Comparision & Order Arrangement
(i) A is taller than B but shorter than C.
(ii) D is taller than E but shorter than F.
(iii) C is shorter than E.
(iv) G is the tallest.
From (i): C > A > B
From (ii): F > D > E
From (iii): E > C. Combining these gives F > D > E > C > A > B.
From (iv): G is the tallest.
Final Order: G > F > D > E > C > A > B.
Clearly, G is the tallest.
Looking at the final order from shortest to tallest: B (1st), A (2nd), C (3rd). Therefore, C is the third shortest.
Based on the combined order (G > F > D > E > C > A > B), B is at the extreme right, making him the shortest.
(i) R is heavier than P but lighter than S.
(ii) Q is the lightest.
(iii) T is heavier than S.
From (i): S > R > P
From (ii): Q is at the bottom.
From (iii): T > S
Final Order: T > S > R > P > Q.
The second heaviest is S.
Based on the combined order (T > S > R > P > Q), T is at the top, making T the heaviest.
F is between B and A.
E is shorter than D but taller than C, who is taller than A.
E and F have two boys between them.
A is not the shortest among them all.
Derivation:
– F is between B and A. (B-F-A or A-F-B).
– D > E > C > A.
– A is not the shortest, so A cannot be at the very front. This means the sub-sequence must be A > F > B.
– Combining: D > E > C > A > F > B.
– Let’s check the condition: “E and F have two boys between them”. In D > E > C > A > F > B, the boys between E and F are C and A (exactly two).
– Final Order: D > E > C > A > F > B.
E is situated directly between D and C.
Based on the established order (D > E > C > A > F > B), the person at the back (the tallest) is D.
1) A weighs twice as much as B.
2) B weighs four-and-a-half times as much as C.
3) C weighs half as much as D.
4) D weighs half as much as E.
5) E weighs less than A but more than C.
Derivation:
Let’s assign a hypothetical weight to C to find the ratios.
Let C = 2 units.
From 3: D = 2C = 4 units.
From 4: E = 2D = 8 units.
From 2: B = 4.5 × C = 4.5 × 2 = 9 units.
From 1: A = 2B = 2 × 9 = 18 units.
Checking condition 5: E(8) < A(18) and E(8) > C(2). This perfectly matches.
Final Order by Weight: A(18) > B(9) > E(8) > D(4) > C(2).
C is the lightest article.
Based on the arrangement: A > B > E > D > C.
E is exactly in the middle position (3rd out of 5).
Given:
1. A + B > C + D
2. A + C = B + D
3. A = (B + D) / 2 ➔ 2A = B + D
Substitute (3) into (2): A + C = 2A ➔ C = A.
Substitute C = A into (1): A + B > A + D ➔ B > D.
From 2A = B + D, we get A – D = B – A. Since B > D, it must be that B > A > D.
Final Order: B > A = C > D.
B has the highest income.
From the established order (B > A = C > D), B earns more than A. Therefore, the statement “A earns more than B” is false.
We know A = 80,000.
From the passage: B + D = 2A. So, B + D = 160,000.
The question states: B – D = A’s income. So, B – D = 80,000.
Add the two equations:
(B + D) + (B – D) = 160,000 + 80,000
2B = 240,000 ➔ B = 120,000.
(Note: Some external answer keys mark ‘C’ for this mistakenly, but mathematical derivation strictly yields 1,20,000.)
Given P = 60,000.
1) P = (Q + S) / 2 ➔ Q + S = 2P = 120,000.
2) P = Q – S ➔ Q – S = 60,000. (Assuming Q > S to make the difference positive like P’s income).
Adding both equations:
(Q + S) + (Q – S) = 120,000 + 60,000
2Q = 180,000 ➔ Q = 90,000.
We know Q + S = 120,000 and Q = 90,000.
90,000 + S = 120,000
S = 120,000 – 90,000 = 30,000.
Preparing for MBA entrance exams like CAT, CET, NMAT, SNAP, and XAT requires strong logical reasoning skills, especially in topics like ordering and ranking puzzles. These questions test your ability to analyze relationships between people, objects, or quantities and arrange them in the correct sequence based on given clues. In exams such as MBA CET, ordering questions frequently appear in Logical Reasoning and can be solved quickly if the right approach is used.
The key to solving ordering problems is to convert every statement into a comparison. For example, clues like “A is taller than B” or “C earns more than D” should immediately be written in sequence form. As you collect multiple clues, you can combine them to create a full order from highest to lowest or fastest to slowest. This structured approach reduces confusion and allows you to eliminate incorrect options quickly.
Another effective technique is to start with the extremes. Statements that indicate the highest, lowest, tallest, or fastest element often help anchor the sequence. Once the extreme positions are identified, the remaining elements can be arranged logically using the remaining clues. Practicing these patterns regularly improves speed and accuracy, which is crucial in time-bound exams.
For MBA aspirants targeting top colleges like JBIMS, SIMSREE, PUMBA, and KJ Somaiya, mastering reasoning topics like ordering, ranking, comparisons, and inequalities can significantly boost the Logical Reasoning score. The more practice sets you solve, the easier it becomes to recognize patterns and avoid common mistakes.
Regular mock tests, daily reasoning practice, and understanding shortcut techniques can help students maximize their performance in competitive exams. With consistent preparation and smart strategies, ordering and ranking questions can become one of the highest-scoring sections in MBA entrance exams.
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