JEE 2023 — Mathematics PYQ

1. Sets ko samajhna:

• $a$ ek even number hai: $A=\{2,4,6,\dots ,100\}$. Isme total $50$ elements hain.

• $b$ ek odd number hai: $B=\{1,3,5,\dots ,99\}$. Isme bhi total $50$ elements hain.

2. Condition:

Hume diya gaya hai ki:

Iska matlab hai $a+b$ ki values $2,25,48,71,94,117,140,163,186$ ho sakti hain.

Lekin, kyunki $a$ (even) aur $b$ (odd) ka sum hamesha odd hona chahiye, isliye hum sirf odd values ko consider karenge:

3. Har case ko calculate karna:

• Case 1: $a+b=25$

  $b$ ki possible odd values: $1,3,5,\dots ,23$ ($12$ values). Har $b$ ke liye ek unique even $a$ milega.

  Ways = 12

• Case 2: $a+b=71$

  $b$ ki possible odd values: $1,3,5,\dots ,69$ ($35$ values). Lekin $a$ maximum $100$ hai, yahan $a$ ki values limit mein hain.

  Ways = 35

• Case 3: $a+b=117$

  Yahan $a$ max $100$ hai, toh $b$ min $17$ hona chahiye ($117-100$).

  $b$ ki values: $17,19,\dots ,99$.

  Number of values $=\frac{99-17}{2}+1=41+1=42$

• Case 4: $a+b=163$

  $a$ max $100$ hai, toh $b$ min $63$ hona chahiye.

  $b$ ki values: $63,65,\dots ,99$.

  Number of values $=\frac{99-63}{2}+1=18+1=19$

4. Total Ways:

Sahi option (2) hai.