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Zorluk: Çok zorBasic Ratios

A forward operating base maintains a stockpile of ammunition, food, and medical supplies in a weight ratio of 5:3:2. An incoming supply convoy delivers an additional 3 tons of food and 8 tons of medical supplies, while the ammunition stockpile remains unchanged. After the delivery, the new weight ratio of food to medical supplies becomes 4:3. What was the total weight of the entire stockpile before the convoy arrived?

  1. A
    20 tons
  2. B
    23 tons
  3. C
    115 tons
  4. 230 tonsCevap

Cevap

The total weight of the entire stockpile before the convoy arrived was 230 tons.
The original weights can be expressed proportionally as 5x5x (ammunition), 3x3x (food), and 2x2x (medical supplies), totaling 10x10x. After adding the newly delivered amounts, the new weights for food and medical supplies become 3x+33x + 3 and 2x+82x + 8. Setting their ratio equal to 4:3 yields the equation 3x+32x+8=43\frac{3x + 3}{2x + 8} = \frac{4}{3}. Cross-multiplying gives 9x+9=8x+329x + 9 = 8x + 32, which simplifies to x=23x = 23. The total original weight is therefore 10x=10(23)=23010x = 10(23) = 230 tons.

Adım Adım Çözüm

1
Define the original weights of the supplies using a common multiplier xx.
Ammunition = 5x5x, Food = 3x3x, Medical = 2x2x. Total original weight = 5x+3x+2x=10x5x + 3x + 2x = 10x.
A ratio represents proportional parts of a whole. Multiplying each ratio term by a constant xx gives expressions for their actual quantities.
2
Set up an algebraic expression for the new weights of food and medical supplies after the delivery.
New Food = 3x+33x + 3; New Medical = 2x+82x + 8.
The convoy delivered 3 tons of food and 8 tons of medical supplies, which are added to their respective original expressions.
3
Create a proportion using the new ratio of food to medical supplies provided in the problem.
3x+32x+8=43\frac{3x + 3}{2x + 8} = \frac{4}{3}
The problem states that after the delivery, the ratio of food to medical supplies becomes 4:3.
4
Solve the proportion for xx using cross-multiplication.
3(3x+3)=4(2x+8)⇒9x+9=8x+32⇒x=233(3x + 3) = 4(2x + 8) \Rightarrow 9x + 9 = 8x + 32 \Rightarrow x = 23.
Cross-multiplying eliminates the fractions, creating a linear equation that can be solved for the common multiplier xx.
5
Calculate the total original weight of the stockpile by substituting the value of xx.
10(23)=23010(23) = 230 tons.
The question asks for the total weight of the entire stockpile before the delivery, which is represented by the expression 10x10x.

Anahtar Kavram

Setting up and solving algebraic proportions from multi-term ratios after quantities undergo additive changes.
Tahmini Süre:1m 30s
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Basic Ratios

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