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M27-11

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Bonnie can paint a stolen car in \(x\) hours, and Clyde can paint the same car in \(y\) hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both \(x\) and \(y\) are odd integers, is \(x=y\)?


(1) \(x^2+y^2 \lt 12\)

(2) Bonnie and Clyde complete the painting of the car at 10:30am
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Bonnie and Clyde, when working together, complete the painting of the car in \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\). From that we can get that \(T=\frac{xy}{x+y}\)). Now, if \(x=y\), then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\), since \(x\) is odd, then this time would be \(\frac{odd}{2}\): 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) \(x^2+y^2 \lt 12\). It's possible that \(x\) and \(y\) are odd and equal each other if \(x=y=1\), but it's also possible that \(x=1\) and \(y=3\) (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am. They complete the job in \(\frac{3}{4}\) of an hour (45 minutes), since it's not \(\frac{odd}{2}\) then \(x\) and \(y\) are not equal. Sufficient.

Answer: B
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Now, if x=y, then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd, then this time would be either an integer or \frac{integer}{2}: 0.5 hours, 1 hour, 1.5 hours, ....


I don't get it: odd/2 can not be integer, so how did we get 1 hour?

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New post 25 Sep 2014, 02:00
Boycot wrote:
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Now, if x=y, then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd, then this time would be either an integer or \frac{integer}{2}: 0.5 hours, 1 hour, 1.5 hours, ....


I don't get it: odd/2 can not be integer, so how did we get 1 hour?


You are right. Edited. Thank you.
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New post 09 Nov 2015, 04:17
(1) x2+y2<12. It's possible that x and y are odd and equal each other if x=y=1, but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.
I don't get this....clearly x=y=1 is not possible since we are given that both of them took 45 minutes and if x=y=1 then, they shud have taken 30 minutes...so the only option is x=1 & y=3 (vice versa) hence, sufficient.
Bunuel ..please correct me .

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New post 09 Nov 2015, 04:26
Sumit124431 wrote:
(1) x2+y2<12. It's possible that x and y are odd and equal each other if x=y=1, but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.
I don't get this....clearly x=y=1 is not possible since we are given that both of them took 45 minutes and if x=y=1 then, they shud have taken 30 minutes...so the only option is x=1 & y=3 (vice versa) hence, sufficient.
Bunuel ..please correct me .


Sorry but where does it say that they took 45 minutes to paint the car? It says that they start working at 9:45am.
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New post 09 Nov 2015, 04:33
Hmm....Thanks...I mixed statement two with the question stem...:(...sorry

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New post 12 Nov 2015, 22:50
"simultaneously" is fine but "independently" has confused me - why should we imply that they work on one and the same car but not on 2 identical cars simultaneously and independently?

the 1st para only sets the individual rates.
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"simultaneously" is fine but "independently" has confused me - why should we imply that they work on one and the same car but not on 2 identical cars simultaneously and independently?

the 1st para only sets the individual rates.


"Work simultaneously and independently" means that the they work on unique parts of the job (in this case, if one person paints the left side of the car, the other person WON'T ALSO paint the left side of the car). This is important to the math because the idea is that the entire car will be painted, but it will be done in the fastest way possible (with no duplicated work). It's the type of "legalese" that has to be included in the question, otherwise the correct answer could be debated.
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New post 08 Nov 2016, 03:11
Given:
1. Bonnie takes x hrs to do a job
2. Clyde takes y hrs to do the same job
3. Both start independently at the same time- 9:45am
4. x and y are odd integers

to find: x=y ?

S1: \(x^2+y^2 \lt 12\)
Since we already know that x and y are Odd integers, there's not much left to check for in the above condition and since adding two squares will very soon pass such a small number as 12, we can do the manual work here-
x=1; y=1 => Yes (both equal and sum of squares less than 12)
x=1; y=3 => No (both unequal and still the sum of squares less than 12)

Hence NS

S2: Bonnie and Clyde complete the painting of the car at 10:30am
Quite straightforward.
Basically, they started at the same time and ended at the same time. Therefore, we can certainly infer that the time they took was same- \(x=y\)
Hence S

B
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New post 08 Nov 2016, 19:31
Bunuel

It is a very good question. But, don't you think it is necessary to mention that they worked simultaneously and independently on the same car?
I understand it is important to mention that they worked independently, but it is worded in such a way that reader can interpret that they are working on two different cars which are same/identical.

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New post 08 Nov 2016, 21:21
sveniga4 wrote:
Bunuel

It is a very good question. But, don't you think it is necessary to mention that they worked simultaneously and independently on the same car?
I understand it is important to mention that they worked independently, but it is worded in such a way that reader can interpret that they are working on two different cars which are same/identical.


Here's my take- Even if you infer that '...they are working on two different cars which are same/identical', it is Okay! Because the main thing that needs to be understood is that both do the same amount of work, and that they started at the same time
At least, this is what i understood from the prompt after the first read, please correct me if im wrong.
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New post 08 Nov 2016, 21:41
arhumsid wrote:
sveniga4 wrote:
Bunuel

It is a very good question. But, don't you think it is necessary to mention that they worked simultaneously and independently on the same car?
I understand it is important to mention that they worked independently, but it is worded in such a way that reader can interpret that they are working on two different cars which are same/identical.


Here's my take- Even if you infer that '...they are working on two different cars which are same/identical', it is Okay! Because the main thing that needs to be understood is that both do the same amount of work, and that they started at the same time
At least, this is what i understood from the prompt after the first read, please correct me if im wrong.


We can not conclude that they both did the same amount of work or their rate of work is equal. For example, A can paint 3/4th of the car whereas B can pain 1/4th of the car.

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New post 08 Nov 2016, 21:51
sveniga4 wrote:
arhumsid wrote:
sveniga4 wrote:
Bunuel

It is a very good question. But, don't you think it is necessary to mention that they worked simultaneously and independently on the same car?
I understand it is important to mention that they worked independently, but it is worded in such a way that reader can interpret that they are working on two different cars which are same/identical.


Here's my take- Even if you infer that '...they are working on two different cars which are same/identical', it is Okay! Because the main thing that needs to be understood is that both do the same amount of work, and that they started at the same time
At least, this is what i understood from the prompt after the first read, please correct me if im wrong.


We can not conclude that they both did the same amount of work or their rate of work is equal. For example, A can paint 3/4th of the car whereas B can pain 1/4th of the car.


Yes indeed, i was wrong. We cannot say that both did the same amount of work. That was a really bad choice of words at my end :) Thanks for correcting me.

But, can we not assume that both worked on the same/identical cars?
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Please explain in detail how b is correct
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New post 06 May 2017, 02:42
I approached the question this way:
1) clear

2)
Rate of Bonnie is 1/b, rate of Clyde 1/c. They work together at rate (b+c)/bc. And to paint a car it will take bc/(b+c) hours
as we know that they work for 3/4 hours, we can say bc/(b+c)=3/4. As we know b,c are odd integers, we can infer that b and c are 3 and 1, hence not the same

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New post 15 Aug 2017, 18:53
Perhaps another approach is via inversion: set x=y and check if question stem constraint that x and y are ODD integers holds.

(1)
Insufficient since x and y could be either same or different odd integers

(2)
T = xy/(y+x) = 3/4 hour

Let x=y=2n+1, i.e. representing an odd integer
Thus we have:

(2n+1)^2 / [2(2n+1)] = 3/4
-- cross multiply, reduce... --
16n^2 + 16n + 4 = 12n + 6
8n^2 + 2n - 1 = 0
(8n + 4)(n - 1/4) = 0
n1 = -1/2
n2 = 1/4

Plugging n1 & n2 back into x=y=2n+1 gives us x = 0 hours (invalid) or x = 1.5 hours (valid)
Therefore, if x=y, then x and y cannot both be odd integers according statement 2's parameters
(said another way, if x and y are both odd integers, then x cannot = y)
=> Sufficient

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New post 16 Sep 2017, 04:25
Explanation of part B:

Given that both completed the work in 3/4 hours, we can write it as
1/x+1/y=1/(3/4)............(A)

Lets assume x=y ...........(B)

Using (A) and (B) we get x=3/2

However, from the question stem, x is an odd integer.

Therefore, our assumtion x=y is incorrect.

Correct answer B.

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New post 16 Sep 2017, 04:33
Thanks kishor! That was helpful :)

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