Question #a3553

Ok, so if Hao's dog ate 40 cans of dog food in 31 days, then he therefore ate `40/31=1.29` cans of dog food per day.

So, there are (`6 times 7 = 42)` days in 6 weeks, so the amount of cans that Hao should buy to feed his dog for 6 weeks can be calculated by:

`1.29 times 42 = 54.18` cans, but then Hao should buy 55 cans as we cannot buy 0.18 of a can. Thus, assuming that Hao's dog eats ~1.29 cans of dog food per day, Hao should buy 55 cans of dog food to feed his dog over the next 6 weeks.

Notice the trap! Over the next 6 `color(red)(weeks)`
The original condition is in `color(red)(days)`

Using ratio in fraction format.

Let the unknown count of cans be `x`

`("consumption in cans")/("time in days")->40/31 =x/(6xx7)`

To get `x` on its own multiply both sides by `color(red)(42)`

`color(green)(40/31 =x/42color(white)("ddd")->color(white)("ddd") 40/31color(red)(xx42)=x/42color(red)(xx42)`

`color(white)("dddddddddd")->color(white)("ddd")color(green)((40color(red)(xx42))/31 =x xx (color(red)(42))/42)`

`color(white)("dddddddddd")->color(white)("ddddd") 1680/31=x`

For 6 weeks the consumption is exactly `color(white)("d")54 6/31 color(white)("d")"cans"`

Approximately `54.1935` cans tp 4 decimal places

Rounding this gives 55 cans: you do not purchase part of a can.