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- Thread starter harpazo
- Start date

This can be solved by referring to the method used here "Original Price 1"`

This can be solved by referring to the method used here "Original Price 1"`

Ok. Will do....

[MATH]P(1+D)=R[/MATH]

Hence:

[MATH]P=\frac{R}{D+1}[/MATH]

Now we have a general formula for all such problems. Applying it to this one, we identify:

[MATH]R=300,\,D=-0.2[/MATH]

Thus:

[MATH]P=\frac{300}{-0.2+1}=\frac{300}{\dfrac{4}{5}}=5\cdot\frac{300}{4}=5\cdot75=375[/MATH]

[MATH]P(1+D)=R[/MATH]

Hence:

[MATH]P=\frac{R}{D+1}[/MATH]

Now we have a general formula for all such problems. Applying it to this one, we identify:

[MATH]R=300,\,D=-0.2[/MATH]

Thus:

[MATH]P=\frac{300}{-0.2+1}=\frac{300}{\dfrac{4}{5}}=5\cdot\frac{300}{4}=5\cdot75=375[/MATH]

I enjoy your ability to create a general formulas for problems. Wish I had such talent back in my college days.

[MATH]P(1+D)=R[/MATH]

Hence:

[MATH]P=\frac{R}{D+1}[/MATH]

Now we have a general formula for all such problems. Applying it to this one, we identify:

[MATH]R=300,\,D=-0.2[/MATH]

Thus:

[MATH]P=\frac{300}{-0.2+1}=\frac{300}{\dfrac{4}{5}}=5\cdot\frac{300}{4}=5\cdot75=375[/MATH]

Nicely done, Mark!(Happy)

@harpazo, please post new question only after you have replied to your previous countless threads, and please don't let Mark to finish up for you. Thanks.

Nicely done, Mark!(Happy)

I occassionally overload the site with math questions for two reasons:

1. To give members extra practice.

2. To increase the popularity of the site.

I have been reviewing trigonometry for a few months and Mark is my online tutor for this purpose.

Last edited:

I should look into creating a thread prefix that can be added to practice questions so we can easily spot which are practice and which aren't.