Mrs. Jordan visited a certain city. She travelled the entire distance

Question

Mrs. Jordan visited a certain city. She traveled the entire distance at an average speed of 45km per hour. If 2/3 of the total distance was traveled at an average speed of 50km per hour, what was the average speed for the remaining distance?

A. 12.5km/hr
B. 25km/hr
C. 37.5km/hr
D. 40km/hr
E. 55km/hr

A. 12.5km/hr
B. 25km/hr
C. 37.5km/hr
D. 40km/hr
E. 55km/hr

huxledotnet · Accepted Answer

You can use the equation approach here. But another method which would be easier for some is to assume a certain distance and the work backwards. Because the fraction 2/3 of distance is mentioned, assume a total distance of 150 km (because it can be easily split into 3 parts of 50 km each). 2/3 of 150 is 100 km. Mrs. Jordan covers 100 km @ 50 km per hour.

Time taken to cover 100 km = 100 km / 50 kmph = 2 hours

Now, the rest of the distance (50 km) is covered in a speed so that the avearge speed is 45 kmph.

Total distance / total time = average speed
150 km / (2 hours + t hours) = 45 kmph
150 km / 45 kmph = 2 hours + t hours
t hours = (150 km/ 45 kmph ) - 2 hours
t hours = (10/3) - 2
t hours = 4/3 hours

Average speed of remaining distance = 50 km / (4/3) hours = 37.5 kmph

eabhgoy · Answer

S =  \frac{D}{T} 
=> 45 =  \frac{x}{(T1+T2)} 
=> T1+T2 =  \frac{x}{45}

Now from the question we kow:
T1= ((2x/3)/50) = x/75
&
T2 = ((x/3)/y) = x/3y

=> (x/75) + (x/3y) = (x/45)
=> (1/3y) = (1/45) - (1/75)

Solving for y we get
y = 37.5

Hence C is the correct answer.

Hadiagh · Answer

i assumed the TD=90
so Total time= 2hrs
 
now for Lap1:
Distnace is 2/3 of the TD= 60
speed= 50 
Time= 6/5

and lap2:
distance: 60-90= 40
time= 2-(6/5)=4/5
speed=50??

please help...

josearchila86 · Answer

How do you get from this => (x/75) + (x/3y) = (x/45)
To this 
=> (1/3y) = (1/45) - (1/75)?

How do you make the X to desappear?

aditijain1507 · Answer

Bunuel  can't we use mixtures formula here? w1/w2 = a2-abg/avg-a1, the answer comes out as 35 this way

stgantayet · Answer

In this question can we use the weighted mean method? If not can anyone please explain why time can be used as weights but not distance?

KB1920 · Answer

Can't we use mixtures formula here? w1/w2 = a2-abg/avg-a1, the answer comes out as 35 this way
My response:
We cannot use the mixture formula directly because that will give Arithmetic Mean (A.M) whereas the average of two speeds will be Harmonic mean (HM) ( Average speed ≠ Average of two speeds). However we can use the weighted average formula along with the concept that AM > HM. So let's take the speed for 1/3 distance as x.
Then AM= (2/3*50+1/3*x)/(2/3+1/3)=45(given) => x=35. However, actually as per the problem, HM=45
Since AM> HM; AM > 45
IF AM > 45, than x > 35. Look into answer choice for the value which is just above 35. So choose 37.5
Ans C This trick will always work in GMAT

luisdicampo · Answer

Deconstructing the Question  
Total Average Speed = 45 km/h.
Leg 1: Distance =  \frac{2}{3}  of Total, Speed = 50 km/h.
Leg 2: Distance =  \frac{1}{3}  of Total (Remaining).
Target: Average Speed for Leg 2.

Method: Smart Numbers  
Choose a Total Distance ( D ) that is easy to work with (a multiple of 45 and 3). Let  D = 450  km.

Step 1: Calculate Total Time 
 T_{total} = \frac{D}{V_{avg}} = \frac{450}{45} = 10  hours.

Step 2: Analyze First Leg 
Distance 1 =  \frac{2}{3} 	imes 450 = 300  km.
Speed 1 = 50 km/h.
Time 1 =  \frac{300}{50} = 6  hours.

Step 3: Analyze Remaining Leg 
Remaining Distance =  450 - 300 = 150  km.
Remaining Time =  T_{total} - T_1 = 10 - 6 = 4  hours.

Step 4: Calculate Speed 
 Speed_{rem} = \frac{	ext{Remaining Distance}}{	ext{Remaining Time}} 
 Speed_{rem} = \frac{150}{4} = 37.5  km/h.

Answer:  C

egmat · Answer

Looking at your doubt about using the weighted mean method with distance as weights - this is a  common misconception  about average speeds!

Key Insight:  Average speed is  NOT the arithmetic mean of speeds . It's always:
 Average Speed = Total Distance ÷ Total Time

Why distance can't be used as weights in arithmetic mean:

If we tried weighted arithmetic mean with distance weights:
- First part:  2/3  weight at  50  km/hr
- Second part:  1/3  weight at  x  km/hr
- Overall average:  45  km/hr

This would give us:  (2/3)  ×  50  +  (1/3)  ×  x  =  45

Solving:  x = 35  km/hr ❌  WRONG!

Why this doesn't work:  When segments have different distances, time spent on each segment varies non-linearly with speed. The arithmetic mean assumes linear relationships, but time = distance/speed is  inversely proportional .

The correct approach:

Let total distance =  D

Step 1:  Find total time
Total time =  D/45  hours

Step 2:  Find time for first  2/3  distance
Time1 =  (2D/3) ÷ 50 = D/75  hours

Step 3:  Find time for remaining  1/3  distance
Time2 = Total time - Time1 =  D/45 - D/75 
Time2 =  D(1/45 - 1/75) = D(2/225) = 2D/225

Step 4:  Find speed for last part
Speed = Distance ÷ Time =  (D/3) ÷ (2D/225) = 225/6 = 37.5  km/hr

Answer: C (37.5 km/hr)

When CAN we use weighted means? 
-  TIME as weights in HARMONIC mean : When different speeds are maintained for different times
-  Equal distances : Use harmonic mean (special case)
-  Never use distance as weights in arithmetic mean for average speed!

Remember:  For average speed problems, always go back to the  fundamental formula : Total Distance ÷ Total Time!

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