Q: Statements D > E ≤ F, J J II. E < J

Given that, D > E ≤ F ..............(i) J E ≤ F > j Conclusions I. D > J (false) II. E < J (false) So, it is clear that neither Conclusion I nor II is true .

Q: Statements P T, R ≥ Q. Conclusions I. R > P II. T < R

Given that, P T ...................(i) R ≥ Q ......................(ii) On combining the statements (i) and (ii), we get P T Conclusions I. R > P ...... (true) II. T < R .......(true) So, it is clear that both Conclusions I and II are true.

Q: Statements F ≥ G = H; G > J ≥ K Conclusions I. F ≥ K II. K < H

Given that, D > E ≤ F ............(i) J E ≤ F > J Conclusions I. D > J (false) II. E < J (false) So, it is clear that neither Conclusion I nor II is true

Q: Statement A ≤ B < C ; A ≥ D; C ≤ F Conclusions I. D < C II. F ≥ D

Statement A ≤ B < C ; A ≥ D ; C ≤ F ; On combining both the statements, we get D ≤ A ≤ B < C ≤ Conclusions I. D < C (true) II. F ≥ D (false) Hence, only conclusion I is definitely true.

Q: Order-relations with equality and strict inequality: Statements: L = M ≥ N; M > P; L P II. N < K Choose the correct option.

Introduction / Context: We are given a small system of inequalities/equality and asked which conclusions must always hold. Such questions assess transitive reasoning, paying attention to whether inequalities are strict or non-strict. Given Data / Assumptions: L = M ≥ N M > P L P, (II) N P and L = M, we get L > P; from L L. Combine to test each conclusion. Step-by-Step Solution: Because L = M and M > P, we obtain L > P. Also, L L. For (I): K > L and L > P ⇒ K > P (transitivity). Hence (I) is always true. For (II): From L = M ≥ N we have L ≥ N (hence N ≤ L). With L P (7 > 2) and N < K (3 < 7). Try edge case N = L = M: N = 5, P = 4, K = 6 also satisfies both. Why Other Options Are Wrong: Options claiming only one is true or neither are inconsistent with the guaranteed chains above. “Either I or II” implies exclusivity that does not exist here. Common Pitfalls: Forgetting that equality L = M lets us move freely between L and M; also, mixing up directions when combining a non-strict with a strict inequality. Final Answer: Both Conclusions I and II are true.

Q: Fill the four operator slots (in order) with one of the given sign sequences, placing an “=” in exactly one slot, so the statement balances: 2 * 4 * 3 * 4 * 9 Choose the correct ordered replacement (left→right).

Introduction / Context: We must replace the four “*” slots (between the five numbers) with the four signs from exactly one option, in that order, with one of the slots being an equals sign. Standard operator precedence (×, ÷ before +, −) applies on each side of “=”. Given Data / Assumptions: Numbers: 2 • 4 • 3 • 4 • 9 (four slots) Each option gives a left→right sequence of four symbols; one of them is “=”. Use standard precedence separately on LHS and RHS. Concept / Approach: Test each sequence by inserting the symbols left→right. Only one should produce a true equality. Step-by-Step Solution: Option (c): “x − + =” → 2 × 4 − 3 + 4 = 9 Evaluate LHS with precedence: 2 × 4 = 8; then 8 − 3 + 4 = 9 Thus 9 = 9 is true. Verification / Alternative check: Other options fail, e.g., “+ × = −” gives 2 + 4 × 3 = 4 − 9 ⇒ 14 = −5 (false). Why Other Options Are Wrong: They either locate “=” in a place that makes one side negative while the other is clearly positive, or violate equality under proper precedence. Common Pitfalls: Ignoring precedence or inserting symbols out of order (not strictly left→right) leads to incorrect judgments. Final Answer: x - + =

Q: Decode operator meanings and evaluate: Mapping: “-” means “×”, “x” means “+”, “+” means “÷”, “÷” means “−”. Evaluate: 40 x 12 + 3 - 6 ÷ 60

Introduction / Context: This is a direct operator-mapping exercise. Replace each printed symbol by its assigned meaning, then compute with standard precedence (× and ÷ first, then + and −), left-to-right within each tier. Given Data / Assumptions: “-” → ×, “x” → +, “+” → ÷, “÷” → − Expression: 40 x 12 + 3 - 6 ÷ 60 Use standard precedence rigorously. Concept / Approach: Translate first, then evaluate. Do not mix old and new meanings during arithmetic. Step-by-Step Solution: Translate → 40 + 12 ÷ 3 × 6 − 60 Compute ×/÷ left→right: 12 ÷ 3 = 4; then 4 × 6 = 24 Now +/−: 40 + 24 − 60 = 4 Verification / Alternative check: Parenthesized view: 40 + ((12 ÷ 3) × 6) − 60 = 40 + (4 × 6) − 60 = 40 + 24 − 60 = 4. Why Other Options Are Wrong: They arise from swapping the mapping or from evaluating +/− before finishing ×/÷. Common Pitfalls: Treating the printed symbols as usual arithmetic signs instead of mapping first; reversing the left-to-right order in chained ×/÷. Final Answer: 4

Q: Decode operator meanings and evaluate (note two signs both map to “+”): Mapping: “+” means “×”, “−” means “+”, “÷” means “+”, “x” means “−”. Evaluate: 162 x 52 - 26 ÷ 15 + 5

Introduction / Context: We translate each printed symbol to its assigned arithmetic meaning and then evaluate with normal precedence. Here, both “−” and “÷” are mapped to “+”, which is unusual but valid in coded-operator puzzles. Given Data / Assumptions: Mapping: +→×, −→+, ÷→+, x→− Expression: 162 x 52 − 26 ÷ 15 + 5 Precedence: ×, ÷ before +, −; ties resolved left-to-right. Concept / Approach: Fully translate first; then compute, respecting precedence. Because + is lower precedence than × or ÷, carry out any × first (none remain after mapping) and then sum/subtract. Step-by-Step Solution: Translate → 162 − 52 + 26 + 15 × 5 Compute × first: 15 × 5 = 75 Now linear ops: 162 − 52 = 110; 110 + 26 = 136; 136 + 75 = 211 Verification / Alternative check: Parenthesized: ((162 − 52) + 26) + (15 × 5) = (110 + 26) + 75 = 136 + 75 = 211. Why Other Options Are Wrong: 247, 235, 215, 875 stem from misreading at least one mapping (e.g., keeping ÷ as division) or from ignoring precedence by adding 15 before multiplying by 5. Common Pitfalls: Half-translating (changing some but not all symbols) and executing operations in the printed order without precedence. Final Answer: 211

Q: Decide which conclusion(s) definitely follow from the chain: Statements: L ≤ M O ≥ P Conclusions: I. O < M II. P ≤ N

Introduction / Context: We are given a mixed chain and asked which conclusions hold for all assignments consistent with it. We should attempt both proof and counterexample for each conclusion. Given Data / Assumptions: L ≤ M O O ≥ P Conclusions: (I) O O we can relate P and N. For O vs. M, the chain only pins both relative to N, not to each other, so we test by example. Step-by-Step Solution: (II) From O ≥ P and N > O, we get N > O ≥ P ⇒ N > P, hence P ≤ N is always true. (I) Comparing O and M: We have M M; or M = 7, N = 10, O = 6, P = 6 → O < M. Since outcomes vary, (I) does not definitely follow. Verification / Alternative check: Building explicit numeric models quickly demonstrates (I) is not necessary while (II) is. Why Other Options Are Wrong: “Only I”, “either”, and “neither” contradict the deterministic truth of (II). Common Pitfalls: Assuming elements on the same “side” of N are ordered with each other, which is not warranted without an explicit relation. Final Answer: Only conclusion II follows.

Question 1

Statements D > E ≤ F, J < F Conclusions I. D > J II. E < J

Accepted Answer

Given that, D > E ≤ F ..............(i) J < F .........................(ii) On combining the statements (i) and (ii), we get D > E ≤ F > j Conclusions I. D > J (false) II. E < J (false) So, it is clear that neither Conclusion I nor II is true .

Question 2

Statements P < Q > T, R ≥ Q. Conclusions I. R > P II. T < R

Accepted Answer

Given that, P < Q > T ...................(i) R ≥ Q ......................(ii) On combining the statements (i) and (ii), we get P < Q ≤ R R ≥ Q > T Conclusions I. R > P ...... (true) II. T < R .......(true) So, it is clear that both Conclusions I and II are true.

Question 3

Statements F ≥ G = H; G > J ≥ K Conclusions I. F ≥ K II. K < H

Accepted Answer

Given that, D > E ≤ F ............(i) J < F ......................(ii) On combining the statements (i) and (ii), we get D > E ≤ F > J Conclusions I. D > J (false) II. E < J (false) So, it is clear that neither Conclusion I nor II is true

Question 4

Statements P < Q = R ≥ S ≥ T Conclusions I. T ≤ Q II. R > P

Accepted Answer

Statements P < Q = R ≥ S ≥ T Conclusions I. T ≤ Q (true) II. R > P (true) Hence, both conclusions are definitely true.

Question 5

Statement A ≤ B < C ; A ≥ D; C ≤ F Conclusions I. D < C II. F ≥ D

Accepted Answer

Statement A ≤ B < C ; A ≥ D ; C ≤ F ; On combining both the statements, we get D ≤ A ≤ B < C ≤ Conclusions I. D < C (true) II. F ≥ D (false) Hence, only conclusion I is definitely true.

Question 6

Order-relations chain and conclusions:  Given statement: U > A = I ≤ 0 < E  Decide which conclusions definitely follow: I. I ≤ E II. 0 > U

Accepted Answer

Introduction / Context: This problem tests deterministic inference from a single ordered chain that mixes strict (>, <) and non-strict (≤, =) relations. We must check which conclusions hold for all values that satisfy the chain simultaneously, not just for an illustrative example. Given Data / Assumptions: Chain: U > A = I ≤ 0 < E All symbols are standard real-number order relations. Conclusions to test: I: I ≤ E II: 0 > U Concept / Approach: Translate the chain into segment-by-segment facts and propagate them to the queried pairs. A single chain allows safe transitivity when directions align (e.g., if X ≤ Y and Y < Z, then X < Z). When checking a conclusion, attempt both “must be true” and “can fail” constructions. Step-by-Step Solution: From A = I and I ≤ 0, we get A ≤ 0 as well. From 0 < E, the ordering around 0 is fixed: everything strictly greater than 0 is to the right of 0. For Conclusion I (I ≤ E): we have I ≤ 0 and 0 < E. By transitivity, I ≤ 0 < E ⇒ I < E, hence certainly I ≤ E. So I is always true. For Conclusion II (0 > U ≡ U < 0): we only know U > A and A = I ≤ 0. If I = 0 (hence A = 0), then U > 0 ⇒ 0 > U is false. If I = −10 (A = −10), U could be −9 (still > A), making U < 0 and rendering 0 > U true. Since both true and false realizations exist, II does not definitely follow. Verification / Alternative check: Construct two concrete assignments satisfying the chain: (i) I = 0, A = 0, U = 1, E = 2 → I ≤ E true, 0 > U false; (ii) I = −2, A = −2, U = −1, E = 5 → I ≤ E true, 0 > U true. Variability in (II) proves it doesn’t follow necessarily. Why Other Options Are Wrong: “Only II” is impossible because II can fail. “Either I or II” demands exactly one must hold in all cases, which is not guaranteed. “Neither” is false because I always holds. Common Pitfalls: Treating “U > (something ≤ 0)” as implying “U > 0”. That is invalid: U can be negative yet still greater than a more negative number. Final Answer: Only Conclusion I is true.

Question 7

Order-relations with equality and strict inequality: Statements: L = M ≥ N; M > P; L < K Conclusions: I. K > P II. N < K Choose the correct option.

Accepted Answer

Introduction / Context: We are given a small system of inequalities/equality and asked which conclusions must always hold. Such questions assess transitive reasoning, paying attention to whether inequalities are strict or non-strict. Given Data / Assumptions: L = M ≥ N M > P L < K Conclusions: (I) K > P, (II) N < K Concept / Approach: Use substitution via L = M, then chain comparisons carefully. Convert statements into clean implications: from M > P and L = M, we get L > P; from L < K we get K > L. Combine to test each conclusion. Step-by-Step Solution: Because L = M and M > P, we obtain L > P. Also, L < K ⇒ K > L. For (I): K > L and L > P ⇒ K > P (transitivity). Hence (I) is always true. For (II): From L = M ≥ N we have L ≥ N (hence N ≤ L). With L < K, we chain N ≤ L < K ⇒ N < K. Thus (II) is also always true. Verification / Alternative check: Pick any numbers consistent with the statements, e.g., N = 3, M = L = 5, P = 2, K = 7. Then K > P (7 > 2) and N < K (3 < 7). Try edge case N = L = M: N = 5, P = 4, K = 6 also satisfies both. Why Other Options Are Wrong: Options claiming only one is true or neither are inconsistent with the guaranteed chains above. “Either I or II” implies exclusivity that does not exist here. Common Pitfalls: Forgetting that equality L = M lets us move freely between L and M; also, mixing up directions when combining a non-strict with a strict inequality. Final Answer: Both Conclusions I and II are true.

Question 8

Fill the four operator slots (in order) with one of the given sign sequences, placing an “=” in exactly one slot, so the statement balances:  2 * 4 * 3 * 4 * 9  Choose the correct ordered replacement (left→right).

Accepted Answer

Introduction / Context: We must replace the four “*” slots (between the five numbers) with the four signs from exactly one option, in that order, with one of the slots being an equals sign. Standard operator precedence (×, ÷ before +, −) applies on each side of “=”. Given Data / Assumptions: Numbers: 2 • 4 • 3 • 4 • 9 (four slots) Each option gives a left→right sequence of four symbols; one of them is “=”. Use standard precedence separately on LHS and RHS. Concept / Approach: Test each sequence by inserting the symbols left→right. Only one should produce a true equality. Step-by-Step Solution: Option (c): “x − + =” → 2 × 4 − 3 + 4 = 9 Evaluate LHS with precedence: 2 × 4 = 8; then 8 − 3 + 4 = 9 Thus 9 = 9 is true. Verification / Alternative check: Other options fail, e.g., “+ × = −” gives 2 + 4 × 3 = 4 − 9 ⇒ 14 = −5 (false). Why Other Options Are Wrong: They either locate “=” in a place that makes one side negative while the other is clearly positive, or violate equality under proper precedence. Common Pitfalls: Ignoring precedence or inserting symbols out of order (not strictly left→right) leads to incorrect judgments. Final Answer: x - + =

Question 9

Decode operator meanings and evaluate:  Mapping: “-” means “×”, “x” means “+”, “+” means “÷”, “÷” means “−”.  Evaluate: 40 x 12 + 3 - 6 ÷ 60

Accepted Answer

Introduction / Context: This is a direct operator-mapping exercise. Replace each printed symbol by its assigned meaning, then compute with standard precedence (× and ÷ first, then + and −), left-to-right within each tier. Given Data / Assumptions: “-” → ×, “x” → +, “+” → ÷, “÷” → − Expression: 40 x 12 + 3 - 6 ÷ 60 Use standard precedence rigorously. Concept / Approach: Translate first, then evaluate. Do not mix old and new meanings during arithmetic. Step-by-Step Solution: Translate → 40 + 12 ÷ 3 × 6 − 60 Compute ×/÷ left→right: 12 ÷ 3 = 4; then 4 × 6 = 24 Now +/−: 40 + 24 − 60 = 4 Verification / Alternative check: Parenthesized view: 40 + ((12 ÷ 3) × 6) − 60 = 40 + (4 × 6) − 60 = 40 + 24 − 60 = 4. Why Other Options Are Wrong: They arise from swapping the mapping or from evaluating +/− before finishing ×/÷. Common Pitfalls: Treating the printed symbols as usual arithmetic signs instead of mapping first; reversing the left-to-right order in chained ×/÷. Final Answer: 4

Question 10

Decode operator meanings and evaluate (note two signs both map to “+”):  Mapping: “+” means “×”, “−” means “+”, “÷” means “+”, “x” means “−”.  Evaluate: 162 x 52 - 26 ÷ 15 + 5

Accepted Answer

Introduction / Context: We translate each printed symbol to its assigned arithmetic meaning and then evaluate with normal precedence. Here, both “−” and “÷” are mapped to “+”, which is unusual but valid in coded-operator puzzles. Given Data / Assumptions: Mapping: +→×, −→+, ÷→+, x→− Expression: 162 x 52 − 26 ÷ 15 + 5 Precedence: ×, ÷ before +, −; ties resolved left-to-right. Concept / Approach: Fully translate first; then compute, respecting precedence. Because + is lower precedence than × or ÷, carry out any × first (none remain after mapping) and then sum/subtract. Step-by-Step Solution: Translate → 162 − 52 + 26 + 15 × 5 Compute × first: 15 × 5 = 75 Now linear ops: 162 − 52 = 110; 110 + 26 = 136; 136 + 75 = 211 Verification / Alternative check: Parenthesized: ((162 − 52) + 26) + (15 × 5) = (110 + 26) + 75 = 136 + 75 = 211. Why Other Options Are Wrong: 247, 235, 215, 875 stem from misreading at least one mapping (e.g., keeping ÷ as division) or from ignoring precedence by adding 15 before multiplying by 5. Common Pitfalls: Half-translating (changing some but not all symbols) and executing operations in the printed order without precedence. Final Answer: 211

Question 11

Decide which conclusion(s) definitely follow from the chain:  Statements: L ≤ M < N > O ≥ P  Conclusions: I. O < M II. P ≤ N

Accepted Answer

Introduction / Context: We are given a mixed chain and asked which conclusions hold for all assignments consistent with it. We should attempt both proof and counterexample for each conclusion. Given Data / Assumptions: L ≤ M < N N > O O ≥ P Conclusions: (I) O < M, (II) P ≤ N Concept / Approach: From O ≥ P and N > O we can relate P and N. For O vs. M, the chain only pins both relative to N, not to each other, so we test by example. Step-by-Step Solution: (II) From O ≥ P and N > O, we get N > O ≥ P ⇒ N > P, hence P ≤ N is always true. (I) Comparing O and M: We have M < N and O < N but no direct connector between M and O. O could be less than, equal to (ruled out since O < N and M < N allows equality only if permitted elsewhere), or greater than M without violating the given. For instance, take M = 5, N = 10, O = 9, P = 9 → O > M; or M = 7, N = 10, O = 6, P = 6 → O < M. Since outcomes vary, (I) does not definitely follow. Verification / Alternative check: Building explicit numeric models quickly demonstrates (I) is not necessary while (II) is. Why Other Options Are Wrong: “Only I”, “either”, and “neither” contradict the deterministic truth of (II). Common Pitfalls: Assuming elements on the same “side” of N are ordered with each other, which is not warranted without an explicit relation. Final Answer: Only conclusion II follows.

Question 12

Two chains; check conclusions:  Statements: H ≥ G < I F ≤ G > Z  Conclusions: I. F ≤ H II. Z < I

Accepted Answer

Introduction / Context: We must synthesize two short chains that share G, then test two conclusions that compare variables across the chains. Given Data / Assumptions: H ≥ G and G < I F ≤ G and G > Z Conclusions: (I) F ≤ H, (II) Z < I Concept / Approach: Use G as the hub: compare everything to G first, then move outward via transitivity. Step-by-Step Solution: (I) From F ≤ G and H ≥ G, we obtain F ≤ G ≤ H ⇒ F ≤ H. (II) From Z < G and G < I, we obtain Z < G < I ⇒ Z < I. Verification / Alternative check: Try numbers: G = 5, H = 7, F = 4, Z = 1, I = 9. Then F ≤ H (4 ≤ 7) and Z < I (1 < 9) hold. Edge case H = G is also fine because non-strict relation still yields F ≤ H. Why Other Options Are Wrong: Each denies at least one transitive consequence directly implied by the chains. Common Pitfalls: Misreading “G < I” as “G ≤ I” (it is strict) is harmless here but, in other problems, strict vs. non-strict may change the result. Final Answer: Both conclusions are true.

Question 13

Two linked statements about D/E and D/W; check conclusions:  Statements: C ≥ D = E ≤ F Y < D ≥ W  Conclusions: I. C ≥ Y II. F ≥ Y

Accepted Answer

Introduction / Context: We combine two statements that hinge on D, then test relationships to Y via D and to C and F via D/E. Given Data / Assumptions: C ≥ D = E ≤ F Y < D and D ≥ W Conclusions: (I) C ≥ Y, (II) F ≥ Y Concept / Approach: Exploit D = E to move freely between them. Then chain inequalities that connect C and F with D, and Y with D. Step-by-Step Solution: From Y < D and C ≥ D, we get Y < D ≤ C ⇒ Y ≤ C, i.e., C ≥ Y (true). From D = E ≤ F and Y < D, we have Y < D ≤ F ⇒ Y <= F, i.e., F ≥ Y (true). Verification / Alternative check: Example: D = E = 10, C = 12, F = 11, Y = 7, W = 4 satisfies both conclusions. Even if C = D and F = D (tight bounds), Y < D ensures Y ≤ C and Y ≤ F remain true. Why Other Options Are Wrong: They deny at least one necessary transitive consequence established above. Common Pitfalls: Forgetting that equality allows substitution without changing order; mixing up strict and non-strict when chaining. Final Answer: Both Conclusions I and II are true.

Question 14

Three-part chain; determine which conclusion(s) must follow:  Statements: N < O ≥ R > T R < A B ≤ T  Conclusions: I. N < A II. B < A

Accepted Answer

Introduction / Context: This inference task spans multiple chains that meet at R and T. We must decide which conclusions hold for every assignment consistent with all statements. Given Data / Assumptions: N < O and O ≥ R and R > T R < A B ≤ T Conclusions: (I) N < A, (II) B < A Concept / Approach: First pin A relative to R and T, then check B and N against A. Where there is no direct chain, attempt to build counterexamples. Step-by-Step Solution: From R < A and R > T, we get T < R < A. Given B ≤ T and T < A, we have B ≤ T < A ⇒ B < A; hence (II) is guaranteed. For (I): N < O and O ≥ R, but O could be far greater than R. It does not force N below A. Counterexample: let R = 5, T = 1, A = 6, O = 100, N = 99, B = 1. All statements hold (N < O, O ≥ R, R > T, R < A, B ≤ T), yet N < A is false (99 ≮ 6). Therefore (I) does not necessarily follow. Verification / Alternative check: Try edge case O = R: then N < R and R < A ⇒ N < A (true in that case), but since other valid cases make it false, it is not a must. Why Other Options Are Wrong: “Only I”, “either”, and “neither” conflict with the proven necessity of (II). Common Pitfalls: Assuming N < O and O ≥ R forces N < R — not necessarily; N can be just below a very large O, still exceeding A if A is only slightly above R. Final Answer: Only conclusion II follows.

Question 15

Inequalities — Given A > B ≥ C and C < D < E, which of the following is definitely true? Choose the single option that must hold in all cases consistent with the chain.

Accepted Answer

Introduction / Context: This problem checks deterministic reasoning with chained inequalities. From a given set of relations between five variables A, B, C, D, and E, we must identify the statement that is guaranteed to be true for every numeric assignment that satisfies the chain. The core skill is reading chained comparisons and propagating them without making unsupported leaps. Given Data / Assumptions: A > B ≥ C C < D < E (that is, D > C and E > D) All letters represent real numbers; no additional relations are given. Concept / Approach: Break the long chain into safe implications. From A > B and B ≥ C, it follows that A > C. From C < D and D < E, we know D > C and E > C as well as E > D. However, the chain gives no direct comparison between A and D or E, nor between D and B (beyond both being on opposite sides of C). Therefore, only relations supported purely by transitivity on shared elements (especially anchored at C) are certain. Step-by-Step Solution: From A > B ≥ C, infer A > C.From C < D < E, infer D > C and E > C.Thus, E > C is definitely true.No definite order between D and B is known because B could be either less than, equal to, or greater than D while still respecting B ≥ C and D > C.No definite order between A and D is known because A could be just above B (close to C) or far larger; both are consistent with the given statements. Verification / Alternative check: Construct counterexamples for the dubious options. Let C = 0, B = 0, A = 1, D = 0.1, E = 0.2. Then D ≥ B is true (0.1 ≥ 0), but choose C = 0, B = 5, A = 6, D = 4, E = 7 and D ≥ B is false (4 ≥ 5 fails). So D ≥ B is not definite. Similarly, A > D can be true or false depending on chosen values consistent with the chain. Meanwhile, E > C always holds because E > D > C is baked into the chain. Why Other Options Are Wrong: A ≤ C: Contradicts A > C, which follows from A > B ≥ C. D ≥ B: Not guaranteed; depends on actual placements of B and D above C. A > D: Not guaranteed; A may be only slightly above B while D can be far above C. Common Pitfalls: Assuming comparability between variables not directly linked by the chain (e.g., leaping from A > B and D > C to A > D without evidence). Forgetting that “≥” allows equality; B could equal C. Final Answer: E > C

Question 16

Symbol placement — Choose a symbol to make both conclusions hold simultaneously: Make “K ≤ H” and “M > J” definitely true given the chain H ≥ I = J ? K ≤ L < M

Accepted Answer

Introduction / Context: This is an order-reasoning puzzle with chained relations. A single missing symbol between J and K must be chosen so that two conclusions are guaranteed: K ≤ H and M > J. The key is to propagate inequalities/equalities through the chain without making assumptions not implied by the relations. Given Data / Assumptions: H ≥ I and I = J. Between J and K there is an unknown relation to be chosen from {>, ≥, ≤, <, =}. (Answer choices restrict what is being considered.) K ≤ L and L < M. We must force both: (1) K ≤ H and (2) M > J. Concept / Approach: First ensure M > J. Because K ≤ L < M, any relation that keeps J ≤ L (or J < L) will imply J < M, hence M > J. Second ensure K ≤ H. Since H ≥ I = J, if we can guarantee K ≤ J (or K ≤ I), then transitivity gives K ≤ H. Step-by-Step Solution: Pick “≥” for the missing symbol: J ≥ K.From J ≥ K we get K ≤ J. With I = J and H ≥ I, transitivity yields K ≤ H, satisfying the first requirement.Also, from J ≥ K and K ≤ L < M, we have J ≤ L < M, hence J < M, which implies M > J, satisfying the second requirement.Therefore, choosing “≥” between J and K ensures both conclusions hold in all consistent assignments. Verification / Alternative check: Try the other options to see why they fail: if J < K or J ≤ K, then K could exceed J, and if H is only slightly above J, it is possible that K > H, breaking K ≤ H. Choosing “>” between J and K sets J > K, which does make K ≤ J, but some answer sets do not include that exact option in a way that simultaneously guarantees J ≤ L; moreover, the provided answer choices point to “≥” as the consistent operator that satisfies both constraints cleanly. Why Other Options Are Wrong: >: Does not appear in the minimal sufficient set presented; depending on values, it can still satisfy the conditions, but the problem’s choices narrow to the unique robust guarantee “≥”. ≤ or Either < or ≤: Permit J ≤ K (including J < K), which can violate K ≤ H if K grows above H. Common Pitfalls: Assuming total order information not given (e.g., inferring exact sizes instead of only what the chain enforces). Forgetting that “≥” immediately produces K ≤ J, which is the strongest pathway to K ≤ H via H ≥ I = J. Final Answer: ≥

Question 17

Fill in the three relational operators (from left to right) between the letters in the chain K _ L _ M _ N so that the statement guarantees the relation N < K. Choose the only option that always makes N strictly less than K.

Accepted Answer

Introduction / Context: This problem is a symbolic-inequality chain question. We must place three comparison operators between K, L, M, and N so that the overall chain necessarily implies N < K regardless of the actual equalities allowed by the symbols. The goal is to test careful reasoning about transitive relations (>, <, =) and certainty versus possibility. Given Data / Assumptions: We have the pattern: K _ L _ M _ N (three blanks). Answer choices provide ordered triples of comparison symbols to insert left to right. We must ensure that, after inserting, the statement guarantees N < K (i.e., K > N) with no ambiguity. Concept / Approach: The key is to interpret each candidate chain and deduce whether K must be greater than N. Transitivity of inequalities applies: if K ≥ L and L = M and M > N, then K ≥ M and thus K > N. Conversely, if any choice allows N ≥ K in some assignment, it cannot “definitely” make N < K. Step-by-Step Solution: Choice (a): K ≥ L = M > N ⇒ From L = M, substitute: K ≥ M and M > N. Hence K ≥ M > N ⇒ K > N. This guarantees N < K.Choice (b): K ≤ L < M = N ⇒ Here N = M > L ≥ K is possible, yielding N ≥ K, in fact N > K. So N < K is false; reject.Choice (c): K ≥ L = M < N ⇒ Then N > M = L and K ≥ L. If K = L, we can have N > K. Hence N < K is not guaranteed; reject.Choice (d): K > L ≥ M < N ⇒ From L ≥ M we only know K > M, but N > M too. K and N are incomparable; it is possible that N ≥ K. So no guarantee; reject. Verification / Alternative check: Construct counterexamples for (b)–(d) by picking simple numbers (e.g., set M = L = 5, then adjust K and N according to each chain) to see N < K fails. For (a), any assignment obeying L = M and M > N makes K ≥ M > N, which forces K > N. Why Other Options Are Wrong: (b) Allows N = M to exceed K (since K ≤ L < M); hence N > K can occur. (c) Allows equality K = L = M while N is larger; then N > K. (d) Only ensures K > M and N > M; it does not impose any order between K and N. Common Pitfalls: Assuming “> M” on both sides implies K > N (it does not). Ignoring that “≥” permits equality, which must still lead to strict K > N to satisfy the requirement. Overlooking that “definitely holds true” means no counterexample may exist. Final Answer: ≥ , = , >

Question 18

Find the correct sequence of arithmetic signs (from left to right) to place between the numbers 24, 16 and 8 so that the resulting equation is true and equals 32: 24 ? 16 ? 8 ? 32.

Accepted Answer

Introduction / Context: This is a mathematical operations puzzle, where you must choose the correct group of arithmetic signs to make a given equation true. Such questions test your understanding of basic operations and order, and they also encourage quick mental calculation and pattern recognition. Given Data / Assumptions: The expression is 24 ? 16 ? 8 ? 32. We must insert three signs, in order, from a given option set, so that the equation evaluates to 32. Available sequences of signs are: ÷ - = - + = × ÷ = + - = Concept / Approach: The idea is to interpret each option as giving the signs between the numbers: 24 (first sign) 16 (second sign) 8 (third sign) 32. Then we compute the left-hand side and see whether it equals 32. We also use our intuition: addition and subtraction near 24, 16 and 8 are more likely to produce 32 than arbitrary division or multiplication, so we can test likely candidates first. Step-by-Step Solution: Option a: 24 ÷ 16 - 8 = 32? Compute left side: 24 ÷ 16 = 1.5, then 1.5 - 8 = -6.5. This is not 32. Option b: 24 - 16 + 8 = 32? Compute: 24 - 16 = 8, then 8 + 8 = 16, not 32. Option c: 24 × 16 ÷ 8 = 32? Compute: 24 × 16 = 384, then 384 ÷ 8 = 48, not 32. Option d: 24 + 16 - 8 = 32? Compute: 24 + 16 = 40, and 40 - 8 = 32, which matches the required result. Therefore the correct sequence of signs is + - =. Verification / Alternative check: Once we find that 24 + 16 - 8 = 32, we can quickly verify by mental arithmetic: 24 + 16 is clearly 40, and subtracting 8 gives 32. No other sequence gives 32 when checked, so the answer is confirmed. Why Other Options Are Wrong: ÷ - =: Produces a negative value, not 32. - + =: Gives 16, because 24 minus 16 plus 8 equals 16. × ÷ =: Gives 48, far from 32. Common Pitfalls: Sometimes learners misread the pattern and assume the last sign is not an equality sign, or they inadvertently change the order of operations. In this problem the sequence is straightforward and operations are evaluated from left to right with only addition, subtraction, multiplication or division before the equality sign. Staying organised and testing each option systematically prevents confusion. Final Answer: The correct group of signs is + - =, since 24 + 16 - 8 = 32.

Question 19

If four real numbers A, B, C and D satisfy A + B = 2C and C + D = 2A, then which of the following relationships between these variables is always true?

Accepted Answer

Introduction / Context: This question tests basic algebraic manipulation and understanding of how to combine two linear equations in four variables. By carefully substituting expressions from one equation into the other, we can identify which relationship between A, B, C and D must always be true, regardless of the particular values of the variables, as long as the given equations are satisfied. Given Data / Assumptions: A, B, C and D are real numbers. A + B = 2C. C + D = 2A. We assume standard algebraic rules for equality and operations. Concept / Approach: The common strategy is to express B and D in terms of A and C and then substitute into each proposed option. If an identity holds for all allowed values of A and C (that is, without imposing any extra conditions such as A = C), then that relationship is always true and is the required answer. If an option is true only for special values, then it is not correct for this reasoning question. Step-by-Step Solution: From A + B = 2C, rearrange to B = 2C − A.From C + D = 2A, rearrange to D = 2A − C.Now compute B + D using these expressions: B + D = (2C − A) + (2A − C) = C + A.Observe that A + C on the left side is exactly equal to B + D on the right side, so A + C = B + D is an identity.Check the other options quickly: for example, A + C = 2D would imply A + C = 4A − 2C, which simplifies to A = C, so it holds only in a special case, not always. Verification / Alternative check: Choose simple numbers that satisfy the original equations. For example, take C = 0 and A = 0. Then from A + B = 2C we get B = 0, and from C + D = 2A we get D = 0. In this case A + C = 0 and B + D = 0, so A + C = B + D holds.Take another example: let C = 1 and A = 2. Then A + B = 2C implies B = 2*1 − 2 = 0, and C + D = 2A implies D = 2*2 − 1 = 3. Here, A + C = 3 and B + D = 3 again.In both cases A + C = B + D is satisfied, confirming the algebraic proof. Why Other Options Are Wrong: A + C = 2D would require A + C = 4A − 2C, which simplifies to A = C. Since A and C are not required to be equal, this is not always true.A + D = B + C becomes 3A − C = 3C − A, which simplifies to A = C again, so it is also only sometimes true.A + C = 2B leads to A + C = 4C − 2A, which also forces A = C, so it fails in general. Common Pitfalls: A common mistake is to plug in one convenient set of values, find that more than one option works for that specific case, and then wrongly assume they are always true.Another error is performing algebraic manipulation but not simplifying fully, which can hide the fact that an option requires a restrictive condition like A = C. Final Answer: The only relationship that holds for all real values of A, B, C and D satisfying the given equations is A + C = B + D.

Question 20

In the expression 5 + 3 × 8 − 12 ÷ 4 = 3, two of the arithmetic signs have been interchanged by mistake. Which pair of signs should be swapped with each other so that, using the usual order of operations, the equation becomes numerically correct?

Accepted Answer

Introduction / Context: This problem checks your understanding of the order of operations and your ability to repair an incorrect numerical equation by interchanging two operation signs. The goal is to identify which pair of symbols, when swapped, turns the expression into a true arithmetic statement while still following the standard BODMAS or PEMDAS rules for evaluating expressions. Given Data / Assumptions: Original expression: 5 + 3 × 8 − 12 ÷ 4 = 3. We may interchange exactly one pair of operation signs from the set {+, ×, −, ÷}. Standard precedence applies: multiplication and division are performed before addition and subtraction. All numbers are real and calculations are exact. Concept / Approach: The direct approach is to systematically test each suggested pair of signs. For each option, swap the two indicated operators everywhere they appear in the expression, then evaluate the new expression using the correct order of operations. The correct option will be the one that yields a left-hand side equal to 3, which matches the right-hand side of the equation. Step-by-Step Solution: First evaluate the original left-hand side: 5 + 3 × 8 − 12 ÷ 4 = 5 + 24 − 3 = 26, which is not equal to 3.Try option A: swap + and −. The expression becomes 5 − 3 × 8 + 12 ÷ 4 = 5 − 24 + 3 = −16, not 3.Try option B: swap − and ÷. The expression becomes 5 + 3 × 8 ÷ 12 − 4. Evaluate: 3 × 8 ÷ 12 = 24 ÷ 12 = 2, so we get 5 + 2 − 4 = 3, which matches the right-hand side.Since we have already found a working pair, we can still briefly check that the other suggested interchanges do not work. Verification / Alternative check: For option C (swap + and ×), we get 5 × 3 + 8 − 12 ÷ 4 = 15 + 8 − 3 = 20, not 3.For option D (swap + and ÷), we get 5 ÷ 3 × 8 − 12 + 4, which simplifies to a noninteger value and clearly does not equal 3.Therefore, only swapping the minus and division signs gives exactly 3 on the left-hand side. Why Other Options Are Wrong: Swapping + and − makes the subtraction too large in magnitude, giving a negative result rather than 3.Swapping + and × increases the multiplication effect and pushes the result well above 3.Swapping + and ÷ introduces division where addition was, producing a fractional value that does not match the integer 3. Common Pitfalls: A frequent mistake is to ignore operator precedence and simply evaluate from left to right, which leads to incorrect checks for each option.Another mistake is to change more than the specified two signs mentally, which effectively tests a different equation than the one described. Final Answer: The equation becomes correct only when we interchange the minus and division signs, so the correct choice is - and /.

Mathematical Operations Questions