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Gary's Beverage

by Andrew
(New York)










































Gary purchased a large beverage, but drank only m/n of this beverage and wasted the rest. If Gary had purchased only half as much and drank twice as much, he would have wasted only 2/9 as much beverage as he did. Find m/n
in reduced form.

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Nov 27, 2011
Word Problem - Gary's Beverage
by: Staff

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Part II


Solve for m/n

2*V₁*(m/n) + (2/9)*V₁*(1 - m/n) = (1/2)* V₁



To remove the variable V₁, divide each side of the equation by V₁


[2*V₁*(m/n) + (2/9)*V₁*(1 - m/n)] / V₁ = [(1/2)* V₁] / V₁

[2*(m/n) + (2/9)*(1 - m/n)] * (V₁ / V₁) = (1/2) * (V₁ / V₁)

[2*(m/n) + (2/9)*(1 - m/n)] * (1) = (1/2) * (1)

[2*(m/n) + (2/9)*(1 - m/n)] = (1/2)

2*(m/n) + (2/9)*(1 - m/n) = ½



Use the distributive law to expand the expression: (2/9)*(1 - m/n)


2*(m/n) + (2/9)*(1 - m/n) = 1/2

2*(m/n) + (2/9)*(1) - (2/9)*(m/n) = 1/2

2*(m/n) + (2/9) - (2/9)*(m/n) = ½



To remove the 2/9 from the left side of the equation, subtract 2/9 from each side of the equation

2*(m/n) + (2/9) - (2/9)*(m/n) = 1/2

2*(m/n) + (2/9) - (2/9) - (2/9)*(m/n) = 1/2 - (2/9)

2*(m/n) + 0 - (2/9)*(m/n) = 1/2 - (2/9)

2*(m/n) - (2/9)*(m/n) = 1/2 - (2/9)

2*(m/n) - (2/9)*(m/n) = (1/2)*(9/9) - (2/9)*(2/2)

2*(m/n) - (2/9)*(m/n) = (9/18) - (4/18)

2*(m/n) - (2/9)*(m/n) = (9 - 4)/18)

2*(m/n) - (2/9)*(m/n) = 5/18



Using the distributive law, factor the fraction (m/n) on the left side of the equation


2*(m/n) - (2/9)*(m/n) = 5/18

(m/n) * [2 - (2/9)] = 5/18

(m/n) * [2*(9/9) - (2/9)] = 5/18

(m/n) * [18/9 - 2/9] = 5/18

(m/n) * (16/9) = 5/18



To remove the fraction 16/9 from the left side of the equation, multiply each side of the equation by its reciprocal: 9/16


(m/n) * (16/9) * (9/16) = (5/18) * (9/16)

(m/n) * (16/16) * (9/9) = (5/18) * (9/16)

(m/n) * (1) * (1) = (5/18) * (9/16)

(m/n) = (5/18) * (9/16)

m/n = (5/18) * (9/16)

m/n = (5 * 9)/(18 * 16)

m/n = (5 * 9)/(9 * 2⁵)

m/n = (5 / 2⁵)* (9/9)

m/n = (5 / 2⁵)* (1)

m/n = (5 / 2⁵)

m/n = (5 / 32)

m/n = 5 / 32

the final answer is: m/n = 5 / 32




Thanks for writing.

Staff
www.solving-math-problems.com


Nov 27, 2011
Word Problem - Gary's Beverage
by: Staff


Part I

Question:


by Andrew
(New York)


Gary purchased a large beverage, but drank only m/n of this beverage and wasted the rest. If Gary had purchased only half as much and drank twice as much, he would have wasted only 2/9 as much beverage as he did. Find m/n in reduced form.



Answer:


1. Gary purchased a large beverage

V₁ = volume Gary purchased

2. Gary drank only m/n of this beverage and wasted the rest

Fraction Consumed = m/n

C₁ = Amount actually consumed

C₁= V₁*(m/n)

Fraction Wasted = 1 - m/n

W₁ = Amount actually wasted

W₁ = V₁*(1 - m/n)


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3. If Gary had purchased only half as much

V₂ = hypothetical volume Gary could have purchased

V₂ = (1/2)* V₁

4. If Gary drank twice as much

C₂ = Amount Gary would have consumed


C₂ = 2*V₁*(m/n)

5. He would have wasted only 2/9 as much beverage as he did.

W₂ = Amount Gary would have wasted

W₂ = (2/9)*V₁*(1 - m/n)




Final Equations

C₁= V₁*(m/n)

W₁ = V₁*(1 - m/n)


V₂ = (1/2)* V₁

C₂ = 2*V₁*(m/n)

W₂ = (2/9)*V₁*(1 - m/n)


C₂ + W₂ = V₂

V₂ = (1/2)* V₁



C₂ + W₂ = (1/2)* V₁


Substituting 2*V₁*(m/n) for C₂ and (2/9)*V₁*(1 - m/n) for W₂

2*V₁*(m/n) + (2/9)*V₁*(1 - m/n) = (1/2)* V₁


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