Solving the Chocolate Costs Logic Puzzle

This problem requires us to determine the specific cost of each of the seven chocolates (A, B, C, D, E, F, G) based on the given clues. All costs are distinct and lie between Rs. 40 and Rs. 50, excluding Rs. 40 and Rs. 50 themselves. This means the possible costs are integers from 41 to 49, with the additional constraint that Rs. 44 is not a possible cost for any chocolate.

Identifying Possible Costs

The range of possible costs is > 40 and < 50. So, the integers are 41, 42, 43, 44, 45, 46, 47, 48, 49. We are told that none of the chocolates costs Rs. 44. Therefore, the set of possible distinct costs for the seven chocolates is:

• 41
• 42
• 43
• 45
• 46
• 47
• 48
• 49

There are 8 possible distinct values, and we need to assign 7 of these to the seven chocolates.

Analyzing the Clues and Deductions

Let's break down each clue and see what we can deduce:

1. The cost of chocolate C is five less than that of chocolate E: $\(C = E - 5\). This implies $\(E = C + 5\). Since both C and E must be within the valid range {41, 42, 43, 45, 46, 47, 48, 49}, we can list possible (E, C) pairs:
   • If $\(C = 41\), then $\(E = 41 + 5 = 46\). Valid pair: (E=46, C=41).
   • If $\(C = 42\), then $\(E = 42 + 5 = 47\). Valid pair: (E=47, C=42).
   • If $\(C = 43\), then $\(E = 43 + 5 = 48\). Valid pair: (E=48, C=43).
   • If $\(C = 45\), then $\(E = 45 + 5 = 50\). Not in range.
   • If $\(C = 46\), then $\(E = 46 + 5 = 51\). Not in range.
   • If $\(C = 47\), then $\(E = 47 + 5 = 52\). Not in range.
   • If $\(C = 48\), then $\(E = 48 + 5 = 53\). Not in range.
   • If $\(C = 49\), then $\(E = 49 + 5 = 54\). Not in range.
   So, the possible (E, C) pairs are (46, 41), (47, 42), and (48, 43).
2. The cost of the chocolate A is a prime number. The prime numbers in the range {41, 42, 43, 45, 46, 47, 48, 49} are 41, 43, and 47. So, A can be 41, 43, or 47.
3. The cost of chocolate F is two more than that of chocolate A: $\(F = A + 2\).
   • If $\(A = 41\), then $\(F = 41 + 2 = 43\). Possible (A, F) = (41, 43).
   • If $\(A = 43\), then $\(F = 43 + 2 = 45\). Possible (A, F) = (43, 45).
   • If $\(A = 47\), then $\(F = 47 + 2 = 49\). Possible (A, F) = (47, 49).
4. The cost of chocolate F is more than that of chocolate E: $\(F > E\). Let's combine this with the possible (A, F) and (E, C) pairs.
   • Case 1: (A, F) = (41, 43). We need $\(43 > E\). Possible E values from the (E, C) pairs are 46, 47, 48. None of these are less than 43. So, (A, F) cannot be (41, 43).
   • Case 2: (A, F) = (43, 45). We need $\(45 > E\). Possible E values from the (E, C) pairs are 46, 47, 48. None of these are less than 45. So, (A, F) cannot be (43, 45).
   • Case 3: (A, F) = (47, 49). We need $\(49 > E\). Possible E values from the (E, C) pairs are 46, 47, 48. All of these are less than 49. This is a valid possibility for A and F.
   So, we know A=47 and F=49. The possible (E, C) pairs are (46, 41), (47, 42), and (48, 43). Since all costs must be distinct, E cannot be 47 (as A is 47). This eliminates (E, C) = (47, 42). The remaining possible (E, C) pairs are (46, 41) and (48, 43).
5. The cost of chocolate D is an odd number. The available costs (not used by A=47, F=49, and a potential (E, C) pair) for B, D, G must include an odd number for D.
6. The cost of chocolate G is three more than the cost of chocolate D: $\(G = D + 3\). Let's check the remaining possibilities from the combined A, F, E, C deduction:
   • Possibility 1: A=47, F=49, E=46, C=41. Used costs: {41, 46, 47, 49}. Remaining costs for B, D, G: {42, 43, 45, 48}. D must be odd from this set: 43 or 45.
     • If $\(D = 43\), then $\(G = 43 + 3 = 46\). But E already costs 46. Costs must be distinct. So, D cannot be 43.
     • If $\(D = 45\), then $\(G = 45 + 3 = 48\). Used costs so far: {41, 45, 46, 47, 48, 49}. All are distinct. This is a valid possibility.
   • Possibility 2: A=47, F=49, E=48, C=43. Used costs: {43, 47, 48, 49}. Remaining costs for B, D, G: {41, 42, 45, 46}. D must be odd from this set: 41 or 45.
     • If $\(D = 41\), then $\(G = 41 + 3 = 44\). But 44 is not allowed. So, D cannot be 41.
     • If $\(D = 45\), then $\(G = 45 + 3 = 48\). But E already costs 48. Costs must be distinct. So, D cannot be 45 in this case.
   The only valid scenario found so far is from Possibility 1 with D=45: A=47, F=49, E=46, C=41, D=45, G=48. The used costs are {41, 45, 46, 47, 48, 49}. The remaining available cost for B is 42.
7. The cost of chocolate B is an even number. In our valid scenario, the remaining cost for B is 42, which is an even number. This fits the condition.

Determining All Chocolate Costs

Based on the deductions, the costs of the chocolates are:

Let's quickly verify all conditions with these costs:

• Costs between 40 and 50 (excl. 40, 50, 44): All costs {41, 42, 43, 45, 46, 47, 48, 49} are in the valid set {41, 42, 43, 45, 46, 47, 48, 49}. Distinct costs are used.
• $\(C = E - 5\): $\(41 = 46 - 5\) (True).
• A is a prime number: 47 is prime (True).
• $\(F = A + 2\): $\(49 = 47 + 2\) (True).
• $\(F > E\): $\(49 > 46\) (True).
• D is an odd number: 45 is odd (True).
• $\(G = D + 3\): $\(48 = 45 + 3\) (True).
• B is an even number: 42 is even (True).

All conditions are satisfied with this set of costs.

Answering the Question: Which chocolate is costlier than E?

The cost of chocolate E is Rs. 46. We need to find which of the given options has a cost greater than 46.

• Cost of D = Rs. 45. Is $\(45 > 46\)? No.
• Cost of G = Rs. 48. Is $\(48 > 46\)? Yes.
• Cost of B = Rs. 42. Is $\(42 > 46\)? No.
• Cost of C = Rs. 41. Is $\(41 > 46\)? No.

Chocolate G is costlier than chocolate E.

Revision Table: Chocolate Costs Summary

Additional Information: Logic Puzzle Solving Tips

Logic puzzles like this require careful reading and systematic deduction. Here are some tips:

• List Possibilities: Start by listing all possible values for the variables based on the constraints.
• Break Down Clues: Analyze each clue separately to understand the relationships between the variables.
• Combine Clues: Look for clues that connect to each other to narrow down possibilities. Use tables or diagrams to keep track of assignments and eliminations.
• Eliminate Impossible Scenarios: If a deduction leads to a contradiction (like costs not being distinct, or a value being outside the allowed range), eliminate that possibility.
• Verify the Solution: Once you have a potential solution, go back and check if it satisfies all the original clues.

This specific puzzle involved number properties like prime, odd, and even numbers, as well as arithmetic relationships and inequality constraints within a defined range, all crucial for successful deduction.