Brain Teasers
Good Guess
Fun: (2.68)
Difficulty: (1.47)
Puzzle ID: #36927
Submitted By: reptile5000 Corrected By: brainglewashed
Submitted By: reptile5000 Corrected By: brainglewashed
In order to win a free laptop computer, students had to guess the exact numbers of gumballs in a jar. The students guessed 45, 41, 55, 50, and 43, but nobody won. The guesses were off by 3, 7, 5, 7, and 2 (given in no particular order). From this information, can you determine the number of gumballs in the jar?
Answer
Forty-Eight gumballs. (48)Since two guessers were off by seven and no guesses were repeated, these values had to refer to numbers at the opposite sides of the spread The two extremes are 41 and 55. If you add 7 to 41, and subtract 7 from 55, you arrive at the answer, 48.
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Comments
you substract the highest margin of error (7) to the highest guess (55), 55 - 7 = 48. then add the highest margin of error (7) to the lowest guess (41), 41 - 7 = 48. the correct answer would be any of the 2 answers. but since its the same, then the correct answer is 48!
yeah- what she said....good one pating (comment nbr 1)
but i did it this way: like the simpleton i am:
45+3 = 48
41+7 = 48
55-7 = 48
43+5 = 48
50-2 = 48
ta daaa
keep em coming - we likey
but i did it this way: like the simpleton i am:
45+3 = 48
41+7 = 48
55-7 = 48
43+5 = 48
50-2 = 48
ta daaa
keep em coming - we likey
lol, thanks, your comments are greatly appreciated. I did notice a spalling error and it is being fixed now.
Yeah I found that too I sent my correction in. oh and good teaser!
Great teaser!
I typically stink at math teasers because I don't know the definitions of technical math terms. This one was right up my alley.
I typically stink at math teasers because I don't know the definitions of technical math terms. This one was right up my alley.
haha i noticed it also.
but on the other hand, u did good!
but on the other hand, u did good!
Oh my goodness that was the easiest teaser I've ever done. there was two sevens so you found the highest and lowest numbers. Then added and subtracted. The other numbers were useless.
Don't forget to send me a message saying what subject you want me to do on a quiz!!!!!!!!!!!!!!!!
I really liked this! I am not a math genie-us, but I was able after some deep thought, to get the answer. It seems like everyone got the same answer by different methods. I reasoned that if there were two sevens, and no repeated wrong guesses, then one seven had to higher and one lower than the correct number. There was no other possibility with that set up. I don't see what the highest and lowest have to do with it though. There could have been wrong guesses higher or lower than seven, but, the answer would still be the same, because: if one has to higher, and one 7 has to be lower, then you just need to find the two wrong guesses that match those requirements, in other words, both wrong guesses are higher or lower than the same number. The other numbers showed me that my answer was correct.
Thanks, Reptile!
Thanks, Reptile!
It just so happened that there are two answers off by 7. You'll see the point of the highest and lowest if the the students guessed 45, 40, 55, 50, and 43, and the guesses were off by 3, 8, 5, 7, and 2. In this case the highest margin of error is 8 (and there is only one . The highest margin of error will always be for either the 2 extreme guesses, that is 40 or 55. We're sure that either the lowest guess or the highest guess is the one off by 8. So,
55 - 8 = 47
40 + 8 = 48
The answer would have to be 47 or 48. You'll just have to check the other guesses to know the correct answer.
55 - 8 = 47
40 + 8 = 48
The answer would have to be 47 or 48. You'll just have to check the other guesses to know the correct answer.
Fun! But, i have to say, you copied! u saw this on another website!
Given any set of guesses and deltas, either the maximum or minimum guess must generate the largest delta (D). Already this reduces the possibilities to two:
min + D
max - D
Also, the delta for the other extreme is
delta = | (max - min) - D |
where |x| is absolute value of x.
min + D
max - D
Also, the delta for the other extreme is
delta = | (max - min) - D |
where |x| is absolute value of x.
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