Math Question

[Jab1113]

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This is from topscore....

Question: How many gallons of pure orange pulp must be added to 30 gallons of a 20% solution of orange pulp to make it a 50% solution?

Answer is 18 gallons. Topscore has the worst explanation. Can somebody help me out?

 

[rockclock]

This is from topscore....

Question: How many gallons of pure orange pulp must be added to 30 gallons of a 20% solution of orange pulp to make it a 50% solution?

Answer is 18 gallons. Topscore has the worst explanation. Can somebody help me out?

[pulp in 30 gal.] + [pulp in additional x gal.] = [pulp in total (30+x) gal.]

(0.20)(30) + (1.00)(x) = (0.50)(30+x)

x = 18 gal.

 

[Josh779]

[pulp in 30 gal.] + [pulp in additional x gal.] = [pulp in total (30+x) gal.]

(0.20)(30) + (1.00)(x) = (0.50)(30+x)

x = 18 gal.

why did you make the pulp additional 1*x? Do you assume it is 100%?

 

[mh0000]

Yes he did. The question said pure pulp.

 

[Jab1113]

[pulp in 30 gal.] + [pulp in additional x gal.] = [pulp in total (30+x) gal.]

(0.20)(30) + (1.00)(x) = (0.50)(30+x)

x = 18 gal.

thank you rockclock 👍

 

[Josh779]

Yes he did. The question said pure pulp.

Oh I see that now, thanks.

 

[rockclock]

thank you rockclock 👍

no prob. shout out to MI...last count we had 5 bachelor's degrees, 2 Ph.D's, an MBA, and an MD from Michigan in the family haha.

 

[Jab1113]

no prob. shout out to MI...last count we had 5 bachelor's degrees, 2 Ph.D's, an MBA, and an MD from Michigan in the family haha.

michigan family full of wolverines or spartans? lol

 

[rockclock]

michigan family full of wolverines or spartans? lol

all U of M wolverines...bleed maize and blue

 

[LetsGo2DSchool]

This is a lot easier if you just take a step back and think about it logically and use high school algebra.

You're starting with 30 gallons of 20% solution so it has already 6 gallons of pulp:

30 gallons * 20% = 6 gallons

You want to know how much more pulp (x) you need to add so the percentage equals 50%:

(6+x) / (30+x) = 50%

Solve for x

 

[Dental2000]

Anyone who has the Math destroyer(2010 ed)

look at TEST 1 #8....They used a simliar formulae but did not have 1*x

how are these 2 questions different?

 

[rockclock]

Anyone who has the Math destroyer(2010 ed)

look at TEST 1 #8....They used a simliar formulae but did not have 1*x

how are these 2 questions different?

i'm guessing you mean something like this...?

(0.20)(30) = (0.10)(30+x)

that would tell you how much water you would need to add to 30 gal. of 20% pulp to dilute it down to 10% pulp. it's m1v1 = m2v2.

the reason I wrote it out the way I did in the original problem, was to keep things consistent with (m)(v) terms:

[original] + [added] = [total]
(0.20)(30) + (1.00)(x) = (0.50)(30+x)

if you're adding something pure, obviously (1.00)(x) is just x.
if you're adding water, obviously (0.00)(x) is 0 and that term goes away and you get m1v1 = m2v2.
if you're mixing in x gal. of another concentration, you have the versatility to use (0.75)(x), etc.

my point was that the (m)(v) thing is more versatile than just m1v1 = m2v2. sure, you can do the same thing step-wise (a la LetsGo) but I don't see how that's any easier than being able to set up a one-line equation once you understand how (m)(v) terms can be additive.