Find the Ratio: 0.0625 : 0.09375
Date: 05/15/2003 at 15:31:38 From: Faisal Subject: Ratio I am in grade 6 and am having problems with ratio. Please help me find the ratio between 0.0625:0.09375 and please tell me how to do it. Thank you, Dr. Maths, Faisal
Date: 05/15/2003 at 17:22:25 From: Doctor Ian Subject: Re: Ratio Hi Faisal, There are a few different ways you might approach this. One is to divide the larger number by the smaller one. For example, if you have a ratio of 12:138 doing the division 1 1.5 ________ 12 ) 1 3 8.0 1 1 --- 1 8 1 2 --- 6 0 6 0 --- 0 shows us that this is the same as the ratio 1:11.5. What do you get when you try this with your numbers? A second thing we can do is scale the values in the ratio by the same amount until they don't look quite as intimidating. For example, a ratio of 2:5 is the same as a ratio of 20:50 or a ratio of 200:500. (Do you see why?) So we can keep multiplying both terms by 10 until we get rid of the decimal point: 0.0625 : 0.09375 0.625 0.9375 6.25 9.375 62.5 93.75 625 937.5 6250 9375 <-- Same ratio, different values This looks easier to reduce, doesn't it? We can look for common factors, and eliminate them, e.g., 6250 : 9375 = 1250 * 5 : 1875 * 5 = 1250 : 1875 Where before we scaled _up_ by a factor of 10, now we're scaling _down_ by a factor of 5. But it's the same idea. Can you find other factors that these values have in common, and eliminate them? A third thing you can do is look for patterns that might work out nicely. In this case, I happen to know that 1/8 is 0.125, which means that 5/8 is 0.625, so that means that 0.0625 is 1/10 of that, or 5/80. So what if I multiply both items by 80? 0.0625 : 0.09375 * 80 80 ------ ------- 5 7.5 That's a lot easier to deal with, isn't it? This third approach requires you to be pretty comfortable with fraction/decimal conversions, so don't feel that this is something that should have jumped out at you. Mostly I was just trying to give you a sense of some of the different ways that you can tackle ratio problems. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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