Confusing math problem?

Question

There are y apples and x baskets. If you put 8 apples in each basket, one of the baskets will contain less than 8 apples. If you put 5 apples in each basket, there will be 12 apples left over. Find the possible values of x and y.

When I try writing the equations out, I get:
8x+(x>8)=y -----(1)
5x+12=y -----(2)
But the first equation is probably incorrect...so please show me how to solve this.

Many thanks!

Answer 1

That means each of the baket cannot contain 8 apples. so all but 1 shall contain 8 and z in other

so y = 8(x-1) +z where z < 8 = 8x -8 + z
5x+12 = y
we need to solve these

5x+12 = 8x - 8 + z

20 = 3x + z

x =6 means z = 2 y = 42
x=5 means z =5 y = 37
x=4 means z = 8 not possible

there are 6 baskets and 42 appples

or 5 baskets and 37 apples

both solutions meet above criteria

Answer 2

You're right with the second equation.

Go back and read the sentence about 8 apples in each basket. If one basket contains less than 8 apples, then it can have at most 7 apples while the others have 8 each. Also, if only one basket has less than 8 apples then the others must be full with 8 each. Can you write an inequality showing the range for how many apples there are? (Hint: there is a minimum and a maximum).

Using this, I found: 8(x-1) <= y < 8x

With that information, can you find the possible values of x and y?

Answer 3

The question doesn't really make much sense - if you put 8 apples in each basket, each basket has 8 apples, so how can a basket have less?
I think what was meant to be said was that you *can't* put 8 apples in each basket, ie y<8x.
So you know that 5x+12<8x, ie 3x>12, ie x>4.
Thus any x>4 will work, with y=5x+12.
(eg x=5, y=37, etc).

As a side note, note that writing '8x+(x>8)=y doesn't actually make sense mathematically.

Answer 4

I do nto know but it is sure the first equation you have written is false
it is 8x+ (x<8) =y

I try but not sure you can write 8x + a = 5x +12 since boyth are equal to y sum of apples
this gives 3x = 12 -a x = 4 -a/3
the number of appels must be an integer
Only two values of a are possible 3 or 6

with a=3 you find x=3 ; with a=6 you have x=2

It works!!!

Answer 5

8(x - 1) + 1 < y < 8x ................ (1)
8x - 7 < y < 8x
y = 5x + 12 ......... (2)
Substitute equation (2) into equation (1)
8x - 7 < 5x + 12 < 8x
3x - 7 < 12 < 3x
4 < x < 6 1/3
32 < y < 43 1/3

So x = 5, y = 37
x = 6, y = 42

So the only possibilities are
5 baskets annd 37 apples or 6 baskets and 42 apples

Answer 6

Don't forget that "less than 8" could be zero. The basic equation is 8x+z=5x+12, where 0<=z<8, hence:

3x = 20-z

which, from the range of z, leads to:

4 <=x < 7

So possible (x, y) values are (4,32), (5,37) and (6,42).

Answer 7

Your Mistake is quite Sily, its just 8(x-1) +(z<8)=y;

Rather than, 8x+(z<8)=y;

So, on sub teh second eqution you will be left with:

3x+z=16;

And Z has to less than 8 so I guess x can take values of 3,4 and 5

Every value of x will have a value for y hence the answer has to 3.