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Deep Dive into the options,
Since Beta takes less time, it will have higher speed. I started with Alpha's speed = 25, which gave k = 16 (400/25) while Beta's speed as 40, which gave k - 2.5 = 10, which is not consistent. Hence I tried Alpha's speed = 32, which gave me k = 12.5 which is consistent when k - 2.5 =10

Hence, Alpha's speed = 32, Beta's speed = 40

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Two cargo ships, Ship Alpha and Ship Beta, transport goods along a 400-nautical-mile channel. Ship Alpha makes the journey in k hours. Ship Beta makes the same journey in (k - 2.5) hours.

In the table, select for Ship Alpha a possible average speed (in nautical miles per hour) and select for Ship Beta a possible average speed (in nautical miles per hour) that would be jointly consistent with the given information. Make only two selections, one in each column.
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We have to select avg speeds consistent with 400 nautical miles journey where ship beta is 2.5 hours faster than ship alpha (k-2.5)

Time = Distance / Speed

If Ship Alpha speed = 32 , Time = 400/32 = 12.5 hours
If Ship Beta speed = 40 , Time = 400/40 = 10 hours.
difference is exactly 2.5 hours

When you will try all other options from the table exact time diff will be something else not 2.5 hours to be exact . So the correct combination is 32 and 40 .
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a: Alpha speed
b: Beta speed

400/a - 400/b = 2.5
ab/(b-a) = 400/2.5 = 160
ab=160(b-a)=160b-160a
a=160b/(b+160)

Testing values: b=40 and a=32

Ship Alpha=32
Ship Beta=40
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Given -> ship A speed = 400/k and ship B speed = 400/(k - 2.5)
looking at the options - let's pick which is will divide 400 and start of with mid -> 40
putting for B speed = 400/(k - 2.5) = 40 => k = 12.5
substituting this in speed of A = 400/12.5 = 32 -- which is one of the options
hence 32, 40 are jointly consistent with the ask.
Note: In similar type of problems, if the combo does not work out, then move up and down to get to the solution.
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Distance D = 400 nautical miles
Using the standard formula for speed,
Speed of alpha ship = 400/k
Speed of beta ship = 400/(k-2.5)

Using trial and error method,
A)
let speed of alpha = 25, then k = 400/25 = 16
hence speed of beta = 400/(13.5) = 29.6 - unavailable

B)
let speed of alpha = 32, then k = 400/32 = 12.5
then speed of beta = 400/(12.5-2.5)=40
available in options

Hence Ship alpha travels at 32 nautical miles per hour and Ship beta travels at 40 nautical miles per hour

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Two cargo ships, Ship Alpha and Ship Beta, transport goods along a 400-nautical-mile channel. Ship Alpha makes the journey in k hours. Ship Beta makes the same journey in (k - 2.5) hours.

In the table, select for Ship Alpha a possible average speed (in nautical miles per hour) and select for Ship Beta a possible average speed (in nautical miles per hour) that would be jointly consistent with the given information. Make only two selections, one in each column.
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We are given,
Distance = 400
Ship Alpha time = k hrs
=>Ship Alpha speed Va= 400/k

Ship Beta time = k -2.5 hrs
=> Ship Beta speed Vb= 400 / k-2.5

From above speeds we can say that,
Va < Vb

After substituting multiple values from the given table, we see that,
For,
Va = 32
k = 400/32 = 12.5

=> Vb = 400 / (12.5 - 2.5) = 400/10 = 40

These values are present in the table and satisfy the requirements

Ship Alpha = 32
Ship Beta = 40
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Try value for A ship: 400/32 = 12.5
Try value for B ship : 400/14 = 10
diff =2.5

logic, as ship a speed is slow compare to B, hence take smaller value in denominator for A ship and greater value for denominator in B, formula of speed=Dist/speed
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400=k*vA -> k=400/vA
400=(k-2.5)*vB

400=(400/vA-2.5)*vB
400/vB = 400/vA - 2.5
400/vA - 400/vB = 2.5

400(vB-VA)/(vA*vB) = 2.5
vA*vB/(vB-vA) = 400/2.5 = 160
vA*vB = 160vB-160vA
vB(160-vA)=160vA
vB=160vA/(160-vA)

It works with vA=32 and vB=40

Ship Alpha=32 and Ship Beta=40
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so for ship alpha equation will be 400/k = a
and for Beta equation will be 400/(k-2.5) = b

We need equation between a and b from above 2 so upon solving it will be

160a/(160-a) = b

Now doing hit and trial on above option

a = 32 and b = 40 solve the equation
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Package/hr = 60*60/50 = 72

Package per day = package/hr*hour per day = 72*8 = 576

Package per week = 576*5 = 2880

Column1: 576
Column2: 2880


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Two cargo ships, Ship Alpha and Ship Beta, transport goods along a 400-nautical-mile channel. Ship Alpha makes the journey in k hours. Ship Beta makes the same journey in (k - 2.5) hours.

In the table, select for Ship Alpha a possible average speed (in nautical miles per hour) and select for Ship Beta a possible average speed (in nautical miles per hour) that would be jointly consistent with the given information. Make only two selections, one in each column.
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time=length/speed

timeAlpha = 400/speedAlpha
timeBeta = 400/speedBeta

timeAlpha - timeBeta = 2.5
400/speedAlpha - 400/speedBeta = 2.5
160 * (1/speedAlpha - 1/speedBeta) = 1

speedAlpha * speedBeta/(speedBeta - speedAlpha) = 160
speedAlpha * speedBeta = 160 * (speedBeta - speedAlpha)

Substituting with the values in the table only speedAlpha=32 and speedBeta=40 fit.

Ship Alpha=32
Ship Beta=40
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Clearly ship beta must be faster than ship aplha.

Time = distance / speed
For alpha time , Ta => 400 / speed a
For Beta speed , Sb = 400/Ta-2.5

using options when speed a = 32 => Ta =12.5
hence sb = 400/10 = 40

Hence speed of :
Ship alpha = 32
Ship beta = 40
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The difference of times is 2.5 hours

400/(Ship Alpha) - 400/(Ship Beta) = 2.5

1/(Ship Alpha) - 1/(Ship Beta) = 2.5/400 = 1/160

((Ship Beta)-(Ship Alpha))/((Ship Alpha)*(Ship Beta)) = 1/160
160*((Ship Beta)-(Ship Alpha)) = (Ship Alpha)*(Ship Beta)

As 160=2^5*5, the product of the two speeds must have 2^5*5 is its factorization. For example, 50 and 25 is not possible.

Tested the values in the table the solution is:

Ship Alpha=32 and Ship Beta=40
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Ship Alpha speed is Va (Time k hour) and Beta is Vb (Time K-2.5 hours)
Distance is 400 m
400/Va - 400/Vb = 2.5
Va = 32 & Vb=40
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