Reasoning Practice: Ineqality Type1 Mock Test – Free Online Practice

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Practice Questions (14 of 14)

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1. Statement: A > B = C ≥ D Conclusions: I. A ≥ C, II. B = D
   कथन: A > B = C ≥ D निष्कर्ष: I. A ≥ C, II. B = D
   (A) (A) Only I follows

   (B) (B) Only II follows

   (C) (C) Both I and II follow

   (D) (D) Neither I nor II follows
   ✅ Answer & Explanation

   Sahi jawab: D) (D) Neither I nor II follows

   Explanation: Method 1 (Inequality Priority Shortcut): Step 1: A se C tak ke chinh dekhein: '>' aur '='. - Priority rule: '>' aur '=' me hamesha '>' ko prathmikta milti hai, atah sahi sambandh $A > C$ hai. Conclusion I ($A \ge C$) galat hai. Step 2: B se D tak ke chinh dekhein: '=' aur '≥'. - Priority rule: '=' aur '≥' me hamesha '≥' ko prathmikta milti hai, atah sahi sambandh $B \ge D$ hai. Conclusion II ($B = D$) galat hai. Atah dono nishkarsh asatya hain. Correct option (D) Neither I nor II follows hai. Method 2 (Definite Relation Check): $A > B = C \implies A > C$. $B = C \ge D \implies B \ge D$. Kisi bhi sthiti me $A \ge C$ ya $B = D$ nishchit nahi hai. Sahi uttar (D) hai.
2. Statement: M ≥ R > T = U Conclusions: I. M > T, II. R > U
   कथन: M ≥ R > T = U निष्कर्ष: I. M > T, II. R > U
   (A) (A) Only I follows

   (B) (B) Only II follows

   (C) (C) Both I and II follow

   (D) (D) Neither I nor II follows
   ✅ Answer & Explanation

   Sahi jawab: C) (C) Both I and II follow

   Explanation: Method 1 (Priority Hierarchy Shortcut): Step 1: Conclusion I ($M > T$): M se T ke beech chinh '≥' aur '>' hain. High priority sign '>' hone ke karan nishkarsh $M > T$ poori tarah satya hai (Follows). Step 2: Conclusion II ($R > U$): R se U ke beech chinh '>' aur '=' hain. High priority sign '>' hone ke karan nishkarsh $R > U$ bhi poori tarah satya hai (Follows). Dono nishkarsh follow karte hain. Correct option (C) Both I and II follow hai. Method 2 (Step-by-Step Traversal): $M \ge R > T \implies M > T$ (True). $R > T = U \implies R > U$ (True). Dono satya hain. Sahi uttar (C) hai.
3. Statement: U < R = T ≤ P = S ≥ Q > U Conclusions: I. U < T, II. R ≤ P, III. S > U
   कथन: U < R = T ≤ P = S ≥ Q > U निष्कर्ष: I. U < T, II. R ≤ P, III. S > U
   (A) (A) Only I follows

   (B) (B) Only II follows

   (C) (C) Only III follows

   (D) (D) All follow
   ✅ Answer & Explanation

   Sahi jawab: D) (D) All follow

   Explanation: Method 1 (Sign Priority Shortcut): Step 1: U se T ke beech: '<' aur '=' $\implies U < T$ satya hai (I follows). Step 2: R se P ke beech: '=' aur '≤' $\implies R \le P$ satya hai (II follows). Step 3: S se U ke beech: '≥' aur '>' $\implies S > U$ satya hai (III follows). Teeno nishkarsh nishchit roop se follow karte hain. Correct option (D) All follow hai. Method 2 (Direct Inspection): $U < R = T \implies U < T$. $R = T \le P \implies R \le P$. $S \ge Q > U \implies S > U$. Sabhi satya hain. Sahi uttar (D) hai.
4. Statement: A = B < C ≤ D Conclusions: I. C > A, II. C < A
   कथन: A = B < C ≤ D निष्कर्ष: I. C > A, II. C < A
   (A) (A) Only I follows

   (B) (B) Only II follows

   (C) (C) Both I and II follow

   (D) (D) Neither I nor II follows
   ✅ Answer & Explanation

   Sahi jawab: A) (A) Only I follows

   Explanation: Method 1 (Reverse Direction Shortcut): Step 1: Statement me A se C ki taraf jane par chinh '=' aur '<' hain, jisse $A < C$ banta hai. Step 2: Jab C se A ki taraf dekhenge (right to left), toh chinh ulat jate hain ($<$ ban jata hai $>$). Atah sahi sambandh $C > A$ hai. Step 3: Conclusion I ($C > A$) satya hai aur Conclusion II ($C < A$) asatya hai. Correct option (A) Only I follows hai. Method 2 (Simple Substitution): $A = B$ hai aur $B < C$ hai $\implies A < C$, jise $C > A$ bhi likha ja sakta hai. Kewal Conclusion I follow karta hai. Sahi uttar (A) hai.
5. Statement: M < R > T > U ≥ S Conclusions: I. T > M, II. T < M, III. S < T, IV. S > T
   कथन: M < R > T > U ≥ S निष्कर्ष: I. T > M, II. T < M, III. S < T, IV. S > T
   (A) (A) Only I and II follow

   (B) (B) Only II and III follow

   (C) (C) Only I and III follow

   (D) (D) Only III follows
   ✅ Answer & Explanation

   Sahi jawab: D) (D) Only III follows

   Explanation: Method 1 (Opposite Family Sign / Block Condition Shortcut): Step 1: T aur M ke beech me dekhein: M < R aur R > T. Yahan dono viprit parivar ke chinh ('<' aur '>') aapas me takra rahe hain (Condition Block). - Condition Block hone par koi definite sambandh nahi banta, isliye Conclusion I aur II dono galat hain. Step 2: S se T ke beech dekhein: T > U ≥ S $\implies T > S$, jise ulat kar likhne par $S < T$ satya banta hai (Conclusion III follows). - Conclusion IV ($S > T$) asatya hai. Atah kewal Conclusion III follow karta hai. Correct option (D) Only III follows hai. Method 2 (Sign Analysis): M aur T ke beech opposite signs hain isliye I aur II galat hain. T se S ki or '>' aur '≥' hain, jisse $T > S \implies S < T$ satya hai. Sahi uttar (D) hai.
6. Statement: A > B < C Conclusions: I. A > C, II. A ≥ C, III. A = C
   कथन: A > B < C निष्कर्ष: I. A > C, II. A ≥ C, III. A = C
   (A) (A) Only I follows

   (B) (B) Only II follows

   (C) (C) Only III follows

   (D) (D) None follows
   ✅ Answer & Explanation

   Sahi jawab: D) (D) None follows

   Explanation: Method 1 (Condition Block Rule Shortcut): Step 1: A se C ke beech jane par viprit parivar ke chinh '>' aur '<' aate hain. Step 2: Golden Rule: Jab bhi do tatvon ke beech viprit chinh ('<' aur '>') aa jayein, toh unke beech Condition Block ban jata hai aur unka koi nishchit sambandh (definite relation) nahi banta. Step 3: Atah A aur C ke beech ka koi bhi nishchit nishkarsh (I, II, III) satya nahi hoga. Correct option (D) None follows hai. Method 2 (Logical Independence Method): A bhi B se bada hai aur C bhi B se bada hai. Lekin A aur C me se kaun bada hai ya dono barabar hain, yeh tay nahi kiya ja sakta. Sahi uttar (D) hai.
7. Statements: A > B ; C ≥ D ; B = C Conclusions: I. A > C, II. B ≥ D
   कथन: A > B ; C ≥ D ; B = C निष्कर्ष: I. A > C, II. B ≥ D
   (A) (A) Only I follows

   (B) (B) Only II follows

   (C) (C) Both I and II follow

   (D) (D) None follows
   ✅ Answer & Explanation

   Sahi jawab: C) (C) Both I and II follow

   Explanation: Method 1 (Common Term Combining Shortcut): Step 1: Common term ke madhyam se sabhi ko ek sath jodein: $A > B = C \ge D$. Step 2: A se C tak dekhein: '>' aur '=' me prathmikta '>' ko milti hai $\implies A > C$ 100% satya hai (I follows). Step 3: B se D tak dekhein: '=' aur '≥' me prathmikta '≥' ko milti hai $\implies B \ge D$ 100% satya hai (II follows). Dono nishkarsh follow karte hain. Correct option (C) Both I and II follow hai. Method 2 (Direct Substitution): $B = C$ hone ke karan $A > B \implies A > C$. Isi tarah $C \ge D \implies B \ge D$. Dono satya hain. Sahi uttar (C) hai.
8. Statements: R > T ; U < T ; U > P Conclusions: I. R > U, II. T > P
   कथन: R > T ; U < T ; U > P निष्कर्ष: I. R > U, II. T > P
   (A) (A) Only I follows

   (B) (B) Only II follows

   (C) (C) Both I and II follow

   (D) (D) None follows
   ✅ Answer & Explanation

   Sahi jawab: C) (C) Both I and II follow

   Explanation: Method 1 (Chain Formation Shortcut): Step 1: Kathano ko ek chain me jodein: $R > T > U > P$. Step 2: Conclusion I ($R > U$): R se U tak sabhi chinh '>' hain, atah $R > U$ satya hai (Follows). Step 3: Conclusion II ($T > P$): T se P tak sabhi chinh '>' hain, atah $T > P$ satya hai (Follows). Dono nishkarsh follow karte hain. Correct option (C) Both I and II follow hai. Method 2 (Direct Inspection): $R > T$ aur $T > U \implies R > U$. $T > U$ aur $U > P \implies T > P$. Dono anivarya roop se satya hain. Sahi uttar (C) hai.
9. Statements: M > N ≤ R ; P ≤ N > U Conclusions: I. M > U, II. M > P, III. R < U
   कथन: M > N ≤ R ; P ≤ N > U निष्कर्ष: I. M > U, II. M > P, III. R < U
   (A) (A) Only I and III follow

   (B) (B) Only I and II follow

   (C) (C) Only III follows

   (D) (D) None follows
   ✅ Answer & Explanation

   Sahi jawab: B) (B) Only I and II follow

   Explanation: Method 1 (Common Term 'N' Connection Shortcut): Step 1: Common term 'N' ke madhyam se sambandh banayein: - M se U tak: $M > N > U \implies$ High priority sign '>' hone se $M > U$ satya hai (Conclusion I follows). - M se P tak: $M > N \ge P \implies$ High priority sign '>' hone se $M > P$ satya hai (Conclusion II follows). - R se U tak: $R \ge N > U \implies R > U$ banta hai, parantu Conclusion III me $R < U$ diya hai (Does not follow). Atah kewal I aur II follow karte hain. Correct option (B) Only I and II follow hai. Method 2 (Step-by-Step Priority Check): $M > N > U \implies M > U$ (True). $M > N \ge P \implies M > P$ (True). $R \ge N > U \implies R > U \ne R < U$ (False). Sahi uttar (B) hai.
10. Statements: M > N ≥ T ; R < N ≤ P Conclusions: I. M > R, II. M > P, III. T ≤ P
    कथन: M > N ≥ T ; R < N ≤ P निष्कर्ष: I. M > R, II. M > P, III. T ≤ P
    (A) (A) Only I and III follow

    (B) (B) Only I and II follow

    (C) (C) Only III follows

    (D) (D) None follows
    ✅ Answer & Explanation

    Sahi jawab: A) (A) Only I and III follow

    Explanation: Method 1 (Sign Alignment & Direction Shortcut): Step 1: M se R tak: $M > N > R \implies M > R$ bilkul sahi hai (Conclusion I follows). Step 2: M se P tak: $M > N$ aur $N \le P$ $\implies M > N \le P$ (Viprit chinh aane se condition block ho gayi, Conclusion II does not follow). Step 3: T se P tak: $T \le N \le P \implies T \le P$ bilkul sahi hai (Conclusion III follows). Atah kewal I aur III follow karte hain. Correct option (A) Only I and III follow hai. Method 2 (Block Condition Identification): M aur P ke beech '>' aur '≤' aate hain jo block banate hain. R se M aur T se P ke beech ek hi parivar ke chinh hain, atah I aur III satya hain. Sahi uttar (A) hai.
11. Statements: R > P ≥ U = S ; P < W = Y Conclusions: I. R > W, II. U < W
    कथन: R > P ≥ U = S ; P < W = Y निष्कर्ष: I. R > W, II. U < W
    (A) (A) Only I follows

    (B) (B) Only II follows

    (C) (C) Both I and II follow

    (D) (D) None follows
    ✅ Answer & Explanation

    Sahi jawab: B) (B) Only II follows

    Explanation: Method 1 (Common Element 'P' Shortcut): Step 1: Conclusion I ($R > W$): R se W tak jane ke liye $R > P < W$ banta hai. Yahan viprit chinh ('>' aur '<') aane ke karan block banta hai, atah Conclusion I galat hai. Step 2: Conclusion II ($U < W$): U se W tak: $U \le P < W \implies U < W$ satya hai (Conclusion II follows). Atah kewal II follow karta hai. Correct option (B) Only II follows hai. Method 2 (Direct Path Method): $W > P \ge U \implies W > U$, jise $U < W$ likha jata hai. $R$ aur $W$ me koi seedha marg nahi hai kyunki $P$ beech me aakar sign flip karta hai. Kewal II sahi hai. Sahi uttar (B) hai.
12. Statements: M > N ≥ R ; N > T ≤ U Conclusions: I. M ≥ R, II. N ≤ U
    कथन: M > N ≥ R ; N > T ≤ U निष्कर्ष: I. M ≥ R, II. N ≤ U
    (A) (A) Only I follows

    (B) (B) Only II follows

    (C) (C) Both I and II follow

    (D) (D) None follows
    ✅ Answer & Explanation

    Sahi jawab: D) (D) None follows

    Explanation: Method 1 (Sign Priority & Block Rule Shortcut): Step 1: Conclusion I ($M \ge R$): M se R tak $M > N \ge R$ hai. Yahan high priority sign '>' hai, isliye sahi nishkarsh $M > R$ hoga. Conclusion I ($M \ge R$) galat hai. Step 2: Conclusion II ($N \le U$): N se U tak $N > T \le U$ hai. Viprit chinh ('>' aur '≤') aane ke karan relation block ho jata hai, atah Conclusion II galat hai. Atah koi bhi follow nahi karta. Correct option (D) None follows hai. Method 2 (Direct Inspection): $M > N \ge R \implies M > R \ne M \ge R$. $N > T \le U$ me opposite signs hone se $N$ aur $U$ me koi nishchit sambandh nahi banta. Sahi uttar (D) hai.
13. Statement: R > U ≥ P ≤ S > T Conclusions: I. R < P, II. P ≤ T, III. P < T
    कथन: R > U ≥ P ≤ S > T निष्कर्ष: I. R < P, II. P ≤ T, III. P < T
    (A) (A) Only I and II follow

    (B) (B) Only II and III follow

    (C) (C) Only II follows

    (D) (D) None follows
    ✅ Answer & Explanation

    Sahi jawab: D) (D) None follows

    Explanation: Method 1 (Direction & Block Evaluation Shortcut): Step 1: R se P tak dekhein: $R > U \ge P \implies R > P$. Conclusion I me $R < P$ diya hai, atah yeh galat hai. Step 2: P se T tak dekhein: $P \le S > T$. Yahan '≤' aur '>' viprit parivar ke chinh aate hain (Block condition), jisse P aur T ke beech koi definite sambandh nahi nikalta. - Atah Conclusion II ($P \le T$) aur Conclusion III ($P < T$) dono asatya hain. Koyi bhi nishkarsh follow nahi karta. Correct option (D) None follows hai. Method 2 (Direct Path Verification): $R > P$ hone ke karan $R < P$ asatya hai. $P$ aur $T$ ke beech opposite signs aane se koi bhi definite conclusion nikalna sambhav nahi hai. Sahi uttar (D) hai.
14. Which of the following options makes 'A > D' and 'C ≤ F' definitely true in the expression: A _ B _ C _ D _ E _ F (A) >, ≥, =, <, < (B) ≥, >, =, <, < (C) >, ≥, =, ≤, ≤ (D) >, <, =, ≤, ≤ (E) >, =, >, ≤, ≤
    निम्नलिखित में से कौन-सा विकल्प अभिव्यक्ति A _ B _ C _ D _ E _ F में 'A > D' और 'C ≤ F' को निश्चित रूप से सत्य बनाता है? (A) >, ≥, =, <, < (B) ≥, >, =, <, < (C) >, ≥, =, ≤, ≤ (D) >, <, =, ≤, ≤ (E) >, =, >, ≤, ≤
    (A) (A) >, ≥, =, <, <

    (B) (B) ≥, >, =, <, <

    (C) (C) >, ≥, =, ≤, ≤

    (D) (D) >, <, =, ≤, ≤

    (E) (E) >, =, >, ≤, ≤
    ✅ Answer & Explanation

    Sahi jawab: C) (C) >, ≥, =, ≤, ≤

    Explanation: Method 1 (Dual Condition Elimination Shortcut): Step 1: 'C ≤ F' ko satya hone ke liye C se F tak kewal '≤' aur '=' chinh aane chahiye, koi '<' nahi aana chahiye. - Options (A) aur (B) me antim do chinh '<, <' hain, isliye yeh eliminate ho gaye. Step 2: 'A > D' ko satya hone ke liye A se D tak greater family ke chinh hone chahiye bina kisi opposite sign ke. - Option (D) me doosra chinh '<' aa jata hai jo condition block banata hai, isliye eliminate. Step 3: Option (C) test karein: $A > B \ge C = D \le E \le F$: - A se D tak: $A > B \ge C = D \implies A > D$ (Satya). - C se F tak: $C = D \le E \le F \implies C \le F$ (Satya). Correct option (C) hai. Method 2 (Direct Priority Verification): Option (C) me A se D tak high priority '>' maujood hai aur C se F tak kewal '=' tatha '≤' hain. Dono maange gaye nishkarsh 100% sidhh hote hain. Sahi uttar (C) hai.

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