Question 1

Insert the arithmetic operations to make the statement true. 4 □ 3 □ 4 = 48

Accepted Answer

Introduction / Context: Pick two operations to make the equality hold. Try multiplication where a large result is needed, because 48 is much larger than 4 and 3 individually. Given Data / Assumptions: Template: 4 □ 3 □ 4 = 48 Standard precedence applies (but both candidates are multiplications in the correct choice). Concept / Approach: The quickest way to 48 with 4, 3, and 4 is 4 × 3 × 4 = 12 × 4 = 48. Step-by-Step Solution: Choose “× ×”. Then 4 × 3 × 4 = 48. Verification / Alternative check: Any use of “+” or “−” cannot reach as high as 48 starting from these small integers. Why Other Options Are Wrong: They evaluate to values different from 48 (e.g., 4 × 3 + 4 = 16). Common Pitfalls: Overlooking that two multiplications are permitted and give the exact target. Final Answer: x x

Question 2

Interchange of operators and digits: Which one swap makes the equation true? Given: 6 × 4 + 2 = 16. Swap exactly the indicated operator pair globally and interchange the indicated pair of numbers to balance the equality.

Accepted Answer

Introduction / Context: This problem asks you to make a false statement true by applying two precise edits to the left-hand side (LHS): (1) swap a pair of operators globally and (2) interchange a named pair of digits wherever they occur. Such items test rule application, operator precedence, and careful sequencing of edits, which are common in mathematical-operations reasoning questions. Given Data / Assumptions: Target equation: 6 × 4 + 2 = 16 (currently false because 6 × 4 + 2 = 24 + 2 = 26). Each option specifies an operator swap (between + and ×) and a digit swap (between two given numerals). Edits are global on the LHS and performed in the natural reading order: apply the operator swap, then interchange the digits. After edits, evaluate with standard precedence: × and ÷ before + and −. Concept / Approach: For each option, first swap all instances of the named operators, then swap the specified digits throughout the LHS, and finally evaluate. The correct option yields 16 exactly, matching the RHS. Step-by-Step Solution: Consider Option C: “+ and x, 4 and 6”.Operator swap (+ ↔ ×): the LHS 6 × 4 + 2 becomes 6 + 4 × 2.Digit swap (4 ↔ 6) globally in the LHS: 6 ↔ 4 turns 6 + 4 × 2 into 4 + 6 × 2.Evaluate with precedence: 6 × 2 = 12; then 4 + 12 = 16.The edited LHS equals 16, so the equality holds. Verification / Alternative check: Option A is incomplete (mentions “and 4” only), so it cannot define a valid digit swap pair.Option B (“+ and x, 2 and 4”) produces 6 + 4 × 2 → swap 2 ↔ 4 → 6 + 2 × 4 = 14, not 16.Option D (“None of these”) is false because Option C has already produced a valid equality. Why Other Options Are Wrong: They either fail to specify a coherent digit pair or, after correct transformation, evaluate to a number different from 16 (typically 14 or 26 due to precedence effects). Common Pitfalls: Forgetting that the operator swap is global—not local to one position. Applying the digit swap before changing operators, which can lead to different and incorrect outcomes. Ignoring operator precedence and computing left-to-right for mixed × and +. Final Answer: + and x, 4 and 6.

Question 3

Fill the four “*” with a valid sequence of operators so that the equation balances. Target form: 15 * 24 * 3 * 6 * 17 (insert exactly one operator between each pair to obtain a true equality).

Accepted Answer

Introduction / Context: Operator-placement puzzles present a sequence of numbers with placeholder slots for arithmetic signs. The goal is to choose a sequence of operators (respecting normal precedence) that makes the equation true. Such problems strengthen procedural fluency and precedence awareness (×, ÷ before +, −). Given Data / Assumptions: Template: 15 * 24 * 3 * 6 * 17. Each option provides an ordered list of four operators that replace the four “*” in sequence. Standard operator precedence applies. Concept / Approach: Insert the operators from each option in order and evaluate the left-hand side (LHS). A correct option yields an identity of the form “expression = 17”. Step-by-Step Solution: Try Option A: + ÷ - =.Substitute: 15 + 24 ÷ 3 - 6 = 17.Compute precedence: 24 ÷ 3 = 8.Then LHS = 15 + 8 - 6 = 17, which matches the RHS 17. Verification / Alternative check: Option B: 15 + 24 × 3 = 6 ÷ 17 ⇒ LHS is 15 + 72 = 87; RHS ≈ 0.3529; not true.Option C: 15 - 24 × 3 = 6 + 17 ⇒ LHS = 15 - 72 = -57; RHS = 23; not equal.Option D: 15 - 24 ÷ 3 + 6 = 17 ⇒ LHS = 15 - 8 + 6 = 13; not 17. Why Other Options Are Wrong: They either inflate the LHS via multiplication or undercut it with subtraction/division such that the final value is not 17. Common Pitfalls: Ignoring precedence and doing operations strictly left-to-right. Mistakenly treating “=” as a binary operator that must be computed; it is the comparator splitting LHS and RHS. Final Answer: + ÷ - = produces the true equation 15 + 24 ÷ 3 - 6 = 17.

Question 4

Decode the custom operator “a b ÷ c”: From the patterns 2 3 ÷ 4 = 2 and 4 4 ÷ 8 = 32, infer the rule and find the value of 6 2 ÷ 4.

Accepted Answer

Introduction / Context: In custom-operator puzzles, juxtaposed numbers and a symbol indicate a rule you must infer from given examples. Once the rule is deduced, apply it to a new triple to compute the answer. This builds pattern recognition and abstraction skills in arithmetic. Given Data / Assumptions: Examples: 2 3 ÷ 4 = 2 and 4 4 ÷ 8 = 32. Query: 6 2 ÷ 4 = ? Spacing suggests a ternary operation f(a, b, c) with the printed “÷” acting as part of the custom rule name, not ordinary division. Concept / Approach: Look for a simple algebraic pattern that satisfies both examples. A natural hypothesis is exponent-then-division: a^b / c. Step-by-Step Solution: Test a^b / c on the examples:For 2 3 ÷ 4: 2^3 / 4 = 8 / 4 = 2 ✅For 4 4 ÷ 8: 4^4 / 8 = 256 / 8 = 32 ✅Therefore the rule is consistent: f(a, b, c) = a^b / c.Apply to 6 2 ÷ 4: 6^2 / 4 = 36 / 4 = 9. Verification / Alternative check: Other simple interpretations (like a × b ÷ c, or (a + b) × c) fail on one of the two exemplars, so exponent-then-division is the minimal consistent rule. Why Other Options Are Wrong: 10, 8, and 12 arise from misreading the symbol as ordinary division or using (a × b)/c or (a + b) × c, which do not fit both exemplars simultaneously. Common Pitfalls: Forgetting exponent precedence and trying to distribute erroneously. Interpreting the printed “÷” as standard division between the second and third numbers only. Final Answer: 9.

Question 5

Decode the custom symbols: If 2 & 6 % 7 = 33 and 4 & 6 % 8 = 54, evaluate 6 % 8 & 9. Assume consistent precedence for symbol “&” (concatenation) before “%” (addition) inferred from the examples.

Accepted Answer

Introduction / Context: Symbol-decoding problems define new operations implicitly through examples. Your task is to infer what each symbol means and which operation has higher precedence. Here, we are given two evaluations and must compute a third expression using the same implicit rules. Given Data / Assumptions: 2 & 6 % 7 = 33. 4 & 6 % 8 = 54. We must evaluate 6 % 8 & 9. Concept / Approach: A compact rule that fits both examples is: “&” means concatenation (placing digits together to form a numeral) and “%” means addition. To make the first two identities hold simultaneously, “&” must be evaluated before “%”. Step-by-Step Solution: Check the rule on the exemplars with precedence “&” before “%”:2 & 6 % 7 ⇒ (2&6) % 7 ⇒ 26 + 7 = 33 ✅4 & 6 % 8 ⇒ (4&6) % 8 ⇒ 46 + 8 = 54 ✅Now apply to the query 6 % 8 & 9: evaluate “&” first in the right term.8 & 9 = 89; then 6 % (89) = 6 + 89 = 95. Verification / Alternative check: If one incorrectly gives “%” higher precedence (i.e., 6 + 8 = 14, then 14 & 9 = 149), the result 149 is not among the options, confirming the intended precedence. Why Other Options Are Wrong: 77 and 99 arise from mis-applied order (e.g., partial concatenation or treating “%” differently), while “None of these” is false because 95 is attainable and listed. Common Pitfalls: Concatenating after addition instead of before (leading to a multi-digit result that is not offered). Treating “&” as multiplication or another arithmetic operator rather than stringing digits together. Final Answer: 95.

Question 6

Infer the rule from examples: 12 (169) 5, 4 (20) 2, and 1 (26) 5. Compute the missing middle value for 7 (?) 3.

Accepted Answer

Introduction / Context: This is a classic “find the rule” puzzle where the middle value in parentheses is derived from the two outer numbers using a consistent operation. Once the pattern is recognized, apply it to the target pair to compute the missing value. Given Data / Assumptions: 12 (169) 5 4 (20) 2 1 (26) 5 Query: 7 (?) 3 Concept / Approach: A compact, natural rule fitting all three examples is: middle = left^2 + right^2. This uses perfect squares, which are common in such reasoning items. Step-by-Step Solution: Check with 12 (169) 5: 12^2 + 5^2 = 144 + 25 = 169 ✅Check with 4 (20) 2: 4^2 + 2^2 = 16 + 4 = 20 ✅Check with 1 (26) 5: 1^2 + 5^2 = 1 + 25 = 26 ✅Apply to 7 (?) 3: 7^2 + 3^2 = 49 + 9 = 58. Verification / Alternative check: Other simple rules (like multiplication or concatenation) fail at least one given example, whereas the sum-of-squares rule satisfies all three, making it the intended pattern. Why Other Options Are Wrong: 78, 68, and 36 correspond to misreading the rule as 7 × (something), 7 concatenated with 8 or 6, or 6^2, none of which matches the consistent square-sum pattern. Common Pitfalls: Using (left + right)^2 instead of left^2 + right^2. Attempting left × right or left^right, which do not fit all exemplars. Final Answer: 58.

Question 7

Fill the four “*” with a valid sequence of operators so that the equation balances. Target form: 5 * 6 * 5 * 8 * 14 (insert exactly one operator between each pair to obtain a true equality).

Accepted Answer

Introduction / Context: We must choose an ordered sequence of operators to make a numeric statement true. This requires both precedence awareness and testing of plausible combinations. Given Data / Assumptions: Template: 5 * 6 * 5 * 8 * 14. Insert the four operators (in the exact order listed in an option) to replace the four asterisks. Standard precedence: × and ÷ before + and −. Concept / Approach: Substitute each option in order and evaluate the left-hand side (LHS). The correct option should yield an identity when compared to the right-most number. Step-by-Step Solution: Try Option B: “× ÷ + =”.Substitute: 5 × 6 ÷ 5 + 8 = 14.Compute precedence: 5 × 6 = 30; 30 ÷ 5 = 6; then 6 + 8 = 14.The equality holds, so Option B is correct. Verification / Alternative check: Option A: 5 + 6 − 5 × 8 = 14 ⇒ 5 + 6 − 40 = −29 ≠ 14.Option C: 5 ÷ 6 × 5 = 8 − 14 ⇒ LHS ≈ 4.166…, RHS = −6; false.Option D: 5 + 6 × 5 − 8 ÷ 14 ⇒ 5 + 30 − 8 ÷ 14 ≈ 34.43; not a clean equality to 14. Why Other Options Are Wrong: They mis-balance the expression through a large product or an ill-placed subtraction/division that prevents the LHS from reaching 14. Common Pitfalls: Executing left-to-right without applying ×/÷ before +/−; or mistakenly trying to “compute” the equality symbol as an operator. Final Answer: x ÷ + = gives the true equation 5 × 6 ÷ 5 + 8 = 14.

Question 8

Symbol remapping puzzle: If “−” means “×”, “×” means “+”, “+” means “÷”, and “÷” means “−”, evaluate 40 × 12 + 3 − 6 ÷ 60 under the remapped meanings.

Accepted Answer

Introduction / Context: In remapping problems, each printed symbol represents a different arithmetic operation. You must translate the expression according to the mapping and then evaluate with standard precedence. Given Data / Assumptions: Mapping: “−” → “×”, “×” → “+”, “+” → “÷”, “÷” → “−”. Expression to evaluate: 40 × 12 + 3 − 6 ÷ 60. Precedence: × and ÷ before + and − (after remapping). Concept / Approach: Translate each symbol to its true meaning, then compute systematically. Step-by-Step Solution: Translate: “×” → “+”, “+” → “÷”, “−” → “×”, “÷” → “−”.Expression becomes: 40 + 12 ÷ 3 × 6 − 60.Precedence: 12 ÷ 3 = 4; 4 × 6 = 24.Now add/subtract: 40 + 24 − 60 = 64 − 60 = 4. Verification / Alternative check: Recomputing left-to-right without precedence would not yield 4, highlighting the necessity of observing ×/÷ precedence. Why Other Options Are Wrong: 44 and 16 typically arise from skipping one of the remaps or mis-ordering operations; 7.95 is a non-integer result from treating a division earlier than multiplication incorrectly. Common Pitfalls: Applying the mapping partially (e.g., forgetting that “+” now means “÷”). Ignoring that “÷” now means subtraction, not division. Final Answer: 4.

Question 9

Choose the correct statement after remapping: If “+” means “÷”, “×” means “+”, “−” means “×” and “÷” means “−”, which of the following becomes a true equation/inequality under the remapped meanings?

Accepted Answer

Introduction / Context: This multiple-choice item asks you to evaluate four candidate statements after remapping each printed operator to a different arithmetic meaning. Your goal is to identify which option becomes a correct numerical relation (equality or inequality) under the new meanings. Given Data / Assumptions: Operator mapping: “+” → “÷”, “×” → “+”, “−” → “×”, “÷” → “−”. Options include both equality “=” and inequalities “<”, “>”, which retain their usual meanings. Standard precedence applies after remapping. Concept / Approach: Translate each option’s expression using the mapping, compute with precedence, and check the stated relation. Step-by-Step Solution: Option B (candidate): 36 + 6 - 3 x 2 = 20.Remap: “+” → “÷”, “−” → “×”, “×” → “+”.Thus, 36 ÷ 6 × 3 + 2 = 20.Compute: 36 ÷ 6 = 6; 6 × 3 = 18; 18 + 2 = 20 ⇒ equality holds ✅Check one more for contrast: Option A → 36 ÷ 6 + 3 ÷ 2 = 20 ⇒ 6 + 1.5 = 7.5 ≠ 20.Options C and D also evaluate to values not satisfying their inequalities under the mapping (both LHS become large due to “×”→“+” and “+”→“÷” interplay). Why Other Options Are Wrong: They produce numerical LHS values that do not satisfy the stated equality/inequality once remapped and evaluated with precedence. Common Pitfalls: Forgetting to remap every occurrence of a symbol. Violating precedence by adding before completing all multiplications/divisions. Final Answer: 36 + 6 - 3 x 2 = 20 (Option B) becomes true under the given remap.

Question 10

Transitivity of strict order: Given L > M, M > N, and N > P, determine which conclusions definitely follow. Conclusions: I) L > P; II) M > P.

Accepted Answer

Introduction / Context: This question checks transitivity in a chain of strict inequalities. When each comparison is strict (">"), we can pass the relation forward to deduce farther-apart elements. The task is to test two conclusions derived from L > M > N > P. Given Data / Assumptions: L > M M > N N > P Conclusions to test: I) L > P; II) M > P. Concept / Approach: Use transitivity of strict order. If a > b and b > c, then a > c. Chaining multiple strict inequalities preserves strictness end-to-end. No equalities or non-strict signs (≥, ≤) are present to weaken the conclusion. Step-by-Step Solution: From L > M and M > N, we obtain L > N (by transitivity).From L > N and N > P, we obtain L > P (I is true).Separately, from M > N and N > P, we obtain M > P (II is true). Verification / Alternative check: Create a numeric model that satisfies the premises: let L=7, M=5, N=3, P=1. Then L > P (7 > 1) and M > P (5 > 1) both hold, confirming the deductions. Why Other Options Are Wrong: Options a/b assert only one conclusion; d says “either”, and e says “neither”. Since both I and II are compelled by transitivity, these alternatives understate the certainty or contradict the logical result. Common Pitfalls: Mistaking strict chains for non-strict and concluding L ≥ P or M ≥ P instead of L > P and M > P. Another error is skipping a link (e.g., comparing L to P without explicitly chaining via M and N), which can obscure the role of transitivity. Final Answer: If both conclusions I and II follow

Question 11

Strict and equality mix: Given A > B, B = H, and H > G, determine which conclusions definitely follow. Conclusions: I) A > G; II) A > H.

Accepted Answer

Introduction / Context: Here we combine equality with strict inequalities. Equalities allow substitution (B = H), which lets us realign the chain and apply transitivity correctly to test A against both G and H. Given Data / Assumptions: A > B B = H H > G Conclusions to test: I) A > G; II) A > H. Concept / Approach: Replace B by H wherever B appears (since B = H). Then apply transitivity: if A > H and H > G, we can deduce A > G. We must verify both conclusions independently. Step-by-Step Solution: From A > B and B = H ⇒ A > H (II true).From A > H and H > G ⇒ A > G (I true). Verification / Alternative check: Let H = B = 10, G = 8, and A = 12. Then A > H (12 > 10) and A > G (12 > 8). Both conclusions hold in this concrete model. Why Other Options Are Wrong: Options a/b assert only one conclusion; option d (“either”) understates the certainty; option e (“neither”) contradicts the established chain. Common Pitfalls: Failing to substitute B = H before chaining; or assuming A ≥ H rather than the required A > H. Here, because A > B and B = H, the comparison remains strict. Final Answer: If both conclusions I and II follow

Question 12

Mixed inequalities with an equality anchor: Given H < J, F < H, and I ≤ J = K, determine which conclusions definitely follow. Conclusions: I) H > I; II) I ≥ F.

Accepted Answer

Introduction / Context: This item explores what can (and cannot) be inferred when an element is bounded above by a value equal to another element. The key is recognizing that knowing I ≤ J tells us nothing about I's position relative to H or F unless additional links exist. Given Data / Assumptions: H < J F < H I ≤ J = K (so I ≤ J and J = K) Conclusions to test: I) H > I; II) I ≥ F. Concept / Approach: Absence of a direct relation between I and H (or F) leaves multiple possibilities. To disprove a “must follow,” it suffices to build a counterexample consistent with all premises where the conclusion fails. Step-by-Step Solution: Counterexample for I (H > I): Let J=10, K=10, H=5 (H < J), I=10 (≤ J). Then H=5 is not > I=10. So I is not forced.Counterexample for II (I ≥ F): Using H=5 and F=1 (F < H), pick I=0 (≤ J). Then I ≥ F is false (0 ≥ 1 fails). Thus II is not forced either. Verification / Alternative check: Because both conclusions fail in some consistent assignments, neither is logically necessary. The premises permit a range of values for I relative to H and F. Why Other Options Are Wrong: a/b/c/e each claims at least one conclusion must hold. Our counterexamples show neither must. Common Pitfalls: Assuming “I ≤ J and H < J” implies I ≥ H or I ≤ H; it implies neither. Without a bridge between I and H, conclusions about their order are indeterminate. Final Answer: If neither conclusion I nor II follows

Question 13

Layered non-strict chain: Given A < B < C ≤ D = E, determine which conclusions definitely follow. Conclusions: I) B ≤ E; II) B < E.

Accepted Answer

Introduction / Context: This question mixes strict and non-strict relations and asks you to compare B against E. Because E equals D and C is no greater than D, the strict part of the chain from B up to C ensures a strict comparison to E as well. Given Data / Assumptions: A < B B < C C ≤ D = E Conclusions: I) B ≤ E; II) B < E. Concept / Approach: From B < C and C ≤ E (since D = E), we have B < E directly. Once B < E is proved, the weaker statement B ≤ E follows automatically. Step-by-Step Solution: Since C ≤ D and D = E, we have C ≤ E.Given B < C and C ≤ E, transitivity yields B < E (II is true).Any strict inequality implies the corresponding non-strict one: from B < E we get B ≤ E (I true). Verification / Alternative check: Example values: let A=1, B=2, C=3, D=4, E=4. Then B < E (2 < 4) and B ≤ E (2 ≤ 4) both hold. Why Other Options Are Wrong: Options a/b assert only one of the two truths; d claims “either,” which undersells certainty; e contradicts the chain. Common Pitfalls: Mixing up ≤ and = when D = E; or thinking C ≤ E prevents a strict B vs E. The strict link is between B and C, which is sufficient to make B < E. Final Answer: If both conclusions I and II follow

Question 14

Two linked chains: Given P > M > Q and Q > Z > N, determine which conclusions definitely follow. Conclusions: I) M ≥ Z; II) N < P.

Accepted Answer

Introduction / Context: Here two strict chains share Q. We must compare M to Z and N to P. Because all relations are strict and connect through Q, both conclusions become straightforward via transitivity. Given Data / Assumptions: P > M > Q Q > Z > N Conclusions: I) M ≥ Z; II) N < P. Concept / Approach: Chaining strict inequalities preserves strictness. From M > Q and Q > Z, deduce M > Z (which implies M ≥ Z). From P > M and M > Q > Z > N, deduce P > N. Step-by-Step Solution: M > Q and Q > Z ⇒ M > Z ⇒ M ≥ Z (I holds).P > M and M > Q > Z > N ⇒ P > N ⇒ N < P (II holds). Verification / Alternative check: Let P=10, M=8, Q=6, Z=4, N=2. Then M ≥ Z (8 ≥ 4) and N < P (2 < 10) both true. Why Other Options Are Wrong: a/b recognize only one of the two truths; d/e are inconsistent with the strict chains. Common Pitfalls: Interpreting ≥ as necessary; note we actually get a stronger result M > Z, which still satisfies ≥. Also beware reversing links through Q incorrectly. Final Answer: If both conclusions I and II follow

Question 15

Determine the definitely true relations from K ≥ G > H ≤ F. Which of the following is/are definitely true?

Accepted Answer

Introduction / Context: The stem provides a mixed chain, K ≥ G > H ≤ F, and asks which relations are guaranteed. Some comparisons are certain (K vs H), others are not (F vs K or F vs G). We minimally repair the options to allow a multi-true outcome consistent with the premise. Given Data / Assumptions: K ≥ G G > H H ≤ F Candidate relations: F ≥ K, H < K, F < G, K ≥ H. Concept / Approach: From K ≥ G and G > H, we get K > H (hence K ≥ H as well). However, because we only know H ≤ F and G > H, F could be less than, equal to, or greater than K and G; those comparisons cannot be fixed. Step-by-Step Solution: Since K ≥ G and G > H ⇒ K > H ⇒ H < K (definitely true).From K > H we immediately have K ≥ H (also definitely true).Regarding F ≥ K: only H ≤ F is known, so F could be smaller or larger than K (not definite).Regarding F < G: again, with only H ≤ F and G > H, F could be below or above G (not definite). Verification / Alternative check: Model 1: H=1, G=3, K=3, F=2 ⇒ H < K and K ≥ H true; F ≥ K false; F < G true (but not guaranteed by all models). Model 2: H=1, G=3, K=4, F=10 ⇒ H < K and K ≥ H true; F ≥ K true; F < G false. Only the K vs H statements persist across models. Why Other Options Are Wrong: Options a/c/d assert single relations; e correctly captures both guaranteed relations. Any option claiming F ≥ K or F < G as definite is overreaching. Common Pitfalls: Assuming H ≤ F and G > H fix F relative to K or G—they do not. Final Answer: Both H < K and K ≥ H are true

Question 16

Identify the pattern where A < D is definitely true. Choose the expression that forces A < D.

Accepted Answer

Introduction / Context: We must pick the single pattern that guarantees A < D in every valid assignment. Options differ in how A compares to B and how C compares to D, with equalities providing anchors. Given Data / Assumptions: (a) A ≥ B = C < D (b) A > B = C ≥ D (c) A < B = C ≤ D (d) A > B < C ≤ D Concept / Approach: We need a chain that ensures A is strictly below D. If we can place A strictly below C and ensure C ≤ D, then A < D follows. Any arrangement that leaves A possibly above D is invalid. Step-by-Step Solution: (c) A < B = C ≤ D ⇒ A < C and C ≤ D ⇒ A < D (definite).(a) A ≥ B = C < D ⇒ A could equal or exceed C; if A is much larger than C, A < D is not guaranteed.(b) A > B = C ≥ D ⇒ A may exceed D; A < D cannot be ensured.(d) A > B < C ≤ D gives B below both A and C, but it fixes nothing about A vs D. Verification / Alternative check: Counterexamples for (a), (b), (d): pick values that satisfy the premises but violate A < D. For (c), any numbers respecting A < C and C ≤ D will always yield A < D. Why Other Options Are Wrong: They permit A ≥ D or leave the A–D relationship underdetermined. Common Pitfalls: Overlooking that “≤” allows equality: with A < C and C ≤ D, A remains strictly less than D regardless of whether D equals C or exceeds it. Final Answer: A < B = C ≤ D

Question 17

Find where “M > R” does NOT necessarily hold true.

Accepted Answer

Introduction / Context: The task is to identify the pattern under which the relation M > R is not guaranteed. Three options force M to be strictly greater than R; one leaves open the possibility that M equals R, breaking strictness. Given Data / Assumptions: (a) M = P > Q > R (b) M > P ≥ Q = R (c) R = P < Q < M (d) R ≤ Q ≤ P = M (punctuation repaired for clarity) Concept / Approach: We check whether each option forces R strictly below M. Any scenario that allows R to equal M invalidates the strict “>” claim. Step-by-Step Solution: (a) With M = P and P > Q > R ⇒ M > R (forced).(b) Q = R and P ≥ R with M > P ⇒ M > R (forced).(c) R = P and P < Q < M ⇒ R < M (forced).(d) R ≤ Q ≤ P = M allows R = M (take R=Q=P=M). Then M > R is false. Hence (d) does not force M > R. Verification / Alternative check: Construct values for (d): let M=P=Q=R=5, which satisfies R ≤ Q ≤ P = M. Then M > R fails (5 > 5 is false). Why Other Options Are Wrong: In (a), (b), and (c), the chains require R to be strictly less than M, so M > R always holds. Common Pitfalls: Confusing “≥” with “>” and overlooking equality cases that break strict inequality. Always check whether equality is permitted somewhere on the path from R to M. Final Answer: R ≤ Q ≤ P = M

Question 18

Mixed non-strict/strict chain: Given H ≥ I = J > K ≤ L, determine which conclusions definitely follow. Conclusions: I) K < H; II) L ≥ I.

Accepted Answer

Introduction / Context: This problem asks what must follow from a chain mixing ≥, =, >, and ≤. We compare K against H and L against I carefully, watching where strictness is preserved or lost. Given Data / Assumptions: H ≥ I = J J > K K ≤ L Conclusions: I) K < H; II) L ≥ I. Concept / Approach: From I = J and J > K, we know I > K. Combining H ≥ I with I > K yields H > K ⇒ K < H (I true). For II, K ≤ L and I > K do not force L ≥ I; L might sit between K and I. Step-by-Step Solution: I) Since I = J and J > K ⇒ I > K. With H ≥ I, we have H > K ⇒ K < H (true).II) From K ≤ L and I > K, L could be ≤ I or < I; nothing mandates L ≥ I. Thus II is not forced. Verification / Alternative check: Counterexample for II: Let K=2, L=3, I=5, H=6 (and J=5). Then L ≥ I is false (3 ≥ 5 fails), while all premises hold. Why Other Options Are Wrong: b/d/c each claim a truth value for II that is not compelled; only I is guaranteed. Common Pitfalls: Assuming K ≤ L and I > K implies L ≥ I—this is not valid without a link between L and I beyond the shared comparison with K. Final Answer: If only Conclusions I is true

Question 19

Sign reasoning around zero: Given S > C ≥ 0 and P < C, determine which conclusions definitely follow. Conclusions: I) 0 < P; II) S > P.

Accepted Answer

Introduction / Context: This item tests careful handling of comparisons with zero. Knowing C is non-negative and S is above C, while P is below C, tells us something definitive about S vs P, but not necessarily about P vs 0. Given Data / Assumptions: S > C C ≥ 0 P < C Conclusions: I) 0 < P; II) S > P. Concept / Approach: If P < C and C ≥ 0, P could still be negative (e.g., C=0, P=-1) or positive (e.g., C=5, P=4). Thus I is not forced. However, S > C and P < C imply S > P strictly by “squeezing” around C. Step-by-Step Solution: I) Not forced: take C=0; then P can be -1 (P < 0), so 0 < P fails.II) Since S > C and P < C, we have S > C > P or at least S > P in all compliant assignments. Hence II is guaranteed. Verification / Alternative check: Example values: C=3, S=5, P=1 ⇒ S > P. Or C=0, S=2, P=-3 ⇒ S > P. In every case consistent with the premises, S > P holds. Why Other Options Are Wrong: a claims I is true—counterexample shown; c claims “either,” which understates certainty; d contradicts the valid deduction in II. Common Pitfalls: Assuming “P < C and C ≥ 0” forces P ≥ 0—false when C=0 or small positive numbers. Final Answer: If only Conclusion II is true

Question 20

Statements A = B ≤ C, A > R Conclusions I. B > R II. R < C

Accepted Answer

Given that, A = B ≤ C ................(i) A > R .........................(II) On combining the statements (i) and (ii), we get R < A = B ≤ C Conclusion I. B > R. .................(true) II. R < C .................(true) So, it is clear that both Conclusions I and II follow from the given statements.

Mathematical Operations Questions