Math Problems and Logic Puzzles

If B is lying, then 'B is older than C' is false, meaning C is older than B. A says A > B, C says C > A. If C > A > B, and B is lying about B > C (which is false since C > B), then C > A > B. B is the youngest, C is the oldest, A is in the middle.

This creates an unsolvable paradox. If A tells truth, B always tells truth, but B says A always lies, contradiction. If A lies, B doesn't always tell truth, but then B's statement might be true or false. No consistent assignment satisfies both statements simultaneously.

The conditional probability formula states P(A|B) = P(A ∩ B) / P(B). Given P(A ∩ B) = 0.32 and P(B) = 0.50, we calculate P(A|B) = 0.32 / 0.50 = 0.64 = 64%. This means there's a 64% chance that event A occurs when we know event B has occurred. Option D would be the joint probability P(A ∩ B), not the conditional probability.

To solve this problem, the user needs to know the capacities of each can and how to divide the milk among the cans to obtain the desired quantities.

Statement 1: It is possible to have 6 litres of milk each in can A and can B.
To obtain 6 litres of milk in can A and can B, we can follow the following steps:

• Pour 6 litres of milk from can C into can A.
• Pour 4 litres of milk from can A into can B.
• Pour 2 litres of milk from can C into can A.
• Pour 4 litres of milk from can A into can B.

After following these steps, can A will have 4 litres of milk, can B will have 6 litres of milk, and can C will have 10 litres of milk. Therefore, statement 1 is true.

Statement 2: It is possible to have 8 litres of milk each in can B and can A.
To obtain 8 litres of milk in can A and can B, we can follow the following steps:

• Pour 10 litres of milk from can C into can B.
• Pour 2 litres of milk from can B into can A.
• Pour 4 litres of milk from can C into can B.
• Pour 2 litres of milk from can B into can A.

After following these steps, can A will have 4 litres of milk, can B will have 8 litres of milk, and can C will have 12 litres of milk. Therefore, statement 2 is true.

Since both statements are true, the answer is:

The Answer is: C. Both 1 and 2.

From the first two statements: Cliff is older than Tanya (C > T), and Tanya is older than Eric (T > E). Therefore Cliff > Eric. The third statement claims Eric is older than Cliff, which directly contradicts this, making it false.

Calvin is directly behind the ball carrier, with Jenkins and Burton side-by-side behind him. Zeller is behind both Jenkins and Burton. When Calvin misses and falls, and Burton trips, Jenkins is the only remaining player in position to make the tackle.

Let the son's current age be x. At the son's birth (x years ago), the father's age was 38-x. The father says he was as old as the son is now, so 38-x = x. Solving gives x = 19. Five years ago, the son was 19-5 = 14 years old.

From Jan 1, 2006 to Jan 1, 2010 is 4 years. Normal years advance the day by 1, while leap years advance by 2. 2008 was a leap year. Total days advanced: 2006→2007 (1), 2007→2008 (1), 2008→2009 (2, leap year), 2009→2010 (1). Sum = 5 days. Sunday + 5 days = Friday.

From 8 AM to 2 PM is 6 hours. The hour hand rotates 360 degrees in 12 hours, so it moves 30 degrees per hour (360/12 = 30). In 6 hours, the hour hand rotates 6 × 30 = 180 degrees.

132 = 4 × 3 × 11. A number is divisible by 132 only if divisible by all three. Checking: 264 (132×2, yes), 396 (132×3, yes), 462 (not by 4, no), 792 (132×6, yes), 968 (not by 3, no), 2178 (not by 4, no), 5184 (not by 11, no), 6336 (132×48, yes). Exactly 4 numbers are divisible.

To be divisible by 132, a number must be divisible by 3, 4, and 11. Testing the numbers reveals that exactly 4 of them (264, 396, 792, and 6336) are divisible by 132.