Question

A car made a run of \[390\] km in ‘ \[x\] ’ hours. If the speed had been \[4\] kmph more it would have taken 2 hours less for the journey. Find ‘ \[x\] ’.

Answer 1

Hint: Here we have to find the time required by car i.e. missing information from the data given. The best way to solve sums involving missing information is to form an equation. A mathematical expression containing the equals symbol is known as an equation. Algebra is often used in equations.

Complete step-by-step answer:
In the given sum, we are provided with the data regarding time, speed and distance. We can apply the following formula to find the missing information:
 \[Speed = \dfrac{{Dis\tan ce}}{{Time}}\]
We have to find the “time” taken by car hence the above formula can be rewritten as:
 \[Time = \dfrac{{Dis\tan ce}}{{Speed}}\]
Since time \[x\] is let speed be \[y\]
From the given information in question, we can form the following equations:
 \[x = \dfrac{{390}}{y}\] …..(1)
 \[x - 2 = \dfrac{{390}}{{y + 4}}\] …..(2)
Now substituting the value of \[x\] in (2) from (1),
 \[\dfrac{{390}}{y} - 2 = \dfrac{{390}}{{y + 4}}\]
Cross-multiplying the terms on both the sides,
 \[(390 - 2(y))(y + 4) = 390(y)\]
Solving the brackets, we get,
 \[390y + 1560 - 2{y^2} - 8y = 390y\]
Forming standard quadratic equation,
 \[ - 2{y^2} + 390y - 390y - 8y + 1560 = 0\]
 \[ - 2{y^2} - 8y + 1560 = 0\]
Dividing by \[2\] ,
 \[ - {y^2} - 4y + 780 = 0\]
Taking the equation on RHS and converting the signs,
 \[{y^2} + 4y - 780 = 0\]
Factorising:
 \[{y^2} + 30y - 26y - 780 = 0\]
Taking the common factors,
 \[y(y + 30) - 26(y + 30) = 0\]
 \[(y + 30)(y - 26) = 0\]
Hence \[y = - 30\] or \[y = 26\] . But since speed cannot be negative, \[y = 26\] kmph is the correct answer.
Now we substitute the value of \[y = 26\] found above in equation (1),
 \[x = \dfrac{{390}}{y} = \dfrac{{390}}{{26}}\]
 \[x = 15\] hours
Hence the time taken by car to travel \[390\] km would have been \[15\] hours.

Note: Remember to denote the unit of measurement. For e.g. speed can be in km/ph, distance can be in “km” and time can be in “hours”.
Linear equations are mostly used to find the missing variables. In such cases, if we have two variables and have to find one variable, graphs can also be used to solve the equation.