One day Andrew goes from his home to his school at a speed of 20 km/hr

Let total distance = 80 km

time from home to school = 80/20 = 4 hrs

time from school to restaurant = 40 / 40 = 1 hrs

Average speed = 120 / 5 = 24 km/hrs

Answer: A

One day Andrew goes from his home to his school at a speed of 20 km/hr but on finding that it was holiday in school, he plans to have party with friends at a restaurant which is situated exactly at midpoint of his way from home to school so he starts back and reaches the restaurant at speed of 40 km/hr without wasting any time at school. Find the average speed of Andrew from home till he reaches restaurant after visiting school.

Let the Distance from Home to School be D km.
Speed of Andrew from Home to School = 20km/hr
Thus Time taken to reach school from home = Distance / speed = \(\frac{D}{20}\)
Restaurant is exactly at midpoint from home to school, therefore Distance from school to restaurant = \(\frac{D}{2}\)km
Speed of Andrew from School to Restaurant = 40km/hr
Time taken to reach restaurant from school = \(\frac{D}{2}\)/40 = \(\frac{D}{80}\)

Average Speed = Total Distance Traveled / Total Time Taken
Average Speed of Andrew from Home till Restaurant = D + \(\frac{D}{2}\) / \(\frac{D}{20}\) + \(\frac{D}{80}\) = \(\frac{2D + D}{2}\) / \(\frac{4D + D}{80}\) = \(\frac{3D}{2}\)/\(\frac{5D}{80}\) = \(\frac{3}{2}\) x \(\frac{80}{5}\) = 24km/hr
Answer A...

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[quote="GMATinsight"]One day Andrew goes from his home to his school at a speed of 20 km/hr but on finding that it was holiday in school, he plans to have party with friends at a restaurant which is situated exactly at midpoint of his way from home to school so he starts back and reaches the restaurant at speed of 40 km/hr without wasting any time at school. Find the average speed of Andrew from home till he reaches restaurant after visiting school.

A) 24
B) 25
C) 30
D) 32
E) 35

let total distance=3 km
total time=2/20+1/40=1/8 hr
average speed=3/(1/8)=24 kmh
A

One day Andrew goes from his home to his school at a speed of 20 km/hr but on finding that it was holiday in school, he plans to have party with friends at a restaurant which is situated exactly at midpoint of his way from home to school so he starts back and reaches the restaurant at speed of 40 km/hr without wasting any time at school. Find the average speed of Andrew from home till he reaches restaurant after visiting school.

R x T = D
There: 20 x 1 = 20
Back: 40 x t = 10

t=\(\frac{1}{4}\) hours = 15 minutes

average speed = \(\frac{Total Distance}{Total Time}\)

average speed = \(\frac{30 km.}{(1 hour + 15 minutes)}\)

average speed = \(\frac{30 km}{75 minutes}\) = \(\frac{10}{25}\) = \(\frac{2}{5}\)

\(\frac{2}{5}\) x 60 minutes = 24 minutes

Answer (A) 24
Visualisation of the problem solving is attached.

Pick a smart number. Let the distance between home and school me 40 kms. [LCM of two speeds, i.e. 20 and 40]. (A number like 100 is not appropriate as it will make the calculations ugly)

We know that speed = \(\frac{distance}{time}\)

Time taken from home to school = \(\frac{40}{20}\) = 2 hrs

Time taken from school to restaurant = \(\frac{(40*0.5)}{40}\) = 0.5 hrs

Average speed = \(\frac{total distance}{total time} = \frac{60}{2.5}\) = 24 km/hrs

Therefore the correct answer is A

Algebraically:-

Let the distance between home and school be x kms.

We know that speed = \(\frac{distance}{time}\)

Time taken from home to school = \(\frac{x}{20}\)

Time taken from school to restaurant = \(\frac{(x*0.5)}{40}\) = \(\frac{x}{80}\)

Average speed = \(\frac{total distance}{total time} = \frac{3x}{2}\) divided by total time [\(\frac{x}{20}\) + \(\frac{x}{80}\)
] = \(\frac{3x}{2}\) divided by \(\frac{5x}{80}\) = \(\frac{24x}{x}\) i.e. = 24 km/hrs

Therefore the correct answer is A

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