Sign Up FREE!  Sign In  Classroom Setup  Common Core Alignment  
02132009  #1 
Join Date: Feb 2009
Posts: 3

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 twocycle 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 cementsand 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, 9karat gold is 9/24 gold). HOw many grams of 9karat gold must be mixed with 18 karat gold to get 200g of 14karat gold? 
02132009  #2  
Join Date: Dec 2008
Posts: 249

Mixture Math!
Quote:
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 cementsand 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. 

02132009  #3 
Join Date: Feb 2009
Posts: 3

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; 02132009 at 07:40 PM.. 
02132009  #4 
Join Date: Feb 2009
Posts: 3

If i'm right about the oil and gas one, i'm lost on the other two.

05292009  #5 
Join Date: May 2009
Posts: 1

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 twocycle 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 cementsand 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, 9karat gold is 9/24 gold). HOw many grams of 9karat gold must be mixed with 18 karat gold to get 200g of 14karat gold? 
Thread Tools  
Display Modes  

