If a car had traveled 20 kmh faster than it actually did, the trip

Question

If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?

A. 35 kmh 
B. 40 kmh 
C. 50 kmh 
D. 60 kmh 
E. 65 kmh

Bunuel · Accepted Answer

If a car had traveled 20 kilometers per hour faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 kilometers, at what speed did it travel ?

A. 35 kilometers per hour 
B. 40 kilometers per hour 
C. 50 kilometers per hour 
D. 60 kilometers per hour 
E. 65 kilometers per hour

Let the car's actual speed be  x  kilometers per hour. Then:
 
At a speed of  x  kilometers per hour, the car would cover 60 kilometers in  time=\frac{distance}{rate}=\frac{60}{x}  hours.
 
At a speed of  x+20  kilometers per hour, the car would cover 60 kilometers in  time'=\frac{distance}{rate'}=\frac{60}{x+20}  hours.
 
We're given that at the faster speed, the car takes 0.5 hours (or 30 minutes) less to cover the 60 kilometers. Therefore,  \frac{60}{x} = \frac{60}{x + 20} + 0.5 .
 
At this stage, plugging in the answer options to back-solve is easier.
 
By doing so, we find that option B fits:  \frac{60}{40} = 1.5  and  \frac{60}{40 + 20} + 0.5 = 1.5 .

Answer: B

Kurtosis · Answer

Time = Distance/Speed

Difference in time = 1/2 hrs

60/x - 60/(x + 20) = 1/2

Substitute the value of x from the options. --> x = 40 --> 60/40 - 60/60 = 3/2 - 1 = 1/2

Answer: B

generis · Answer

Using answer choices, it is much easier if you use 30 minutes =  \frac{1}{2}  hour

Start with Answer C to get a benchmark
If original speed was 50, then time, t = D/r: 
Time at 50 mph:  \frac{60}{50}=\frac{6}{5}  hours
20 mph faster? (50 + 20) = 70 mph. t = D/r
Time at 70 mph:  \frac{60}{70}=\frac{6}{7}  hours

That hour fraction in minutes will not yield an integer (30 minutes), but find the difference in times to get a benchmark
 (\frac{6}{5} - \frac{6}{7})=\frac{(42-30)}{35}=\frac{12}{35}  hours
 \frac{12}{35}\approx{\frac{1}{3}}  hour
 \frac{1}{3} < \frac{1}{2}  hr

We need time to be longer. That means speed must be slower (decreased speed = increased travel time). Eliminate answers D and E.

Try B) 40
Time at 40 mph:  \frac{60}{40} = \frac{3}{2}  hour
20 mph faster: (40 + 20) = 60 mph
Time at 60 mph:  \frac{60}{60}=1  hour
Time difference?  \frac{3}{2} - 1 = \frac{1}{2}  hour = 30 minutes

Answer B

EgmatQuantExpert · Answer

Let the actual speed be "S" kmh

The increased speed will be "S + 20"

The difference in timing has been given as 30 minutes or 1/2 hour

Actual distance travelled = 60 km

Thus, we can write :

\frac{60}{S} - \frac{60}{(S+20)} = 0.5

Though we can solve the equation, a quicker way would be to look at the options, which either gives us an integer when  \frac{60}{S}  or when  \frac{60}{(S+20)} .

35, 50 and 65 would not help get integer values of time.

A good option would be 40, because  \frac{60}{40} = 1.5  and  \frac{60}{(40+20)} = 1  and the difference is 1.

Thus, the correct answer would be    Option B   .

ScottTargetTestPrep · Answer

We can let rate = r and create the equation:

60/(r + 20) + 1/2 = 60/r

Multiplying by 2r(r+20) we have:

120r + r^2 + 20r = 120r + 2400

r^2 + 20r - 2400 = 0

(r + 60)(r - 40) = 0

r = -60 or r = 40

Since r can’t be negative, r = 40 kmh.

Answer: B

SarthakSharma8 · Answer

Let the initial speed be 100 km/h
A 20 km/h increase is 1/5th increase of the initial speed and  since distance is constant therefore we can use  :-

If 1 variable increases by n/d the other variable decreases by n/d+n 
So if S increases by 1/5, Time decreases by 1/6
1/6th of time is given as 30 minutes or 1/2 hr

Let 'x' be initial time taken at initial speed to cover 60 km
so,
1/6th of x = 1/2
x = 3 hrs

now putting time and distance in speed equation we can find the actual initial speed :-

S*T = D
S*3 = 60
S = 20 km/h ----  Actual initial speed

But since the speed was increased by 20 km/h so the final speed becomes = 20+20 = 40 km/h

Shan0110 · Answer

I tried equating the distance with the new speed and time. Speed = S+20 and time = 60/S-0.5, if we multiply these two the distance would be 60. But, I'm unable to solve that equation. I don't see anything wrong with my logic. Can anyone explain why equating it with the distance isn't working for me?

Bunuel · Answer

(S + 20) * (60/S - 0.5) = 60 gives the correct solution, S = 40. However, expanding the equation results in a quadratic, so as shown in the solution above, it's easier to back-solve by plugging in the answer choices.

If a car had traveled 20 kmh faster than it actually did, the trip

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If a car had traveled 20 kmh faster than it actually did, the trip

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