Question 14 Marks

Three quantities $x, y$ and $z$ are such that $x$ and $y$ are directly proportional to each other while $y$ and $z$ are inversely proportional to each other. If $x=12$ then $y=72$ and $z=8$.

(1) If $y=42$ then $x$ is equal to
(a) 6
(b) 7
(c) 8
(d) 10

(2) If $z=12$ then $y$ is equal to
(a) 36
(b) 48
(c) 60
(d) 64

(3) Which of the following cannot be a possible combination of the values of $x$ and $y$ ?
(a) 6.5 and 39
(b) 9 and 54
(c) 47 and 282
(d) 52 and 260

(4) If $x=24$ then $z$ is equal to
(a) 4
(b) 6
(c) 8
(d) 12

Answer

$x$ and $y$ are directly proportional
$\Rightarrow \frac{x}{y}=\text { constant }$
$y$ and $z$ are inversely proportional
$\Rightarrow y z=\text { constant } .$

(1) (B) 7
We have
$
\frac{x}{y}=\frac{12}{72} \Rightarrow \frac{x}{42}=\frac{12}{72} \Rightarrow x=\left(\frac{12}{72} \times 42\right)=7
$

(2) (B) 48
$y z=72 \times 8 \Rightarrow y \times 12=72 \times 8 \Rightarrow y=\frac{72 \times 8}{12}=48$

(3) (D) 52 and 260
$\frac{x}{y}=\frac{12}{72}=\frac{1}{6} .$
Now, $\frac{6.5}{39}=\frac{1}{6} ; \frac{9}{54}=\frac{1}{6} ; \frac{47}{282}=\frac{1}{6}$. But $\frac{52}{260}=\frac{1}{5} \neq \frac{1}{6}$.

(4) (A) 4
$\frac{x}{y}=k_1$ and $y z=k_2$, where $k_1$ and $k_2$ are constants. Putting $y=\frac{x}{k_1}$ in $y z=k_2$, we have
$x z=k_1 k_2=$ constant $\Rightarrow 12 \times 8=24 \times z \Rightarrow z=\frac{12 \times 8}{24}=4$.

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