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Anne and Beth will participate in a sack race (In a sack race, people

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Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 04 Oct 2014, 00:59
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Question Stats:

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Anne and Beth will participate in a sack race (In a sack race, people hop to reach the finish line). In the time that Anne takes 3 hops, Beth takes 4 hops but the distance covered by Anne in 4 hops is equal to distance covered by Beth in 5 hops. What is the ratio of Anne’s speed: Beth’s speed?

A. 3 : 5
B. 12 : 20
C. 15 : 16
D. 1 : 1
E. 5 : 3

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Re: Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 04 Oct 2014, 04:46
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aadikamagic wrote:
Anne and Beth will participate in a sack race (In a sack race, people hop to reach the finish line). In the time that Anne takes 3 hops, Beth takes 4 hops but the distance covered by Anne in 4 hops is equal to distance covered by Beth in 5 hops. What is the ratio of Anne’s speed: Beth’s speed?

A. 3 : 5
B. 12 : 20
C. 15 : 16
D. 1 : 1
E. 5 : 3


The distance covered by Anne in 4 hops is equal to distance covered by Beth in 5 hops --> 4*(distance of Ann's 1 hop) = 5*(distance of Beth's 1 hop).

The time that Anne takes 3 hops, Beth takes 4 hops --> time for Ann to take 12 hops --> 3*(time for Ann's 1 hop) = 4*(time for Beth's 1 hop).

Divide one by another:

4/3*(distance of Ann's 1 hop)/(time for Ann's 1 hop) = 5/4*(distance of Beth's 1 hop)/(time for Beth's 1 hop);

4/3*(Anne's speed) = 5/4*(Beth's speed);

4/3*(Anne's speed)/(Beth's speed) = 15/16.

Answer: C.
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Re: Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 04 Oct 2014, 10:01
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aadikamagic wrote:
Anne and Beth will participate in a sack race (In a sack race, people hop to reach the finish line). In the time that Anne takes 3 hops, Beth takes 4 hops but the distance covered by Anne in 4 hops is equal to distance covered by Beth in 5 hops. What is the ratio of Anne’s speed: Beth’s speed?

A. 3 : 5
B. 12 : 20
C. 15 : 16
D. 1 : 1
E. 5 : 3


Let time taken by Anne and beth to complete 3 and 4 hops respectively be 't'

Therefore time taken by Anne & Beth to do 1 hop is 't/3' & 't/4'

Let distance covered in Anne & beth's 4 & 5 hops respectively be 'd'

There distance covered by Anne & beth in 1 hop is 'd/4' & 'd/5'

Anne's speed: Beth's speed = [d/4] / [t/3] : [d/5] / [t/4] = [3/4]*[d/t] : [4/5] * [d/t] = 15:16
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Re: Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 04 Oct 2014, 11:32
aadikamagic wrote:
Anne and Beth will participate in a sack race (In a sack race, people hop to reach the finish line). In the time that Anne takes 3 hops, Beth takes 4 hops but the distance covered by Anne in 4 hops is equal to distance covered by Beth in 5 hops. What is the ratio of Anne’s speed: Beth’s speed?

A. 3 : 5
B. 12 : 20
C. 15 : 16
D. 1 : 1
E. 5 : 3


Since A & B cover the same distance in 4 and 5 hops respectively, we can say that distance covered in 4 hops by A = distance covered in 5 hops by B = x
Distance covered by A in 1 hop = x / 4
Distance covered by B in 1 hop = x / 5
Now, A and B make hops in the ratio of 3 : 4 for a given time interval.

Thus distance covered by A in 3 hops = 3x/4
Distance covered by B in 4 hops = 4x/5

Since these above mentioned distances are covered in a stipulated time, so their ratios will be equal to the ratios of their speed

Ratio of A's speed to B's Speed = 3x/4 : 4x/5 = 15 / 16
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Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 05 Oct 2014, 02:50
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aadikamagic wrote:
Anne and Beth will participate in a sack race (In a sack race, people hop to reach the finish line). In the time that Anne takes 3 hops, Beth takes 4 hops but the distance covered by Anne in 4 hops is equal to distance covered by Beth in 5 hops. What is the ratio of Anne’s speed: Beth’s speed?

A. 3 : 5
B. 12 : 20
C. 15 : 16
D. 1 : 1
E. 5 : 3


My approach

Distance covered by Anne in 4 hops = distance covered by Beth in 5 hops

Premise #1: Equal distance

Assume Distance = 10 (Plug in always helps)
=> 1 hop of Anne = 10/4 = 2.5
=> 1 hop of Beth = 10/5 = 2

Premise #2: Equal time

Time Anne took 3 hops = t
=> \(V1 = D1/t = (2.5*3)/t\)

Time Beth took 4 hops = t
=> \(V2 = D2/t = (2*4)/t\)

=> \(V1/V2 = 7.5/8 = 15/16\)

Ans: C
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Re: Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 16 Jun 2016, 06:21
1
aadikamagic wrote:
Anne and Beth will participate in a sack race (In a sack race, people hop to reach the finish line). In the time that Anne takes 3 hops, Beth takes 4 hops but the distance covered by Anne in 4 hops is equal to distance covered by Beth in 5 hops. What is the ratio of Anne’s speed: Beth’s speed?

A. 3 : 5
B. 12 : 20
C. 15 : 16
D. 1 : 1
E. 5 : 3


Let the time Anne takes for 3 hops and Beth takes for 4 hops = x

Anne takes x mins for 3 hops, so she will take 4x/3 mins for 4 hops

Beth takes x mins for 4 hops, so she will take 5x/4 mins for 5 hops.

We know distance= Speed *time (distance covered by Anne and Beth is same)

S1T1=S2T2
S1*4x/3= S2*5x/4

S1/S2= 15/16

C is the answer
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Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 17 Jun 2016, 18:10
anne's time per hop=4/3 of beth's
anne's distance per hop=5/4 of beth's
anne's speed=(5/4)/(4/3)=15/16 of beth's
ratio of anne's speed to beth's=15:16
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Re: Anne and Beth will participate in a sack race (In a sack race, people  [#permalink]

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New post 30 Sep 2018, 04:51
Attachment:
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Attachment:
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AB2.jpg [ 79.02 KiB | Viewed 250 times ]


Speed = \(\frac{Distance}{time}\)

Speed of Anne from aforesaid is \(\frac{20 Miles}{16 minutes}\) = \(\frac{5}{4}\)

Similarly Speed of Beth from aforesaid is \(\frac{20 Miles}{15 minutes}\) = \(\frac{4}{3}\)

Ratio of speed of ANNE:BETH = \(\frac{5}{4}\) Divided by \(\frac{4}{3}\)

or \(\frac{5}{4}\) * \(\frac{3}{4}\) = \(\frac{15}{16}\)

Therefore correct answer is C
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Re: Anne and Beth will participate in a sack race (In a sack race, people &nbs [#permalink] 30 Sep 2018, 04:51
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