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

Statement 1.
15t= S*0.6t
S= 25
Statement 2
15*5t = S*3t
S=25

Hence OPTION D

Let the distance from Home to School be x and hence the distance from school to home will also be x. Hence, we know that
Total distance = 2x
and
Time taken from Home to school = total distance/ average speed = x/15

We know that Average Speed = Total Distance/ Total time

With this in mind let's evaluate the statements -

From statement 1 we can infer that Time taken from school to home = 0.6*time taken from home to school = 0.6*(x/15) = x/25. Plugging this in the formula for average speed we can calculate average speed = 2x/((x/15) +(x/25))
Sufficient.

Statement 2 tells us that the ratio of the two times from which we can again calculate that time taken from school to home = x/25. Hence sufficient

Answer - D

d be the distance between home and school. t be the time taken to travel the distance.
d = 15*t
From Statement (1)
let v be the speed from school to home. Given time taken is 0.6t
d = v*0.6t
15t = v * 0.6t => v = 25mph
avg speed for round trip = total distance/ total time = 15t+25*0.6t/1.6t = 30/1.6 ----(Sufficient)
From Statement (2)
let t1,t2 be the times taken from school to home and home to school. v be the speed from school to home.
t1/t2 = 5/3
15 * t1 = v * t2 => v = 25mph
avg speed for round trip = 15*t1 + 25*t2/ t1+t2 = 75/4 --- (Sufficient)

So either statement is sufficient.

Option D

lets take the distance between school and home as 'd' and time taken from school to home as 't'
the average speed from home to school is d/t

I) As per first condition the time taken from school to home is 40% less which means its 0.6t
Hence avg speed is (1/0.6) d/t
using this we can calculate the overall avg speed

II) As per second condition t1/t2 is 5/3
using this same as before avg speed from school to home is 5/3(d/t)
using this we can calculate overall avg speed

Hence option D

Let d = distance (same for both journeys)
t1 = time for first journey
t2 = time for second journey

Total average speed = total distance/total time => 2d/t1+t2
Given, first journey: 15 = d/t1
2d/t1+t2 multiply divide by t1 we get => 2(d/t1)/1+t2/t1 => 30/1+t2/t1
Thus we only need the ratio of t2 and t1.

Statement 1:

t2 = 0.6t1 => t2/t1 = 0.6 => Sufficient

Statement 2:

t1/t2 = 5/3 → t2 = 0.6t1 => t2/t1 = 0.6 => Sufficient

Both statements give the same information => either statement alone is sufficient.

Answer: D

The correct answer is D

We are given - Clair bikes back home along the same route - which means the distance is constant.

When the distance is constant, Speed is inversely proportional to time. We can use the formula:
Speed1/Speed2 = Time2/Time1 or S1/S2 = T2/T1

Given: For home to school
Average speed (S1) = 15 miles per hour
Distance = d miles
Time taken be denoted by T1

For school to home:
Average speed (S2) = S2 miles per hour
Distance = d miles
Time taken be denoted by T2

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

So, if Clair took x hours from home to school, she took 0.6x hours from school to home

Using the above formula,
15/S2 = 0.6x/x
S2 = 25 miles per hour

Since we now have both S1 and S2, we can find Avg Speed - This statement is sufficient

Statement 2:
T1:T2 = 5:3
Let T1 be 5x and T2 be 3x

Again, using the same formula,
15/S2 = 3x/5x
S2 = 25 miles per hour

Since we now have both S1 and S2, we can find Avg Speed - This statement is sufficient

Here Distance 'D' is equal

Speed*Time = Distance

15*T = X*T1

Statement 1

15*T = 0.6T*Y(Speed while coming back)
Y= 25, Sufficient
Statement 2
T/T1 = 5/3

15*T = 3/5(T)*Y

Y = 25, Sufficient

IMO D

Average speed = Total dist covered/total time taken

Let the distance by d, we need 's' with which he returned home

Distance	Speed	Time
d	15	d/15
d	s	d/s
2d		d/15 + d/s

St 1) d/s = (1-40%)*d/15
We can get 's' SUFFICIENT

St 2) (d/15)/(d/s)=5/3
=> s/15=5/3 again we get s SUFFICIENT

So correct answer is option D

Initial Statement

	Distance	Speed	Time
To School	D	15 mph	D/15
From School	D

Average speed = ?

Above is the initial statement captured as table (Time alone was computed)
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S1:

	Distance	Speed	Time
To School	D	15 mph	D/15
From School	D	30 mph	0.6 * D / 15

Since Speed = Distance / time,

We can compute speed Speed From school as 30mph

THerefore average speed can be computed (since the distance is same, the average speed is harmonic mean of the two speeds)

Sufficient

-----------------------------------------------------------------------------------------
S2 :

	Distance	Speed	Time
To School	D	15 mph	D/15
From School	D	25 mph	D/25

Let t be the time taken from school
D/15 : t = 5/3

=> t = 3D/75 = D/25

speed is D / D/ 25 = 25mph

THerefore average speed can be computed (since the distance is same, the average speed is harmonic mean of the two speeds)

Sufficient
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Option D

Overall Avg. Speed = Total Distance / Total Time

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

Overall Avg. speed = 2*d/(d/15 + d/s) = 30*s/(s+15)

We'll get overall Avg. Speed if we find s

Statement 1.

d/s = 0.6*(d/15)
s = 25 SUFFICIENT

Statement 2.

(d/15)/(d/s) = 5/3
s/15 = 5/3
s = 25 SUFFICIENT

Answer D.

(1)
Let D be the distance
Let t be the time
Let V be the velocity

	D	t	V
Home to School	D	t	15
School to Home	D	0.6t	15*0.6 = 9

Average speed for the trip = 15+9/(1+0.6) (Sufficient)

(2) The ratio of 5 to 3 is equivalent to t vs 0.6t above (Sufficient)

Therefore the answer is D

So here distance is same , Let it be d.

Stmt (1) The journey from school to home took her 40% less time than the journey from home to school.
Let call 100t Home to School. So School to home is 60t.

We know home to school 15m/hr so d=1500t.

We know 2d = 3000t , Total time =160t.

Avg = 3000/160 => 150/8

Hence sufficient.

So AD.

Stmt (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.
If time from home to school = 100t. then time from school to home is 60t.
This is exactly same as stmt 1. Hence this is also sufficient.

IMO D

To solve the problem, we calculate Clair's average speed for the round trip using the formula for average speed when the distances are the same:

Average speed= 2*(Speed going)*(Speed returning)/(Speed going +Speed returning)
​
We need the speeds for the journey to school and back. Let’s analyze the statements.

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

Let the time taken to bike from home to school be t1
​
, and the time taken to bike from school to home be t2
​
The statement tells us: t2 = 0.6*t1
The speed going to school is 15mph.
Since
Speed= Distance/Time
​
, the distance is the same both ways, and we can express the speed returning (S return ) as:

S return = Distance/t2 = Distance / (0.6*t1) = 15/0.6 = 25 mph
Thus, the speeds are:
Speed going=15mph,Speed returning=25mph.

The average speed is:
Average speed= 2⋅15⋅25 / (15+25) = 750/40 =18.75mph.
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 5:3."

Let the time taken to go to school be 5x, and the time taken to return be 3x.
The speed going to school is 15 mph

Since distance is the same both ways, we can calculate the speed returning (S return ) as:

S return = Distance /Time returning = 15⋅5x / 3x =25mph.
Thus, the speeds are:
Speed going= 15mph, Speed returning=25mph.

The average speed is:

Average speed= 2*15*25 / (15+25) = 750/40 =18.75mph.
Statement (2) alone is sufficient.

Final Answer: D
​

Let the distance be x miles between school and home.
Therefore home -> school time = x/15
Assume school -> home time = y
Avg speed = 2x/(x/15 + y) = 2/(1/15 + y/x)

Case 1: y = x/15 * 2/5 => y/x is some constant which is sufficient to find the avg speed
Case 2: Again its evident that it provides a direct correlation between y and x which is again suff

Hence (C)

Let the distance to school be d and time taken from home to school be t1, then 15 = d/t1. Let the time taken from school to home be t2.
Calculate avg speed = Total distance/Total time = 2d/(t1+t2)

Statement 1 - t2 = 0.6t1

avg speed = 2d/(t1 + 0.6t1)= (2/1.6)(d/t) = (2/1.6)*15 = 18.75

Hence, this statement is sufficient.

Statement 2 - t1/t2 = 5/3 => 3t1=5t2 => t2 = 0.6t1

avg speed = 2d/(t1 + 0.6t1)= (2/1.6)(d/t) = (2/1.6)*15 = 18.75

Hence, this statement is sufficient.

Answer: D

Distance = d miles

Time (home to school) = t1 hours
Avg. Speed (home to school) = v1 = 15 miles/hour

Time (school to home) = t2 hours
Avg. Speed (school to home) = v2 = x miles/hour

Total Avg. Speed = Total distance / Total time taken = 2/(1/15 + 1/x)
=> We need the value of x

(1) The journey from school to home took her 40% less time than the journey from home to school.
t2 = 60% of t1 = 3/5 * t1
=> x = 25
Statement (1) alone is sufficient.

(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
=> x = 25
Statement (2) alone is sufficient.

Answer: D.

The stem tells us that;

• 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. (Let that speed be x mph)

Then it asks us;

• What was her average speed for the entire round trip?

Average speed = \(\frac{Total distance D}{Total time T}\)

Let the time of journey from school to home =t'
and, the time of journey from home to school = t

Let's look at the statements;

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

t'=0.6t and t=\(\frac{D}{15}\)

Average speed = \(\frac{D+D}{t+t'}\)
or, Average speed = \(\frac{2D}{1.6t}\)
or, Average speed = \(\frac{2D*15}{1.6D}\)
or Average speed = \(\frac{300}{16}\)

Statement 1 is sufficient

(Eliminate BCE)

Let's look at statement 2;

(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 is same as statement 1 and gives us \(\frac{t}{t'}\)=\(\frac{5}{3}\)

Statement 2 is sufficient.

(Eliminate A)

Hence the answer is D