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Given and assumptions:

let there be three legs A, B, C
Averages speeds are given, Sa = 40 kmph, Sb = 60 kmph, Sc = 80 kmph

S1
Ts ratios are given let total Time (T) be T = 3t + 2t + 1t = 6t
then Da = Sa*Ta = 40 (3t) = 120t
similarly Db = 120t
Dc = 80t
total distance = 320t
total avg speed S = D/T = 320t/6t = 160/3 some value but is possible
S1 suff

S2 -
Ds ratios are given
Da = 3d, Db = 3d. Dc = 2d
Total D = 8d
each T can again be Ta = Da/Sa = 3d/40
solving for S, d from numerator (distance) and d from denominator (time) gets cancelled out again is solvable giving us a value of S

So answer - D each alone are sufficient
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In the question stem we are provided with the information of the speeds for the three legs. To get to the average speed for the entire trip we need to either the relative time or the relative distance.

S1: Relative time is given. Through which we can find relative distance and thus we get average speed.
S2: Relative distance is given. Through which we can find relative time and thus we get average speed.

(D) Each statement alone is sufficient
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Answer: D) each statement alone is sufficient

Because we can express the distance in terms of time, and vice versa, both statements alone are sufficient.

Average speed = total distance / total time

Statement (1):
3t:2t:t
Average speed = (40*3t + 60*2t + 80t) / 6t = 160/3

Statement (2):
3d:3d:2d
Average speed = 8d / (3d/40 + 3d/60 + 2d/80) = 160/3
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t=3:2:1
v=40(3)+60(2)+80(1)/3+2+1=160/3 suff

d=3:3:2
v=8/3/40+3/60+2/80=160/3 suff
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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?


total 3 legs and avg speed in each leg
leg 1= 40 kmph
leg 2= 60 kmph
leg 3= 80 kmph

find avg speed of entire trip


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

distance = speed * time
time for each leg 40*3x ; 60*2x ; 80*x
total time = 120x+120x+80x ; 320x
net time is 6x
avg speed = 320/6 ; 160/3 ; sufficient


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

with length ratio given time = distance / speed
time for each leg will be
3x/40 ; 3x/60 ; 2x/80
sum of time = 3x/40 + 3x/60+ 2x/80 ; 3x/20
avg speed will be 3x+3x+2x ; 8x*20/3x ; 160/3 ; sufficient

both statements alone are sufficient ; OPTION D is correct
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Average speed = total dist/total time
statement 1:
ratio of time = 3:2:1
let it be 3 hr, 2hrs, 1 hr
so distance becomes=
40x3, 60x2, 80x1
= 120, 120, 80
total dist= 320
ttoal time = 6 hrs
average speed = 320/6= 160/3
Statement 1 sufficient

Statement 2:
Well I didnt actually had to solve.
Distance ratios were given so I knew I will be able to find time and then like i did with statement 1, we can find average speed.
So statement 2 was sufficient

Hence D
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We are given the speeds for the 3 legs 40, 60, 80, need to find Avg speed

1) = time is in ratio 3:2:1

let times be 3x, 2x , x

d = 120x, 120x, 80x = Avg speed = 320x / 6x SUFFICIENT

2) = d is in ratio 3:3:2

T = 3X / 40, 3X / 60, 2X / 80

8X / (3X/20) SUFFICIENT

IMO D
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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?
Avg speed= (40a+60b+80c)/(a+b+c)

(1) The ratio of the times the courier spent on the first, second, and third legs, respectively, was 3:2:1.
a:b:c= 3:2:1
Avg speed= 40(3x)+60(2x)+80(x)/6x =160/3 kmph
Sufficient
(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.
D1:D2:D3= 3:3:2
Total distance= 8y
a=3y/40 ; b=3y/60 ; c=2y/80
Total time= (3y/40)+(3y/60)+(2y/80)= 3y/20
Avg speed= 8y/(3y/20) = 160/3 kmph
Sufficient

D
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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.
Let the times the courier spent on 1st, 2nd and 3rd legs be 3k, 2k & k hours respectively
Total distance travelled = 40*3k + 60*2k + 80*k = 120k + 120k + 80k = 320k km
Total time taken = 3k + 2k + k = 6k
The courier's average speed for the entire trip = 320k/6k = 160/3 = 53 1/3 km/hr
SUFFICIENT

(2) The ratio of the lengths of the first, second, and third legs, respectively, was 3:3:2.
Let the lengths of 1st, 2nd and 3rd legs be 3k, 3k & 2k respectively
Total distance travelled = 3k + 3k + 2k = 8k km
Total time taken = 3k/40 + 3k/60 + 2k/80 = 3k/40 + k/20 + k/40 = 6k/40 = 3k/20
The courier's average speed for the entire trip = 8k/(3k/20) = 160/3 = 53 1/3 km/hour
SUFFICIENT

IMO D
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Statement I. We are given ratios of time. Let them be 3x, 2x and x

Average Speed for the entire trip= Total Distance / Total Time

Total time= 6x
Total distance = 40*3x + 60*2x + 80x

We can get the speed because x will get cancelled out

Statement II. Distance ratio- 3x, 3x and 2x

Total D= 8x

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

When we do D/T for average speed, x will get cancelled out and get the speed

Option (D)- Each statement alone is sufficient
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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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d1/t1 = 40

d2/t2 = 60

d3/t3 = 80

average of entire trip = (d1 +d2+ d3)/(t1 + t2 + t3)

1) t1 : t2 : t3 = 3 : 2: 1

From the base equations we can represent each of the distances in terms of their time.

From the relationship in statement 1, we can represent each time in one term.

As its a ratio, the term will get cancelled.

2) d1:d2:d3 = 3:3:2

Each of the distance can be represented in one common term.

From the base equations they can be represented in terms of the time.

Hence, the common term will get cancelled out.

This statement is also sufficient.

Option D
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Statement 1

Time ratios are 3:2:1. Let times be 3t, 2t and 1t

Distances are

First 40 x 3t
Second 60 x 2t
Third is 80 x t

Total distance is 120t + 120 t + 80 t = 320 t

Total time. is 3t + 2t + t = 6t

Average speed is 320 t / 6t = 53.3333 km/h

Statement 1 is sufficient


Statement 2

Distance ratio is 3:3:2 so 3d : 3d : 2d

Times are distance over time.
First. 3d/40.
Second 3d/60 = d/20
Third 2d/80 = d/40

total time
3d/40 + d/20 + d/40 = (3d + 2d +d)/40 = 6d/40 = 3d/20

total distance
3d + 3d + 2d = 8d

average speed
8d/(3d/20) = 8 x 20/3 = 160/3 = 53.3333 km/hour

Sufficient

ANSWER. D. BOTH alone are sufficient
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I generally write the average speed formula every time I encounter a question related to it viz:
Total Distance travelled/total time taken = avg speed

S1: the ratio of times is there, let it be 3t, 2t, t for leg 1, 2, 3 respectively.

Speed * time = distance,
So distance of: Leg 1 = 120t , Leg 2 = 120t, leg 3 = 80t
total distance = 320t
and total time is = 6t

So using avg speed formula we will get 320t/6 = 53.33 km/hr (i did not calculate it as I knew it would give me a definitive answer)
SUFFICIENT

S2: using the info given, let 3d, 3d and 2d be leg 1, 2 and 3 distance respectively now we could calculate total time by:
Total time would be time taken by(leg 1 + leg 2 + leg 3)

3d/40 + 3d/60 + 2d/80

and total distance is 8d, hence using average speed formula we could compute a definitive answer of average speed.

Sufficient

IMO D
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Avg. Speed = Total Distance / Total Time Taken

Can also be represented as 40x + 60y + 80z / x + y + z where x,y,z represents the time taken in each of the stretches by the courier.

Now moving to Statement 1:

Ratio of time taken between legs: 3:2:1. Assume total Time taken = 6Hours
Leg 1 - 3Hrs; Leg 2 - 2Hrs; Leg 3 - 1 Hour
Total Distance = 40X3 + 60X3 + 80X1 / 3 + 2 + 1 [Total Time Taken] = 320/6 or 160/3 Kms/Hr -> Hence Statement 1 is sufficient

Statement 2 which provides the Lengths are in ratio 3:3:2

Let's assume Total Distance = 40L1 + 60L2 +80L3 where L1, L2 and L3 are the Distance of each leg. The common factor divisible between 40,60,80 is 480Kms. Assume 480Kms divided in 3:3:2 ratio

L1 = 3/8 X 480= 180 Kms
L2 = 3/8 X 480 = 180 Kms
L3 = 2/8 X 480 = 120 Kms

Thus Avg Speed = Total Distance / Total Time
Total Time = 180/40 + 180/60 + 120/80 = 4.5Hrs + 3Hrs + 1.5Hrs = 9 Hours Total

Again avg. speed = 480/9 = 160/3 Km/Hr

Hence Statement 2 is also sufficient
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Avg Speed = Total distance/ total time.

Statement 1: Courier spent on 1st, 2nd and 3rd leg: 3:2:1
Speed already given 40:60:80
So 40x3 : 60x2: 80x1 = 120 : 120 : 80.
Total dist = 320 in 3 + 2 + 1 hours total so we can calculate the avg speed.
Sufficient.

Statement 2:
Distance ratio given, Similar to statement 1 we can fnd the total time like 3/40:3/80:2/80
reducing to something similar to statement 1 which can be solved.
Sufficient

IMO Ans 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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Statement 1:
Ratio of t1:t2:t3=3:2:1. Let's assume t1=3x, t2=2x, t3=x
That gives us d1=120x, d2=130x, d3=80x
Since we have all the individual time and distances in terms of x, total distance and total time can both be computed in terms of x. On dividing, this would give us a fixed fraction, which would be our average speed. Hence, SUFFICIENT.

Statement 2:
Similar to Statement 1, we can assume distances in terms of x and can then compute the individual lap time in terms of x as well. On summing total distances and total time (which would both be in terms of x), and dividing, we will get a fixed fraction, which will be our average speed. Hence, SUFFICIENT.

Since both Statement 1 and 2 are sufficient on their own, 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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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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(1) Time ratio = 3 : 2 : 1

First leg: 3 hours at 40 km/h = 120 km
Second leg: 2 hours at 60 km/h =120 km
Third leg: 1 hour at 80 km/h = 80 km

Total distance = 320 km
Total time = 6 hours

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

A unique value is found, so (1) is sufficient.

(2) Distance ratio = 3 : 3 : 2

First leg: 3 km at 40 km/h = time = 3/40
Second leg: 3 km at 60 km/h = time = 3/60
Third leg: 2 km at 80 km/h = time = 2/80

Total distance = 8 km

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

= 3/40 + 1/20 + 1/40

= 6/40

= 3/20

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

Again, a unique value is found, so (2) is sufficient.

Option D
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