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Retired Moderator
Joined: 16 Apr 2020
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WE:Education (Education)
Circumference of the front wheel of a cart is 40ft long and that of
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26 Mar 2021, 08:56
26 Mar 2021, 08:56
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Question Stats:
43% (02:20) correct
57% (02:24) wrong based on 200 sessions
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Circumference of the front wheel of a cart is \(40ft\) long and that of the rear wheel is \(48ft\) long. The front wheel makes \(5\) more revolutions than the rear wheel.
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Distance travelled by the cart |
\(500yds\) |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Circumference of the front wheel of a cart is 40ft long and that of
[#permalink]
27 Mar 2021, 07:11
27 Mar 2021, 07:11
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Explanation:
Both the wheels will travel the same distance irrespective of the number of revolutions
We know,
\(2πrn = d\)
\(Cn = d\)
\(n = \frac{d}{C}\)
Where, \(C\) is circumference of the wheel, \(r\) is the radius of the wheel, \(d\) is the distance travelled by the cart, and \(n\) is number of revolutions.
We have been given, \(n_f - n_r = 5\)
\(\frac{d}{40} - \frac{d}{48} = 5\)
\(48d - 40d = 5(40)(48)\)
\(8d = 5(40)(48)\)
\(d = 1200ft\)
Col. A: \(1200ft\)
Col. B: \(500yds\)
Col. A: \(1200ft\)
Col. B: \((500)(3) = 1500ft\)
Hence, option B
Both the wheels will travel the same distance irrespective of the number of revolutions
We know,
\(2πrn = d\)
\(Cn = d\)
\(n = \frac{d}{C}\)
Where, \(C\) is circumference of the wheel, \(r\) is the radius of the wheel, \(d\) is the distance travelled by the cart, and \(n\) is number of revolutions.
We have been given, \(n_f - n_r = 5\)
\(\frac{d}{40} - \frac{d}{48} = 5\)
\(48d - 40d = 5(40)(48)\)
\(8d = 5(40)(48)\)
\(d = 1200ft\)
Col. A: \(1200ft\)
Col. B: \(500yds\)
Col. A: \(1200ft\)
Col. B: \((500)(3) = 1500ft\)
Hence, option B
General Discussion
Re: Circumference of the front wheel of a cart is 40ft long and that of
[#permalink]
26 Mar 2021, 10:48
26 Mar 2021, 10:48
1
48/40 = 1.2. Thus, for every one rotation of the larger wheel, the smaller one will rotate 1.2 times. If the cart travels 1500 ft (aka 500 yards), the large wheel will rotate 1500/48 = 31.25 times. We can then find the number of rotations of the smaller wheel by calculating (1.2)(31.25) = 37.5 times. Since the smaller wheel would need to rotate 6.25 times more than the larger one to move the cart forward 1500 feet and we know it only rotated 5 times more, we know the cart must have moved less than 1500 feet.
Imo, the answer is B.
Imo, the answer is B.
