# Solutions for the first year

### Logical Problems

1. “Being a champion $\Rightarrow$ being good at math” but Being good at math need not necessarily imply being a champion.

2. Condition: *If an even number is written on one side of the card, then a vowel is written on the other side.*

Note that the only vowels (disregarding caps) are ‘a’,‘e’,‘i’,‘o’,‘u’. We need to turn two cards, the cards with ‘B’ and ‘4’ written on their visible sides.We trun:

B to check whether the number on the other side is even or odd

4 to check whether the alphabet written on the other side is vowel or not.

We do not turn the cards with ‘A’ and ‘5’ (written on their visible sides) as their others sides can have number of any parity or alphabet of any kind respectively.

3. % to be done

4. Yes.

Let U be the set of all people, T be the set of all people with a TV, $T_m$ be the set of mathematicians with a TV and S be the set of all non mathematicians who swim. Condition a) claims that $A:=T\setminus T_m\ne \emptyset$. Condition b) claims that $B:=T \cap S=\emptyset$. In order to show that there is one person who has a TV set and does not swim everyday: we prove $T\setminus S \ne \emptyset$. Suppose, $T\setminus S=\emptyset$, then T must be empty, as B is empty. But since $A\ne \emptyset$, T is non empty. contradiction!

5. Darmouse stole the cookies.

Suppose, March Hare is the guilty. Then in his testimony he should have claimed that he stole the cookies. Hence, he is not guilty. Since he gave false testimony, neither is Mad Hatter guilty. That leaves us with Darmouse.

6. Let $A_1,\dots ,A_n$ be the sets of pencils of (pairwise) different colors. Suppose we find $a\in A_i$ and $b\in A_j$ ($i\ne j,\{i,j\}\in \{1,\dots ,n\}^2$) such that a and b differ in size, then we are done. Else, wlog let $A_1$ be the set with at least two pencils with different sizes say c and d. Pick any pencil from any $A_i$ with $i\ne 1$. Then it differs from at least one among c and d in size.

7. No.

# Contributors

talegari