Question

Mr. Joshi has $430$ cabbage-plants which he wants to plant out. Some 25 to a row and the rest 20 to a row. If there are to be 18 rows in all how many rows of 25 will be there?
A. 10
B. 14
C. 8
D. 12

Answer 1

Hint: First we have to assume the number of rows of 25 be $x$ and number of rows of 20 to be $y$. Then, we form two equations by using the information given in the question that there are total $430$ cabbage-plants which are planted out in 5 to a row and the rest 20 to a row and there are to be 18 rows in all. Then, by solving these equations we get our answer.

Complete step-by-step answer:
We have been given that Mr. Joshi has $430$ cabbage-plants which he wants to plant out, some 25 to a row and the rest 20 to a row.
We have to find the number of rows of 25.
Now, let us assume that there are $x$ number of rows of 25 and $y$ number of rows of 20.
Now, as given there are total $430$ cabbage-plants, so we have
$25x+20y=430...........(i)$
Now, we have 18 rows in all, so, we have
$x+y=18.........(ii)$
Now, to solve these equations we use elimination method. For which we have to multiply equation (ii) with 25, we get
$\begin{align}
  & 25x+25y=18\times 25 \\
 & \Rightarrow 25x+25y=450 \\
\end{align}$
Now, subtract the obtained equation from equation (i), we get
$\begin{align}
  & \left( 25x+20y-430 \right)-\left( 25x+25y-450 \right)=0 \\
 & \Rightarrow 25x+20y-430-25x-25y+450=0 \\
\end{align}$
Now, cancel out the terms, we get
$\begin{align}
  & \Rightarrow 20y-25y-430+450=0 \\
 & \Rightarrow -5y+20=0 \\
 & \Rightarrow -5y=-20 \\
 & \Rightarrow y=\dfrac{-20}{-5} \\
 & y=4 \\
\end{align}$
So, we have $4$ rows of $20$.
Now, substituting the value in equation (ii), we get
$\begin{align}
  & x+y=18 \\
 & x+4=18 \\
 & x=18-4 \\
 & x=14 \\
\end{align}$
So, there are $14$ rows of $25$.

Note: As there are two unknown variables, so we have two equations in two variables. We can also use a substitution method to solve these equations. As we have two equations
$25x+20y=430...........(i)$
$x+y=18.........(ii)$
Substituting $y=18-x$ in equation (i), we get
$\begin{align}
  & 25x+20\left( 18-x \right)=430 \\
 & \Rightarrow 25x+360-20x=430 \\
 & \Rightarrow 5x=430-360 \\
 & \Rightarrow 5x=70 \\
 & \Rightarrow x=\dfrac{70}{5} \\
 & x=14 \\
\end{align}$