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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 10:14
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t = t1, t2, t3
d = 40t1, 60t2, 80t3
total d/total t??
S1: t1:t2:t3 = 3:2:1; 3x, 2x, x
do total d can be written as k.x, t will be 6x, x will cancel, solvable ACD remains
S2: d1:d2:d3 = 3:3:2, so we get t1:t2:t3, as we solved above, solvable. Answer (D)
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 10:28
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Since for 3 consecutive legs, avg speed is given : 40kmph, 60kmph, 80kmph
Avg speed for entire trip = Total dist/Total time
St1 : Let the time resp be 3x, 2x, x
Then Avg speed = (40.3x+ 60.2x+ 80. x)/ 6x = Sufficient
St2 : Let resp lengths = 3x, 3x, 2x
Then Avg speed = {8x/ (3x/40)+(3x/60) + (2x/80)} = Sufficient

Hence Each statement suficient alone .D is the ans
Bunuel
A courier traveled from Hillford to Lakeshore in three consecutive legs, with no stops between the legs. On the first leg, the courier’s average (arithmetic mean) speed was 40 kilometers per hour; on the second leg, the courier’s average (arithmetic mean) speed was 60 kilometers per hour; and on the third leg, the courier’s average (arithmetic mean) speed was 80 kilometers per hour. What was the courier’s average speed for the entire trip?

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.


 


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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 10:33
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Average speed = (40t1 + 60t2 + 80t3)/(t1+t2+t3)

Statement 1 : Ratio of times spent on the first, second, and third legs, respectively, was 3:2:1 = 3x:2x: x
hence avg speed = ( 40*3x + 60*2x+80*x)/(3x + 2x + x) = (120 +120+80)/(3+2+1)
Hence, statement 1 is sufficient

Statement 2 : Ratio of the lengths of the first, second and third legs, respectively, was 3:3:2

i.e. 40t1:60t2:80t3 = 3:2:2 = 3x : 2x : 2x
i.e. t1 = 3x/40
t2 = 2x/60
t3 = 2x/80

avg speed = (3x + 2x + 2x)/(3x/40 + 2x/60 + 2x/80) = (3+2+2)/(3/40 + 2/60 +2/80)
Hence, Statement 2 is sufficient

Hence, each statement alone is sufficient. Answer = D
Bunuel
A courier traveled from Hillford to Lakeshore in three consecutive legs, with no stops between the legs. On the first leg, the courier’s average (arithmetic mean) speed was 40 kilometers per hour; on the second leg, the courier’s average (arithmetic mean) speed was 60 kilometers per hour; and on the third leg, the courier’s average (arithmetic mean) speed was 80 kilometers per hour. What was the courier’s average speed for the entire trip?

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.


 


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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 10:36
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Avg speed is total distance/total time

Statement 1: Let the times be 3x, 2x and x

Hence avg speed is (40 *3x + 60 * 2x + 80*x)/(3x + 2x + x). All the x's will get cancelled. Sufficient

Statement 2:

Let distances equal 3x + 3x + 2x

Time would be 3x/40, 3x/60 and 2x/80

Again the x's will get cancelled Hence sufficient

I ll go with D


Bunuel
A courier traveled from Hillford to Lakeshore in three consecutive legs, with no stops between the legs. On the first leg, the courier’s average (arithmetic mean) speed was 40 kilometers per hour; on the second leg, the courier’s average (arithmetic mean) speed was 60 kilometers per hour; and on the third leg, the courier’s average (arithmetic mean) speed was 80 kilometers per hour. What was the courier’s average speed for the entire trip?

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.


 


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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 10:59
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Here's my solution for this question
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Screenshot_2026-07-16-23-28-34-40_40deb401b9ffe8e1df2f1cc5ba480b12.jpg
Screenshot_2026-07-16-23-28-34-40_40deb401b9ffe8e1df2f1cc5ba480b12.jpg [ 1.27 MiB | Viewed 135 times ]

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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 11:10
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St1: Since the ratio of time taken is 3:2:1, we can assume the time to be 3x:2x:1x (total 6x) respectively (as shown in the attached photo). The distance will then be 120x, 120x and 80x (total 320x). Average speed = 320x / 6x = 53.33 kmph. SUFFICIENT

St2: Here, the ratio of distance travelled is given as 3:3:2. We can assume the distance to be 120x:120x:80x. This will give us speeds same as St 1 and final answer will be 53.33 kmph. SUFFICIENT

Final answer D
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 11:15
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i) treat as weighted average. when finding average of rate such as speed, the denominator of the rate is the weightage. which means the denominator of speed i.e. time is the weightage. and the given speeds of each leg is the statistic. we can easily find the average speed for entire trip.
40(3)+60(2)+80(1)/(3+2+1) = 320/6 = 160/3 = 53.3 km/h sufficient.

ii) ratio of distances = 3:3:2 = 8
average speed = total distance/total time
Let total distance = d
if we can get total time in terms of d, we can find the average speed.
leg 1 time = (3d/8)(1/40) = 3d/320
leg 2 time= (3d/8)(1/60) = d/160
leg 3 time= (d/4)(1/80) = d/320
total time = 6d/320 = 3d/160

average speed = d/(3d/160) = 160/3 = 53.3 km/h sufficient.
D
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 11:22
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speedtimeDistance
L140
L260
L380
Statement-1
STD
L1403t120t
L2602t120t
L380t80t
Avg speed = 320t/6t = 320/6
sufficient
Statement II
Ratio of Distance = 3:3:2 , So taking the same values as the previous statement, d1=120t, d2=120t, d3=80t
Avg speed = 320t/6t = 320/6
sufficient
Ans D
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 11:28
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given: courier travelled in 3 legs without stop
1st leg- avg speed 40km/h
2nd leg- avg speed 60 km/h
3rd leg- avg speed 80 km/h

We have to find if we can find courier's average speed for entire trip?

Lets check both statement one by one.

Statement-1: Ratio of times courier spent on 1st, 2nd and 3rd leg are 3:2:1

Say total time is t

then, 1st leg time is 3/6(t) ; distance travelled for 1st leg=3/6(t)* 40= 20t
2nd leg time is 2/6(t) ; distance travelled for 2nd leg= 2/6(t)*60= 20t
3rd leg time is 1/6(t) ; distance travelled for 3rd leg= 1/6(t)*80= 40/3(t)

total distance travelled = 20t+20t+ 40/3(t)= 160/3(t)

Average speed foe entire trip = total distance travelled/ total time taken = 160/3(t)/t= 160/3 km/h

Statement found sufficient

Statement-2: Ratio of lengths of first, 2nd and 3rd leg are 3:3:2

lets assume total distance as d.

distance for 1st leg= 3/8(d) ; time for 1st leg = (3/8)*(d)/40 = (3/320)*(d)
distance for 2nd leg= 3/8(d) ; time for 2nd leg= (3/8)* (d)/60 = (1/160)*(d)
distance for 3rd leg= 2/8(d) ; time for 3rd leg= (2/8)*(d)/80 = (1/320)*(d)

total time taken = (3/320)*(d)+ (1/160)*(d)+ (1/320)*(d)= (6/320)*(d)

average speed for entire journey= total distance travelled/total time taken= d/(6/320)*(d)= 320/6= 160/3 km/h

Statement-2 is also sufficient

so each statement alone seems sufficient.

My answer choice is -D
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 11:31
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1. t1:t2:t3=3:2:1
t1=3x, t2=2x, t3=x
d1=40*3x, d2=60*2x, d3=80*x
D=d1+d2+d3=320x
T=t1+t2+t3=6x

D/T=320/6....SUFFICIENT

2. d1:d2:d3=3:3:2
d1=3x, d2=3x, d3=2x
t1=3x/40, t2=3x/60, t3=2x/80
T=t1+t2+t3= 3x/40 + 3x/60 + 2x/80
D=d1+d2+d3=8x

both D & T in terms of x..so, D/T is unique value.. SUFFICIENT

Ans D
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 12:00
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Let the total length is L and total time is T. Then average speed = L/T
Now, let leg 1,2 & 3 distance and times be L1, L2, L3 and T1, T2 and T3 respectively.
So, Av Speed Leg 1 = L1/T1 = 40, or L1 = 40T1, or T1 = L1/40. Similarly for leg 2 and 3, L2=60T2, T2=L2/60 & L3=80T3, T3=L3/80. --- eq 1


Since, average speed = (L1+L2+L3)/ (T1+T2+T3) --- eq -2
now consider the statements
A) T1:T2:T3 = 3:2:1
so, T1 = 3x, T2= 2x and T3=x
putting the values in our eq 1 and 2, we get av speed = (40T1+60T2+80T3)/(T1+T2+T3)
we can solve it to get a definitive answer. So A is sufficient.

B) L1:L2:L3 = 3:3:2
so, L1 = 3y, L2 = 3y, L3 = 2y
again, putting the values in eq 1 & 2, we can get a final answer wihtout any variable. Hence this is also sufficient.

Since both A & B are sufficient, hence answer is D.


Bunuel
A courier traveled from Hillford to Lakeshore in three consecutive legs, with no stops between the legs. On the first leg, the courier’s average (arithmetic mean) speed was 40 kilometers per hour; on the second leg, the courier’s average (arithmetic mean) speed was 60 kilometers per hour; and on the third leg, the courier’s average (arithmetic mean) speed was 80 kilometers per hour. What was the courier’s average speed for the entire trip?

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.


 


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Average speed=total distance/total time
three legs speed= 40km/h, 60km/h, 80km/h
Statement 1
time ratio=3:2:1
time= 3t, 2t, t
distance
1st leg, 40(3t) =120t
2nd leg, 60(2t) =120t
3rd leg 80(t)=80t
ttl d=120t+120t+80t=320t
ttl t=3t+2t+t=6t
320t/6t=160/3=53.3333 km/h sufficient

Statement 2
distance ratio 3:3:2
d=3d, 3d, 2d
time 3d/40, 3d/60, 2d/80
ttl time: 3d/40+3d/60+2d/80=(9d+6d+3d)/120=18d/120=3d/20
ttl d: 3d+3d+2d=8d
average speed= 8d/(3d/20)=160/3=53.3333km/h sufficient
ANS: D.
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 12:21
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Bunuel
A courier traveled from Hillford to Lakeshore in three consecutive legs, with no stops between the legs. On the first leg, the courier’s average (arithmetic mean) speed was 40 kilometers per hour; on the second leg, the courier’s average (arithmetic mean) speed was 60 kilometers per hour; and on the third leg, the courier’s average (arithmetic mean) speed was 80 kilometers per hour. What was the courier’s average speed for the entire trip?

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.


 


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Statement 1 - t1:t2:t3 = 3:2:1
let total time of the courier = 6t,
t1 = 3t, t2 = 2t and t3 = t
speed = dist/time
40 = d1/3t => d1 = 120t
60 = d2/2t => d2 = 120t
80 = d3/t => d3 = 80t
Average speed = total dist / total time
avg speed = d1 + d2 + d3 / t1 + t2 + t3
= 120t + 120t + 80t / 6t
= 320t /6t = 320/6 km/hr
SUFFICENT

Staement 2 - d1:d2:d3 = 3:3:2
Total distance = 8d
d1 = 3d, d2=3d, d3 =2d
40 = 3d/t1 => t1 = 3d/40
60 = 3d/t2 => t2 = 3d/60
80 = 2d/t3 => t3 = 2d/80

Avg speed = 8d / (3d/40) + (3d/60) + (2d/80)
SUFFICIENT

Answer - D
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 13:17
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Bunuel
A courier traveled from Hillford to Lakeshore in three consecutive legs, with no stops between the legs. On the first leg, the courier’s average (arithmetic mean) speed was 40 kilometers per hour; on the second leg, the courier’s average (arithmetic mean) speed was 60 kilometers per hour; and on the third leg, the courier’s average (arithmetic mean) speed was 80 kilometers per hour. What was the courier’s average speed for the entire trip?

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.


 


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Correct Answer (D)

Avg Speed = Distance/ Time

Statement 1: We have ratio of time - 3t, 2t, t.

Avg Speed= (40*3T+ 60*2T + 80*T) / (3T + 2T + T) which equals 53.33 km/hr. Sufficient.

Statement 2: We have ratio of distance 3d, 3d, 2d

Avg Speed = (3d + 3d + 2d) / (3d/40 + 3d/60 + 2d/80) which equals to 53.33 km/hr. Sufficient.
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 14:23
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Answer = Each statement alone is sufficient

Given
Hillford 40km/hr60 km/hr80 km/hrLakeshore

Average speed = Total distance (D) / Total Time (T)

Three separate distances T(1)*40=D1, T(2)*60=D2, T(3)*80=D3

Statement 1:
Use ratio multiplier (3x:2x:1x)
40*3x=D1
60*2x=D2
80*1x=D3

Total Distance D1+D2+D3= 320x
Total Time=6x
Average speed 320x/6x

Sufficient

Statement: 2
Use ratio multiplier (3y:3y:2y)
40(T1)=3y
T1=(3y)/40

60(T2)=3y
T2=(3y)/60

80(T3)=2y
T3=(2y)//80

Total Time= T1+T2+T3
Total Distance= 8y
Average speed= 8y / ((3y)/40+(3y)/60+2y/(80)) = 320y/6y

Sufficient
Bunuel
A courier traveled from Hillford to Lakeshore in three consecutive legs, with no stops between the legs. On the first leg, the courier’s average (arithmetic mean) speed was 40 kilometers per hour; on the second leg, the courier’s average (arithmetic mean) speed was 60 kilometers per hour; and on the third leg, the courier’s average (arithmetic mean) speed was 80 kilometers per hour. What was the courier’s average speed for the entire trip?

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.


 


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Answer is D - each statement alone.

This is a question testing our conceptual understanding of rate, distance and time. Let's use a logical approach.

1) We know the rates / speeds for each segment. What we do not know is the distance of each segment, nor do we know how much time he spent on each segment. If we know the ratio of the time spent for each of the segments, even though we do not know the actual amount of time, it doesn't matter because we know how long was spent at each speed per segment relative to each other segment. This allows us to know the total average speed. Another way to see it is if we know the avg. speed formula, that is:

Avg. Speed = Total Distance / Total Time
and; Distance = Time * Rate

We would have (T1*R1+T2*R2+T3*R3)/T1+T2+T3
Through knowing the ratio we essentially know the weight average and can find the answer.

S2) Since we are given the ratios of the distance of each segment, this is also enough to answer the question because we know the rates at which each segment is traveled, and that combined with the ratio of distance is enough to calculate average speed. This is because no matter the scale of the distance the same proportion of time would have to be spent traveling each one, i.e., the same average speed.

Answer A.
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mehtyas
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 18:02
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IMO - Answer should be Option D

The speeds for the 3 legs of the trip are 4km/h, 60km/h and 80km/h, respectively.

We need the overall average speed, which is = Total Distance / Total Time

S1 -> The time being spent on the 2 legs is in the ratio = 3:2:1

Let the times be 3t, 2t and t.

Then the distances are:
  • 40 * 3t = 120t
  • 60 * 2t = 120t
  • 80 * t = 80t

Total distance = 320t and Total Time = 6t

Average speed = 320t/6t =53 1/3 km/h


S1 is sufficient

S2 -> The distances are in the ratio = 3:3:2. Let the actual distances be 3d, 3d and 2d.

The corresponding times are:
  • 3d/40
  • 3d/60
  • 2d/80 = d/40

Total time = 3d/40 +3d/60 + d/40

Total distance = 8d

Average speed = 8d / (3d/20) = 160/3 = 53 1/3 km/h

S2 is sufficient

Answer: Option D - Each statement is alone sufficient.
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 18:27
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Let the first leg distance be d1 and speed traveled is 40 kmph
The second leg distance is d2 and the speed with which it traveled is 60 kmph
The last leg distance is d3 and the speed with which it traveled is 80 kmph
If the avg speed of the travel is = total distance / total time
St1: Time spent in each leg is 3:2:1
d1/40=3t; d2/60=2t and d3/80=t => d1=120t d2=120t and d3= 80t
so Avg speed= d1d2+d3/(3t+2t+t)=120t+120t+80t / 6t = 53.33 kmph
So sufficient
St2:
d1:d2:d3=3:3:2
d1+d2+d3/ tot time= 3d+3d+2d/(3d/40+3d/60+2d/80)=> 8d/(3d/40+d/20+d/40) =53.33 kmph
St 2 is sufficient
So option D
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 19:10
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We have to find the average speed for the entire trip, so we need total distance/total time.
Leg 1 average speed = 40 kmph
Leg 2 average speed = 60 kmph
Leg 3 average speed = 80 kmph

(1)
Let the time spent on Leg 1, 2, 3 be 3x, 2x, x hours respectively.
Leg 1 distance = 40*3x = 120x km
Leg 2 distance = 60*2x = 120x km
Leg 3 distance = 80*x = 80x km
Total distance = 320x km
Total time = 6x hours
Average speed = 320x/6x kmph = 160/3 kmph
Sufficient.

(2)
Let the distance of Leg 1, 2, 3 be 3x, 3x, 2x km respectively.
Leg 1 time = 3x/40 hours
Leg 2 time = 3x/60 = x/20 hours
Leg 3 time = 2x/80 = x/40 hours
Total distance = 8x km
Total time = 3x/20 hours
Average speed = 8x/(3x/20) = 160/3 kmph
Sufficient.

(D)
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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to 16 Jul 2026, 21:35
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Average speed of the courier = d/t

St. 1
The given ratio = 3:2:1 = 3t:2t:1t = 6t (total time)
distance covered = 40*3t + 60*2t + 80*t = 320t
Average speed = 320t/6t
Sufficient

St. 2
length ratio = 3:3:2 = 3d:3d:2d = 8d (total distance)
Time generated = 3d/40 + 3d/60 + 2d/80
= d(3/40 + 3/60 + 1/40)
= d*(6/40) = 3d/20

Speed = 8d/(3d/20) = 160/3 Sufficient

Answer (D) Each statement alone is sufficient.
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