Question

(i) If d varies directly as t, and if d = 4 when t = 9, find d when t = 21.
(ii) The mass of a uniform copper bar varies directly as its length. If a bar 40cm long has a mass approximately 420g, find the mass of a bar 136cm long.

Answer 1

Hint: The two given questions are based on direct proportionality or sequentiality. In both the questions, 2 quantities are sequential. So, by the rule of sequentiality, if x and y are two quantities, both are sequential to each other, so, $x\propto y$ and let ${{y}_{1}}$ be corresponding to ${{x}_{1}}$ and ${{y}_{2}}$ corresponding to ${{x}_{2}}$. Then, $\dfrac{{{x}_{1}}}{{{y}_{1}}}=\dfrac{{{x}_{2}}}{{{y}_{2}}}$. In both the questions, we have to use this rule and solve the equations, so that we can get the quantity which is unknown.

Complete step-by-step answer:
In order to solve this question, we need to understand the concept of sequentiality. It is the linear unidirectional succession of elements or events, either reversible or irreversible. So, if 2 or more quantities are directly proportional or increase or decrease in the same ratio of each other, then those quantities are said to be sequential to each other.
Let x and y be sequential. So, $x\propto y$.
Now, if we convert this into an equality, then there will be a proportionality constant, say k. So, we can write,
x = ky
Now, if we assume that at ${{y}_{1}}$, x is ${{x}_{1}}$ and at ${{y}_{2}}$, x is ${{x}_{2}}$, we get,
$\begin{align}
  & {{x}_{1}}=k{{y}_{1}}.........(i) \\
 & {{x}_{2}}=k{{y}_{2}}.........(ii) \\
\end{align}$
On dividing equation (i) by equation (ii), we get,
$\begin{align}
  & \dfrac{{{x}_{1}}}{{{y}_{1}}}=\dfrac{k{{x}_{2}}}{k{{y}_{2}}} \\
 & \Rightarrow \dfrac{{{x}_{1}}}{{{y}_{1}}}=\dfrac{{{x}_{2}}}{{{y}_{2}}} \\
\end{align}$
Now, in the first question, we have been given that d varies directly as t. So, we can say, $d\propto t$.
Now, let us assume that when ${{d}_{1}}=4,\text{then }{{t}_{1}}=9$ and let us consider ${{t}_{2}}=21\text{ and }{{d}_{2}}={{d}_{2}}$. So, we have to find ${{d}_{2}}$.
Now, according to the rule of sequentiality, we can write,
$\begin{align}
  & \dfrac{{{d}_{1}}}{{{t}_{1}}}=\dfrac{{{d}_{2}}}{{{t}_{2}}} \\
 & \dfrac{4}{9}=\dfrac{{{d}_{2}}}{21} \\
\end{align}$
On cross multiplying we get,
\[\begin{align}
  & 4\times 21=9\times {{d}_{2}} \\
 & {{d}_{2}}=\dfrac{4\times 21}{9} \\
 & {{d}_{2}}=\dfrac{4\times 7}{3} \\
 & {{d}_{2}}=\dfrac{28}{3} \\
\end{align}\]
Therefore, when $t=21$, we get $d=\dfrac{28}{3}$.
Now, in the second question, we have been given that, the mass of a uniform copper bar varies directly as its length varies. So, the length and mass are sequential.
Let l and m be the length and the mass of the copper bar respectively. So, we can write, $l\propto m$.
Now, let us assume that when ${{l}_{1}}=40,\text{then }{{\text{m}}_{1}}=420$ and let us consider ${{l}_{2}}=136\text{ and }{{\text{m}}_{2}}={{m}_{2}}$. So, we have to find ${{m}_{2}}$.
Now, according to the rule of sequentiality, we can write,
\[\begin{align}
  & \dfrac{{{l}_{1}}}{{{m}_{1}}}=\dfrac{{{l}_{2}}}{{{m}_{2}}} \\
 & \dfrac{40}{420}=\dfrac{136}{{{m}_{2}}} \\
\end{align}\]
On cross multiplying we get,
\[\begin{align}
  & 40\times {{m}_{2}}=420\times 136 \\
 & {{m}_{2}}=\dfrac{420\times 136}{40} \\
 & {{m}_{2}}=42\times 34 \\
 & {{m}_{2}}=1428 \\
\end{align}\]
Therefore, when the length of the copper bar is 136 cm, we get its mass as 1428g.

Note: While solving these types of questions, students may make mistakes in the proportionality and hence, the students must check whether the quantities are directly or inversely proportional. Also, the students must take care of the multiplications and the ratios in order to avoid any kind of mistakes.