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Quant Question of the Day (7/25/09)

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Manager
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Joined: 16 Feb 2009
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Schools: Anderson FEMBA
Quant Question of the Day (7/25/09) [#permalink]

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New post 27 Jul 2009, 19:56
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The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the rear wheel?
A. 20 ft
B. 25 ft
C. 750 ft
D. 900 ft
E. 1000 ft

Kudos [?]: 141 [0], given: 1

Senior Manager
Senior Manager
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Joined: 18 Jun 2009
Posts: 356

Kudos [?]: 78 [0], given: 15

Location: San Francisco
Schools: Duke,Oxford,IMD,INSEAD
Re: Quant Question of the Day (7/25/09) [#permalink]

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New post 27 Jul 2009, 21:41
Let x be the revolutions made by the rear wheel

therefore 30(x+5) = 36x ==> 30x+150 - 36x

== > x = 25

therefore 36x = 36*25 = 900


Answer is D

Kudos [?]: 78 [0], given: 15

Senior Manager
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Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 150 [0], given: 6

Re: Quant Question of the Day (7/25/09) [#permalink]

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New post 27 Jul 2009, 23:25
AndersonBound wrote:
The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the rear wheel?
A. 20 ft
B. 25 ft
C. 750 ft
D. 900 ft
E. 1000 ft



Its D,

Let X be no. of the revolutions made by rear wheel, then X+ 5 are the no. of revolutions made by front wheel.

So X x36 = (X+ 5)30
on solving we get X =25
Hence Distance travelled by cart = 36 x 25 = 900 ft

Kudos [?]: 150 [0], given: 6

Manager
Manager
avatar
Joined: 16 Feb 2009
Posts: 57

Kudos [?]: 141 [0], given: 1

Schools: Anderson FEMBA
Re: Quant Question of the Day (7/25/09) [#permalink]

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New post 28 Jul 2009, 13:36
Ahhhhh, I see my mistake. I was multiplying 25 revolutions by 30 (vs. 36). Duh. Yet again, another silly mistake.

And yes, OA: D.

Kudos [?]: 141 [0], given: 1

Re: Quant Question of the Day (7/25/09)   [#permalink] 28 Jul 2009, 13:36
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Quant Question of the Day (7/25/09)

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