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D.

lets Say V1 is vol of small cargo and v2 is large cargo
r1 = v1/2.5
r2 = 2r1

x = v2/r2
x= v2/2v1/2.5

x= 5v2/4v1
v2/v1 = 4x/5 SO D
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Manaswita
At the Marseille port, a small cargo ship is loaded in 2.5 hours and a large cargo ship is loaded at double the rate in X hours. In terms of X, how many times does the load of a small cargo ship fit into a large cargo ship?

A. 2X/5
B. 5X/2
C. 5/4X
D. 4X/5
E. 5X/4


Sol:

work done by small ship = A (rate ) * 5/2 (time)= 5A/2

Work done by large ship = B(rate)* T = 2A*X

ratio = Large/small ship = 2AX/(5A/2) = 4X/5
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Rate Time Work

1/2.5 2.5 1
2/2.5 X (2/2.5)X

1* What = 2/2.5 X

What = (2/2.5)X

Hence (D)
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Work = rate *time;

r1 t1 = 2.5 * ra;
r2t2 = X*2 rb;

2.5 * ra = X * rb;

ratio = 4X/5;
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Let Small ship capacity be 'S'
Large ship capacity be 'L'

Form question -- small cargo ship is loaded in 2.5 hours, the rate at which the small ship loaded is S/2.5
-- large cargo ship is loaded at double the rate in X hours, the rate at which the large ship is loaded is L/x

Also, from question -- a large cargo ship is loaded at double the rate , so L/x = S/2.5 *2
L = S * X* 2/5 *2
L= S *(4X/5)
Hence 'D'
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Volume (Small): R*T =2.5r
Volume (Big): R*T = X*2r
Big/Small = 20Xr/25r =4X/5. Answer is D.
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