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

Using 1) If the journey from school to home is 40% less than the journey from home to school then ratio becomes 60:100 ~ 3:5

Since the distance is equal, the Speed ratio will be inverse of time ratio, Hence Avg speed school to home:Avg speed home to school= 5:3
We can find the speed accordingly as if 3x=15 then 5x=25.

We can find the avg speed as well since we have speed and time, hence 1 is sufficient


Using 2) Sufficient as we have been given the ratio of speed, and we can find the time ratio by taking its inverse.


Answer (D) IMO

Distance from Home to School is the same as the Distance from School to Home = 'd'
Let's assume d = 90 for convenience (multiple of 15 for easier calculations).
Average speed from Home to School = As1, Time = t1
Average speed from School to Home = As2, Time = t2

(As1 + As2)/2 = ?

Statement 1: t2 = 0.6 * t1
Speed = dist/time
t2 < t1

In the first case As1 = d/t1
15 = 90/t1
t1 = 90/15 = 6

0.6 * t1 = t2
0.6 * 6 = t2
t2 = 3.6

In the second case As2 = d/t2
As2 = 90/3.6 = 25 mph

(As1 + As2)/2 = (15 + 25)/2 = 40/2 = 20

Statement (1) is sufficient.

Statement 2: Ratio of t1 : t2 is 5:3

We can find the values of As2 using the same steps as above.

Statement (2) is sufficient.

[D] is the correct answer. Both statements are individually sufficient.

I would say answer D

(1) The journey from school to home took her 40% less time than the journey from home to school. -> We have the duration of the journey, the distance being equal, we can determine the global average speed

(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 -> -> We have the ratio duration between Home to school and school to home, the distance being equal, we can determine the global average speed

St 1 and st 2 both give the same info that T1:T2 =5:3

And speed ratio is inversely proportional to time hence we can determine average speed

And D

Hi All,

Statement 1 ,

Speed 2= D/ t2 --> D/ 0.6 t1

t1=D/15, applying this in above statement,

S2 = D/0.6*(D/15) --> 15/0.6-->25 m/h

Average speed = 2*15*25/(15+25)

Sufficient


Statement 2,

Given t1/t2=5/3,

Speed 2 = D/t2, apppying above equation in this

S2 = D/(3/5t1)--> (5/3)*(D/t1)

as already given in D/t1=S1=15

there fore S2=(5/3)*15 --> 25 m/h

Average speed = 2*15*25/(15+25)

Sufficient


Therefore D is the answer

Answer is D. Each statement is independently sufficient

1) Let's assume it took her 1 hour to bike to school, which means the route was 15 miles. If she took 0.6 * 1 hour = 36 minutes to bike home on the 15 mile route, her speed was 25 mph. Her average speed was (15 + 25)/2 = 20 mph. Now let's assume it took her 2 hours to bike to school, which means the route was 30 miles. If it took her 0.6 * 2 hours = 72 minutes to bike home on the 30 mile route, her speed was still 25 mph, and her average speed was still 20 mph. Statement 1 is sufficient as her speed home will always be 15 mph / 0.6.
2) Statement 2 is essentially giving the same information as Statement 1, since a ratio of 5:3 is the same as 40% less time. Statement 2 is sufficient for the same reasons as Statement 1.

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

We're given that the speed from home to school is 15 mph
Let the speed from school to home be s mph
The distance be d, which will be the same in both cases
Formula for time = Distance/Speed
Therefore, d/s = 0.6(d/15)
s = 25

We can now find the average speed using the formula (2*15*25)/(15+25) = 18.75mph

Therefore statement (1) is sufficient

Statement (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.

Time from home to shool/Time from school to home = 5/3
(d/15)/(d/s) = 5/3
s = 25

We can now find the average speed using the formula (2*15*25)/(15+25) = 18.75mph

Therefore statement (2) is sufficient

Therefore, (D) Each statement alone is sufficient

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

Time ∝ Speed - > S

Statement 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.

Assuming Distance = 15, ratio can be used to find speed when returning - > S

Answer D.

Statement A: The return trip took 40% less time than the first trip.

This implies that the time ratio is 10:6 or 5:3. Since speed is inversely proportional to time, return speed = 15 × (5/3) = 25 mph.
Using average speed formula, 2(15)(25) / (15 + 25) = 18.75 mph. Sufficient.

Statement B: Time ratio is directly given as 5:3. Sufficient.

Ans: Option D. Each statement alone is sufficient.

Let the speed and time of the 1st trip be v1 & t1. Let the speed and time of the 2nd trip be v2 & t2. Total distance = d. So v1 = 15 miles/hour.

The question is: 2d / (t1 + t2) = ?

(1) The journey from school to home took her 40% less time than the journey from home to school.
t2 = 0.6t1
d/v2 = 0.6 x d/v1
v2 = v1/0.6 = 15/0.6 = 25

Answer: 2d /(t1+t2) = 2d / (d/v1+ d/v2) = 18.75


(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.
t1/t2 = 5/3
t2 = 3/5 x t1 = 0.6t1
This is similar to (1). So we can calculate the answer.

Answer: D

Average speed when the distance is same or equal involves just the time taken in both the journeys.

Let d be the distance between school and home
Let s be the average speed from school to home.

total Average speed = (2*t1*t2)/(t1+t2)

Where,

t1 = time taken from home to school

t2 = time taken from school to home.

Statement 1 -

60% of her time from home to school is equal to time from school to home

(60/100)* t1 = t2

(60/100)* (d/15) = (d/s)

s = 25.

Now we have speed for the return as well. We can easily calculate total average speed.SUFFICIENT

STATEMENT 2 -

(t1)/(t2) = 5/3

(d/15)/(d/s) = 5/3
s/15 = 5/3
s = 25
We can easily calculate total average speed.
SUFFICIENT

FINAL ANSWER - Option D

let the distance be d and time(home to school)be t. given: d/t=15

average speed= total distance/total time

Statement 1: time(school to home) that is t' is 0.6t
avg speed = d+d/t+0.6t = 2d/1.6t =1.25 d/t = 18.75

Statement 2: t/t'= 5/3 so t'=3/5t=0.6 t
avg speed = 2d/1.6t =1.25 d/t = 18.75

So both statements are independently sufficient
Answer is D

Wait, where are the options to mark?

Problem Recap:

• Outbound Trip (Home to School):
  
  • Speed: 15 mph
• Return Trip (School to Home):
  
  • Speed: Different (unknown)
• Goal: Find the average speed for the entire round trip.

Understanding Average Speed:
The average speed for a round trip is calculated using the formula:
Where:

• Speed1 = 15 mph (Outbound)
• Speed2 = Return speed (unknown)

Analyzing the Statements:
Statement (1):
"The journey from school to home took her 40% less time than the journey from home to school."

1. Let’s Assume:
   
   • Time to go (Home to School) = T hours
   • Time to return (School to Home) = 0.6T hours (40% less)
2. Distance is the Same Both Ways:
   Distance = Speed × Time
   So,
   15 × T = Return Speed × 0.6T
   Return Speed = 15/0.6 = 25 mph
3. Calculate Average Speed:
   Average Speed = 2 × 15 × 25/15 + 25 = 750/40 = 18.75 mph

Conclusion: Statement (1) alone is sufficient.
Statement (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. Let’s Assume:
   
   • Time to go (Home to School) = 5k hours
   • Time to return (School to Home) = 3k hours
2. Distance is the Same Both Ways:
   15 × 5k = Return Speed × 3k
   Return Speed = 75k/3k = 25 mph
3. Calculate Average Speed:
   Average Speed = 2 × 15 × 25/15 + 25 = 750/40 = 18.75 mph

Conclusion: Statement (2) alone is sufficient.

Final Answer:
D) EACH statement ALONE is sufficient to answer the question asked.

Evaluating Statement 1 alone:
(1) The journey from school to home took her 40% less time than the journey from home to school.
Avg Speed = Total Distance / Total Time
If the distance between home to school is x, for the round trip it would be 2x
Total time = time taken to school + Time taken from school
Time = Distance/ Speed. Hence, Total time = x/15 + x/15*60%
Since all of this is expressed in one variable, this data is sufficient
Answer would be A

s1=15
Average Speed = 2(s1*s2)/(s1+s2) = 30s2/(15+s2)
Knowing s2 we can calculate the average speed.

Distance = D

(1) 40% less means 60% of the time

0.6D/15=D/s2
s2=25

Condition (1) is sufficient

(2)
T1=D/15
T2=D/s2

T1/T2=5/3=(D/15)*(s2/D)=s2/15
s2=25

Condition (2) is sufficient

Answer D

IMO D

Statement (1):
The journey from school to home took her 40% less time than the journey from home to school.
Let's denote:
• t1as the time taken to bike from home to school.
• t2 as the time taken to bike from school to home.
• D as the distance from home to school.
From the statement,
t2=t1−0.40t1 =0.6t1

Since speed is distance divided by time, we have:
• Speed from home to school: 15 miles per hour.
• Speed from school to home: v2=d/t2=d/0.6t1=15/0.6=25 mph

Now, we can calculate the average speed for the round trip

So, statement (1) is sufficient to determine the average speed.

Statement (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.
Let's denote:
• t1as the time taken to bike from home to school.
• t2 as the time taken to bike from school to home.

From the statement, t1/t2=5/3
which means t1=5/3t2.

Since speed is distance divided by time, we have:
• Speed from home to school:15 miles per hour.
• Speed from school to home: v2=d/t2
Using the ratio v2= 25 miles per hour
Now, we can calculate the average speed for the round trip
So, statement (2) is also sufficient to determine the average speed.
Conclusion:
Both statements (1) and (2) are individually sufficient to determine the average speed for the entire round trip. Therefore, the correct answer is:
(D) Each statement alone is sufficient.