📌 Section 1 is Logical reasoning specially designed for CAT

Teams like Parallel, Turbopuffer, and Wordsmith are already setting the pace on Attio. Get an always-on revenue engine, with agents and workflows that build pipeline, chase every buying signal, and move deals forward with your team. Whether you're working in your browser, inbox, or favorite agent, connect to your customer data in real-time through Attio's web app, MCP, API, and SDK.

Try Attio now

A team is to be selected from among 10 students A through J. The selection is to take place according to the following conditions:

(1)If A is selected, either D or F but not both should be selected

(2)G must be selected. At least one between E and F must be selected.

(3)Only one of I and J has to be selected.

(4)B and H are not together. Size of the team is the number of students selected. In each of pairs (E, H) and (F, C), if one student is selected, then the other student must be selected.

“If I were a farmer, I would absolutely use AI.” 

That’s Jensen Huang, founder and CEO of NVIDIA. 

He’s pointing to one of AI’s biggest untapped opportunities: farming. 

DIT AgTech is already putting AI to work with 500+ units deployed and 370,000 head on-platform.

Invest in DIT AgTech Today

^{𝘐𝘯 𝘮𝘢𝘬𝘪𝘯𝘨 𝘢𝘯 𝘪𝘯𝘷𝘦𝘴𝘵𝘮𝘦𝘯𝘵 𝘥𝘦𝘤𝘪𝘴𝘪𝘰𝘯, 𝘪𝘯𝘷𝘦𝘴𝘵𝘰𝘳𝘴 𝘮𝘶𝘴𝘵 𝘳𝘦𝘭𝘺 𝘰𝘯 𝘵𝘩𝘦𝘪𝘳 𝘰𝘸𝘯 𝘦𝘹𝘢𝘮𝘪𝘯𝘢𝘵𝘪𝘰𝘯 𝘰𝘧 𝘵𝘩𝘦 𝘪𝘴𝘴𝘶𝘦𝘳 𝘢𝘯𝘥 𝘵𝘩𝘦 𝘵𝘦𝘳𝘮𝘴 𝘰𝘧 𝘵𝘩𝘦 𝘰𝘧𝘧𝘦𝘳𝘪𝘯𝘨, 𝘪𝘯𝘤𝘭𝘶𝘥𝘪𝘯𝘨 𝘵𝘩𝘦 𝘮𝘦𝘳𝘪𝘵𝘴 𝘢𝘯𝘥 𝘳𝘪𝘴𝘬𝘴 𝘪𝘯𝘷𝘰𝘭𝘷𝘦𝘥. 𝘋𝘐𝘛 𝘈𝘨𝘛𝘦𝘤𝘩 𝘩𝘢𝘴 𝘧𝘪𝘭𝘦𝘥 𝘢 𝘍𝘰𝘳𝘮 𝘊 𝘸𝘪𝘵𝘩 𝘵𝘩𝘦 𝘚𝘦𝘤𝘶𝘳𝘪𝘵𝘪𝘦𝘴 𝘢𝘯𝘥 𝘌𝘹𝘤𝘩𝘢𝘯𝘨𝘦 𝘊𝘰𝘮𝘮𝘪𝘴𝘴𝘪𝘰𝘯 𝘪𝘯 𝘤𝘰𝘯𝘯𝘦𝘤𝘵𝘪𝘰𝘯 𝘸𝘪𝘵𝘩 𝘪𝘵𝘴 𝘰𝘧𝘧𝘦𝘳𝘪𝘯𝘨, 𝘢 𝘤𝘰𝘱𝘺 𝘰𝘧 𝘸𝘩𝘪𝘤𝘩 𝘮𝘢𝘺 𝘣𝘦 𝘰𝘣𝘵𝘢𝘪𝘯𝘦𝘥 𝘩𝘦𝘳𝘦: https://bit.ly/4bzuWCi​}

A, B, C, D, E, F, G, H, I, J

First convert the conditions into simpler rules:

To maximize the team, consider whether we can take both E and F.

If E is selected, H must also be selected.

If F is selected, C must also be selected.

So we can have:

E, H, F, C

Since H is selected, B cannot be selected.

G is compulsory, and exactly one of I/J can be selected.

If we don't select A, there is no restriction involving D and F, so D can also be selected.

Thus a maximum team can be:

C, D, E, F, G, H, I

That's 7 students.

Could we get 8 by adding A?

No. If A is selected, exactly one of D and F can be selected. Since F is already selected, D would have to be excluded.

So adding A simply replaces D:

A, C, E, F, G, H, I = 7

Therefore:

Maximum size = 7

✅ Answer: (b) 7

If A is selected, exactly one of D and F must be selected.

So there are two major cases.

Since F is selected:

C must be selected.

And because A requires exactly one of D/F:

D cannot be selected.

So compulsory members so far:

A, F, C, G

Now consider E.

Then H must be selected.

Since H is selected, B cannot be selected.

We also choose exactly one of I/J.

So there are:

2 possibilities
(one with I and one with J)

Then H is also not selected.

B can either be:

That's 2 possibilities for B.

And there are 2 possibilities for I/J.

Therefore:

2 × 2 = 4 possibilities

Total for Case 1:

2 + 4 = 6

Since exactly one of D/F must be selected:

F cannot be selected.

If F is absent, C must also be absent.

But at least one of E/F must be selected.

Since F is absent:

E must be selected.

Therefore H must also be selected.

Since H is selected:

B cannot be selected.

The fixed members are therefore:

A, D, E, H, G

We only have to choose exactly one of I/J.

Hence:

2 possibilities

Case 1 = 6

Case 2 = 2

Therefore:

6 + 2 = 8

✅ Answer: (b) 8

We are given:

D and F are selected.

Since F is selected:

C must be selected.

G is compulsory.

Exactly one of I/J must be selected.

So we definitely have:

D, F, C, G + one of {I, J}

Now notice something important about A.

If A were selected, exactly one of D/F should be selected.

But both D and F are selected.

Therefore:

A cannot be selected.

Now consider E.

Then H is also not selected.

Since H is absent, B is unrestricted.

So B can be:

= 2 possibilities

For each possibility, we can choose I or J:

2 possibilities

Therefore:

2 × 2 = 4 ways

Then H must be selected.

Since H is selected:

B cannot be selected.

We only choose between I and J:

= 2 ways

Therefore total number of teams:

4 + 2 = 6

✅ Answer: (c) 6

We continue with the condition:

D and F are both selected.

We established that the compulsory students are:

D, F, C, G + exactly one of I/J

That's already:

5 students

Don't select E/H and don't select B.

For example:

C, D, F, G, I

Size = 5

So:

Minimum = 5

If E is selected, H must also be selected.

Then we can have:

C, D, E, F, G, H, I

That's:

7 students

B cannot be added because H is present, and A cannot be added because D and F are both present.

Therefore:

Maximum = 7

So:

Minimum = 5, Maximum = 7

✅ Answer: (c) 5, 7

⚡ Brain teasers are not just about getting the right answer — they train you to think differently under pressure.

Stay consistent with your daily practice, challenge your assumptions, and keep sharpening your reasoning skills.

See you in tomorrow’s challenge!