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MSC #23: Proportionately: Partitive Proportion

MSC #23: Proportionately: Partitive Proportion

“A rectangle is divided into two rectangles with areas shown.

The length of the whole rectangle is 49cm. What is the length marked “x”?

A) 49 cm; B) 51 cm; C) 54 cm; D) 55 cm; E) 57 cm”

This problem from the Junior level of the 2nd International Vedic Mathematics Olympiad using the topic for the 23rd of our 30 MATH-Inic Specials for Christmas series.

Our 23rd MATH-Inic Special for Christmas is one of my favorites. I originally titled it the “See Saw Solution” the reason for which will be clear later. We use it to mentally solve not only average speed problems, but also combination and mixture problems and even problems in engineering mechanics(statics). We can just “feel’ the answer.

When we apply “the product of the means is equal to the product of the extremes” to a proportion such as a:b:: c:d, we have bc = ad which is a balanced “see-saw”.

Imagine a 5-meter-long weightless beam unevenly placed in a fulcrum with 2 meters on one side and 3 meters on the other. How can we place 20 1-newton weights on the two ends of this “see-saw” so that they will balance?

We can use “trial and error” here or we could simply partition the 20 into two groups in the ratio of 2 : 3 which is 8 : 12 and place 8 pieces at the end of the 3 meter side while placing 12 pieces at the end of the 2 meter side. We will have a balanced see-saw with 24 newton-meters per side.

The IVMO question can be solved by portioning 89 into x and 89 -x and using the proportion, 480:588 :: (89 – x):x, leading to 588 (89-x) = 480x.

 However, looking at the areas 480 and 588 of the two triangles, an easier solution can be found. Both figures are divisible by 3 (the sum of digits are divisible by 3) and by 4 (the last two digits are divisible by 4). So they are both divisible by 12. 480 divided by 12 is 40. If we add 12 to 588 we will get 600 which is 12 x 50. So, 588 divided by 12 is (50 -1) or 49. Now we have 40 + 49 = 89. Thus, we have x = 49.

Let us now consider the featured problem with the motorist going up the mountain resort at 40 kph and returning at the rate of 60 kph. Since the distance is not given, the most common solution to this type of problem is to take the numeric least common multiple of the speeds, which is 120 in this case, and assume this to be the distance.

This means that the motorist will need 120 km/ 40kph or 3 hours to travel uphill and 120 km/60kph or 2 hours on the return trip. Then we compute the total distance traveled and divide it by the total time to get the average speed.

(2x 120)/(3 + 2) = (240)/5 =  48 kph.

But this is time consuming and totally, unnecessary. Given the same distance the speed is inversely proportional to the time. So, if the ratio of the speed is 40 : 60 or 2 : 3, then their time to traverse the same distance would be in the ratio of 3 : 2

Thus, the lower speed will contribute to 3/5 of the average speed while the faster speed will account for the other 2/5. If we use 40 kph as the base, then we need to compute only 2/5 of the difference in their speed which is (60 -40) or 20. This is also the same as partitioning 20 in to a ratio of 2 : 3. For 20 it is 8:12. So The average speed is 40 + 8 = 48 kph.

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