Question

if the ratio of \\(\frac{2}{3} : \frac{3}{2}\\) is equivalent to the ratio of \\(\frac{b}{9} : \frac{a}{6}\\), what is the value of \\(a + b\\)?\\(\bigcirc\\) a. 6\\(\bigcirc\\) b. 9\\(\bigcirc\\) c. 12\\(\bigcirc\\) d. 15

Explanation:

Step1: Convert ratios to fractions

A ratio $x:y$ equals $\frac{x}{y}$. So:
$$\frac{\frac{2}{3}}{\frac{3}{2}} = \frac{\frac{b}{9}}{\frac{a}{6}}$$

Step2: Simplify left-hand side

Multiply by reciprocal of denominator:
$$\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$$

Step3: Simplify right-hand side

Multiply by reciprocal of denominator:
$$\frac{b}{9} \times \frac{6}{a} = \frac{6b}{9a} = \frac{2b}{3a}$$

Step4: Set equal and cross-multiply

Equate the two simplified fractions:
$$\frac{4}{9} = \frac{2b}{3a}$$
$$4 \times 3a = 9 \times 2b$$
$$12a = 18b$$

Step5: Simplify the equation

Divide both sides by 6:
$$2a = 3b$$
We can take the simplest integer values: $a=3$, $b=2$ (since this satisfies $2*3=3*2$)

Step6: Calculate $a+b$

$$a+b = 3+2=5$$
Wait, this is not an option. Let's re-express the ratio equivalence correctly:
Original ratio $\frac{2}{3}:\frac{3}{2}$ is equivalent to $\frac{b}{9}:\frac{a}{6}$, which means $\frac{\frac{2}{3}}{\frac{3}{2}} = \frac{\frac{b}{9}}{\frac{a}{6}}$
Wait, alternatively, ratios are equivalent if $\frac{2}{3} \div \frac{3}{2} = \frac{b}{9} \div \frac{a}{6}$
$\frac{4}{9} = \frac{b}{9} \times \frac{6}{a} = \frac{6b}{9a} = \frac{2b}{3a}$
Cross multiply: $4*3a = 9*2b$ → $12a=18b$ → $2a=3b$ → $\frac{a}{b}=\frac{3}{2}$
So $a=3k$, $b=2k$ for any non-zero $k$. Then $a+b=5k$. Now check the options:
If $k=3$, $a=9$, $b=6$, $a+b=15$; $k=2$, $a=6$, $b=4$, $a+b=10$ (not option); $k=1.8$, $a=5.4$, $b=3.6$, $a+b=9$; $k=1.2$, $a=3.6$, $b=2.4$, $a+b=6$; $k=2.4$, $a=7.2$, $b=4.8$, $a+b=12$
Wait, the ratio $\frac{2}{3}:\frac{3}{2} = 4:9$ (multiply both by 6). The ratio $\frac{b}{9}:\frac{a}{6} = \frac{2b}{18}:\frac{3a}{18} = 2b:3a$
Set $4:9 = 2b:3a$ → $\frac{4}{9}=\frac{2b}{3a}$ → $12a=18b$ → $2a=3b$ → $a=\frac{3b}{2}$
Then $a+b = \frac{3b}{2}+b = \frac{5b}{2}$
If $b=6$, $a=9$, $a+b=15$; if $b=3.6$, $a=5.4$, $a+b=9$; if $b=2.4$, $a=3.6$, $a+b=6$; if $b=4.8$, $a=7.2$, $a+b=12$
But the question implies a unique answer, so we need to find the ratio of $a$ to $b$ from the equivalent ratios:
$\frac{2}{3}:\frac{3}{2} = \frac{b}{9}:\frac{a}{6}$
Write as fractions: $\frac{\frac{2}{3}}{\frac{3}{2}} = \frac{\frac{b}{9}}{\frac{a}{6}}$
$\frac{4}{9} = \frac{6b}{9a}$ → $\frac{4}{9} = \frac{2b}{3a}$ → $12a=18b$ → $2a=3b$ → $\frac{a}{b}=\frac{3}{2}$
So the simplest form is $a=3$, $b=2$, $a+b=5$, but 5 is not an option. Wait, maybe I mixed up the ratio:
Wait, the problem says "the ratio of $\frac{2}{3}:\frac{3}{2}$ is equivalent to the ratio of $\frac{b}{9}:\frac{a}{6}$"
Equivalent ratios mean $\frac{2}{3} \times \frac{a}{6} = \frac{3}{2} \times \frac{b}{9}$ (product of extremes = product of means)
$\frac{2a}{18} = \frac{3b}{18}$ → $2a=3b$ → same as before.
Ah, wait, maybe the question is asking for $a+b$ where $a$ and $b$ are integers such that the ratios are equivalent, and the answer is 15? No, wait, 2a=3b, so a=3, b=2, sum 5; a=6, b=4, sum 10; a=9, b=6, sum 15; a=12, b=8, sum 20. 15 is an option. Wait, maybe the question assumes that $\frac{b}{9}$ and $\frac{a}{6}$ are in the same ratio as $\frac{2}{3}$ and $\frac{3}{2}$, so $\frac{2}{3} = \frac{b}{9}$ → $b=6$, and $\frac{3}{2} = \frac{a}{6}$ → $a=9$. Then $a+b=9+6=15$. That makes sense! Because equivalent ratios mean each term is scaled by the same factor.
$\frac{2}{3} = \frac{b}{9}$ → $b= \frac{2*9}{3}=6$
$\frac{3}{2} = \frac{a}{6}$ → $a= \frac{3*6}{2}=9$
Then $a+b=9+6=15$

Answer:

D. 15