Ratios and Proportions
A ratio is the comparison of two quantities of the same unit by division, where at least one of the quantities is non-zero.
Example:
\[\frac{10 \;\; \text{kg} }{3\;\; \text{kg} } \] is a ratio, but \[\frac{10 \;\;\text{kg}}{3 \;\; \text{cm}} \] is not a ratio.
PROPORTION:
A proportion is an equation stating that two or more ratios are equal.
\[\displaystyle\frac{a}{b} = \displaystyle\frac{c}{d} = k \quad \text{(Simple Proportion)} \qquad \displaystyle\frac{a}{b} = \displaystyle\frac{c}{d} =\displaystyle\frac{e}{f} =k \quad \text{(Extended Proportion)} \]
Here, \(k\) is called the constant of proportionality.
Example:
\( \displaystyle{\frac{10}{5} = \frac{6}{3} = \frac{-14}{-7} }\) \(= k \), where \( k = 2 \).
Note:
The proportion \(\displaystyle\frac{a}{b} =\displaystyle \frac{c}{d} = \displaystyle \frac{e}{f} \) can also be written in the form \( a\;:\;c\;:\;e = b\;:\;d\;:\;f\).
Properties of Proportions:
\( 1. \quad \displaystyle\frac{a}{b} = \displaystyle\frac{c}{d} \) or \( \overbrace{a : \underbrace{ b = c}_{\text{means}} : d }^{\text{extremes}} \)
In a proportion, the product of the means equals the product of the extremes (Cross-Multiplication).
\[ \displaystyle\frac{a}{b} = \displaystyle\frac{c}{d} \quad \text{means} \quad b \cdot c = a \cdot d \]
Since this equality always holds true, the proportion \[ \frac{a}{b} = \frac{c}{d} \] can be rearranged as follows:
\[ \frac{a}{c} = \frac{b}{d} \] (the means can swap places),
\[ \frac{d}{b} = \frac{c}{a} \] (the extremes can swap places),
\[ \frac{b}{a} = \frac{d}{c} \] (taking the reciprocal of both sides).
\( 2. \quad \) Ratios in a proportion can be simplified or expanded by a non-zero multiplier:
\[ \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k \quad \Rightarrow \quad \frac{m \cdot a }{m \cdot b} = \frac{n \cdot c }{n \cdot d}= \frac{r \cdot e}{r \cdot f} = k \]
\(\quad a) \;\;\) When the numerators are added together and the denominators are added together, the constant of proportionality remains unchanged.
\[\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k \quad \Rightarrow \quad \frac{a + c + e }{b + d + f } = k \]
or with linear combinations:
\[\frac{m \cdot a}{m \cdot b} = \frac{n \cdot c}{n \cdot d} = \frac{r \cdot e}{r \cdot f} = k \quad \Rightarrow \quad \frac{m \cdot a + n \cdot c + r \cdot e }{m \cdot b + n \cdot d + r \cdot f } = k \]
Example:
\[ \bullet \quad \frac{3}{6}= \frac{-2}{-4} = \frac{7}{14} =\frac{1}{2} \quad \text{and} \quad \frac{3-2+7}{6-4+14} = \frac{8}{16} = \frac{1}{2} \]
\(\quad b) \;\;\) If \( {\displaystyle \frac{a}{b} = \displaystyle\frac{c}{d} =\displaystyle \frac{e}{f} }=k \), then \( { \displaystyle\frac{a^n}{b^n} = \displaystyle\frac{c^n}{d^n} = \displaystyle\frac{e^n}{f^n} }=k^n \).
Consequently:
\[ \displaystyle\frac{a^n + c^n + e^n}{b^n + d^n + f^n}= k^n \]
\( 3. \quad \)
If \[ \frac{a}{b}= \frac{c}{d} = \frac{e}{f} = k \], then multiplying the ratios gives: \[ \frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f} = k \cdot k \cdot k = k^3 \]
\( 4. \quad \) If the same linear operations are applied uniformly across all components of a proportion, the proportionality relationships are preserved.
If \[ \frac{a}{b} = \frac{c}{d} = \frac{e}{f}\], then:
\[\frac{a+b}{a-b} = \frac{c+d}{c-d} = \frac{e+f}{e-f} = \frac{2a+3b}{a+b} = \frac{2c+3d}{c+d} = \frac{2e+3f}{e+f} \cdots \]
Question 1
If \[ \frac{x+ 3y + z}{x+2y-z} = \frac{3}{2} \], what is the value of the fraction ? \[ \frac{x+z}{2x-z} \]
\[
\text{A)} \frac{2}{3} \quad
\text{B) } 2 \quad
\text{C) } \frac{5}{2} \quad
\text{D) } 3 \quad
\text{E) } \frac{7}{2}
\]
Solution:
\[ \frac{x+ 3y + z}{x+2y-z} = \frac{3}{2} \Rightarrow 3(x + 2y – z) = 2(x + 3y + z)\]
\[ \Rightarrow 3x + 6y – 3z = 2x + 6y + 2z\]
Subtracting \(6y\) from both sides:
\[ \Rightarrow 3x – 3z = 2x + 2z\]
\[ \Rightarrow x = 5z \]
Substituting \(x = 5z\) into the target expression:
\[ \frac{x+z}{2x-z} = \frac{5z + z }{2(5z) – z } = \frac{6z}{9z} = \frac{2}{3} \]
\(\textbf{Answer: A} \)
Question 2
If \[ \frac{a}{2} = \frac{b}{3} = \frac{c}{4} \], what is the value of the expression ? \[ \frac{a^3 + b^3 + c^3 }{abc} \]
\[
\text{A)} 2 \quad
\text{B) } \frac{1}{2} \quad
\text{C) } \frac{10}{3} \quad
\text{D) } \frac{11}{2} \quad
\text{E) } \frac{33}{8}
\]
Solution:
Set the proportions equal to a constant \(k\):
\[ \frac{a}{2} = \frac{b}{3} =\frac{c}{4} = k \Rightarrow a= 2k, \;\; b= 3k, \;\; c= 4k \]
Substituting these into the equation:
\[ \frac{a^3 + b^3 + c^3 }{abc} = \frac{(2k)^3 + (3k)^3 + (4k)^3}{(2k)(3k)(4k)} \]
\[ = \frac{8k^3 + 27k^3 + 64k^3}{24k^3} = \frac{99k^3 }{24k^3} = \frac{33}{8} \]
\(\textbf{Answer: E} \)
Question 3
Given \[ ax= by = cz = 3 \quad \text{and} \quad \frac{1}{x } + \frac{1}{y } + \frac{1}{z } = 6 \], find the sum \(a + b + c \).
\[
\text{A)} 12 \quad
\text{B) } 14 \quad
\text{C) } 16 \quad
\text{D) } 18 \quad
\text{E) } 20
\]
Solution:
\[ ax = by = cz = 3 \Rightarrow x = \frac{3}{a}, \;\; y = \frac{3}{b}, \;\; z= \frac{3}{c} \]
Taking the reciprocal values:
\[ \frac{1}{x} = \frac{a}{3}, \;\; \frac{1}{y} = \frac{b}{3}, \;\; \frac{1}{z} = \frac{c}{3} \]
Substitute these into the given sum:
\[ \frac{a}{3} + \frac{b}{3} + \frac{c}{3} = 6 \]
\[ \Rightarrow \frac{a + b + c}{3} = 6 \Rightarrow a + b + c = 18\]
\(\textbf{Answer: D} \)
Question 4
A special paint mixture consists of three compounds with weights x, y, and z. In a total mixture batch of 430 grams, the weight ratios of these substances are given as:
\[\frac{y}{x} =\frac{3}{2} \quad \text{and} \quad \frac{z}{y} = \frac{6}{5} \]
Based on this, what is the weight difference \( z – x\) in grams?
\[
\text{A)} 70 \quad
\text{B) } 80 \quad
\text{C) } 90 \quad
\text{D) } 100 \quad
\text{E) } 110
\]
Solution:
To combine the ratios, make the parts for \(y\) equivalent. The least common multiple for \(y\)’s ratio markers (3 and 5) is 15:
\[ \frac{y}{x} = \frac{3 \cdot 5}{2 \cdot 5} = \frac{15}{10} \Rightarrow \frac{x}{10} = \frac{y}{15} \]
\[ \frac{z}{y} = \frac{6 \cdot 3}{5 \cdot 3} = \frac{18}{15} \Rightarrow \frac{y}{15} = \frac{z}{18} \]
Now we can write the extended proportion:
\[ \frac{x}{10} = \frac{y}{15} = \frac{z}{18} = k \Rightarrow \frac{x + y + z }{10 + 15 + 18 } = k \]
Since the total weight is \(x + y + z = 430\) grams:
\[ \frac{430}{43} = k \Rightarrow k = 10 \]
We want to find \(z – x\):
\[ z = 18k, \;\; x = 10k \Rightarrow z – x = 8k \]
\[ z – x = 8(10) = 80 \text{ grams} \]
\(\textbf{Answer: B} \)
Question 5
Given that x and y are positive integers and \[ \frac{x + y }{ 2x – y } = \frac{3}{2} \], which of the following values could be the sum ? \(3x + y \)
\[
\text{A)} 9 \quad
\text{B) } 27 \quad
\text{C) } 57 \quad
\text{D) } 64 \quad
\text{E) } 72
\]
Solution:
\[ \frac{x + y }{2x – y} = \frac{3}{2} \Rightarrow 3(2x – y) = 2(x + y)\]
\[ \Rightarrow 6x – 3y = 2x + 2y \]
\[ \Rightarrow 4x = 5y \Rightarrow \frac{x}{y} = \frac{5}{4} \]
Thus, we can define \(x = 5k\) and \(y = 4k\). Substitute these into the expression:
\[ 3x + y = 3(5k) + 4k = 19k \]
Since x and y are positive integers, \( k \in Z^+\). Therefore, the sum \(3x + y\) must be a positive multiple of 19. Looking at the options, 57 is the only multiple of 19 (\(19 \cdot 3 = 57\)).
\(\textbf{Answer: C} \)
Question 6
On a farm, there are chickens, rabbits, and birds. Their respective quantities are x, y, and z. If \( x:y:z = 0.3 : 0.6 : 0.9\), what is the minimum possible value for the total sum \( x + y + z \)?
\[
\text{A)} 3 \quad
\text{B) } 6 \quad
\text{C) } 18 \quad
\text{D) } 21 \quad
\text{E) } 27
\]
Solution:
\[x:y:z = 0.3 : 0.6 : 0.9 \Rightarrow \frac{x}{0.3} = \frac{y}{0.6} = \frac{z}{0.9} \]
Multiply the denominators by 10 to eliminate decimals, then simplify by dividing by 3:
\[ \frac{x}{3} = \frac{y}{6} = \frac{z}{9} \Rightarrow \frac{x}{1} = \frac{y}{2} = \frac{z}{3} = k \]
To minimize the total population count of live animals, choose the smallest possible positive integer for the constant, \(k = 1\):
\[x = 1, \;\; y = 2, \;\; z = 3 \Rightarrow x + y + z = 1 + 2 + 3 = 6 \]
\(\textbf{Answer: B} \)
Question 7
\[ \text{If } \quad \frac{a}{b } = \frac{c}{d} = \frac{e}{f} = \frac{1}{2}, \quad 3a – c + e = 40, \quad \text{and} \quad d – f = 1, \quad \text{find the value of } b. \]
\[
\text{A)} 9 \quad
\text{B) } 15 \quad
\text{C) } 21 \quad
\text{D) } 27 \quad
\text{E) } 33
\]
Solution:
Apply multipliers to the component ratios to match the given equation:
\[ \frac{3a}{3b} = \frac{-c}{-d} = \frac{e}{f} = \frac{1}{2} \]
Summing the numerators and denominators preserves the ratio constant \(k\):
\[ \frac{3a – c + e }{3b – d + f} = \frac{1}{2} \]
Group the denominator terms to substitute \(d – f = 1\):
\[ \frac{40}{3b – (d – f)} = \frac{1}{2} \Rightarrow \frac{40}{3b – 1} = \frac{1}{2} \]
\[ 80 = 3b – 1 \Rightarrow 3b = 81 \Rightarrow b = 27 \]
\(\textbf{Answer: D} \)
Question 8
\[ \text{If } \frac{x-y}{a} = x + 2y = \frac{y+2z}{b}, \quad x + y + z = 6, \quad \text{and} \quad a + b = 2, \text{ what is the value of } x + 2y ? \]
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
We can view \(x+2y\) as a fraction with a denominator of 1:
\[ \frac{x-y}{a} = \frac{x + 2y}{1} = \frac{y+2z}{b} = k \]
Sum the numerators and denominators:
\[ \frac{(x-y) + (x + 2y) + (y + 2z)}{a + 1 + b} = k \]
Simplify the numerator and use \(a + b = 2\):
\[ \frac{2x + 2y + 2z}{(a + b) + 1} = k \Rightarrow \frac{2(x+y+z)}{2+1} = k \]
Substitute \(x + y + z = 6\):
\[ \frac{2 \cdot 6 }{3} = k \Rightarrow k = 4 \]
Since \(x + 2y = k\), we get:
\[ x + 2y = 4 \]
\(\textbf{Answer: D} \)
Question 9
\[ \text{If } \frac{a}{b} = \frac{b}{c} = k, \text{ which of the following is equivalent to } \left( \frac{a+b}{c} \right) \cdot \left(\frac{a}{b + c} \right) ? \]
\[
\text{A)} 1 \quad
\text{B) } k \quad
\text{C) } k^2 \quad
\text{D) } k^3 \quad
\text{E) } k^4
\]
Solution:
From the proportions, write variables in terms of \(c\) and \(k\):
\[ b = ck, \quad a = bk = (ck)k = ck^2 \]
Substitute these values into the given expression:
\[ \left(\frac{a+b}{c}\right) \cdot \left(\frac{a}{b+c}\right) = \left(\frac{ck^2 + ck}{c}\right) \cdot \left(\frac{ck^2}{ck + c}\right) \]
Factor out common terms:
\[ = \frac{ck(k + 1)}{c} \cdot \frac{ck^2}{c(k + 1)} = k \cdot k^2 = k^3 \]
\(\textbf{Answer: D} \)
Question 10
\[ \text{If } \frac{a}{b} = \frac{c}{d} = \frac{e}{f}, \text{ find the equivalent algebraic form of } (b-d) \cdot (b^2+bd+ d^2) \cdot e^3. \]
\[
\text{A)} (a-c)f \quad
\text{B) } (a+c)f \quad
\text{C) } acf \quad
\text{D) } (a^3 + c^3 )f^3 \quad
\text{E) } (a^3- c^3) f^3
\]
Solution:
Notice that the product of the first two terms forms a difference of cubes:
\[ (b-d) \cdot (b^2+bd+d^2) = b^3 – d^3 \]
So the target expression is: \((b^3 – d^3)e^3\). Now let’s cube our structural proportion terms:
\[ \frac{a}{b} = \frac{c}{d} = \frac{e}{f} \Rightarrow \frac{a^3}{b^3} = \frac{c^3}{d^3} = \frac{e^3}{f^3} \]
Subtracting component ratios keeps the proportionality constant same:
\[ \frac{a^3 – c^3}{b^3 – d^3} = \frac{e^3}{f^3} \]
Cross-multiplying gives us:
\[ (b^3 – d^3)e^3 = (a^3 – c^3)f^3 \]
\(\textbf{Answer: E} \)
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