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Bunuel
calreg11
Abby and Bobby type at constant rates of 80 words per minute and 60 words per minute, respectively. Bobby begins typing before Abby and has typed 600 words when Abby begins typing at 1:30 pm. If they continue typing at their respective rates, at what time will Abby have typed exactly 200 more words than Bobby?

A. 1:40 PM
B. 1:50 PM
C. 2:00 PM
D. 2:10 PM
E. 2:20 PM

Say time needed for Abby to type 200 more words than Bobby is t. In that time she would type 80t words and Bobby would type 60t words.

Now, total words typed by Bobby would be 600+60t and we want that number to be 200 less than 80t: 600+60t=80t-200 --> t=40.

1:30 PM + 40 minutes = 2:10 PM.

Answer: D.


Nice explanation Bunuel... ur methods are most of the time the simplest and easier-to-follow..thanks for that once again
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Lets take the time required as 't'

So, words typed by Abby-words typed by Bobby = 200
ie. 80t - (60t+600)=200

80t-60t-600=200
20t=800
t=40 min

1:30 PM +40 Min = 2:10 PM

D
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This question can be solved easily using relative speed
Relative speed of Abby over booby = 20 words/min
Extra work needed to be done by abby = 600+ 200 = 800 words
Time take by abby to write extra 800 words = 800/20 = 40 min
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Output of B = Output of A - 200

60t + 600 = 80t - 200
20t =800

t = 40 minutes
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B's elapsed time for typing 600 words: 600/60 = 10 min (No. of words typed by B/Rate of B)
A can type 800 words every 10 min; ( Rate of A x Time = 80 x 10)

At 1:30 if A also starts typing with 800 words/min.


Number of words typed by A and B in 10 min intervals
At 1:20
A=0; B=0

At 1:30
A=0; B=600

At 1:40
A=800; B=1200

At 1:50
A=1600; B=1800

At 2:00
A= 2400; B= 2400

At 2:10
A=3200; B=3000
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This one is a classic Rate work and Rate distance time problem.

Rate(A) = 80
Rate(B) = 60

B starts at 01:30 and completes 600 words.

To be 200 words ahead of B, A will have to cover the distance of 600 + 200.

Total Distance = 800
Total speed = Relative speed = 80-60

800/20 = 40 minutes

01:30 + 40 = 02:10
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Let T be the time after 1:30 PM.

In this time if Bobby typed 'x' words, Abby should have typed 600+x+200 i.e. '800+x' words.

Time is same so Words_bobby/Rate_Bobby = W_Abby/Rate_Abby.

x/60 = 800+x/80 => x = 2400 words.

Therefore T = 2400/60 or 3200/80 which gives T = 40 mins = 2:10 PM
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Hey,


Difference between their reading speeds: 20 Words p/m

Abby already typed 600 words
So to cover abby + type 200 more words
Bobby will have to type 800 words

800/20 = 40 mins
1:30 +40 mins = 2:10 pm
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If Abby types at 80wpm then in 'x' mins words typed => 80x
Similarly, Bobby will type => 60x words in 'x' mins

so if you want Abby to type 200 more than Bobby in a given time where Bobby had already started with a head start of 600 words then the equation would look as below:

80x = 600 + 60x + 200 (600 words HeadStart & 200 more words to be typed by Abby in the future)
=> 20x = 800
=> x = 40mins

if started at 1:30, thenit should conclude by 2:10pm!

D. 2:10 PM
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80x = 600+ 200 + 60x
20x = 800
x= 40 min after 1:30 i.e. 2:10
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