We received several incorrect solutions like the ones below:
Combining paints A (1:3) and B (1:7):
|
Required
Ratio
|
Amount of
paint A |
Amount of paint B |
| 1:4 | 3 | 1 |
| 1:5 | 1 | 1 |
| 1:6 | 1 | 3 |
Combining paints C (1:4) and D (1:9):
|
Required
Ratio
|
Amount of
paint C
|
Amount of
paint D
|
| 1:5 | 4 | 1 |
| 1:6 | 3 | 2 |
| 1:7 | 2 | 3 |
| 1:8 | 1 | 4 |
They are based on the misconception that you
can add the ratios to work out the necessary combinations. The
solutions given have assumed that the 'parts' in the ratios are of
equal size so that a can in the ratio 1:3 contains half the amount
of the one in the ratio 1:7.
However, one can of paint A and one can of paint B does
not produce paint
in the ratio 2:10 (or 1:5) as suggested above, since that would
assume that the one part red in can A has the same volume as the
one part red in can B.
This can't be the case since there are 4 parts in can A and 8 parts
in can B,
so 1/4 of can A is red and 1/8 of can B is red..
To compare equal quantitities we will need to express the ratio of
the colours in can A as 2:6, so we have:
in can A: 2/8 red and 6/8 white
in can B: 1/8 red and 7/8 white
Combining one can of each paint will now give us
3/8 red and 13/8 white,
that is, paint in the ratio 3:13
Can anyone now solve this tough nut?
Yes!
We've received two correct solutions from Year 8 students at St
Albans High School.
Click here
to see Anjali's solution and here to see Lydia's
solution.
Well done Anjali and Lydia, and all the other students in Ms
Chapman's class who managed to crack this tricky problem.