How do I solve this? Easy but I can't remember
- Thread starter SLSCoder
- Start date
I understand what you wrote, but you really should not abuse the language of mathematics like that: 41 does NOT equal 33
If you use the notation correctly, all becomes quite straightforward
[math]\dfrac{33}{41} = \dfrac{40}{x} \implies \dfrac{41x}{33} * \dfrac{33}{41} = \dfrac{41x}{33} * \dfrac{40}{x} \implies x = \dfrac{41 * 40}{33} \approx 50.[/math]
If you use the notation correctly, all becomes quite straightforward
[math]\dfrac{33}{41} = \dfrac{40}{x} \implies \dfrac{41x}{33} * \dfrac{33}{41} = \dfrac{41x}{33} * \dfrac{40}{x} \implies x = \dfrac{41 * 40}{33} \approx 50.[/math]
D
Deleted member 4993
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Let's write it out!
We want to FIND amount of cereal.
That is:
To get 40 grams of carbohydrate - how many grams of cereal I would need?
To get 33 grams of carbohydrate I need 41 grams of cereal
To get 1 grams of carbohydrate I need (41/33) grams of cereal............................. (see now 33 is below 41)
To get 40 grams of carbohydrate I need (41/33 * 40 = 49.69) ~ 50 grams of cereal ............................. however we get the same answer because we use correct logic.
In essence you are asking what percentage of the cereal is carbohydrate.
How do we calculate that percentage? We divide 33 by 44, which can be shown in math notation as [imath]\dfrac{33}{41} \approx 80.5\%.[/imath]
Then you want to find the amount of cereal that has 40 grams of carbohydrate. But the percentage of the cereal that will be carbohydrate will be the same. So
[math]\dfrac{33}{41} = \dfrac{40}{x} \implies 0.8 \approx \dfrac{40}{x} \implies x \approx \dfrac{40}{0.8} = 50.[/math]
A hundred years ago this was taught as the rule of three as part of arithmeti.
D
Deleted member 4993
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Not hundred years ago - only ~60 years - when I was a mere child
As it was explained to me, it was called rule of three because it required 3 lines to complete the calculations.
Otis
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And, of course, we have the grade school proportion rule, "The product of the means is equal to the product of the extremes", or, as Denis liked to say, "criss-cross multiplication".
For anyone interested, here's how that rule works.
Given a proportion [imath]\quad \dfrac{a}{b} = \dfrac{c}{d}[/imath]
Then [imath]\quad a \cdot d = b \cdot c[/imath]
That rule is handy for clearing fractions in algebra, and there's also a shortcut for solving proportions that I rather like, and it's based on the rule above:
"Multiply on the diagonal and divide by the number not used."
Solve the proportion [imath]\quad \dfrac{33}{41} = \dfrac{40}{x}[/imath]
41 and 40 are on the diagonal.
[imath]x = \dfrac{41 \cdot 40}{33}[/imath]
?
Steven G
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I like the number 1! Why? Because I know what to do to any number to make it 1 and I know what to multiply 1 by to get any number.
33 grams of carbohydrates corresponds to 41grams of cereal
Divide both numbers by 33 (since that is what you do to 33 to get 1)
Now 1 gram of carbohydrates corresponds to 41/33 grams of cereal
Now multiply both numbers by 40 (since that is what you do to 1 to get 40)
Now 40 gram of carbohydrates corresponds to (41/33)40 grams of cereal
If you realized that you need to multiply 40/33 (since that is what you do to 33 to get 40), then you just multiply both numbers by 40/33 and you can be done in one step..
33 grams of carbohydrates corresponds to 41grams of cereal
Divide both numbers by 33 (since that is what you do to 33 to get 1)
Now 1 gram of carbohydrates corresponds to 41/33 grams of cereal
Now multiply both numbers by 40 (since that is what you do to 1 to get 40)
Now 40 gram of carbohydrates corresponds to (41/33)40 grams of cereal
If you realized that you need to multiply 40/33 (since that is what you do to 33 to get 40), then you just multiply both numbers by 40/33 and you can be done in one step..