A driver completed the first 20 miles of a 40-mile trip at an average

Average speed when a body covers m part of the trip at speed x and the second n part of the trip at speed y:

average speed = (m+n)xy / (my+nx)

60 = (20+20)50x / (20*50 + 20x)

60 = 2000x / 1000 + 20x

60 000 = 800x

x = 75

Questions of speed, distance, work and time are the few types of questions where I really prefer to use memorized formulas.

tot dis/tot time= 60(avg speed)
40/(20/50+20/x)= 60 *(x is the average speed for the second 20 miles)
why does this not give the answer?

Bunuel VeritasKarishma, This looks like a weighted average question. But when I am applying it it gives me different answer why?

Hi Karishma,
I have a doubt regarding using the weighted average method in this question -->> Why are we taking the ratio of time taken for the 2 trips as the ratio of weights in the w.avg question ?? why not the distance.
When we use the ratio of the distance as the ratio of weights we get the ans as 70 as inverse of the ratio of weights(1:1) == the ratio of the distance of data points away from the main average.
When we use the ratio of the TIME as the ratio of weights we get the ans as 75 as inverse of the ratio of weights(2:3) == the ratio of the distance of data points away from the main average.

I prefer to reject few options before getting into calculation.
We can reject A,B and C by weighted average concept.
If answer is A or B, the weightage of 65 or 68 would be more because 60 is closer to 65 or 68 than 50. Logically it means time taken would be more at 65 or 68 than 50 m/h which is not possible.

If one speed is 50 m/h and avg speed is 60, then answer must be greater than 70 because weightage of time to cover 50 m/h is more than weightage of time at other speed.

parth2424

With similar logic, we can reject C (70 m/h) because if other speed is 70, then time taken for first 20 miles is same as time taken for next 20 miles. Therefore answer must be D or E.

Now check D and E using the formula
2ab/(a+b) = 2*50*75/50+70 = 75

To find the average speed for the entire 40-mile trip, we can use the concept of harmonic mean.

Let's denote the average speed for the remaining 20 miles as V. The harmonic mean of the two speeds (50 mph and V mph) will be equal to the reciprocal of the average speed:

Harmonic mean = 2 / ((1/50) + (1/V))

We know that the desired average speed for the entire 40-mile trip is 60 mph. So, we can set up the equation:

60 = 2 / ((1/50) + (1/V)) = 2 / ((V + 50) / (50V)) = 2 * (50V) / (V + 50)

Multiplying both sides by (V + 50) to eliminate the denominator:

60 * (V + 50) = 2 * 50V

Dividing both sides by 2:

30 * (V + 50) = 50V

30V + 1500 = 50V

1500 = 50V - 30V = 20V

V = 1500 / 20 = 75 mph

Therefore, the average speed for the remaining 20 miles must be 75 miles per hour. Hence, the answer is (D) 75 mph.

I used a formula that I picked up from the thread of a question similar to the question here.
If m and n are distances covered in the complete journey and if p and q are the avg speed in which dist m and n are covered respectively,

then, Avg Speed = ((m + n) * p * q)/(mq + np)

Using this formula, 60 = ((20 + 20) * 50 * q)/(20q + 20*50)
therefore, q = 75 mph.