David covers a total of 240 miles journey, of which 120 miles are covered at a constant speed and remaining 120 miles are covered at 10 mph more than first half speed. The Time took to cover the second half of 240 miles is 1 hour less than the first half of 240 miles. At what speed did David cover his second half of the journey?
(A) 30 mph
(B) 35 mph
(C) 40 mph
(D) 45 mph
(E) 50 mph
The OA is the option C.
Experts, may you help me here? I don't know what are the equations I should set here.
Hi Vincen,
Let's take a look at your question.
Let the first half of 120 miles journey is covered at a constant speed of x miles/hr.
Then the second half of 120 miles journey will be covered in (x+10) miles/hr.
Let the time taken to cover the first half of journey is y hours.
Then the time taken to cover the second half of journey will be (y-1) hours.
We know that:
$$Distance\ =\ Speed\ \times\ Time$$
Using Distance formula, we can write two equations for first half and second half of the journey.
$$120=xy...\left(i\right)$$
$$120=\left(x+10\right)\left(y-1\right)...\left(ii\right)$$
LHS of both equations is the same i.e. 120, so we will equate the RHS of the equations.
$$xy=\left(x+10\right)\left(y-1\right)$$
$$xy=xy-x+10y-10$$
$$0=-x+10y-10$$
$$x=10y-10 ... (iii)$$
Substitute the value of x from equation (iii) in eq(i), we get:
$$\left(10y-10\right)y=120$$
$$10\left(y-1\right)y=120$$
$$\left(y-1\right)y=12$$
$$y^2-y=12$$
$$y^2-y-12=0$$
Factorize to find y:
$$y^2-4y+3y-12=0$$
$$y\left(y-4\right)+3\left(y-4\right)=0$$
$$\left(y+3\right)\left(y-4\right)=0$$
$$y=-3,\ 4$$
Since, y represents time and time can not be negative, therefore, we will discard y = -3 and consider only:
$$y=4$$
Plugin y = 4 in eq(i) to find y to find y:
$$xy=120$$
$$x\left(4\right)=120$$
$$x=\frac{120}{4}=30$$
We are asked to find the speed of the second half of the journey i.e. represented by x+10.
$$Speed\ of\ second\ half=x+10=30+10=40mph$$
Therefore, Option
C is correct.
Hope this helps.
I am available if you'd like any follow up.
