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26 Apr 2026, 12:52
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A warehouse tug can operate either with no carts attached or with several identical carts attached. Attaching carts reduces the tug’s speed, and the speed reduction is directly proportional to the square root of the number of carts attached. The tug’s speed is 18 km/h when 4 carts are attached and 12 km/h when 9 carts are attached. What is the greatest number of carts that can be attached so that the tug can still move?
A. 14
B. 24
C. 25
D. 30
E. 36
A. 14
B. 24
C. 25
D. 30
E. 36
26 Apr 2026, 12:52
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Official Solution:
A warehouse tug can operate either with no carts attached or with several identical carts attached. Attaching carts reduces the tug’s speed, and the speed reduction is directly proportional to the square root of the number of carts attached. The tug’s speed is 18 km/h when 4 carts are attached and 12 km/h when 9 carts are attached. What is the greatest number of carts that can be attached so that the tug can still move?
A. 14
B. 24
C. 25
D. 30
E. 36
Let the tug’s speed with no carts attached be \(V\).
Since the speed reduction is proportional to \(\sqrt{n}\), we can write the tug’s speed with \(n\) carts as speed = \(V - k\sqrt{n}\), where \(k\) is a positive constant.
Using the two given cases:
When \(n = 4\), \(\sqrt{n} = 2\), so \(V - 2k = 18\).
When \(n = 9\), \(\sqrt{n} = 3\), so \(V - 3k = 12\).
Subtract the second equation from the first:
\((V - 2k) - (V - 3k) = 18 - 12\)
\(k = 6\).
Plug back into \(V - 2k = 18\):
\(V - 12 = 18\)
\(V = 30\).
So the speed with \(n\) carts is:
speed = \(30 - 6\sqrt{n}\).
For the tug to still move, its speed must be positive:
\(30 - 6\sqrt{n} > 0\)
\(\sqrt{n} < 5\)
\(n < 25\).
The greatest integer less than 25 is 24, so the maximum number of carts is 24.
Answer: B
A warehouse tug can operate either with no carts attached or with several identical carts attached. Attaching carts reduces the tug’s speed, and the speed reduction is directly proportional to the square root of the number of carts attached. The tug’s speed is 18 km/h when 4 carts are attached and 12 km/h when 9 carts are attached. What is the greatest number of carts that can be attached so that the tug can still move?
A. 14
B. 24
C. 25
D. 30
E. 36
Let the tug’s speed with no carts attached be \(V\).
Since the speed reduction is proportional to \(\sqrt{n}\), we can write the tug’s speed with \(n\) carts as speed = \(V - k\sqrt{n}\), where \(k\) is a positive constant.
Using the two given cases:
When \(n = 4\), \(\sqrt{n} = 2\), so \(V - 2k = 18\).
When \(n = 9\), \(\sqrt{n} = 3\), so \(V - 3k = 12\).
Subtract the second equation from the first:
\((V - 2k) - (V - 3k) = 18 - 12\)
\(k = 6\).
Plug back into \(V - 2k = 18\):
\(V - 12 = 18\)
\(V = 30\).
So the speed with \(n\) carts is:
speed = \(30 - 6\sqrt{n}\).
For the tug to still move, its speed must be positive:
\(30 - 6\sqrt{n} > 0\)
\(\sqrt{n} < 5\)
\(n < 25\).
The greatest integer less than 25 is 24, so the maximum number of carts is 24.
Answer: B
29 Apr 2026, 02:50
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original speed=S
reduced speed=S1
speed ruduction= S-S1
Acc to Question,
S-S1=k*(root(n))
statement 1; S-18=k*(root(4)) i.e S-18=2k
statement 2; S-12=k*(root(9)) i.e S-12=3k
solving above equations we get k=6, S=30
30-S1=6*(root(n))
for (root(n)) to have highest value , S1 should have smallest value
lets assume S1=0
30=6 *(root(n))
5= (root(n))
25=n
but s1 cant be zero as the question says tug should still be moving so the next value smaller than 25 is 24. And we have our answer.
reduced speed=S1
speed ruduction= S-S1
Acc to Question,
S-S1=k*(root(n))
statement 1; S-18=k*(root(4)) i.e S-18=2k
statement 2; S-12=k*(root(9)) i.e S-12=3k
solving above equations we get k=6, S=30
30-S1=6*(root(n))
for (root(n)) to have highest value , S1 should have smallest value
lets assume S1=0
30=6 *(root(n))
5= (root(n))
25=n
but s1 cant be zero as the question says tug should still be moving so the next value smaller than 25 is 24. And we have our answer.
