An equation for an algebra word problem

[mech-eng]

Homework Statement

Suppose that b is 20% more than a, c is 25% more than b, and d is k% less than c. Find k such that a=d

Homework Equations

a=d; b=1.2a; c=1.25b=1.5a

The Attempt at a Solution

1. a=##\frac{c}{15 }##=d=##\frac{100c-k}{100}##
2. By multiplication of extremes and means we get 100c=150c-15.k
3. 50c=1.5k
4. Again simplifying k=33.3c

I have tried above solution but I think something is incorrect because k should be a number not a ratio in terms of c.
I don't latex well enough and it seems that step 1 is not seeming properly, too small symbols.
Source: Algebra and Trigonometry by Keedy Bittinger.
Thank you.

 

[SammyS]

Homework Statement

Suppose that b is 20% more than a, c is 25% more than b, and d is k% less than c. Find k such that a=d

Homework Equations

a=d; b=1.2a; c=1.25b=1.5a

The Attempt at a Solution

1. a=##\frac{c}{15 }##=d=##\frac{100c-k}{100}##
2. By multiplication of extremes and means we get 100c=150c-15.k
3. 50c=1.5k
4. Again simplifying k=33.3c

I have tried above solution but I think something is incorrect because k should be a number not a ratio in terms of c.
I don't latex well enough and it seems that step 1 is not seeming properly, too small symbols.
Source: Algebra and Trigonometry by Keedy Bittinger.
Thank you.

You have a typo and an error in multi-equation #1.

For one, ##\ \displaystyle a=\frac{c}{1.5} \ ##. (Missed the decimal point.)

More importantly the last part is in error.

##\ \displaystyle \frac{100c-k}{100} \ ##​

is incorrect. ##\ k\ ## must also multiply ##\ c\ ##.

 

[mech-eng]

But if ##d## is ##k%## less than ##c##, does not this mean ##\frac{c}-k}{100} ## and isn't it ##\frac{100c-k}{100}##?

And would you please explain why percent sign not appear in this code?

Thank you.

 

[Ray Vickson]

Homework Statement

Suppose that b is 20% more than a, c is 25% more than b, and d is k% less than c. Find k such that a=d

Homework Equations

a=d; b=1.2a; c=1.25b=1.5a

The Attempt at a Solution

1. a=##\frac{c}{15 }##=d=##\frac{100c-k}{100}##
2. By multiplication of extremes and means we get 100c=150c-15.k
3. 50c=1.5k
4. Again simplifying k=33.3c

I have tried above solution but I think something is incorrect because k should be a number not a ratio in terms of c.
I don't latex well enough and it seems that step 1 is not seeming properly, too small symbols.
Source: Algebra and Trigonometry by Keedy Bittinger.
Thank you.

You need
$$d =\left( 1 - \frac{k}{100} \right) c$$

 

[mech-eng]

$$d =\left( 1 - \frac{k}{100} \right) c$$=##\frac{c}{1.5}##

##50c=15ck##

and k=3.333

Would you please write above in the proper latex form?

Thank you.

 

[UsableThought]

@mech-eng, I suggest you check your answer of ##k = 3.333\%## against your initial set of equations ##a=d; b=1.2a; c=1.25b##.

Also, although percentages proper are expressed as ##x/100##, it's easier & faster during calculation to assume 1 = 100%; thus instead of $$d = \left(1 - \frac{k}{100}\right) c$$ you can work more simply with ##d = (1 - k)c##. For that matter you can substitute a temporary value for that expression and do the subtraction later on.

 

[SammyS]

$$d =\left( 1 - \frac{k}{100} \right) c$$=##\frac{c}{1.5}##

##50c=15ck##

and k=3.333

Would you please write above in the proper latex form?

Thank you.

That looks proper to me.

Mathematically, it's equivalent to

##\ \displaystyle d = \left(\frac{100-k}{100}\right) c ##

 

[UsableThought]

I think the OP was referring to the part of his comment that wasn't already formatted in LaTeX.

 

[mech-eng]

##\ \displaystyle d = \left(\frac{100-k}{100}\right) c=\frac c {15} ##

I added the right side but is it proper. I didn't used symbols such as \left and \right.
The following is what I added there. But it is not the same when it is on the right side. Would you please explain this?
## \frac c {15}##

Thank you.

 

[SammyS]

##\ \displaystyle d = \left(\frac{100-k}{100}\right) c=\frac c {15} ##

I added the right side but is it proper. I didn't used symbols such as \left and \right.
The following is what I added there. But it is not the same when it is on the right side. Would you please explain this?
## \frac c {15}##

Thank you.

Proper in what way?

You dropped the decimal point from 1.5 again, otherwise, it looks right math-wise .

By the way: for % symbol in LaTeX, use \ suffix as in \%.

## 7\% ## or ##k\%##

 

[mech-eng]

Proper in what way?

You dropped the decimal point from 1.5 again, otherwise, it looks right math-wise .

Sorry. Question is solved. But look the Latex. Their size of the c/15 parts are different.

Thank you.