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12 Days of Christmas GMAT Competition - Day 6: Clair bikes from home

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GMATNextStep

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24 Dec 2024, 06:33

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Clair bikes from home to school to drop off her daughter at an average (arithmetic mean) speed of 15 miles per hour. She then immediately bikes back home along the same route at a different average speed. What was her average speed for the entire round trip?

(1) The journey from school to home took her 40% less time than the journey from home to school.

(2) The ratio of the time she spent biking from home to school to the time she spent biking from school to home was 5 to 3.

1) Call t the time she takes from Home to School. d the distance between home and school
average speed for the entire trip = distance/ time.
- Distance: 2d, since it is a round trip (1)
- Time: I'll calculate v for return trip. Ratio of t1/t2 = 5/3 => ratio of speed = 3/5. Since first leg of the trip is 15mph, return trip speed is 15*5/3 = 25 mph
=> Total time = d/15 + d/25 (2)

From (1) and (2), we will be able to plug in to distance/time formula to eliminate d and calculate average speed. Sufficient
2) Gives the same information as (1). We can use the same methodology to arrive at the average speed. Sufficient as well.

Choice D

HansikaSachdeva

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24 Dec 2024, 07:33

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Statement 1

Time taken from home to school = t hours
Time taken from school to home is t−0.4t = 0.6t

The distances for both trips are the same, so the speed for the return journey is:
Speed = Distance / Time = d / 0.6t

The total time for the round trip is:
t + 0.6t = 1.6t

Average speed = 2d / (t + 0.6t) = 2d / 1.6t

We know that d/t = 15

Therefore, Speed = 2 * 15 / 1.6 = 18.75

Statement 1 alone is sufficient

Statement 2

Time from home to school = 5t
Time from school to home = 3t
Total time = 3t + 5t = 8t

Average speed = 2d / 8t

Speed from home to school = 15 = d / 5t
Therefore, d = 75t

Substituting d in average speed,

Average speed = (2 * 75t) / 8t = 18.75

Statement 2 alone is sufficient

Each statement alone is sufficient

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24 Dec 2024, 07:56

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Speed from home to school S1=15mph.
Distance for each side of the trip = d
Speed from school to home: S2

1 ->
time taken from home to school = t1
=> t2 = 0.6 t1

T = t1+t2 = t1 + 0.6t1 = 1.6t1
Distance from home to school = d = S1.t1 = 15t1

Total distance = d+d = 2d = 30t1
Average Speed = Total Distance/Total Time = 30t1/1.6t1 = 18.75

2->
The ratio of the time spent biking from home to school to the time spent biking from school to home was 5:3
=> T = 3x + 5x = 8x
d= S1*t1 = 15*5x = 75x
Average Speed = Total Distance / Total Time = 75x + 75x / 8x = 150x/8x = 18.75

D. EACH statement ALONE is sufficient.

prashantthawrani

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24 Dec 2024, 08:09

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Let v1=15 mph.

Let the distance from home to school be d

Time from home to school: t1=d/15

Time from school to home: t2=d/v2 (Let v2 be the speed back home)

Total distance: D=2d

Total time: T=t1+t2=(d/15+d/v2)

Average speed:
Vavg=D/T=2d/(d/15+d/v2)
Vavg=2/(1/15+1/v2), So we just need the value to V2 to find average speed.

Q: V2=?

Statement (1) : t2=0.6 t1
=> d/v2=0.6d/v1
=> v2=v1/0.6= 15/0.6 = 25
sufficient

Statement (2)
t1/t2=5/3
(d/v1)/(d/v2)=5/3
v2=5(15)/3 = 9
sufficient

Answer is D.

BillyButcher0956

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24 Dec 2024, 08:22

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Let the distance be D b/w home and school
T = time for the trip from home to school

Home to School => D/T = 15 miles/hr ---- (eqn 1)

A) School to Home => Time = 0.4 * T
Then Speed = D/(0.4*T) = 15*T/ (0.6*T) => 25 miles/hr
Average Speed = Total Distance/Total Time = 2D/1.6T = (2/1.6)*(D/T) = 5/4 * 15 miles/hr

A is sufficient alone

B) T = time for the trip from home to school
15 miles/hr = speed for the trip from home to school
t = time from school to home
s = speed from school to home

T/t = 5/3
(D/15) / (D/s) = 5/3 => s/15 = 5/3 => s = 25 miles/hr
Average Speed = Total Distance/Total Time = 2D/1.6 T = 5/4 * 15 miles/hr

B is sufficient alone

Bunuel

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

Clair bikes from home to school to drop off her daughter at an average (arithmetic mean) speed of 15 miles per hour. She then immediately bikes back home along the same route at a different average speed. What was her average speed for the entire round trip?

(1) The journey from school to home took her 40% less time than the journey from home to school.

(2) The ratio of the time she spent biking from home to school to the time she spent biking from school to home was 5 to 3.

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

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24 Dec 2024, 08:35

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Distance $S=15t_1=v_2t_2$
Average speed = $\frac{2s}{t_1+t_2}$

(1) $t_2=0.6t_1$
Therefore, $t_1=\frac{S}{15}$ and $ t_2=\frac{0.6S}{15}$

So, average speed $= \frac{ 2S}{t_1+t_2}=\frac{2S}{(S/15+0.6S/15)}$
S cancels out and we get the ratio. Sufficient.

(2) $ t_1:t_2=5:3$
=>$ t_2=\frac{3t_1}{5}=0.6t_1$
Same info as in point 1, so sufficient as well.

Therefore, the answer is D.

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24 Dec 2024, 08:38

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To find : Average speed for entire round trip => we need to know both trips speed
To school speed = 15 miles/hr
From school speed = x

Assume distance is d

Statement 1 - x = 15/ 0.4
Sufficient alone

Statement 2 -
Time to school = d/15
Time from school = d/x

d/15 : d/x = 5 : 3
Speeds in the ratio : 3:5

Speed from school = 25
Sufficient alone

OPTION D

Bunuel