How much time did it take a certain car to travel 400 kilometers?
(1) The car traveled the first 200 kilometers in 2.5 hours.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
OA B
Source: GMAT Prep
How much time did it take a certain car to travel 400
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Question: How much time did it take a certain car to travel 400 kilometers?BTGmoderatorDC wrote:How much time did it take a certain car to travel 400 kilometers?
(1) The car traveled the first 200 kilometers in 2.5 hours.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
OA B
Source: GMAT Prep
Let's take each statement one by one.
(1) The car traveled the first 200 kilometers in 2.5 hours.
Certainly insufficient since we do not know how much it took to travel the remaining 200 kms. Insufficient.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
Say the average speed is x kmps.
Thus, time taken to travel 400 kms @x kmph = 400/x hours
Time taken to travel 400 kms @(x + 20) kmph = 400/(x + 20) hours
=> 400/x - 400/(x + 20) = 1
=> 8000 = x(x + 20)
We see that 8000 is a product of two positive integers that are 20 away from each other. Upon hit and trial, we see that if we write 8000 as 80*100, we have 80*(80 + 20). Thus, x = 80 kmph. Sufficient.
The correct answer: B
Hope this helps!
-Jay
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$$400\,{\rm{km}}\,\,\,\,\,\left\{ \matrix{BTGmoderatorDC wrote:How much time did it take a certain car to travel 400 kilometers?
(1) The car traveled the first 200 kilometers in 2.5 hours.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
Source: GMAT Prep
\,\,\left( {{\rm{real}}\,\,{\rm{speed}}\,,\,\,{\rm{real}}\,\,{\rm{time}}} \right)\,\,\,{\rm{ = }}\,\,\,\left( {{V_R}\,\,,\,\,{T_R}} \right) \hfill \cr
\,\,\left( {{\rm{hypothetical}}\,\,{\rm{speed}}\,,\,\,{\rm{hypothetical}}\,\,{\rm{time}}} \right)\,\,\,{\rm{ = }}\,\,\,\left( {{V_H}\,\,,\,\,{T_H}} \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\,\,\left[ {\,{{{\rm{km}}} \over {\rm{h}}}\,} \right]\,\,\,,\,\,\,\,\left[ {\,{\rm{h}}\,} \right]\,\,\,} \right)$$
$$? = {T_R}\,\,\,\left[ {\rm{h}} \right]$$
$$\left( 1 \right)\,\,\left\{ \matrix{
\,{\rm{If}}\,\,{\rm{it}}\,\,{\rm{took}}\,\,0.5\,{\rm{h}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{last}}\,\,200\,{\rm{km}}\,\,\,\,\, \Rightarrow \,\,\,\,?\,\, = \,\,3\, \hfill \cr
\,{\rm{If}}\,\,{\rm{it}}\,\,{\rm{took}}\,\,1\,{\rm{h}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{last}}\,\,200\,{\rm{km}}\,\,\,\,\, \Rightarrow \,\,\,\,?\,\, = \,\,3.5\,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,{V_H} = {V_R} + 20\,\,\,\left( * \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{T_R} - {T_H} = 1\,\,\,\,\,\,\left[ {\rm{h}} \right]$$
Now it´s time for UNITS CONTROL, one of the most powerful tools of our method:
$${{{\rm{km}}} \over {\,\,\,{{{\rm{km}}} \over {\rm{h}}}\,\,}} = {\rm{h}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{
{T_R} = {{400} \over {{V_R}}} \hfill \cr
{T_H}\,\mathop = \limits^{\left( * \right)} \,\,{{400} \over {{V_R} + 20}} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{{400} \over {{V_R}}} - {{400} \over {{V_R} + 20}} = 1$$
$${{400\left( {{V_R} + 20} \right)} \over {{V_R}\,\,\left( {{V_R} + 20} \right)}} - {{400\,\,{V_R}} \over {\left( {{V_R} + 20} \right)\,\,{V_R}}} = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,400 \cdot 20 = {V_R}\,\,\left( {{V_R} + 20} \right)$$
$${V_R}^2 + 20{V_R} - 400 \cdot 20 = 0\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{of}}\,\,{\rm{roots}}}^{{\rm{product}}} \,\,\,\,\,\,\,\left( {{c \over a} = } \right)\,\,\,{{ - 400 \cdot 20} \over 1} < 0\,\,\,\, \Rightarrow \,\,\,\,\,{V_R}\, > 0\,\,\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\,\,$$
$$\left( {\,\,\,\left. \matrix{
{V_R}\,\,{\rm{unique}}\,\, \hfill \cr
{\rm{400}}\,{\rm{km}} \hfill \cr} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,{T_R}\,\,\,{\rm{unique}}\,\,\,} \right)$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Target question: How long did it take to travel 400kmBTGmoderatorDC wrote:How much time did it take a certain car to travel 400 kilometers?
(1) The car traveled the first 200 kilometers in 2.5 hours.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
To find the travel time, we need to know the average speed traveled.
Let x = the average speed traveled.
REPHRASED target question: What is the value of x?
Statement 1: The car traveled the first 200 km in 2.5 hrs
No info about the 2nd half of the trip, so we can't determine the overall average speed (aka x).
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: If the car's average speed had been 20 km/h faster, it would have traveled the 400 km in 1 hour less time.
Let's start with a word equation:
(travel time at x km/h) - 1 = (travel time at x+20 km/h)
Since time = distance/speed, we can now write:
(400/x) - 1 = 400/(x+20)
IMPORTANT: At this point, we need only determine whether this equation will yield 1 or 2 valid values of x. If it yields only 1 valid value, then statement 2 is sufficient. If it yields 2 valid values, then statement 2 is not sufficient.
Rewrite as: (400-x)/x = 400/(x+20)
Cross multiply: (400)(x) = (400-x)(x+20)
Simplify: 400x = 8000 + 380x - x²
Rewrite: x² + 20x - 8000 = 0
Factor: (x + 100)(x - 80) = 0
So, x = -100 or 80
Since x cannot be negative, it must be the case that x = 80
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent