Vic can beat Harold by $1/10$ of a mile in a $2$ mile race. Harold can beat Charlie by $1/5$ of a mile in a $2$ mile race. Very Confused.

Question

Vic can beat Harold by $1/10$ of a mile in a $2$ mile race. Harold can beat Charlie by $1/5$ of a mile in a $2$ mile race. If Vic races Charlie how far ahead will he finish?

Now I don't know the correct answer but I've done the problem in $2$ different ways and got $2$ different answers‍♂️

Let V be Vic, H be Harold and C be Charlie. In V and H race V covers $2$ miles while H covers $2 - 1/10 = 1.9$ miles. In H and C race H covers $2$ miles while C covers $2 - 1/5 = 1.8$ miles

If H covers $1.9$ miles then C covers $(1.8)*(1.9)/2 = 1.71$ miles. Hence V is $2-1.71=0.29$ miles ahead.

Second method:

$V/H=2/1.9$

$H/C=2/1.8$

$V/C=200/171$

Since $V$ runs for 2 miles, $C$ runs $2(170)/200=1.7$ miles Hence V is ahead by $2-1.7=0.3$ miles.

Now there were four options after I've searched the problem online but in book there wasn't any options. (A) $0.15$ miles (B) $0.22$ miles (C) $0.25$ miles (D) $0.29$ miles (E) $0.33$ miles

But I don't understand If my second method is wrong or not since I've used it successfully to solve problems like this before.

Answer 1

It looks like your error was changing $171$ to $170$ when you inverted $V/C=200/171$ to obtain what should have been $2(171)/200=1.71$. The subtraction would have then given the correct answer, $2-1.71=0.29$, again.

Answer 2

Notice, the speeds of all three racers remain constant from one race to the other.

Race-1 Let $t_1$ be the time taken to complete race-1. Then the speeds of Vic & Harold $V$ & $H$ respectively are given as $$V=\frac{2}{t_1}, \ \ \ H=\frac{2-0.1}{t_1}=\frac{1.9}{t_1}$$

Race-2 Let $t_2$ be the time taken to complete race-2. Then the speeds of Harold & Charlie $H$ & $C$ respectively are given as $$H=\frac{2}{t_2}, \ \ \ C=\frac{2-0.2}{t_2}=\frac{1.8}{t_2}$$ Equating the speeds of Harold, $$\frac{1.9}{t_1}=\frac{2}{t_2}\implies \frac{t_1}{t_2}=\frac{1.9}{2}$$

Race-3 Vic will cover a distance of $2$ miles in time $t_1$ (which is also equal to the time taken in race-1) to finish race-3. Then the distance covered by Charlie in the same time $t_1$ of race-3 $$=\text{speed of Charlie}\times \text{time} t_1=\frac{1.8}{t_2}\times t_1$$ $$=1.8\left(\frac{t_1}{t_2}\right)=1.8\left(\frac{1.9}{2}\right)=1.71\ \text{miles}$$ Hence, in race-3, Vic beats Charlie by $2-1.71=\color{blue}{0.29\ \text{miles}}$

Answer 3

If Vic beats Harold by 1/10 of a mile in a 2 mile race (i.e. he completes 2 miles when Harold is at 1.9 miles), his velocity $V$ is $20/19$ than that of Harold. Similarly, if Harold can beat Charlie by 1/5 of a mile in a 2 mile race (i.e. he completes 2 miles when Harold is at 1.8 miles), his velocity $H$ is $20/18=10/9$ than that of Charlie. If we call $C$ the velocity of Charlie, we have $$V=\frac{20}{19}\cdot H=\frac{20}{19} \cdot \frac{10}{9} \cdot C =\frac{200}{171} C$$

and then

$$C=\frac{171}{200} \,V$$

Therefore, in the same time in which Vic completes 2 miles, Charlie completes $171/200$ of the course, that is to say $2\cdot 171/200=1.71$ miles. So, when Vic finishes, the difference with Charlie is

$$2 -1.71=0.29 \text{ miles} $$