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# Answers to Classic & Logic Puzzles

1. This answer is for 2 weighings. The problem I originally had posted here was for 3 weighings, but since you can complete the problem with 2 weighings, I list only the most efficient method. Place 3 marbles on each pan. If they balance, remove the marbles from the pans, & place 2 of the remaining unweighed marbles on the pans, 1 on each pan. If 1 is heavier, it is the heavier marble, but if they balance, the remaining unweighed marble is the heavier 1. If your first weighing does not balance, remove the marbles from the lighter pan, & place 1 marble on each pan from the heavier pan. The heavier 1 is the 1.5 ounce marble, but if they balance, then the marble from the heavy pan from the first weighing that was not weighed in the second weighing is the heavy 1.

2. There are 2 possible answers, both of which are almost the same:

Answer 1: Andy & Brenda walk together to the house (2 minutes), & Andy returns (1 minute). Carl & Dana walk to the house (10 minutes), & Brenda returns with the umbrella (2 minutes). Andy & Brenda walk to the house again (2 minutes).

Answer 2: Andy & Brenda walk together to the house (2 minutes), & Brenda returns (2 minutes). Carl & Dana walk to the house (10 minutes), & Andy returns with the umbrella (1 minute). Andy & Brenda walk to the house again (2 minutes).

3. There are 2 possible answers for this puzzle:

Answer 1: You ask 1 of the individuals what the other 1 would say if you asked him or her which door you should open to get to the coffer. In this case, you would open the other door.

Answer 2: You ask 1 of the individuals what the other 1 would say if you asked him or her which door is holding back the hungry lion. In this case, you would open this door.

4. For the chart below, assume A = knave, I = knight, & O = normal.

A B C
A I O
A O I
I A O
I O A
O A I
O I A

The first condition & the last 2 conditions cannot occur because contradictions would occur. In 2 of the second, third, & fourth possibilities, the knight is A, the person who claims to love cats. You would be wisest to bet on A, as there is better than a 66% chance that that individual is a knight.

5. All 4 individuals are knaves. The primary colors are red, yellow & blue for pigments, but for the spectrum they are red, green & blue. This was a tough puzzle if you didn't use reference aids - not many people know these things!

6. 1-333. The reason why is because 200-299 each begins with a 2!

7. There are 64 different possible outcomes, & in 9 of these, exactly 2 of the balls will be red. There is thus a slightly better than 14% chance that exactly 2 balls will be red. A much faster way to solve the problem is to look at it this way. There are 3 scenarios where exactly 3 balls are red:

1 2 3
R R X
R X R
X R R

X is any ball that is not red. There is a 4.6875% chance that each of these situations will occur. Take the first 1, for example: 25% chance the first ball is red, multiplied by a 25% chance the second ball is red, multiplied by a 75% chance the third ball is not red. Because there are 3 scenarios where this outcome occurs, you multiply the 4.6875% chance of any 1 occurring by 3, & you get 14.0625%.

8. This problem can be solved algebraically: 10 (8k + 14l) = 510 and 12 (13k + 6l) = 492. You then isolate 1 variable in 1 equation, & substitute it in the other equation. Kips build 2 tors an hour, & ligs build 2.5 tors an hour.

9. Pretend that A & B are the objects in the juggler's first hand, & C & D are the objects in the juggler's second hand. You thus start out as AB CD, & each step shows the result after each throw (note: some of the objects would not actually reach their destination in the steps in which they are shown arriving):

1. B ACD
3. C ABD
4. CD AB
5. D ABC
7. A BCD
8. AB CD

It would take a total of 8 throws to get the objects back into their original positions.

A. AROUSE (synonym of EXCITE).

B. KANT.

C. OUNCE.

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