Target Test Prep GMAT Quant Challenge Problem 7

Solution:

This is an if/then rate question. We're given a hypothetical “if” scenario, and we need to use this to determine the actual scenario. We are told that the distance traveled was 60 miles. Most importantly, we are told that the hypothetical speed was 2 mph faster than the actual speed. Since we do not have any values for the actual speed, we can express the actual speed as r and the faster speed as (r+2).

Now we have enough information to determine the faster and slower times:

$\begin{array}{l}\Rightarrow {\text{time}}_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{ miles}}{\mathrm{r}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{ miles}}{1}\times \frac{\text{1 }\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\mathrm{r}\text{ miles}}=\frac{60}{\mathrm{r}}\text{ hours}\\ \Rightarrow {\text{time}}_{\text{faster }\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{ miles}}{\left(\mathrm{r}+2\right)\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{ miles}}{1}\times \frac{\text{1 }\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\left(\mathrm{r}+2\right)\text{ miles}}=\frac{60}{\left(\mathrm{r}+2\right)}\text{ hours}\end{array}$

We can now set up an equation using the travel time. We know that if Thomas had ridden at the faster speed, he would have arrived 1 hour earlier. Hence:

$\begin{array}{l}\text{⇒ Faster Time}+\text{1 Hour}=\text{ Slower Time}\\ \text{⇒}\frac{60}{\mathrm{r}+2}+1=\frac{60}{\mathrm{r}}\\ \text{⇒}\mathrm{r}(\mathrm{r}+2)(\frac{60}{\mathrm{r}+2}+1)=\frac{60}{\mathrm{r}}\\ \text{⇒}60\mathrm{r}+\mathrm{r}(\mathrm{r}+2)=60(\mathrm{r}+2)\\ \text{⇒}60\mathrm{r}+{\mathrm{r}}^2+2\mathrm{r}=60\mathrm{r}+120\\ \text{⇒}{\mathrm{r}}^2+2\mathrm{r}-120=0\\ \text{⇒}(\mathrm{r}+12)(\mathrm{r}-10)=0\\ \text{⇒}\mathrm{r}=-12,\text{ r}=10\end{array}$

Since we cannot have a negative rate, Thomas's actual rate was 10 mph.

(Note: If you thought that factoring this quadratic was difficult, keep in mind that once you have r² + 2r - 120 = 0, you can look at the answer choices to help you determine r. Notice that 10 and 12 are 2 units apart and also multiply to 120. Thus, we can quickly see that r² + 2r - 120 = 0= 0 will factor down to (r + 12)(r - 10). This is a useful method to use anytime you have a tricky quadratic in a multiple choice question.)