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A and B ran, at their respective constant rates, a race of 480 m. In

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A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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New post 11 Oct 2009, 18:14
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A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?

(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-and-b-ran-at-their-respective-constant-rates-a-race-of-480-m-in-106921.html
[Reveal] Spoiler: OA

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Kudos [?]: 10 [0], given: 0

Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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New post 12 Oct 2009, 03:55
solve these two simultaneous equations

432/b - 480/a = 6 - I
480/a - 336/b = 2 - II

I - II gives you
so we get 96/b = 8
therefore b = 12

Hence answer is A
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Kudos [?]: 71 [0], given: 17

Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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New post 12 Oct 2009, 07:37
Come again. WHere'd you get the 6 and the 2 from? Please help me out.
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Kudos [?]: 79421 [0], given: 10016

Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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New post 12 Oct 2009, 09:26
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Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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New post 12 Oct 2009, 09:35
1/10th of a minute = 6 seconds
Similarly 1/30th of a minute is 2 seconds
Also in first race A reaches before B whereas in second race B reaches before. Hence the order of equations
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Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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New post 09 May 2011, 03:39
432/Sb -480/Sa = 6

480/Sa - 336/Sb = 2

gives Sb = 12
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Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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Kudos [?]: 79421 [0], given: 10016

Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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New post 24 Jul 2016, 08:18
Bunuel wrote:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?

(A) 12
(B) 14
(C) 16
(D) 18
(E) 20


Let \(x\) be the speed of B.

Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

\(\frac{480-48}{x}-6=\frac{480-144}{x}+2\) --> \(x=12\).

Answer: A.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-and-b-ran-at-their-respective-constant-rates-a-race-of-480-m-in-106921.html
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A and B ran, at their respective constant rates, a race of 480 m. In   [#permalink] 24 Jul 2016, 08:18
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