Marbles problem to solve
Here's a nice "brain teaser" problem sent to my way... probably a student who wants to know the solution. I'll post it for you all and the solution in a few days.
(I do not have a clue where the problem originates, but I DO LIKE it!)
Solution:
Now, we need to find three numbers.
The one hint tells us that one number is double the other, and the third is less than the first (the doubled one). This won't yet get us started.
It is actually the latter hints that provide a starting point.
We learn the numbers are divisible by 4. They're also divisible by 6, which means they're divisible by 2 and 3. But we already knew they're divisible by 2 (since thye were divisible by 4), so the new information in this hint really is that the numbers are divisible by 3.
Divisible by 4 AND divisible by 3 means.... DIVISIBLE by 12!
Also all numbers are less than 100.
This really now restricts our "search space". We're looking at multiples of 12 that are less than 100.
Now, one was double the other. Since the 8th multiple of 12 (96) is the largest we can use, then two of the numbers COULD be the 4th and 8th multiples of 12 (48 and 96). They could also be the 3rd and 6th multiples of 12 (36 and 72).
The first hints will now "lock in" the solution... the sum has to be 204, which is 17 x 12. So, choosing 4 x 12, 8 x 12, and 5 x 12 - or 48, 96, and 60 - as our numbers will work. And, there are therefore 60 red marbles.
204 marbles are divided into 3 groups according to colour. Ahmad found that there are twice as many blue marbles as white marbles and there are fewer red marbles than blue marbles. Ben found that the number of marbles in each group are divisible by 4 and 6. Cally found that the number of marbles in each group is less than 100.
How many red marbles are there?
(I do not have a clue where the problem originates, but I DO LIKE it!)
Solution:
Now, we need to find three numbers.
The one hint tells us that one number is double the other, and the third is less than the first (the doubled one). This won't yet get us started.
It is actually the latter hints that provide a starting point.
We learn the numbers are divisible by 4. They're also divisible by 6, which means they're divisible by 2 and 3. But we already knew they're divisible by 2 (since thye were divisible by 4), so the new information in this hint really is that the numbers are divisible by 3.
Divisible by 4 AND divisible by 3 means.... DIVISIBLE by 12!
Also all numbers are less than 100.
This really now restricts our "search space". We're looking at multiples of 12 that are less than 100.
Now, one was double the other. Since the 8th multiple of 12 (96) is the largest we can use, then two of the numbers COULD be the 4th and 8th multiples of 12 (48 and 96). They could also be the 3rd and 6th multiples of 12 (36 and 72).
The first hints will now "lock in" the solution... the sum has to be 204, which is 17 x 12. So, choosing 4 x 12, 8 x 12, and 5 x 12 - or 48, 96, and 60 - as our numbers will work. And, there are therefore 60 red marbles.
Comments
Thanks for posting it!
~Laura
1st B; nb of blue ones
R: nd of red ones
W: nd of white ones
clearly b+r+w=204 divide by 12 the equation
b1+r1+w1=17 where
b1=b/12 and r1=r/12 and w1=w/12 and b1, r1 and w1 are natural since b, r and w divisble by 12
now we know b1=2w1 since blure double that white
so we got 3w1+r1=17 then
w1=(17-r1)/3
so we have 2 possible casses for r1 either 2 or 5 in order to get w1 natural number
1st case
r1=2 then w1=5 and b1=10
so r=24
w= 60
blue=100 which is rejected since all numbers are less than 100
2nd case:
r1 =5 then w1=4 and b1=8
so r=60 nb of red marbles
w=48 number of white marbles
b= 96 number of blue marbles
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i would thank u alot for the nice problems and site really
hope 4 u the best
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