Reasoning Practice: Alphabaticalseries Evenseries Mock Test – Free Online Practice

Reasoning Practice: Alphabaticalseries Evenseries ke liye free mock test: 4 questions, timer, negative marking aur explanation ke saath. Koi login nahi, koi payment nahi — turant shuru karein.

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Reasoning Practice: Alphabaticalseries Evenseries ke liye ye 4 questions ka practice set General Knowledge pattern par banaya gaya hai. Test me 45 second per question ka timer hai aur har galat answer par 0.25 marks kate jaate hain — bilkul asli exam jaisa. Submit karne ke baad aapko exact score, accuracy aur har question ka explanation milta hai. Bihar ke competitive exams (BPSC, BSSC, Bihar Police, BTSC) ki tayari karne wale students ke liye.

Practice Questions (4 of 4)

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1. If every alternative letter starting from A is deleted then which letter will be 4th to the right of 5th from the left end?
   यदि A से शुरू करते हुए प्रत्येक एकांतर (alternative) अक्षर को हटा दिया जाए, तो बाएँ छोर से 5वें अक्षर के दाएँ चौथा अक्षर कौन-सा होगा?
   (A) (A) P

   (B) (B) R

   (C) (C) Q

   (D) (D) T
   ✅ Answer & Explanation

   Sahi jawab: B) (B) R

   Explanation: Method 1 (Even Series $2n$ Shortcut Trick): Step 1: Series form samjhein: - A se shuru karke alternative letters (A, C, E, G, ...) hatane par kewal sam sthanon (even positions) wale letters bachenge: $$\text{Series}: \text{B, D, F, H, J, L, N, P, R, T, V, X, Z (kul 13 letters)}$$ Step 2: Position calculate karein (Left + Right): $$\text{Position from Left} = 5 + 4 = 9\text{th letter}$$ Step 3: Even series me Left se $n$-th letter ki real alphabetical position $2 \times n$ hoti hai: $$\text{Original Position} = 9 \times 2 = 18$$ 18th letter 'R' hota hai. Correct option (B) R hai. Method 2 (Manual Even Counting Method): Even series: $B(1), D(2), F(3), H(4), J(5), L(6), N(7), P(8), R(9)$. 9th position par letter 'R' prapt hota hai. Sahi uttar (B) hai.
2. If every alternative letter starting from A is deleted then which letter will be 5th to the right of 5th from the left end?
   यदि A से शुरू करते हुए प्रत्येक एकांतर अक्षर को हटा दिया जाए, तो बाएँ छोर से 5वें अक्षर के दाएँ 5वाँ अक्षर कौन-सा होगा?
   (A) (A) S

   (B) (B) U

   (C) (C) T

   (D) (D) V
   ✅ Answer & Explanation

   Sahi jawab: C) (C) T

   Explanation: Method 1 (Even Series $2n$ Shortcut Trick): Step 1: Position from Left end nikalein (Opposite directions = Addition): $$\text{Position from Left} = 5 + 5 = 10\text{th letter}$$ Step 2: Even series me baayein se $n$-th letter ki original place value $= 2 \times n$: $$\text{Original Position} = 10 \times 2 = 20$$ Standard alphabet me 20वाँ letter 'T' hota hai. Correct option (C) T hai. Method 2 (Step-by-Step Counting Method): Even series: $B, D, F, H, J, L, N, P, R, T$. 10th sthan par aane wala letter 'T' hai. Sahi uttar (C) hai.
3. If every alternative letter starting from A is deleted then which letter will be 3rd to the left of 5th from the right end?
   यदि A से शुरू करते हुए प्रत्येक एकांतर अक्षर को हटा दिया जाए, तो दाएँ छोर से 5वें अक्षर के बाएँ तीसरा अक्षर कौन-सा होगा?
   (A) (A) J

   (B) (B) L

   (C) (C) N

   (D) (D) K
   ✅ Answer & Explanation

   Sahi jawab: B) (B) L

   Explanation: Method 1 (Even Series Right-to-Left Conversion Shortcut): Step 1: Position from Right end nikalte hain (Right + Left): $$\text{Position from Right} = 5 + 3 = 8\text{th letter}$$ Step 2: Even series me kul 13 letters hote hain, isliye Left end me badalne ke liye $(13 + 1) = 14$ me se ghataate hain: $$\text{Position from Left} = 14 - 8 = 6\text{th letter}$$ Step 3: Original alphabet me position $= 2 \times n$: $$\text{Original Position} = 6 \times 2 = 12$$ 12th letter 'L' hota hai. Correct option (B) L hai. Method 2 (Direct Backward Counting Method): Even series right se: $Z(1), X(2), V(3), T(4), R(5), P(6), N(7), L(8)$. Right end se 8th letter 'L' prapt hota hai. Sahi uttar (B) hai.
4. If every alternative letter starting from A is deleted then which letter will be 4th to the right of 10th from the right end?
   यदि A से शुरू करते हुए प्रत्येक एकांतर अक्षर को हटा दिया जाए, तो दाएँ छोर से 10वें अक्षर के दाएँ चौथा अक्षर कौन-सा होगा?
   (A) (A) P

   (B) (B) R

   (C) (C) N

   (D) (D) O
   ✅ Answer & Explanation

   Sahi jawab: A) (A) P

   Explanation: Method 1 (Even Series Conversion Shortcut): Step 1: Samaan dishaon (Right aur Right) ke liye ghataav (subtraction) karein: $$\text{Position from Right} = 10 - 4 = 6\text{th letter}$$ Step 2: Kul 13 letters hone ke karan isse Left end me convert karein: $$\text{Position from Left} = (13 + 1) - 6 = 14 - 6 = 8\text{th letter}$$ Step 3: Real alphabetical position ke liye 2 se multiply karein: $$\text{Original Position} = 8 \times 2 = 16$$ 16वाँ letter 'P' hota hai. Correct option (A) P hai. Method 2 (Right End Inspection Method): Even series right se: $Z(1), X(2), V(3), T(4), R(5), P(6)$. Daayein छोर se 6th letter 'P' hai. Sahi uttar (A) hai.

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