target = 6000/7000 = 85.71%

→ Incentive = Rs. 10,500

Common Calculation

Vendor P

Basic pay = 3200 → Sales target = 8000

Units sold = 250 → Sales achieved = 250 × 20 = 5000

% target = 5000/8000 = 62.5% → Incentive = Rs. 7,500

Vendor Q

Sales target = 7000 → Basic pay = 7000/2.5 = 2800

Units sold = 300 → Sales achieved = 300 × 20 = 6000

% target = 6000/7000 = 85.71% → Incentive = Rs. 10,500

Vendor R

Basic pay = 3600 → Sales target = 9000

Sales achieved = 4500 → Units sold = 4500/20 = 225

% target = 4500/9000 = 50% → Incentive = Rs. 6,000

Vendor S

Basic pay = 4000 → Sales target = 10,000

Units sold = 330 → Sales achieved = 330 × 20 = 6600

% target = 6600/10000 = 66% → Incentive = Rs. 7,500

Calculation

Incentive earned by Vendor Q = Rs. 10,500

Sales of P as % of Q = (5000/6000) × 100 = 83.33%

Common Calculation

Let's work it out step-by-step.

Solve the quadratic x2 − 14x + 48 = 0

Discriminant =142 – 4 × 48 = 196 – 192 = 4

<br>

Roots =14±22={8,6}

The highest root =8=8=8.

Let n be the difference between the roots: n= 8 – 6 = 2

1st term = highest root = 8

2nd term = (n+4) − 2n = 4 – n = 4 – 2 = 2.

3rd term = (n + 1)2−(n + 1) = n(n + 1) =2 × 3=6.

4th term = (n+2)2−(n + 2) = (n+2)(n+1)=4 × 3=12.

5th term = (n+1)3 − (n + 5) = 27 − 7=20.

6th term = given = 30.

So the 6-number list (in the order requested) is:

 8,   2,   6,   12,  20,  30 

Calculation

Sequence: 8,  2,  6,  12,  20,  30

And n difference of roots =8 – 6 = 2

Now check each statement.

I. Product of 2nd and 5th numbers = 2 × 20 = 40

40 is not a perfect square (since 62 = 36 and 72 =49).

So I is false.

II. Last four numbers are 6, 12, 20, 30.

Sum =6 + 12 + 20 + 30 = 68

Average = 68/4 =17

So II is true.

III. 7th term is 6th term+n3= 30 + 8= 38

Check if 38 is a factor of 114: 114÷38=3

So III is true.

Final answer

I: False

II: True

III: True

Common Calculation

Let's work it out step-by-step.

Solve the quadratic x2 − 14x + 48 = 0

Discriminant =142 – 4 × 48 = 196 – 192 = 4

Roots =14±22={8,6}

The highest root =8=8=8.

Let n be the difference between the roots: n= 8 – 6 = 2

1st term = highest root = 8

2nd term = (n+4) − 2n = 4 – n = 4 – 2 = 2.

3rd term = (n + 1)2−(n + 1) = n(n + 1) =2 × 3=6.

4th term = (n+2)2−(n + 2) = (n+2)(n+1)=4 × 3=12.

5th term = (n+1)3 − (n + 5) = 27 − 7=20.

6th term = given = 30.

So the 6-number list (in the order requested) is:

 8,   2,   6,   12,  20,  30 

Calculation

So, required HCF is 6

Common Calculation

1. Lateral surface area of cuboid = 2h(l+b) = 140 ⇒ h(l+b) = 70
   
2. Volume of cuboid = l × b × h = 120
   
3. We need integer solutions (l, b, h) with l > b.

Checking integer triples, we get two valid cuboids:

• (l, b, h) = (4, 3, 10)

• (l, b, h) = (12, 2, 5)

Case 1: (l, b, h) = (4, 3, 10)

Quantity I = 2(4) - 3 - (10/2) = 0

Quantity II = (3×3×10)/4 = 90/4 = 22.5

Quantity III = (2×3 - 4)/10 = 2/10 = 0.2

<br>

Ordering: Quantity II > Quantity III > Quantity I

Case 2: (l, b, h) = (12, 2, 5)

Quantity I = 2(12) - 2 - (5/2) = 19.5

Quantity II = (3×2×5)/12 = 30/12 = 2.5

Quantity III = (2×2 - 12)/5 = -8/5 = -1.6

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Ordering: Quantity I > Quantity II > Quantity III

Conclusion

Since the two valid cuboids give different orderings of the three quantities, we cannot determine a single consistent comparison. Hence, the data are insufficient to establish a unique order among Quantity I, II, and III.

Shortcut Trick Total M + N = 56 kg with ratio 3:5 ⇒ M = 21 kg, N = 35 kg. Condition for O: M < O < N ⇒ 21 < O < 35 ⇒ Max O = 34 kg (Statement a is True). Condition for P and Q: N < P < Q. Total weight = 200 kg ⇒ 56 + O + P + Q = 200 ⇒ O + P + Q = 144. For Max Q, O and P must be minimum: O = 22, P = 36 ⇒ Q = 144 − 22 − 36 = 86 kg (Statement b is False). For Max P, O is min and Q is just P + 1: 22 + P + (P + 1) = 144 ⇒ 2P = 121 ⇒ P = 60 kg (Statement c is False). ∴ The correct answer is Only a true. Alternate Method

Given: Total = 200 kg; M + N = 56 kg; M:N = 3:5; M < O < N; N < P < Q. ⇒ 3x + 5x = 56 ⇒ 8x = 56 ⇒ x = 7. ⇒ Weight M = 21 kg, Weight N = 35 kg. ⇒ For O: 21 < O < 35. Integers range [22, 34]. Max O = 34 kg. ⇒ For P and Q: P + Q = 200 − (M + N + O) = 200 − 56 − O = 144 − O. ⇒ Since N < P < Q, minimum P = 36 kg and minimum Q = 37 kg. ⇒ Max Q occurs when O and P are min: O = 22, P = 36. Q = 144 − 22 − 36 = 86 kg. ⇒ Max P occurs when O is min and Q is P + 1: 22 + P + P + 1 = 144 ⇒ 2P = 121 ⇒ P = 60.5 (integer P = 60 kg). ∴ The correct answer is Only a true. Additional Information Ratio and Proportion To divide a quantity S into ratio a:b, the parts are (a × S)/(a+b) and (b × S)/(a+b). Optimization with Constraints To maximize one variable in a sum, all other variables must be kept at their minimum possible values allowed by the constraints. Integer Constraints In discrete weight problems, strictly greater than (>) for an integer x implies the next value is at least x + 1.

Shortcut Trick Roots of x2 − 130x + 4200 = 0 are 70 and 60 (since 70+60=130 and 70×60=4200). Given SA > SB ⇒ SA = 70 km/hr and SB = 60 km/hr. When speeds are swapped, Time A increases by 2: D/60 − D/70 = 2 ⇒ D(10/4200) = 2 ⇒ D = 840 km. TA = 840/70 = 12 hours and TB = 840/60 = 14 hours. Sum of times = 12 + 14 = 26 hours. ∴ The correct answer is 26 hours. Alternate Method Given: Equation x2 − 130x + 4200 = 0; SA > SB; Swap condition: A takes TA+2 at SB. Formula Used: Time = Distance ÷ Speed; Quadratic roots: x = [−b ± √(b2 − 4ac)] / 2a

⇒ x2 − 70x − 60x + 4200 = 0 ⇒ (x − 70)(x − 60) = 0 ⇒ x = 70, 60. Since SA > SB, SA = 70 km/hr, SB = 60 km/hr. ⇒ TA = D/70 and TB = D/60. ⇒ Swap condition for Train A: D/SB = TA + 2 ⇒ D/60 = D/70 + 2 ⇒ D/60 − D/70 = 2 ⇒ (7D − 6D)/420 = 2 ⇒ D/420 = 2 ⇒ D = 840 km. ⇒ TA = 840 / 70 = 12 hours. ⇒ TB = 840 / 60 = 14 hours. ⇒ Sum = 12 + 14 = 26 hours. ∴ The correct answer is 26 hours. Additional Information Quadratic Roots Property For ax2 + bx + c = 0, sum of roots is −b/a and product of roots is c/a. Here 70+60=130 and 70×60=4200. Inverse Proportion of Speed and Time For a constant distance, speed is inversely proportional to time (S1/S2 = T2/T1). Swapping speeds results in a direct swap of time values. Average Speed If a distance is covered at two different speeds S1 and S2, the average speed for the whole journey is 2S1S2 / (S1 + S2).

Shortcut Trick

Ratio of Lengths (P:Q) = 100:75 = 4:3. Ratio of Heights (P:Q) = 120:100 = 6:5.

Volume Ratio = (L_P × B_P × H_P) / (L_Q × B_Q × H_Q) = 8/3.

⇒ (4 × B_P × 6) / (3 × B_Q × 5) = 8/3 ⇒ 24B_P / 15B_Q = 8/3 ⇒ B_P / B_Q = (8 × 15) / (3 × 24) = 5/3.

Given B_P − B_Q = z. Since B_P:B_Q = 5:3, let B_P = 5k and B_Q = 3k, then z = 2k.

From Statement II, Side a = B_P = 5k. Surface Area 6a² = 6(5k)² = 150k² = 225z = 225(2k) = 450k.

⇒ 150k = 450 ⇒ k = 3. Therefore, z = 2k = 2 × 3 = 6.

∴ The correct answer is 6.

Alternate Method

Given: L_Q = 0.75 L_P, H_P = 1.2 H_Q, B_P − B_Q = z, V_P : V_Q = 8 : 3, Cube Area = 225z, Side = B_P.

⇒ V_P / V_Q = (L_P × B_P × H_P) / (L_Q × B_Q × H_Q) = 8/3

⇒ (L_P × B_P × 1.2H_Q) / (0.75L_P × B_Q × H_Q) = 8/3

⇒ (1.2 × B_P) / (0.75 × B_Q) = 8/3

⇒ B_P / B_Q = (8 × 0.75) / (3 × 1.2) = 6 / 3.6 = 5 / 3

⇒ Let B_P = 5k and B_Q = 3k. Then z = 5k − 3k = 2k

⇒ Surface Area of Cube = 6 × (Side)² = 6 × (B_P)² = 6 × (5k)² = 150k²

⇒ 150k² = 225z ⇒ 150k² = 225(2k) = 450k

⇒ 150k = 450 ⇒ k = 3

⇒ z = 2k = 2 × 3 = 6

∴ The correct answer is 6.

Additional Information

Volume of a Cuboid

The volume of a cuboid is given by the formula V = Length × Breadth × Height.

Surface Area of a Cube

The total surface area of a cube with side 'a' is 6a². If the lateral surface area is mentioned, it is 4a².

Percentage and Ratio Conversion

75% is equivalent to 3/4 and 20% more is equivalent to a multiplier of 1.2 or 6/5.

Given:

The first quadratic equation is x² - Px + 32 = 0.

One root is 4 and the other root is 4A.

⇒ Product of roots = 4 × 4A = 32

⇒ 16A = 32

⇒ A = 2

The other root is 4A = 4 × 2 = 8.

⇒ Sum of roots = P = 4 + 8 = 12

∴ P = 12 and A = 2

Now, from the second equation:

2 × cube root of Y = Square root of Y

Let u = Y^(1/6), then:

⇒ 2u² = u³

⇒ u²(u - 2) = 0

⇒ u = 0 or u = 2

If u = 2 ⇒ Y = 2⁶ = 64

If u = 0 ⇒ Y = 0

Calculations for Quantities:

For Quantity I:

Equation is Z² - KZ + 990 = 0, where K = 5P + 3.

⇒ K = 5(12) + 3 = 63

So, the equation becomes Z² - 63Z + 990 = 0.

Let the roots be a and b:

Sum of roots (a + b) = 63

Product of roots (a × b) = 990

Finding factors of 990 that add up to 63: 33 × 30 = 990 and 33 + 30 = 63.

⇒ The roots are 33 and 30.

Taking a as the larger root, a = 33.

⇒ Quantity I = 2a = 2 × 33 = 66

For Quantity II:

Quantity II = Value of Y + A

Case 1: If Y = 64

⇒ Quantity II = 64 + 2 = 66

Case 2: If Y = 0

⇒ Quantity II = 0 + 2 = 2

For Quantity III:

Quantity III = Value of (Y + 2) - Highest root of the given equation (Z2 - 63Z + 990 = 0)

The highest root of the equation is 33.

Case 1: If Y = 64

⇒ Quantity III = (64 + 2) - 33 = 66 - 33 = 33

Case 2: If Y = 0

⇒ Quantity III = (0 + 2) - 33 = -31

Conclusion:

Comparing the values:

If Y = 64 ⇒ Quantity I (66) = Quantity II (66) > Quantity III (33)

If Y = 0 ⇒ Quantity I (66) > Quantity II (2) > Quantity III (-31)

∴ Quantity I ≥ Quantity II > Quantity III

Common Calculation

Percentage distribution of number of students who like Tea, Coffee, Milk and only coffee is 35%, 30%, 40% and 20% respectively.

Let the Venn regions be:

Tea only = a

Coffee only = given = 40

Milk only = b

Tea & Milk only = x

Tea & Coffee only = y

Coffee & Milk only = z

None (like no drink) = n

(There are no students who like all three.)

Given percentages → absolute counts (out of 200):

Tea total = 35% of 200 = 70

Coffee total = 30% of 200 = 60

Milk total = 40% of 200 = 80

Only coffee = 40

Given relations:

n=2 × (Tea & Milk only) = 2x

n= (Tea & Coffee only) + 20 = y + 20

From coffee total: 40 + y + z = 60 

 ⟹ y + z = 20

From tea total: a + x + y = 70 

 ⟹   a = 70 – x − y

From milk total: b + x + z = 80 

 ⟹ b = 80 – x − z

Total students: a + 40 + b + x + y + z + n = 200

Solving this system gives a unique integer solution.

Tea only (a) = 60

Coffee only = 40 (given)

Milk only (b) = 50

Tea & Milk only (x) = 10

Tea & Coffee only (y) = 0

Coffee & Milk only (z) = 20

None (n) = 20

Calculation

Number of girls students who like only Tea is 60 × 60 / 100 = 36

Common Calculation

Percentage distribution of number of students who like Tea, Coffee, Milk and only coffee is 35%, 30%, 40% and 20% respectively.

Let the Venn regions be:

Tea only = a

Coffee only = given = 40

Milk only = b

Tea & Milk only = x

Tea & Coffee only = y

Coffee & Milk only = z

None (like no drink) = n

(There are no students who like all three.)

Given percentages → absolute counts (out of 200):

Tea total = 35% of 200 = 70

Coffee total = 30% of 200 = 60

Milk total = 40% of 200 = 80

Only coffee = 40

Given relations:

n=2 × (Tea & Milk only) = 2x

n= (Tea & Coffee only) + 20 = y + 20

From coffee total: 40 + y + z = 60 

 ⟹ y + z = 20

From tea total: a + x + y = 70 

 ⟹   a = 70 – x − y

From milk total: b + x + z = 80 

 ⟹ b = 80 – x − z

Total students: a + 40 + b + x + y + z + n = 200

Solving this system gives a unique integer solution.

Tea only (a) = 60

Coffee only = 40 (given)

Milk only (b) = 50

Tea & Milk only (x) = 10

Tea & Coffee only (y) = 0

Coffee & Milk only (z) = 20

None (n) = 20

Calculation

Required percentage = [0/20] × 100 = 0

Common Calculation

Percentage distribution of number of students who like Tea, Coffee, Milk and only coffee is 35%, 30%, 40% and 20% respectively.

Let the Venn regions be:

Tea only = a

Coffee only = given = 40

Milk only = b

Tea & Milk only = x

Tea & Coffee only = y

Coffee & Milk only = z

None (like no drink) = n

(There are no students who like all three.)

Given percentages → absolute counts (out of 200):

Tea total = 35% of 200 = 70

Coffee total = 30% of 200 = 60

Milk total = 40% of 200 = 80

Only coffee = 40

Given relations:

n=2 × (Tea & Milk only) = 2x

n= (Tea & Coffee only) + 20 = y + 20

From coffee total: 40 + y + z = 60 

 ⟹ y + z = 20

From tea total: a + x + y = 70 

 ⟹   a = 70 – x − y

From milk total: b + x + z = 80 

 ⟹ b = 80 – x − z

Total students: a + 40 + b + x + y + z + n = 200

Solving this system gives a unique integer solution.

Tea only (a) = 60

Coffee only = 40 (given)

Milk only (b) = 50

Tea & Milk only (x) = 10

Tea & Coffee only (y) = 0

Coffee & Milk only (z) = 20

None (n) = 20

Calculation

Required ratio = 60 : 50 = 6:5

Shortcut Trick

For Boat P: Total Time = D/15 + D/12 = 27 hours.

LCM(15, 12) = 60 ⇒ (4D + 5D)/60 = 27 ⇒ 9D/60 = 27 ⇒ Distance (D) = 180 km.

For Boat Q: Total Time = 180/18 + 180/15.

Total Time = 10 + 12 = 22 hours.

∴ The correct answer is 22 hours.

Alternate Method

Given: Downstream speed of P (D_p) = 15 km/hr, Upstream speed of P (U_p) = 12 km/hr, Total time for P = 27 hours. Downstream speed of Q (D_q) = 18 km/hr, Upstream speed of Q (U_q) = 15 km/hr.

Formula Used: Distance = Speed × Time | Total Time = (Distance ÷ Downstream Speed) + (Distance ÷ Upstream Speed)

Calculations:

⇒ Let the distance be x km.

⇒ For boat P: (x ÷ 15) + (x ÷ 12) = 27

⇒ (4x + 5x) ÷ 60 = 27

⇒ 9x ÷ 60 = 27

⇒ 9x = 27 × 60

⇒ x = 3 × 60 = 180 km

⇒ For boat Q, distance is 180 km.

⇒ Time taken downstream = 180 ÷ 18 = 10 hours

⇒ Time taken upstream = 180 ÷ 15 = 12 hours

⇒ Total time for Q = 10 + 12 = 22 hours

∴ The correct answer is 22 hours.

Additional Information

Speed in Still Water and Stream

Speed of Boat (B) = (Downstream + Upstream) ÷ 2. Speed of Stream (S) = (Downstream − Upstream) ÷ 2.

Average Speed of Boat

When a boat covers equal distance upstream and downstream, Average Speed = (2 × U × D) ÷ (U + D), where U = Upstream and D = Downstream speeds.

Relative Speed in Water

Downstream speed = Speed of boat + Speed of current. Upstream speed = Speed of boat − Speed of current.

Calculation

1st term is x0 = 1 [ x>1] [20 = 1 , 30 = 1 ….]

2nd term is smallest even prime number is 2.

3rd term of the series is cube of second term , so 3rd term of series = 23 = 8

The square root of 4th term of the series is equals to third term of series. So, 4th term of the series is 64.

The square root of the 5th term of the series is 25.

So, 5th term of the series is 32 × 32 = 1024

So, series is 1, 2 , 8, 64, 1024,

1 × 2 = 2, 2 × 22 = 8, …….6th term = 1024 × 25

 Calculation

So, required value = [ 1024 × 25 ] / 64 = 512

Calculation

1st term is x0 = 1 [ x>1] [20 = 1 , 30 = 1 ….]

2nd term is smallest even prime number is 2.

3rd term of the series is cube of second term , so 3rd term of series = 23 = 8

The square root of 4th term of the series is equals to third term of series. So, 4th term of the series is 64.

The square root of the 5th term of the series is 25.

So, 5th term of the series is 32 × 32 = 1024

So, series is 1, 2 , 8, 64, 1024,

1 × 2 = 2, 2 × 22 = 8, …….6th term = 1024 × 25

Calculation

So, required difference = [5/8] × 1024 - 83 = 640 - 512 = 128

Calculation

1st term is x0 = 1 [ x>1] [20 = 1 , 30 = 1 ….]

2nd term is smallest even prime number is 2.

3rd term of the series is cube of second term , so 3rd term of series = 23 = 8

The square root of 4th term of the series is equals to third term of series. So, 4th term of the series is 64.

The square root of the 5th term of the series is 25.

So, 5th term of the series is 32 × 32 = 1024

So, series is 1, 2 , 8, 64, 1024,

1 × 2 = 2, 2 × 22 = 8, …….6th term = 1024 × 25

Calculation

Sum is 1024 + 1 = 1025

So, Multiply of 5.

Sum of digit is 1+0+2+5=8

So, it is wrong.

It is not perfect square.

Shortcut Trick

Days ratio: P = 75, Q = 75(1 + 2x/100), R = 75(1 + 2x/100)(1 + x/100).

Given R = P + Q ⇒ 75(1 + 2x/100)(1 + x/100) = 75 + 75(1 + 2x/100).

Let (1 + 2x/100) = k, then k(1 + x/100) = 1 + k. On solving for x = 50%.

Days: P = 75, Q = 75 × 2 = 150, R = 150 × 1.5 = 225.

Required time = 50% × (Q × R) ÷ (Q + R) = 0.5 × (150 × 225) ÷ 375 = 45 days.

∴ The correct answer is 45 days.

Alternate Method

Given: Time taken by P = 75 days. Let x/100 = a.

Formula Used: Time taken = Total Work ÷ Efficiency.

Time for Q = 75(1 + 2a)

Time for R = 75(1 + 2a)(1 + a)

According to the question: R = P + Q

⇒ 75(1 + 2a)(1 + a) = 75 + 75(1 + 2a)

⇒ (1 + 2a)(1 + a) = 1 + 1 + 2a

⇒ 1 + a + 2a + 2a² = 2 + 2a

⇒ 2a² + a − 1 = 0

⇒ (2a − 1)(a + 1) = 0 ⇒ a = 1/2 (as time cannot be negative)

⇒ x/100 = 0.5 ⇒ x = 50

⇒ 50% of Total Work = 450 × 0.5 = 225 units

⇒ Combined Efficiency of Q and R = 3 + 2 = 5 units/day

⇒ Time taken = 225 ÷ 5 = 45 days

∴ The correct answer is 45 days.

Additional Information

Work and Time Inverse Relation

Efficiency is inversely proportional to the time taken to complete the work. Efficiency = Work / Time.

Combined Work Formula

If two people take A and B days respectively, time taken together is (A × B) / (A + B) days.

Percentage Increase in Time

If time increases by r%, New Time = Old Time × (1 + r/100).

​Common Calculation

Let Set M = {m, m+2, m+4, m+6}. Since the HCF of the smallest numbers of both sets is 6, m must be divisible by 6. Also, the 2nd lowest of M must be a single digit ⇒ m+2 ≤ 9. Thus m = 6. So M = {6, 8, 10, 12}.

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2. Let Set N = {7n, 7(n+1), 7(n+2), 7(n+3)}. The gcd(smallest of M=6, smallest of N=7n)=6 ⇒ n must be a multiple of 6. So smallest of N = 42k, with k an integer.
   

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3. One element of N must be a perfect square. This condition allows many possibilities for k, but the smallest valid solution occurs at k=1, giving N = {42, 49, 56, 63} (49 is 7²).

Calculation

Evaluation of Statements (for k=1)

I. Largest of N = 63, which is divisible by 9 ⇒ True.

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II. Smallest of M = 6, 2nd smallest of N = 49. Difference = 49 - 6 = 43, which is prime ⇒ False.

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III. Sum of N = 42 + 49 + 56 + 63 = 210, which is less than 250 ⇒ True.

​Common Calculation

Let Set M = {m, m+2, m+4, m+6}. Since the HCF of the smallest numbers of both sets is 6, m must be divisible by 6. Also, the 2nd lowest of M must be a single digit ⇒ m+2 ≤ 9. Thus m = 6. So M = {6, 8, 10, 12}.

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2. Let Set N = {7n, 7(n+1), 7(n+2), 7(n+3)}. The gcd(smallest of M=6, smallest of N=7n)=6 ⇒ n must be a multiple of 6. So smallest of N = 42k, with k an integer.
   

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3. One element of N must be a perfect square. This condition allows many possibilities for k, but the smallest valid solution occurs at k=1, giving N = {42, 49, 56, 63} (49 is 7²).

Calculation

Second Largest Numbers

• Set M = {6, 8, 10, 12} ⇒ 2nd largest = 10.

• Set N (for k=1) = {42, 49, 56, 63} ⇒ 2nd largest = 56.

So, Sum is = 56 +10 = 66

Common Calculation

1. Let sizes be p, q, r for sets P, Q, R. Given p = q - 1 and q = r + 1 ⇒ p = r. Total elements p + q + r = 13 ⇒ 3p + 1 = 13 ⇒ p = 4. So |P| = 4, |Q| = 5, |R| = 4.
   

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2. For set P: smallest × largest = 27 ⇒ possibilities (1,27) or (3,9). Since R uses 1 as smallest, P must use (3,9).
   

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3. For set Q: product = 27 + 58 = 85 ⇒ possibilities (1,85) or (5,17). Smallest cannot be 1 (already in R), so Q uses (5,17).
   

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4. For set R: product = 23 ⇒ (1,23). So R has {1, ..., 23} with two primes only.

Calculation

Possible products in Q

All three values are possible:

• Example for 70: Q = {5, 7, 14, 16, 17} ⇒ 5 × 14 = 70

• Example for 72: Q = {5, 6, 12, 13, 17} ⇒ 6 × 12 = 72

• Example for 120: Q = {5, 10, 12, 14, 17} ⇒ 10 × 12 = 120

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Answer: i), ii), and iii) — all possible.

Common Calculation

1. Let sizes be p, q, r for sets P, Q, R. Given p = q - 1 and q = r + 1 ⇒ p = r. Total elements p + q + r = 13 ⇒ 3p + 1 = 13 ⇒ p = 4. So |P| = 4, |Q| = 5, |R| = 4.
   

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2. For set P: smallest × largest = 27 ⇒ possibilities (1,27) or (3,9). Since R uses 1 as smallest, P must use (3,9).
   

<br>

3. For set Q: product = 27 + 58 = 85 ⇒ possibilities (1,85) or (5,17). Smallest cannot be 1 (already in R), so Q uses (5,17).
   

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4. For set R: product = 23 ⇒ (1,23). So R has {1, ..., 23} with two primes only.

Calculation

Maximum sum of set R

R = {1, a, b, 23} with exactly two primes. Since 23 is prime, only one among a and b can be prime. Choose the largest available prime (19) and the largest composite (22).

Sum = 1 + 19 + 22 + 23 = 65.

Common Calculation

1. Let sizes be p, q, r for sets P, Q, R. Given p = q - 1 and q = r + 1 ⇒ p = r. Total elements p + q + r = 13 ⇒ 3p + 1 = 13 ⇒ p = 4. So |P| = 4, |Q| = 5, |R| = 4.
   

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2. For set P: smallest × largest = 27 ⇒ possibilities (1,27) or (3,9). Since R uses 1 as smallest, P must use (3,9).
   

<br>

3. For set Q: product = 27 + 58 = 85 ⇒ possibilities (1,85) or (5,17). Smallest cannot be 1 (already in R), so Q uses (5,17).
   

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4. For set R: product = 23 ⇒ (1,23). So R has {1, ..., 23} with two primes only.

Calculation

Set P has {3, ..., 9}. The 2nd smallest must be just above 3. If 2nd smallest = 4, difference = 9 - 4 = 5.

<br>

Answer: 5.

Common Calculation

1. Let sizes be p, q, r for sets P, Q, R. Given p = q - 1 and q = r + 1 ⇒ p = r. Total elements p + q + r = 13 ⇒ 3p + 1 = 13 ⇒ p = 4. So |P| = 4, |Q| = 5, |R| = 4.
   

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2. For set P: smallest × largest = 27 ⇒ possibilities (1,27) or (3,9). Since R uses 1 as smallest, P must use (3,9).
   

<br>

3. For set Q: product = 27 + 58 = 85 ⇒ possibilities (1,85) or (5,17). Smallest cannot be 1 (already in R), so Q uses (5,17).
   

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4. For set R: product = 23 ⇒ (1,23). So R has {1, ..., 23} with two primes only.

Calculation

​(I) The root of equation x2 - 18x + 81 = 0

(II) 162 – 82

(III) Cube of smallest prime number

Smallest(P) = 3, Smallest(Q) = 5, Smallest(R) = 1.

Sum = 3 + 5 + 1 = 9.

Options:

(I) Root of x² - 18x + 81 = 0 ⇒ root = 9.

(II) 162 – 82 = 80.

(III) Cube of smallest prime = 8.​

Shortcut Trick

Let Total Marbles be T. Probability of Yellow is p ⇒ Yellow = p × T.

Purple marbles = 1/4 of T = 0.25T. Remaining Orange = T − pT − 0.25T.

When half of Orange (0.5 × Orange) is removed, new Total = T − 0.5 × Orange.

New Prob = Yellow / New Total = pT / (T − 0.5(T − pT − 0.25T)) = 1.5p.

Solving 1 / (1 − 0.5 + 0.5p + 0.125) = 1.5 ⇒ 1 / (0.625 + 0.5p) = 1.5 ⇒ 1 = 0.9375 + 0.75p ⇒ 0.0625 = 0.75p ⇒ p = 1/12.

∴ The correct answer is 1/12.

Alternate Method

Given: P(Yellow) = p, P(Purple) = 1/4 = 0.25.

Let Total Marbles = T. Then Yellow = pT, Purple = 0.25T, Orange = O.

⇒ T = pT + 0.25T + O ⇒ O = 0.75T − pT.

New Total after removing O/2 = T − O/2 = T − (0.75T − pT)/2.

⇒ New Total = T − 0.375T + 0.5pT = 0.625T + 0.5pT.

New P(Yellow) = (pT) / (0.625T + 0.5pT) = 1.5p.

⇒ 1 / (0.625 + 0.5p) = 1.5.

⇒ 1 = 1.5 × (0.625 + 0.5p) ⇒ 1 = 0.9375 + 0.75p.

⇒ 0.75p = 0.0625 ⇒ p = 625 / 7500 = 1/12.

∴ The correct answer is 1/12.

Additional Information

Definition of Probability

Probability of an event E is P(E) = (Number of favorable outcomes) ÷ (Total number of outcomes).

Sum of Probabilities

The sum of probabilities of all mutually exclusive and exhaustive events in a sample space is always 1.

Effect of Removing Items

When items are removed, both the count of that specific category and the Total Count change, affecting the probability of all other categories.

Shortcut Trick

Total Initial Quantity = 150 litres. Final Quantity of honey = 150 − 88.56 = 61.44 litres.

Using the formula: Final Qty / Initial Qty = (1 − Z/Total)ⁿ, where n = 4 (total operations).

61.44 / 150 = (1 − Z/150)⁴ ⇒ 6144 / 15000 = (1 − Z/150)⁴ ⇒ 256 / 625 = (1 − Z/150)⁴.

Taking 4^th root: 4 / 5 = 1 − Z / 150 ⇒ Z / 150 = 1/5 ⇒ Z = 30.

∴ The correct answer is 30 litres.

Alternate Method

Given: Total initial honey = 150 litres. Honey reduced by = 88.56 litres. Total process done = 1 + 3 = 4 times.

Formula Used: Final Quantity = Initial Quantity × (1 − Z/V)ⁿ

⇒ Honey left = 150 − 88.56 = 61.44 litres

⇒ 61.44 = 150 × (1 − Z/150)⁴

⇒ 61.44 / 150 = (1 − Z/150)⁴

⇒ 6144 / 15000 = (1 − Z/150)⁴

⇒ 256 / 625 = (1 − Z/150)⁴

⇒ (4 / 5)⁴ = (1 − Z/150)⁴

⇒ 4 / 5 = 1 − Z / 150

⇒ Z / 150 = 1 − 4 / 5 = 1 / 5

⇒ Z = 150 / 5 = 30

∴ The correct answer is 30 litres.

Additional Information

Mixture Replacement Formula

For n repeated replacements, Remaining Qty = Initial Qty × (1 − Outgoing/Total)ⁿ.

Ratio Method for Mixtures

If 1/k of the mixture is replaced with water, the concentration of the original liquid becomes (1 − 1/k)ⁿ times the original.

Successive Percentage Change

This process is an application of successive percentage decrease, similar to reducing a value by x% repeatedly.

The correct answer is Operation Brahma.

In News

• India launched 'Operation Brahma' to assist Myanmar after the 7.7 magnitude earthquake in March 2025.

Key Points

• A 7.7 magnitude earthquake struck Myanmar on 28 March 2025, causing extensive casualties and infrastructure damage.
• Government of India launched dedicated HADR mission named Operation Brahma to support quake-hit areas.
• National Disaster Response Force teams and Indian Army field hospital units deployed for rescue and medical support.
• Indian Air Force C-130J and C-17 aircraft airlifted personnel, medical equipment, and relief material to Nay Pyi Taw and Yangon.
• Specialized medical task forces delivered trauma care, emergency surgeries, and essential health services to civilians.
• Indian Navy ships sailed with relief consignments, reinforcing India's Neighbourhood First and HADR commitment.
• Ministry of External Affairs coordinated multi-agency response with Integrated Defence Staff, NDRF, Army, Air Force, and Navy.

Additional Information

• HADR operations involve rapid deployment of military and civilian assets for humanitarian assistance and disaster relief.
• India conducted previous HADR missions like Operation Maitri for Nepal earthquake 2015 and Operation Dost for Türkiye-Syria earthquake 2023.
• MEA acts as nodal body coordinating with foreign governments during overseas disaster relief efforts.
• Such operations deepen strategic and people-to-people ties with neighbouring countries beyond immediate relief.