Brain Chase Brain Teaser
Help our team solve weekly conundrums.
February 6, 2018
In his European travels, Tate Grayson meets the bell ringer of an ancient cathedral. He finds that it takes the bell ringer three seconds to ring four o’clock. How long does it take him to ring midday?
Solution: The three seconds that go by in ringing four o’clock correspond to the intervals between bell tolls, not the number of tolls. To ring the twelve tolls of midday, the ringer will take eleven seconds, corresponding to the eleven intervals separating the twelve tolls of the bell.
January 25, 2018
Mae invited the following people to a Merriweather family reunion:
One grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, two sisters, one brother, two daughters, two sons, one father-in-law, one mother-in-law, and one daughter-in-law.
What is the fewest number of people who could have attended the reuniion, and who were they?
Solution: Seven people are at the reunion: two girls and a boy, their parents, and their father’s parents.
December 27, 2017
In his journeys, Grayson Academy member Sean Drake meets an old farmer who raises chickens and rabbits. When Sean asks how many he owns, the farmer replies, “When I count heads, I find eight. When I count feet, I find twenty-eight.”
How many rabbits does the farmer have? And how many chickens?
Solution: Let x be the number of chickens and y be the number of rabits. The equation for number of heads is x + y = 8. The equation for number of feet would be 2x + 4y = 28.
Solving for x in the first equation, we get x = 8 – y. If we use 8 – y to replace x in the second equation, we get (8 – y) + 4y = 28. Next, 16y-2y + 4y = 28, or 2y = 12, so y = 6. Plugging that into our x = 8 – y equation, we get x = 2.
The farmer has two chickens and six rabbits.
December 20, 2017
Riddle: A Professor at the Grayson Academy of Antiquities asks her student, “I am four times as old as you were when I was the same age as you are now. I am forty years old. How old are you?”
Solution: Let’s write out what we know:
- Professor: age before = x; age now = 40
- Student: age before = y; age now = z
We know that 40 = 4 * y, because “I am four times as old as you were.” So y = 10.
We also know that z = x, because “I was the same age as you are now.” Let’s update:
- Professor: age before = x; age now = 40
- Student: age before = 10; age now = x
Since we know that the same number of years have passed for each person, we can say x – 40 = 10 – x. Or 2 * x = 50. So x = 25.
The student is 25 years old.
October 18, 2017
Riddle: In his journey through the Lost World, Max Merriweather comes across a vault with three doors. One door contains the lost dinosaur fossil and a sizable treasure while the other two doors each contain a dangerous snake.
Max is allowed to pick a door which he thinks is the treasure. After he makes his choice, a different door will open revealing either a snake or a treasure. If this new open door reveals a snake, then Max is allowed to keep his choice or pick the other door.
Max chooses door number one. Door number three opens up revealing a snake. Should Max stay with his first choice of door number one, or should be change it to door number two?
Solution: Max should change his choice from door one to door two. Here’s the reason. When Max chose door number one he had a 1 in 3 chance of guessing correctly, giving the treasure a 2 in 3 chance of being behind door number two or three. With the new information of the treasure NOT being behind door number three, the new odds are this: door number one is still a one in three chance, while door number two is a 2 in 3 chance. Door number two is a 2 in 3 chance because doors two and three were previously a 2 in 3 chance before we knew what was behind door number 3. Once door number three revealed a snake, its chances became 0 in 3. If doors two and three have a 2 in 3 chance, and if door number three becomes 0, then door number two becomes the sole 2 in 3 chance of winning. For a lengthier explanation of this, see the Monty Hall Problem entry on Wikipedia.
October 15, 2017
Riddle: Grayson Academy of Antiquities members Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates:
May 15, May 16, May 19
June 17 June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert and Bernard separately the month and the day of her birthday, respectively.
Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard also does not know.
Bernard: At first I didn’t know when Cheryl’s birthday is, but now I know.
Albert: Then I also know when Cheryl’s birthday is.
When is Cheryl’s birthday?
Special thanks to Khan Academy, who introduced us to this logical puzzle. Click here and click “Cheryl’s Birthday” to see the entire solution.
October 6, 2017
Riddle: When Tate Grayson and two rival explorers discover a lost civilization, the tribal chief offers one of them a great treasure. To determine which is worthy, the chief has them close their eyes while he paints a colored dot on each of their foreheads. They are told that the dot is either red or blue. As they open their eyes, they are told to raise a hand if they see a red dot on another, and that the first to stand and correctly announce the color on his own forehead will win.
While their eyes are closed, the chief puts a red dot on all three explorers. All three raise a hand when they open their eyes. And then, after a minute, Tate stands and correctly announces that he has a red dot on his forehead. How did he figure it out?
Solution: Let’s call them Explorers A, B, and C (Tate Grayson). When he opened his eyes, Explorer A saw that both other men had red dots on their foreheads – and he knew that they could each be raising hands because of the other, and not because of him. If Explorer A had seen that Explorer B was raising a hand and that Explorer C had a blue dot, it would be easy – he would immediately realize the his own dot was red, and he would have immediately stood up to claim the prize. The same logic would apply to Explorer B.
Tate Grayson quickly noticed that both other explorers were raising their hands but not standing up, and surmised that each was as confused as he. He therefore knew that he had a red dot on his own forehead, and quickly announced as much. He is rumored to have thoroughly enjoyed his new riches.
September 26, 2017
Riddle: Famed adventurer Tate Grayson stands at the intersection of two roads – one which leads to a treasure, and one which leads to a perilous trap. Each path is guarded by a tribal sentry – one who ALWAYS tells the truth, and one who ALWAYS lies. Tate doesn’t know which is which. He is only permitted to ask one question. What question should he ask in order to find the treasure?
Solution: Tate should choose one of the two sentries and ask, “Which way would the other soldier tell me to go?” The lying sentry would answer by lying about the direction that the truthful sentry would advise, and the truthful sentry would correctly point to the false direction which the liar would recommend. Asking either sentry will produce the perilous trail – so Tate should pose his question, and then select the opposite path.
September 19, 2017
Riddle: Savannah Brice, a shady antiquities dealer, travels between Eastern and Western Europe regularly. The authorities know that she smuggles large amounts of money through customs each time, but they are unable to catch her. Each time they search her car, they only find a few coins in the ashtray. How is Savannah eluding the customs officers?
Solution: Savannah Brice smuggles extremely rare coins worth hundreds of thousands of dollars.
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