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GMAT Club World Cup 2026 (Day 9): A courier traveled from Hillford to

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LibrarianTurtle

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17 Jul 2026, 06:50

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First Leg: $\frac{D1}{T1} =\frac{40km}{h}$
Second Leg:$\frac{ D2}{T2} =\frac{ 60 km}{h}$
Third Leg: $\frac{D3}{T3} = \frac{80Km}{H}$

Statement 1: T1 = 3K , T2 =2k , T3 =1K total time of travel = 6K
$D1=40*3k D2=60*2k D3=80*1k $
$Total distance= \frac{320K}{6K}$
Statement 1 = Sufficient

Statement 2: D1= 3K D2= 3K D3= 2K
$T1=\frac{3K}{40} T2=\frac{3K}{60 }T3=\frac{2K}{80}$We know total distance and total time of travel so we can calculate avg. speed no need to do whole calculationStatement 2 = Sufficient

Ans. (D)
IMO

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first leg: 40 km/h
second leg: 60 km/h
third leg: 80 km/h

average = (total length)/(total time)

(1)
time ratio 3:2:1
total time = 3k+2k+k = 6k

total length = 40*3k + 60*2k + 80*k = 120k + 120k + 80k = 320k

average = 320k/6k = 160/3

Sufficient

(2)
length ratio 3:3:2
total length = 3k+3k+2k = 8k

total time = 3k/40 + 3k/60 + 2k/80 = (18k + 12k + 6k)/240 = 36k/240 = 3k/20

average = 8k/(3k/20) = 160/3

Sufficient

IMO D

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IMO D
both statements are sufficient
I )
1st leg == 40 km/ hr
2nd = 60
3rd = 80
d1+d2+d3 / t1+t2+t3 = avg speed

t1:t2: t3 = 3:2:1
d1 40. 3k
d2 = 60 .2k
d3 = 80 . k
avg = 320 k / 6k --- unique

statement 2
d1: d2:d3 = 3:3:2
t1 = 3x/40
t2= 3x/ 60
t3 = 2x / 80
avg speed = 8x / 3x/20 = 160 / 3
unique

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Speeds: 40, 60, 80

(1) Times 3:2:1 -> Total = 3a+2a+a = 6a

Distance: 40*3a+60*2a+80*a = 320a

Mean = 320a/6a = 320/6 = 160/3 km/h

Condition sufficient

(2) Distances 3:3:2 -> Total = 3a+3a+2a = 8a

Time: 3a/40+3a/60+2a/80 = 3a/40+a/20+a/40 = 6a/40 = 3a/20

Mean = 8a/(3a/20) = 160/3 km/h

Condition sufficient

Answer D

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Average speed total trip?

(1) Ratio of the times is 3:2:1, so times are 3i, 2i and i with i=integer

t = t1+t2+t3 = 3i+2i+i = 6i

l = l1+l2+l3 = 3i*40 + 2i*60 + i*80 = 320i

average speed = l/t = 320i/6i = 160/3

Condition (1) is sufficient

(2) Ratio of the lengths is 3:3:2, so lengths are 3i, 3i and 2i with i=integer

l = l1+l2+l3 = 3i+3i+2i = 8i

t = t1+t2+t3 = 3i/40 + 3i/60 + 2i/80 = 6i/40 = 3i/20

average speed = l/t = 8i/3i * 20 = 160/3

Condition (2) is sufficient

The answer is D

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(1) if ratio of times is 3:2:1 then times=3t+2t+t=6t

distances = 40*3t+60*2t+80t = 320t

Speed = 320t/6t = 160/3 km/h

Sufficient

(2) if ratio of distances is 3:3:2 then distances=3d+3d+2d=8d

times = 3d/40+3d/60+2d/80 = 3d/40+d/20+d/40 = 6d/40 = 3d/20

Speed = 8d/(3d/20) = 160/3 km/h

Sufficient

The correct answer is D

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3 legs of journey

First leg, avg speed = 40, distance = d1, time = t1
Second leg, avg speed = 60, distance = d2, time = t2
Third leg, avg speed = 80, distance = d3, time = t3

Total avg speed for trip = total distance/ total time = (d1+d2+d3)/(t1+t2+t3)

Statement A : t1:t2:t3 = 3:2:1 => t1 = 3x, t2 = 2x, t3 = x (let x be a constant)

D = S x T => d1 = 40x t1 = 60 x 3x = 120x
d2 = 60 x t2 = 60 x 2x = 120x
d3 = 80 x t3 = 80 x x = 80x

Avg speed = (120x + 120x +80x)/(3x+2x+x). X cancels out so avg speed can be calculated. Statement A alone is sufficient

Statement B : d1:d2:d3 = 3:3:2 =>d1 =3x, d2 =3x, d3 =2x (let x be a constant)

D = S x T = > T = D/S

T1 = d1/s1 = 3x / 40, similarly t2 = 3x / 60, t3 = 2x/80

Average speed = (d1+d2+d3)/(t1+t2+t3) = (3x+3x+2x)/(3x/40+3x/60+2x/80) X gets cancelled out and average speed can be calculated.

Statement B alone is sufficient

Each statement alone is sufficient. Answer D

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My ans is D)
We know the speeds at 3 legs, we need either the time or distance at each leg to find the average speed
1) Stat 1 gives ratio of time 3:2:1

For each leg assuming time taken as 3hr,2hr and 1 hr will give us the distance at each leg and thus the total distance. We have total distance and total time (6hrs)- SUFFICIENT

2) Stat 2 gives ratio of distance, assuming the distances as 300km, 300km and 200km we can find the time and total time- SUFFICIENT

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17 Jul 2026, 09:59

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avg speed = total dist / total time

(1) time ratio = 3:2:1

take times as 3h, 2h, 1h.

dist = 40(3) + 60(2) + 80(1)

= 120 + 120 + 80 = 320

total time = 6h

avg speed = 320/6

can find exact ans.

suff

(2) dist ratio = 3:3:2

take dists as 3x, 3x, 2x.

time = 3x/40 + 3x/60 + 2x/80

total dist = 8x

so avg speed can also be found.

suff

each stmt alone is sufficient.

ans: 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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17 Jul 2026, 10:13

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In average speed formula, time are weights

(A) time ratio is given, we can derive average speed which is based on using time ratio as weights. Hence, Sufficient

(B) distance ratio is given and actual speeds too. Using D/S formula we can derive time ratio - similar reasoning as A from here. This is sufficient too.

(D) is the answer

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31 Jul 2026, 11:07

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V(avg) = (d1+d2+d3)/(t1+t2+t3)

We know v1, v2, v3
We know V=d/t
Since we know V for each leg, to find out the d or t for each leg we need information on one or the other.

I) This gives us time information for each leg which means we can solve for distances of each leg. This means we have all we need to solve for the average.
SUFF.

II) This gives us distance information which can be used to solve for t of each leg.
SUFF.

MrMaxMan

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This one got me but I see it now in this way

I don’t need the exact distances or times because they can scale up or down together, while the fixed speed keeps each distance tied to its corresponding time. So the ratio is enough because it tells me the relative amount of distance and time for each leg, and any scaling of the whole trip leaves the overall average unchanged. Therefore, either the time ratio or the distance ratio is sufficient, so the answer is D. There are an infinite number of distances and times that satisfy the speed ratio, but if you want average speed, the distances and times all come to the same total average because they all move up in scale with each other.