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On planet Larky, 7 ligs = 4 lags, and 9 lags = 20 lugs. How many ligs are equivalent to 80 lugs?

Logic Aug 20, 2018

#1**+2 **

This is one way to think of it:

We know...

9 lags = 20 lugs Multiply both sides of this equation by 4 .

36 lags = 80 lugs

And we know...

7 ligs = 4 lags Multiply both sides of this equation by 9 .

63 ligs = 36 lags

63 ligs = 36 lags and 36 lags = 80 lugs So...

63 ligs = 80 lugs

hectictar Aug 20, 2018

#2**+2 **

Excellent hectictar!

Here is another way:

1 lag = 7/4 ligs

7/4 ligs x 9 lags = 20 lugs

15.75 ligs = 20 lugs. Multiply both sides by 4

63 ligs = 80 lugs.

Guest Aug 20, 2018

#3**+3 **

On planet Larky, 7 ligs = 4 lags, and 9 lags = 20 lugs. How many ligs are equivalent to 80 lugs?

Here is a slightly different variation. Once you get the hang of this method you do not have to understand very much.

It all just falls out.

Think of them all as rates and write them as fractions. Since they have units it is perfectly valid to turn the fractions upside down if that helps. We have.

\(\frac{7\;ligs}{4\; lags}\qquad \frac{9\; lags}{20\;lugs}\qquad \frac{80\;lugs}{1}\qquad ?ligs\)

Ligs is what you want so put if first and on the top. \(\frac{7\;ligs}{4\; lags}\)

Now you need to get rid of the lags on the bottom so find a fraction with lags and make sure lags is on the top, then put a mult sign between them.

\(\frac{7\;ligs}{4\; lags}\times \frac{9\; lags}{20\;lugs}\)

Now the lags will cancel but you need to get rid of the lugs so multiply with lugs on the top.

\(\frac{7\;ligs}{4\; lags}\times \frac{9\; lags}{20\;lugs}\times \frac{80\;lugs}{1}\)

Now the lags cancel and the lugs cancel leaving you with just ligs on the top which is what you want.

\(\frac{7\;ligs}{4\; lags}\times \frac{9\; lags}{20\;lugs}\times \frac{80\;lugs}{1}\\ =\frac{7\;ligs}{4}\times \frac{9}{20}\times \frac{80}{1}\\ =7\times9 \;ligs\\ =63 \; ligs \)

.Melody Aug 21, 2018

#5**0 **

Is it?

I had to teach maths for nurses one time and they had all these super horrible questions involving rates.

I 'invented' this method all by myself. LOL

I have never seen anyone else use it in mathematics although it is very similar to methods regularly used in chemistry I think.

It makes difficult rates questions super easy!

Melody
Aug 22, 2018