Original Price / Partial Solution

[Guest]

The price of cookies was increased by 10 cents per dozen. As a result, 5 fewer dozen could be purchased for $6. Determine the original price per dozen.

A partial solution was provided with problem:

Let P = original price per dozen in cents.

600/(P + 10) + 5 = 600/P

How do I proceed ?

[Guest]

You should always try to understand the problem logically first, then solve it mathematically.
You are letting P = the original price per dozen.
And you are considering 2 purchases of cookies....one before the price increase and one after the price increase.
Before the price increase the price was P cents per dozen so the number of dozen able to be bought was 600/P
After the price increase the price per dozen has increased to (P + 10) cents per dozen so the number of dozen able to be bought after the increase is (600/(P + 10))
The difference between the numbers of cookies purchased on each purchase is 5 fewer after the increase. So the difference (original - increased) is 5.
We setup an equation to show this mathematically, ie. (600/P) - (600/(P + 10)) = 5
OR....Re-arranging the terms to match your equation .... (600/(P + 10)) + 5 = 600/P ... (what I buy now) + 5 = (what I used to buy)

Now all you have to do is solve for P.
You should see there is a common factor of 5 .....
So divide across by 5 ....may simplify with smaller numbers...
There are 2 terms on the LHS ....find a common denominator and combine them as one fraction...
That leaves a single fraction on each side so you can cross multiply....
Then collect the like terms and move to one side of equation and you end up with a quadratic equal to Zero, which factorises easily to solve for P.

[Guest]

"Now all you have to do is solve for P.
You should see there is a common factor of 5 .....
So divide across by 5 ....may simplify with smaller numbers...
There are 2 terms on the LHS ....find a common denominator and combine them as one fraction...
That leaves a single fraction on each side so you can cross multiply....
Then collect the like terms and move to one side of equation and you end up with a quadratic equal to Zero, which factorises easily to solve for P. "


Please post a similar problem with a step by step solution. I am not very good at solving quadratics. Thanks.

[Guest]

I will now do a similar problem to show steps.........You then post your own question solution.....

This time 440 cents spent and price increase 2 cents...and 2 less items bought after increase....

It will be a similar equation...

(440/(A + 2)) + 2 = 440/A

Now all you have to do is solve for A.

You should see there is a common factor of 2 .....

So divide across by 2 ....may simplify with smaller numbers...but may not...but we will do it to keep same method as original question....

This gives (220/(A + 2)) + 1 = 220/A

There are 2 terms on the LHS ....find a common denominator and combine them as one fraction...consider the 1 as a fraction of 1/1.

((220) + 1(A + 2))/(A + 2) = 220/A

Read this as 220 plus 1 times (A + 2) all over (A + 2).....(A + 2) is the common denominator of (A + 2) and 1.

then simplify... (220 + A + 2)/(A + 2) = 220/A

this gives..... (222 + A)/(A + 2) = 220/A

That leaves a single fraction on each side so you can cross multiply....(top one side x bottom on other)

That gives...222A + A^2 = 220A + 440

Then collect the like terms ....gives...A^2 + 222A - 220A - 440 = 0

and move all to one side of equation
and you end up with a quadratic equal to Zero...

A^2 + 2A - 440 = 0

which factorises easily to solve for A.

The factors of 440..... 20 and 22 when multiplied together give 440 and when subtracted give 2 ...so that will work.
We want a minus 440 so one is plus and other is minus and we want a plus 2A so the biggest .... 22 is the positive one

So the factors are ... (A - 20)(A + 22) = 0

Therefore either (A - 20) = 0 giving A = 20 OR (A + 22) = 0 giving A = -22

We are wanting the original price so take the +ve one .... So.....A = 20.

Check....when price was 20 then 440/20 = 22 items could be bought.

after increase then 440/(20 + 2) = 440/22 = 20 items could be bought and this is 2 less.

..........................Now its your turn to post the original question solution..........

[Guest]

600/(P + 10) + 5 = 600/P

((600/(P +10)) + 5 = 600/P

(120/(P + 10)) = 120/P

((120) (P+10)) / (P+10) = 120/P

(120 (P+10) / (P +10) = 120/P

(130 + P) / (P + 10) = 120/P

130P + P^2 = 120P + 1200

P^2 + 130P - 120P - 1200 = 0

P^2 + 10P - 1200 = 0

I don't know the factors of 1200.

[Guest]

I think I have one solution:

P^2 + 10P - 1200 = 0

(P + 40) (P - 30) = 0

P = 40

600/40 = 15

I am not sure how to check.

[Guest]

Your solution attempt...with my notes in between......

600/(P + 10) + 5 = 600/P

((600/(P +10)) + 5 = 600/P

(120/(P + 10)) = 120/P...wrong..you forgot about the 5....divided by 5 = 1
you need to divide each term by 5 as in example last post...

((120) (P+10)) / (P+10) = 120/P....reads wrong because.... what is this..((120) (P+10))

(120 (P+10) / (P +10) = 120/P....reads wrong same reason over again....and what is this..((120) (P+10))

xxxxxxxxxxx....The above should be....((120) + (P+10)) / (P +10) = 120/P

(130 + P) / (P + 10) = 120/P.......Correct even though errors made above...???

130P + P^2 = 120P + 1200....correct..this is after cross multiplying

P^2 + 130P - 120P - 1200 = 0.....correct

P^2 + 10P - 1200 = 0......correct

I don't know the factors of 1200......you said this...


Then your next post....you said...???

I think I have one solution:

P^2 + 10P - 1200 = 0

(P + 40) (P - 30) = 0 .....correct...these are the 2 factors of the quadratic eqn.
If either bracket has value Zero then the eqn. is Zero.

So... (P + 40) = 0 when P = -40 .... also....(P - 30) = 0 when P = +30.

So....either P + 40 = 0 which gives P = - 40

OR......... P - 30 = 0 which gives P = + 30

Our price needs to be positive so we take the +ve value .....so P = 30 cents per dozen original price


You took P = 40 .....Wrong.... you just took the figures from the brackets....you need to work it out as shown above

Check.... 600/30 = 20 dozen cookies bought at original price.
and.......600/(30 + 10) = 600/40 = 15 dozen cookies bought at increased price.....and this is 5 dozen fewer.........

[Guest]

"We want a minus 440 so one is plus and other is minus and we want a plus 2A so the biggest .... 22 is the positive one"


"So... (P + 40) = 0 when P = -40 .... also....(P - 30) = 0 when P = +30.

So....either P + 40 = 0 which gives P = - 40

OR......... P - 30 = 0 which gives P = + 30"

I am not clear on this part.

[Guest]

Your first query is from the earlier post.....
The factors of 440..... 20 and 22 when multiplied together give 440 and when subtracted give 2 ...so that will work.
We want a minus 440 so one is plus and other is minus and we want a plus 2A so the biggest .... 22 is the positive one.

What we need to satisfy the equation is 2 factors that multiply to give -440
20 x 22 = 440 .....this is plus 440 ....positive
-20 x -22 = 440.....this is plus 440 ....positive
+20 x -22 = -440,,,,,this is minus 440 ....negative
-20 x +22 = -440,,,,,this is minus 440 ....negative

The only ones that give -440 are the last 2...

The difference between the 2 factors needs to be 2....plus 2
The only way I can get that is with +22 - 20 = +2.......plus 2
So 22 needs to be the positive one and take 20 from it, OR add -20 to it which ever way you look at it.

To check you can multiply out the factors.....
(A + 22)(A - 20)
gives... A^2 - 20A + 22A -440
gives.... A^2 +2A - 440
...................................................................

Your next query is just basic algebra......

"So... (P + 40) = 0 when P = -40 .... also....(P - 30) = 0 when P = +30.

So....either P + 40 = 0 which gives P = - 40

OR......... P - 30 = 0 which gives P = + 30"

.................

Basically ... if P + 40 = 0 then P must be minus 40
Because..... -40 + 40 = 0

and similarly for.. P - 30 = 0 then P must be plus 30
Because........... +30 - 30 = 0

And the equation will be 0 for either of these conditions.....
So we get 2 roots for the equation.....
But the only one making sense to the problem is the positive value for P because the price has to be a real positive number.

[Guest]

I understand that. Thanks again.