Ratio and Proportion


Ratio and Proportion.

Yr9 asked, 20130318, about ratio questions.

In general the problem can expressed as in this example::

1.  A tank is filled with two fluids, A and B in the ratio a:b. After some use, the level in the tank is reduced to fraction or percentage p. The tank is then filled up again with only material B. Find the ratio of A to B in the newly filled tank.

The technique I recommend is to draw a picture of the tank, divided with a horizontal line into a and b. Forget about mixing, draw it the way you need to picture it. When some is taken out, draw the tank again, but divide it again with a vertical line to show the newly empty part. Now fill the ‘empty’ part as directed by the question. You should have created a fractions problem that you can see well enough to tackle and answer. Reading the question carefully, find the requested ratio.

We had a series of questions in class of which this was the hardest:

2. A vessel is filled with mixed liquids, say oil and vinegar, in the ratio 4:3. Later, 2/3 has been used and the vessel is topped up with a mixture which is half oil and half vinegar. What is the new ratio of oil to vinegar?

                                  Answer: a ratio of 4:3 means working in sevenths. the left-hand diagram shows teh container divided 4:3, the 4 at the bottom. the right-hand diagram is divided by a vertical line one third of the way across, showing the two-thirds used. that space is then divided horizontally by a horizontal supoosedly dividing the space in two halves, though it doesn’t look it on my diagram.

Bottom left corner of the right hand diagram is 4/7*1/3 = 4/21 oil
Bottom right corner is 2/3 * 1/2 = ⅓
Top left corner is 3/7*1/3 = 1/7 vinegar
Top right corner is ½*2/3 = 1/3 vinegar.
Total vinegar is 1/7+1/3 = 10/21
Total Oil is 4/21+1/3 = 11/21
So the ratio of oil to vinegar is 11:10


Assuming you understand this, here are some problems of the same type:

3.   A cylindrical drum is filled with a well-mixed 2:1 mixture of cement and sand in preparation for making concrete. After the cylinder has been half-emptied, it is filled with only sand. What ratio of cement : sand is now in the drum?

4.   A food manufacturer has a large container filled with a thorough mixture of two fruit juices, apple and blackcurrant, in the ration three parts apple to two parts blackcurrant. After filling a large number of bottles for sale the tank is 80% empty and is filled (in error) only with blackcurrant. What is the % of blackcurrant in the new mixture?

5.   On another day the same manufacturer has a similar problem, but what is added is equal parts of apple and blackcurrant when the tank was 15% full. What percentage of apple is left and what is the ratio of apple to blackcurrant?


These questions become practical (‘real’, believable problems) only when the ‘system’ needs to rectify a problem, so a Y9 or Y10 student should be able to provide an answer that restores the balance:

6.   In Q4, if the error is discovered when the tank is 32% full (having been filled with only blackcurrant as the question said, so we have to assume you have answer Q4 correct), what ratio of mix is required to be added to rectify the problem?

7.  In Q5, assume the problem is found less quickly and that it is just under half full, at 45%; this time add just enough (and find out how much that is) of one material (presumably, apple juice) to put the balance back to its proper 3:2 ratio. 


A different class of problem deals with truncating shapes. The difficulty generated is to do with perception of area and volume when instructions are given about length. A classic problem follows:


8.   I discover that I need only one 5ml tin of paint for my model plane. How much do I need to paint a full size plane is my model is built at 1:80?

9.   A vertical square-based pyramid of height 6m is cut horizontally so that the top 3m is discarded. What % volume remains?


Answer 8: the 1:80 is a linear scale, and paint goes on an area, so I need 80x80 times as much, or 32,000ml, 32 litres. That’s before we worry about paint thicknesses and appropriateness of the model.

Answer 9: We don’t care how tall it is, the top half in length is what is cut off, so this is an eighth (a half cubed) of the volume and 7/8 is left, 87.5%.

10.  I build a model of a car at 1:20 scale. (i) You recognise the size of a steering wheel, so what sort of diameter do you think the model wheel will be? (ii) If the full size car has a 10 gallon tank, what size should the model tank be? (You don’t need to know that a gallon is 4.54 litres.) (iii) If I used 8ml to paint all the black bits on the model, how much should I need to do the same paint job on the full size car?

11.   I have a cylindrical cone and cut off the top two thirds by height. (i) What is the volume of the remainder? (ii) What about the remaining surface area?

12.    I have two spheres (think planets) of similar density and one is twice the mass of the other. What is the ratio of their diameters?





Q3 Cement was 1/3, reduced to 1/6th of total volume, so new ratio is 1:5

Q4 Apple content was 60%, reduced to 20% of that, i.e. 12%, so new mix is 88% blackcurrant.

Q5 Apple content was 60%, reduced to 60%*15% = 9% apple. What is added is 50/50, and 85% of total volume, so apple content goes to 60%x15%+85%x50% = 51.5%, while the blackcurrant goes to 40%x15%+85%x50% = 48.5%, which checks nicely.

Q6 60% apple went down to 9% while the blackcurrant went down to 6%.  17% more blackcurrant has been added and the current ratio is 9%:23%, filling 32% of the tank. To fill the tank we need 68% and to end with 60:40, so what is needed is 51:17, or 3:1 in favour of the apple.

Q7 Go back to the point where there is 15% left, 9%:6%. Add 30% blackcurrant and 36% of the total volume is blackcurrant. We need the apple volume to rise to 36x1.5=54%, so we need to add 45%. That makes lots of sense, adding 50% more than the ‘wrong’ 30% of blackcurrant, and leaving the tank now at 15+30+45 = 90% full.


Answer 10: if a wheel is 50cm, the scale model would be 25mm. 10 gallons needs to be divided by 203, so 10/8000 of a gallon,  5.675ml. 8ml x 202 (area) = 3200ml = 3.2 litres.

Answer 11: (⅔)³ = 8/27 so 19/27 remains.
Area is hard, since there is a new surface to add. There are three parts, the base (A, area πr
²), the newly revealed cut surface (which must be 4/9 of A) and the reduced curved surface area (the remaining 5/9 of the previous curved part, πrɩ, for slant height ɩ on the originalWhat there was in the original complete cone is A + πrɩ and the newly truncated cone has surface area of 13/9 A + 5/9 πrɩ  (where ɩ²=r²+h²)
As a ratio, this might be written as (13r+5ɩ) / 9(r+ɩ). if you want a nicer answer, you have to choose some suitable expression for ɩ in terms of r.

Answer 12: 1 : 1.26    (1.26 = 2 to 3sf ).



Q3 Cement was 1/3, reduced to 1/6th of total volume, so new ration is 1:5

Q4 Apple content was 60%, reduced to 20% of that, i.e. 12%, so new mix is 88% blackcurrant.

Q5 Apple content was 60%, reduced to 60%*15% = 9% apple. What is added is 50/50, and 85% of total volume, so apple content goes to 60%x15%+85%x50% = 51.5%, while the blackcurrant goes to 40%x15%+85%x50% = 48.5%, which checks nicely.

Q6 60% apple went down to 9% while the blackcurrant went down to 6%.  17% more blackcurrant has been added and the current ratio is 9%:23%, filling 32% of the tank. To fill the tank we need 68% and to end with 60:40, so what is needed is 51:17, or 3:1 in favour of the apple.

Q7 Go back to the point where there is 15% left, 9%:6%. Add 30% blackcurrant and 36% of the total volume is blackcurrant. We need the apple volume to rise to 36x1.5=54%, so to add 45%. that makes lots of sense, adding 50% more than the ‘wrong’ 30% of blackcurrant, and leaving the tank now at 15+30+45 = 90% full.


Answer 10: if a wheel is 50cm, the scale model would be 25mm. 10 gallons needs to be divided by 203, so 10/8000 of a gallon,  5.675ml. 8ml x 202 (area) = 3200ml = 3.2 litres.

Answer 11: (2/3)3 = 8/27 so 19/27 remains. Area is harder, since there is a new surface to add. What there was in the complete cone volume is 1/3Ah (A is the base area, πr2, and h the height). The removal adds a new surface of 4/9 πr2, while making the whole area 4A/9 and the area of the truncated cone is 4/9(A+ πr2).

Answer 12: 1 : 1.26 = 21/3  .


© David Scoins 2017