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The Discreet Charm of the DS 25 Feb 2012, 05:27
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All,

7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange?
(1) None of the customers bought more than 4 oranges
(2) The difference between the number of oranges bought by any two customers is even

In respect to the question above, I assumed that any two of those 19 customer might have bought 5 & 3 oranges and hence I, marked the option insufficient. Bunnel have equated and treated the option in totally different way. I ,lack the skill to convert these sort of condition in to equation.

please can some post or point to the list of similar Word translation sentences and how to convert them in to equation. Im very new to GMAT club so please forgie me if this is the repeated posting.

Thanks,
Vids
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The Discreet Charm of the DS 25 Feb 2012, 05:53
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vidhya16
All,

7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange?
(1) None of the customers bought more than 4 oranges
(2) The difference between the number of oranges bought by any two customers is even

In respect to the question above, I assumed that any two of those 19 customer might have bought 5 & 3 oranges and hence I, marked the option insufficient. Bunnel have equated and treated the option in totally different way. I ,lack the skill to convert these sort of condition in to equation.

please can some post or point to the list of similar Word translation sentences and how to convert them in to equation. Im very new to GMAT club so please forgie me if this is the repeated posting.

Thanks,
Vids
Show more

I did not use any equation for this question.

Statement (2) says: the difference between the number of oranges bought by ANY two customers is even --> in order the difference between ANY number of oranges bought to be even, either all customers must have bought odd number of oranges or all customers must have bough even number of oranges.

Now, the sum of 19 odd integers is odd and we have that fruit stand sold total of 76, so even number of oranges, which means that the case where all customers buy odd number of oranges is not possible. And since 1 is odd then no one bought only one orange. Sufficient.

As for word translation check this: word-problems-made-easy-87346.html

Hope it helps.
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The Discreet Charm of the DS 16 May 2012, 01:39
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9. If x and y are negative numbers, is x<y?

(1) 3x + 4 < 2y + 3 --> \(3x<2y-1\). \(x\) can be some very small number for instance -100 and \(y\) some large enough number for instance -3 and the answer would be YES, \(x<y\) BUT if \(x=-2\) and \(y=-2.1\) then the answer would be NO, \(x>y\). Not sufficient.

(2) 2x - 3 < 3y - 4 --> \(x<1.5y-\frac{1}{2}\) --> \(x<y+(0.5y-\frac{1}{2})=y+negative\) --> \(x<y\) (as y+negative is "more negative" than y). Sufficient.

Answer: B.


Hi bunuel,
Did not got how u solved option 2.Could you please explain in detail.
thanks.
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(2) 2x - 3 < 3y - 4 --> \(x<1.5y-\frac{1}{2}\) --> \(x<y+(0.5y-\frac{1}{2})\). Now, since \(y\) is a negative number then \(0.5y-\frac{1}{2}=negative\) so, we have that: \(x<y+negative\). \(y+negative\) is less then \(y\) and if \(x\) is less than \(y+negative\) then it must also be less than \(y\) itself: \(x<y\).

Hope it's clear.
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The Discreet Charm of the DS 27 Jun 2012, 08:03
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Hi Bunuel,

Can we solve this question using graphical approach.

Since (x-a)^2+(y-b)^2 = r^2 ----(A)

Equation 1 represents equation of circle with radius 1.

Now, if we can maximize the value of both x and y, we can get the maximum value of xy and that could let us know whether statement (1) is sufficient or not.

Here, I require your help. Intuitively, I find that x=y will maximize the equation x^2+y^2=1. But I am not sure about my reasoning as well as the Mathematics behind. Could you please look into this.

Thanks

Bunuel


Please read the thread. Solutions to ALL the questions are given on the previous pages.

2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

(2) x^2-y^2=0 --> \(|x|=|y|\). Clearly insufficient.

Answer: A.
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The Discreet Charm of the DS 27 Jun 2012, 08:17
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imhimanshu
Hi Bunuel,

Can we solve this question using graphical approach.

Since (x-a)^2+(y-b)^2 = r^2 ----(A)

Equation 1 represents equation of circle with radius 1.

Now, if we can maximize the value of both x and y, we can get the maximum value of xy and that could let us know whether statement (1) is sufficient or not.

Here, I require your help. Intuitively, I find that x=y will maximize the equation x^2+y^2=1. But I am not sure about my reasoning as well as the Mathematics behind. Could you please look into this.

Thanks

Bunuel


Please read the thread. Solutions to ALL the questions are given on the previous pages.

2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

(2) x^2-y^2=0 --> \(|x|=|y|\). Clearly insufficient.

Answer: A.
Show more

Useful property: For given sum of two numbers, their product is maximized when they are equal.

(1) says that \(x^2+y^2=1\). So, \(x^2y^2\) will be maximized when \(x^2=y^2\): \(x^2+x^2=1\) --> \(x^2=\frac{1}{2}\) --> the maximum value of \(x^2y^2\) thus is \(\frac{1}{4}\) and the maixmum value of \(xy\) is \(\frac{1}{2}\).

Hope it's clear.
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The Discreet Charm of the DS 06 Dec 2012, 23:19
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Bunuel

1. 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<12

(2) Bonnie and Clyde complete the painting of the car at 10:30am
Show more

Responding to a pm:

Time taken by Bonnie to complete one work = x hrs
Time taken by Clyde to complete one work = y hrs
x and y are odd integers i.e. they could take values such as 1/3/5/7/9/11...

Question: Is x = y? i.e. is the time taken by Bonnie equal to time taken by Clyde? i.e. is the speed of Bonnie equal to the speed of Clyde?

(1) x^2+y^2<12
This info is not related to work concepts. It's just number properties. x and y are odd integers.
If x = y = 1, this inequality is satisfied.
If x = 1 and y = 3, this inequality is satisfied.

This means x may or may not be equal to y. Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30 am.r
Together, they take 45 mins to complete the painting of the car. This means, if their rate of work were the same, each one of them would have taken 1.5 hrs working alone. But their time taken is an integer value. We can say that they do not take the same time i.e. x is not equal to y. Hence this statement is sufficient to say \(x \neq y\)

Answer (B)
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The Discreet Charm of the DS 14 Feb 2013, 01:24
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Bunuel
SOLUTIONS:

1. 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?

Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) --> \(T=\frac{xy}{x+y}\)). 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 odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible \(x\) and \(y\) to be odd and equal to 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 3/4 of an hour (45 minutes), since it's not odd/2 then \(x\) and \(y\) are not equal. Sufficient.

Answer: B.
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Bunuel,
I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?
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The Discreet Charm of the DS 14 Feb 2013, 02:16
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thinktank
Bunuel
SOLUTIONS:

1. 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?

Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) --> \(T=\frac{xy}{x+y}\)). 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 odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible \(x\) and \(y\) to be odd and equal to 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 3/4 of an hour (45 minutes), since it's not odd/2 then \(x\) and \(y\) are not equal. Sufficient.

Answer: B.

Bunuel,
I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?
Show more

From the stem we got that 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 odd/2: 0.5 hours (1/2 hours = 0.5 hours), 1.5 hours (3/2 hours = 1.5 hours), 2.5 hours (5/2 hours = 2.5 hours), 3.5 hours (7/2 hours = 3.5 hours), 4.5 hours (9/2 hours = 4.5 hours), ....

Now, from the second statement we got that they complete the job in 0.75 hours, since the total time (0.75 hours) is NOT odd/2 (0.5 hours, 1.5 hours, 2.5 hours, 3.5 hours, 4.5 hours, ....), then \(x\) and \(y\) are not equal.

Hope it's clear.
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The Discreet Charm of the DS 22 Jul 2013, 23:53
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Bunuel
12. If 6a=3b=7c, what is the value of a+b+c?

Given: \(6a=3b=7c\) --> least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) --> \(a=7x\), \(b=14x\) and \(c=6x\).

(1) ac=6b --> \(7x*6x=6*14x\) --> \(x^2=2x\) --> \(x=0\) or \(x=2\). Not sufficient.

(2) 5b=8a+4c --> \(5*14x=8*7x+4*14x\) --> \(70x=80x\) --> \(10x=0\) --> \(x=0\) --> \(a=b=c=0\) --> \(a+b+c=0\). Sufficient.

Answer: B.
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Hi Bunuel,

In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c.

Now substituting 12a=ac=14c

I get 12a=ac --> c = 12 and similarly a = 14 and hence b= 28.

Can you please point where am I going wrong?

Regards
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The Discreet Charm of the DS 22 Jul 2013, 23:56
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Spaniard
Bunuel
12. If 6a=3b=7c, what is the value of a+b+c?

Given: \(6a=3b=7c\) --> least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) --> \(a=7x\), \(b=14x\) and \(c=6x\).

(1) ac=6b --> \(7x*6x=6*14x\) --> \(x^2=2x\) --> \(x=0\) or \(x=2\). Not sufficient.

(2) 5b=8a+4c --> \(5*14x=8*7x+4*14x\) --> \(70x=80x\) --> \(10x=0\) --> \(x=0\) --> \(a=b=c=0\) --> \(a+b+c=0\). Sufficient.

Answer: B.

Hi Bunuel,

In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c.

Now substituting 12a=ac=14c

I get 12a=ac --> c = 12 and similarly a = 14 and hence b= 28.

Can you please point where am I going wrong?

Regards
Show more

ac=12a (here you cannot reduce by a and write c=12 as you exclude possibility of a=0) --> a(c-12)=0 --> either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12.

Does this make sense?
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The Discreet Charm of the DS 01 Jun 2014, 21:18
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Bunuel
8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?

First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: math-number-theory-88376.html

(1) a+b>14 --> the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.

(2) a-c>6 --> the least value of a is 7 (7-0=7>6), but we don't know the value of b. Not sufficient.

(1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient.

Answer: C.
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Hi Bunnel,

One question here.

in st2 i can consider 8-0 also then why we are considering only least value. in st1 we are considering least value as well as other value. why this is different in st2?

Thanks
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The Discreet Charm of the DS 02 Jun 2014, 01:23
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PathFinder007
Bunuel
8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?

First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: math-number-theory-88376.html

(1) a+b>14 --> the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.

(2) a-c>6 --> the least value of a is 7 (7-0=7>6), but we don't know the value of b. Not sufficient.

(1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient.

Answer: C.

Hi Bunnel,

One question here.

in st2 i can consider 8-0 also then why we are considering only least value. in st1 we are considering least value as well as other value. why this is different in st2?

Thanks
Show more

Because considering the least value of a (7), while knowing nothing about b is enough to say that the second statement is not sufficient. If b=8, then the answer is yes but if b is 0, then the answer is no.
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The Discreet Charm of the DS 03 Jun 2014, 00:39
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Hi Bunuel,
Referring to below:
4. How many numbers of 5 consecutive positive integers is divisible by 4?
(1) The median of these numbers is odd
(2) The average (arithmetic mean) of these numbers is a prime number

The average (arithmetic mean) of these numbers is a prime number --> in any evenly spaced set the arithmetic mean (average) is equal to the median --> mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.

I didn't quite get the explanation on the second statement. Do you mind explaining with some numbers?

Thanks,
Mitesh
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The Discreet Charm of the DS 03 Jun 2014, 01:49
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mrvora
Hi Bunuel,
Referring to below:
4. How many numbers of 5 consecutive positive integers is divisible by 4?
(1) The median of these numbers is odd
(2) The average (arithmetic mean) of these numbers is a prime number

The average (arithmetic mean) of these numbers is a prime number --> in any evenly spaced set the arithmetic mean (average) is equal to the median --> mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.

I didn't quite get the explanation on the second statement. Do you mind explaining with some numbers?

Thanks,
Mitesh
Show more

If mean=median=prime=2, then the 5 consecutive integers would be {0, 1, 2, 3, 4}. But this set is not possible since we are told that the set consists of 5 positive integers and 0 is neither positive nor negative integer.

Hence, mean=median=odd prime --> {Odd, Even, Odd prime, Even, Odd}. So, the set includes 2 consecutive even numbers. Out of 2 consecutive even integers only one is a multiple of 4.

Hope it's clear.
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The Discreet Charm of the DS 10 Jun 2017, 11:50
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Bunuel
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

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Hi Bunuel, a small doubt instead of \(2x+8\geq{0}\) --> \(x\geq{-4}\) this logic if we take |x+10| to be both positive and negative we get 2 vale by solving equation x=2 and x=-6. however by putting values back in eqtn we can see that only for x=2 equation is satisfying. Is this correct approach?
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The Discreet Charm of the DS 10 Jun 2017, 12:47
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Bunuel
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.


Hi Bunuel, a small doubt instead of \(2x+8\geq{0}\) --> \(x\geq{-4}\) this logic if we take |x+10| to be both positive and negative we get 2 vale by solving equation x=2 and x=-6. however by putting values back in eqtn we can see that only for x=2 equation is satisfying. Is this correct approach?
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Yes, that's also a correct way of solving. Good thing you did is that you did not forget to test both values after you got them.
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The Discreet Charm of the DS 06 Sep 2020, 04:13
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Bunuel
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2. Is xy<=1/2?
(1) x^2+y^2=1
(2) x^2-y^2=0

If we take (X+Y)^2 instead of (X-Y)^2 for analyzing statement 1, why we get two different answers.
Also, the detailed explanation link is not working perhaps.
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The Discreet Charm of the DS 06 Sep 2020, 04:44
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Bunuel
In
2. Is xy<=1/2?
(1) x^2+y^2=1
(2) x^2-y^2=0

If we take (X+Y)^2 instead of (X-Y)^2 for analyzing statement 1, why we get two different answers.
Also, the detailed explanation link is not working perhaps.
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The link to the solution is here: https://gmatclub.com/forum/the-discreet ... l#p1039634


The point is that if you consider \((x+y)^2 \ge 0\), you'll get \(xy\geq{-\frac{1}{2}}\), which is not helpful at all. Actually since both \(xy\geq{-\frac{1}{2}}\) (from \((x+y)^2 \ge 0\)) and \(xy \le \frac{1}{2}\) (from \((x-y)^2 \ge 0\)) are true, then we get that \(-\frac{1}{2} \leq{xy} \leq{\frac{1}{2}}\). But only the approach given in the solution above gives you the answer we are looking for, while another one gives you an inequality which is not helpful.

P.S. You posted a question IN the thread, why use report along with it? Pleas do NOT use report button that way. It's for reporting problems, not for questions.,
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The Discreet Charm of the DS 29 Jul 2022, 10:59
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SOLUTIONS:

1. 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?

Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) --> \(T=\frac{xy}{x+y}\)). 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 odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible \(x\) and \(y\) to be odd and equal to 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 3/4 of an hour (45 minutes), since it's not odd/2 then \(x\) and \(y\) are not equal. Sufficient.

Answer: B.
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Bunuel in Statement 2, does this mean that because t = 3/4, which is a fraction, it is not odd and hence 3/4 =x and 3/4= y?

Separate question: I was confused by the wording of St2. For a second it made it sound like Bonnie and Clyde were working together on painting, but question says independent. Which statement to trust more here?
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The Discreet Charm of the DS 30 Jul 2022, 04:24
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Bunuel
SOLUTIONS:

1. 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?

Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) --> \(T=\frac{xy}{x+y}\)). 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 odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible \(x\) and \(y\) to be odd and equal to 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 3/4 of an hour (45 minutes), since it's not odd/2 then \(x\) and \(y\) are not equal. Sufficient.

Answer: B.

Bunuel in Statement 2, does this mean that because t = 3/4, which is a fraction, it is not odd and hence 3/4 =x and 3/4= y?

Separate question: I was confused by the wording of St2. For a second it made it sound like Bonnie and Clyde were working together on painting, but question says independent. Which statement to trust more here?
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1. From the stem we concluded that IF x = y, then the total time would be (x hours)/2 = (odd hours)/2 (since given that x is odd). (odd hours)/2 means the total time would be 0.5 hours, 1.5 hours, 2.5 hours, ... (2) says that the total time is 3/4 hours. 3/4 hours is NOT in the form of (odd hours)/2, so x does not equal to y.

2. Working simultaneously and independently, means that they work on different parts of the job. For example, one is painting the front and another the back of the car. This ensure that their works does not overlap and they do the job in fastest way possible.
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