Question

An airplane takes $1$ hour less for a journey of $1200\,km$ if its speed is increased by $100\,{km}/{hr}\;$ from its usual speed. Its usual speed is ________ ${km}/{hr}\;$ .

Answer 1

Hint: The given question is related to linear equations in two variables. Consider speed $x\,{km}/{hr}\;$ and time as $y\,hr$ . Form two equations using the information given in the problem and solve them simultaneously to get the values of $x$ and $y$ .
Complete step-by-step answer:
We are given that the distance of the journey is $1200\,km$ . Let the usual speed of the airplane be $x\,{km}/{hr}\;$ and the time taken to complete the journey at usual speed be $y\,hr$ . We know that the relation between the speed, distance travelled and time taken is given as $speed=\dfrac{\text{distance travelled}}{\text{time taken}}$ . So, for a journey of $1200\,km$ at a speed of $x\,{km}/{hr}\;$ taking time $y\,hr$ , the relation can be written as $x=\dfrac{1200}{y}......(i)$ .
Now, it is given that if the speed of the airplane is increased by $100\,{km}/{hr}\;$ , then it takes $1\,hr$ less for the same journey. Since the speed is increased by $100\,{km}/{hr}\;$ , so, the new speed of the airplane is $\left( x+100 \right){km}/{hr}\;$ . Also, the time taken for the journey is $1\,hr$ less. So, the time taken will be $\left( y-1 \right)\,hr$ . So, the relation between new speed, distance travelled and time taken is given as $\left( x+100 \right)=\dfrac{1200}{y-1}......(ii)$ .
Now, substituting $x=\dfrac{1200}{y}$ from equation $(i)$ in equation $(ii)$ , we get:
$\dfrac{1200}{y}+100=\dfrac{1200}{y-1}$
$\Rightarrow \dfrac{1200}{y}+\dfrac{100y}{y}=\dfrac{1200}{y-1}$
$\Rightarrow \left( y-1 \right)\left( 1200+100y \right)=1200y$
$\Rightarrow 1200y+100{{y}^{2}}-1200-100y=1200y$
$\Rightarrow 100{{y}^{2}}-1200-100y=0$
$\Rightarrow {{y}^{2}}-y-12=0$
Now, we have a quadratic in $y$ . To solve the quadratic, we will use middle term splitting. We have to split $-y$ into two terms, such as their sum is equal to $-y$ and their product is equal to $-12{{y}^{2}}$ .
We can write $-y$ as $-4y+3y$ . Here, the sum of $-4y$ and $3y$ is $-y$ and the product of $-4y$ and $3y$ is $-12{{y}^{2}}$ .
So, ${{y}^{2}}-y-12=0$ can be written as ${{y}^{2}}-4y+3y-12=0$ .
$\Rightarrow y\left( y-4 \right)+3\left( y-4 \right)=0$
$\Rightarrow \left( y-4 \right)\left( y+3 \right)=0$
So, either $y=4$ or $y=-3$ . Since, time cannot be negative, so, $y=4\,hr$.
Now, substituting $y=4$ in equation $(i)$ , we get $x=\dfrac{1200}{4}=300$ . So, $x=300\,{km}/{hr}\;$ .
So, the usual speed of the airplane is $300\,{km}/{hr}\;$ .

Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong. We can find the value of x first as well by substituting the value of y.