Fruit Frenzy
The ratio of the number of apples to oranges to pears is 7:11:9. Timmy ate 21 fruits. As a result, the ratio of the number of apples to oranges to pears became 2:3:3. How many fruits were left ?
The answer is 168.
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8 solutions
there can be more than 1 possiblity... even 27(15) -21 = 384
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@bharanithara
yes u r right,that x=15 and y=48 can also be a solution of the equation.But,in that case,we have
no. of pears in initial state- 9 x= 9 15 =135 and, no. of pears in final state - 3 y= 3 48 = 144
But,its common sense,that if some fruits are eaten,final no. cant be greater than the initial no..so, u have to keep in mind that , no. of any of the fruits in the final state dont go higher than that of the initial state.
u can ask for further clarification!!!....
no, that case you mentioned is not possible, try substituting values of x and y
exactly. the answer is 27x-21 where x is an integer with remainder 7 when divided by 8. hence, there are infinitely many possible answers.
i got the same answer!!
Phew! very long process. Isn't there a shorter way ?
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Well, I don't really know... The only other way, I suppose, would be to construct a graph and see when it crosses are certain point. So far, though, I haven't been able to find a correct way to graph them for an answer clearly.
idk, I think this is simple enough : /
You can use Excel, it helps saving a lot of time for cheking cases
Yes. There is a much shorter way. We essentially need to solve for n where 2 7 n − 2 1 ≡ 0 ( m o d 8 )
We realize that 2 7 ≡ 3 ( m o d 8 ) So, if we want our value to be 21 greater than something divisible by 8, and for each value of 27 we add the "surplus" over something divisible by 8 increases by 3, this means the number of values of 27 we need to add = 21/3 = 7. This means the original number = 27 7, so the secondary number equals 27 7-21 = 168
typical Diophantine equation
idol
Did it exactly the same way!
I did the same way, too! :))
The original ratio was 7:11:9, so 7x+11x+9x=27x.
The new ratio was 2:3:3, so 2y+3y+3y=8y.
We write 27x-21=8y, or 27x-8y=21
Here we see that 2 7 ( 1 ) − 8 ( 3 ) = 3
Multiplying both sides by 7 , 2 7 ( 1 ∗ 7 ) − 8 ( 3 ∗ 7 ) = 3 ∗ 7
27(7)-8(21)=21
Therefore, the final number of fruits is 8 ( 2 1 ) = 1 6 8
Instead of solving diophontaine equations, i tried the following approach :
Let the original total number of fruits be x.
Then apples = 7x/27 Oranges = 11x/27 Pears = 9x/27
Now we know that 21 fruits were eaten. So the total number of fruits remaining = x -21
Proportion of remaining apples = 2 (x-21)/8 Proportion of remaining oranges = 3 (x-21)/8 Proportion of remaining pears = 3*(x-21)/8
Therefore, if we let amount of apples, oranges and pears be a1, a2 and a3 respectively,
a1 = 7x/27 - 2 (x-21)/8 a2 = 11x/27 - 3 (x-21)/8 a3 = 9x/27 - 3*(x-21)/8
But in the last equation, we know that a3 = 21-a1-a2
Thus substituting the values on the RHS in the first two equations, we get
21- (7x/27 - 2 (x-21)/8) - ( 11x/27 - 3 (x-21)/8) = 9x/27 - 3*(x-21)/8
which can be solved for x but i seem to have gone wrong somewhere
Of course, the diophontaine approach is much more elegant and computationally simpler
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a1,a2 and a3 are the amount of apples, oranges and pears that were eaten
This solution will work regardless of how big the number:
Let, a= no. of apples
o= no. of oranges
p= no. of pears
a' = no. of apples eaten
o'= no. of oranges eaten
p'= no. of pears eaten
a : o : p
7u:11u:9u
Total at first = 27u
After:
2p:3p:3p
Total later: 8p
27u = 8p +21 --(1)
2p + a' =7u --(2)
3p + o' = 11u --(3)
3p + p' = 9u --(4)
o' - p' = 2u ((3)-(4))
o' = 2u +p' --(5) ; p' = o' -2u --(7)
a' +0' +p' = 21--(6)
(5) put into (6)
a' +2u +2p' = 21--(8)
(7) into (8)
a' +2o' - 2u = 21 --(9)
(8) - (9):
4u +2p' +2o' = 0
p' +o' = 2u
p' = 2u - o'
o' = 2u - p' --(10)
(10) put into (5)
2u +p' = 2u - p'..............................??!!?!?!?!
p' = 0 --(11)....................................??!!?!?!!!?
(11) add into (4)
3p = 9u
p = 3u --(12)
(12) put into (1)
27u = 24u + 21
21 = 3u
u = 7
Therefore, 24u = 168
Im quite lucky that p' = 0, or the solution would be much longer
I figured a simple way by logical sense . Firstly the ratio 7:11:9 , the nearest way to make it equivalent to 2:3:3 is to get 6:9:9 in which i minus 3 from the total . 3 could also be equivalent to 21(the number of fruits eaten) if you multiply it by 7 . So i multiplied the whole left ratio 6:9:9 to 7 = 42:63:63 (the total of fruits left) . which sums up to 168 :) I hope its easy and understandable
The ratio of the fruits was, 7:11:9
total of the numbers of ratio= 7+11+9 = 27 which means , it should a multiple of 27. so we can denote the number of fruits as, 27x.
After 21 fruits have been eaten, the new ratio is 2:3:3.
total of the numbers of ratio= 2+3+3 = 8 which means , it should a multiple of 8. so we can denote the number of fruits remained as, 8y.
So we have the equation as: 27x - 21 = 8y
We need to find an integer x such that 27x - 21 is a multiple of 8 .
When x = 7, 27x - 21 = 168 which is divisible by 8. Therefore, the final number of fruits is 168.
i also could make the equation.... but didnt get the solution.
11+7+9=27, , 2+3+3=8, 8- =24, 27-24=3, 21/3=7, 24 7=168
Multiply the original ratio by 7 & becomes 49:77:63 meaning 49 apples, 77 oranges & 63 pears a total of 189 fruits. Subtracting21 results to 168 fruits which is proportional to the ratio 2:3:3. Using ratio 2:3:3 yields to the no.of apples= 42, oranges=63, & pears= 63 a total of 168 fruits.
get the lcm of (27x-21) & 8:(x E integer.)
The original ratio was 7:11:9 at total of 7x+11x+9x = 27x After 21 fruits have been eaten, the new ratio is 2:3:3 = 2y+3y+3y = 8y So we have 27x - 21 = 8y We need to find an integer x such that 27x - 21 is a multiple of 8 and another integer y. Or 27x - 21 must be divisible by 8. When x = 7, 27x - 21 = 168 which is divisible by 8. Therefore, the final number of fruits is 168