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Old 02-13-2009   #1
milk123
 
Join Date: Feb 2009
Posts: 3
Default word problem help?

Hi everyone, i'm new here. Seems like a great educational site.

Its been a while for me and i really need some help with the formula and explaination.



The fuel for a two-cycle motorboat engine is a mixture of gasoline and oil
in the ratio of 15 to 1. How many liters of each are in 6.6L of mixture?


Fifty kilograms of a cement-sand mixture is 40% sand. How many kilograms
of sand must be added for the resulting mixture to be 60% sand?


A karat equals 1/24 part of gold in an alloy (for example, 9-karat gold is
9/24 gold). HOw many grams of 9-karat gold must be mixed with 18 karat
gold to get 200g of 14-karat gold?
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Old 02-13-2009   #2
MAS1

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Quote:
Originally Posted by milk123 View Post
Hi everyone, i'm new here. Seems like a great educational site.

Its been a while for me and i really need some help with the formula and explaination.



The fuel for a two-cycle motorboat engine is a mixture of gasoline and oil
in the ratio of 15 to 1. How many liters of each are in 6.6L of mixture?


Fifty kilograms of a cement-sand mixture is 40% sand. How many kilograms
of sand must be added for the resulting mixture to be 60% sand?


A karat equals 1/24 part of gold in an alloy (for example, 9-karat gold is
9/24 gold). HOw many grams of 9-karat gold must be mixed with 18 karat
gold to get 200g of 14-karat gold?
1. If the ratio of gasoline to oil in the mixture is 15 to 1, then in a 16 L sample of the mixture there are 15 L of gasoline and 1 L of oil. In other words, 15/16 of the mixture is gasoline, and 1/16 is oil.

So if the total is 6.6L then 15/16 of it is gasoline and 1/16 is oil.
(15/16)*6.6 = 6.1875 L gasoline
(1/16)*6.6 = 0.4125 L oil

2. If 50 kg of a cement-sand mixture is 40% sand then:
(0.40)*50kg = 20 kg sand (S) and (50 - 20) = 30 kg cement (C).

The ratio of sand in the mixture is given by:
S / (S + C).

Therefore if we want 60% sand then we have:
S / (S + C) = 0.60
Since the amount of cement does not change then C = 30 and we have:
S / (S + 30) = 0.60
Solving for S gives S = 45 kg. Since we already have 20 kg of sand, then we need to add 25 kg of sand.

3. 9 karat gold = 9/24 gold
18 karat gold = 18/24 gold
X = grams of 9 karat gold
Y = grams of 18 karat gold

(9/24)*X + (18/24)*Y = (14/24)*200
Also by weight we have the equation:
X + Y = 200

Since we have two equations and two unknowns set them up like:

(9/24)*X + (18/24)*Y = 2800/24
X + Y = 200
Multiplying the bottom equation by 9/24 gives:

(9/24)*X + (18/24)*Y = 2800/24
(9/24)*X + (9/24)* Y = 1800/24

Subtracting the bottom equation from the top gives:

(9/24)*Y = 1000/24. Solving for Y gives:

Y = 1000/9 = 111 1/9 = 111.11 grams of 18 karat gold

Plugging the value for Y back into one of the equations and solving for X gives:

X = 800/9 = 88 8/9 = 88.89 grams of 9 karat gold.

Another way to set up the equations is:

(9/24)*X + (18/24)*Y = (14/24)*200 (Amount of gold in the alloy)
(15/24)*X + (6/24)*Y = (10/24)*200 (Amount of other metal in the alloy).

Hope this helps.
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Old 02-13-2009   #3
milk123
 
Join Date: Feb 2009
Posts: 3
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Exellent explaination but i'm looking for the algbratic equation.

So for the sand and cement one could i say;
15a+1a = 6.6
16a = 6.6
a = 6.6/16
a=.4125
So therefore 15a = 6.1875L of gasoline
1a = .4125L of oil


Does this one look correct?

Last edited by milk123; 02-13-2009 at 07:40 PM..
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Old 02-13-2009   #4
milk123
 
Join Date: Feb 2009
Posts: 3
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If i'm right about the oil and gas one, i'm lost on the other two.
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Old 05-29-2009   #5
shae marks
 
Join Date: May 2009
Posts: 1
Default word problem help?

Hi everyone, i'm new here. Seems like a great educational site.

Its been a while for me and i really need some help with the formula and explaination.



The fuel for a two-cycle motorboat engine is a mixture of gasoline and oil
in the ratio of 15 to 1. How many liters of each are in 6.6L of mixture?


Fifty kilograms of a cement-sand mixture is 40% sand. How many kilograms
of sand must be added for the resulting mixture to be 60% sand?


A karat equals 1/24 part of gold in an alloy (for example, 9-karat gold is
9/24 gold). HOw many grams of 9-karat gold must be mixed with 18 karat
gold to get 200g of 14-karat gold?
shae marks is offline   Reply With Quote
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