15 Examples of Compound Rule of Three

It is known as compound rule of three a particular case of calls ‘rules of three’, which are those that simplify the resolution of mathematical problems in which a proportionality relationship governs on the basis of three known data and an unknown data (unknown).

The simplest case of a ‘rule of three’ is that of ‘Rule of three simple direct’, which is the one that describes the direct or positive proportionality between two quantities and is the one that governs many situations daily: if I want to buy two kilos of bread, for example, I need twice as much money as if I go to buy only one kilo. In other cases there is also a relationship of proportionality but negative: these relationships conform to the ‘rule of three simple inverse’.

In the call ‘compound rule of three ’, there is also a unknown data but the known data that allow us to solve this unknown are more than three (usually five), and there are two proportional relationships at the same time. What must be done in these cases to calculate the unknown value is to combine in a single expression the relationship between the two proportionalities, which implies reducing everything to the unitary expression minimal.

Example explained:

If 7 miners dig 49 meters in 21 days, how many meters will 14 miners dig in 35 days?

To solve this we first try to find out how much each miner digs in a single day (assuming all miners can work at the same rate).

To do this, the 49 meters are divided between the 21 days (with that we are assuming that all days are equally suitable for work) and among the 7 miners, thus reaching a value of ‘meters per day per miner'. Then it is enough to multiply it by the number of days and by the number of miners to arrive at the desired result. In short, the result will arise from doing 49*14*35/21*7.

By reducing the global problem to the smallest unit of the relation, the compound rule of three is transformed into a new simple rule of three.

Examples of compound rule of three

Here, as an example, twelve cases of application of the compound rule of three are listed, with the corresponding explanation:

1. Thirteen horses in 4 days consume 30 kg of feed. How many days can 8 horses be fed 60 kg of feed?
   We must put together an equation that in the numerator has the value of the kg of food that these will consume animals in the hypothetical problem situation (60 kg), the number of animals for said hypothetical situation (8 animals) and the number of days that is known as known data in the known situation (4 days), and in the denominator the number of animals and the amount of food that is kept in the known situation (13 and 30, respectively). In short: (60 kilos * 8 horses * 4 days) / (13 horses * 30 kilos) = 4,923. Horses may be fed for four full days (Arguably for 'almost' five).
2. Eleven workers can do a job in twenty days, but after eight days of work 6 workers retire. What day will they actually deliver the finished work?
   Days with eleven horses: (1 work * 11 workers * 8 days) / (11 workers * 20 days) = 0.4 work. Days with five workers: (0.6 works –the rest- * 5 workers * 20 days) / (5 workers * 0.4545 –proportional construction of five workers) = 26.4 days. In total, it will take 26.4 + 8 = 34.4 days.
3. For twelve days a family of 6 people has spent € 900 on food. How much would a couple spend in 20 days?
   Again and as in all other cases, the equation is assembled with what corresponds in each case; here: (900 euros * 20 days * 2 people) / (12 days * 6 people) = 500. They will spend € 500
4. To make a 40 m wall²12 workers have worked 6 days at a rate of 12 hours a day. How many days will 15 workers work at 9 hours a day to make a 100 m trench?² Wide?
   Again (6 days * 100 m² * 135 hours -15 workers with 9 hours a day-) / (40 m² * 144 hours -12 workers with 12 hours a day-) = 14,062. They will work 15 days to make that trench.
5. To feed 40 workers of a business 192 loaves of bread are needed. How many loaves of bread will one have to buy to feed 65 people for 80 days?
   (192 bars * 80 days * 65 people) / (24 days * 40 workers) = 1040. 1040 loaves of bread will have to be bought.
6. Five artisans make 60 rings in 15 days. If you want to make 150 rings in 25 days. How many artisans should be hired?
   (5 artisans * 150 rings * 25 days) / (60 rings * 15 days) = 20,833. 21 artisans must be hired.
7. A group of 20 workers must milk six cows in 10 days. After 4 days, 5 doubly efficient people join them. How many days will it take to milk all the cows?
   Part with 20 workers: (6 cows * 20 workers * 4 days) / (20 workers * 10 days) = 2.4 cows. Part with 25 workers: (10 days * 25 workers * 3.6 cows –remaining part) / (25 workers * 9 cows –double productivity) = 4 days. 4 + 4 = It will take 8 days in total.
8. For sending 5 kg to a town that is 60 km away, a company has charged me € 9. How much will it cost me to send an 8 kg package 200 km away?
   (9 euros * 8 kilos * 200 kilometers) / (5 kilos * 60 kilometers) = It will cost € 48.
9. In 9 days four workers, working 5 hours each day, have earned a total of $ 1200. How much will ten workers earn in 10 days, working 6 hours a day?
   ($ 1200 * 10 workers * 60 hours of work) / (4 workers * 45 hours of work) = They will win $ 4000.
10. For depositing $ 260 in a bank they give me $ 140 per year. How much money will I get if I deposit $ 10 for twice as long?
    ($ 140 * $ 10 * 2 years) / ($ 260 * 1 year) = They will give me $ 10.76.
11. Four tractors can remove 800 m³ of land in 3 hours. How long will it take six tractors to remove 1200 m³ of Earth?
    (3 hours * 6 tractors * 1200 m³) / (4 tractors * 800 m³) = 6,75. It will take 6 hours and 45 minutes.
12. Three people can live in a hotel for 9 days for $ 695. How much will the 15-person hotel cost for eight days?
    ($ 695 * 15 people * 8 days) / (3 people * 9 days) = It will cost $ 3,088.88.