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I'm posting the next set of medium/hard DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

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

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

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

10. The function f is defined for all positive integers a and b by the following rule: f(a,b)=(a+b)/GCF(a,b), where GCF(a,b) is the greatest common factor of a and b. If f(10,x)=11, what is the value of x? (1) x is a square of an integer (2) The sum of the distinct prime factors of x is a prime number.

Hi Bunuel, Where can I find the AO and explainations. I can see them anywhere in the thread ?

Rgds

Switch view mode of the topic from "Best Reply" to "Oldest" and the links from the initial post (the-discreet-charm-of-the-ds-126962.html) will lead you to the posts with solutions.

For 12. If 6a=3b=7c, what is the value of a+b+c? (1) ac=6b (2) 5b=8a+4c

statement 2 i used 5b=8a+4c => 5x3b= 3x(8a +4c) ==> replace 3b with 6a given 5x6a=3x(8a+4c) solved and got a=2c (now we know from question step that 6a=7c), so the only number for which this is posible is zero (hence the solution is B

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 wrote:

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.

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 wrote:

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.

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 \(\frac{1}{2}\).

(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.

Bunuel, while solving (1), how do you know which nos to plugin in and test? Can you suggest some approach please? I got this one wrong because for the values I plugged in, I was always getting \(x<y\)

(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.

(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.

Because \((x-y)^2\geq{0}\) expands to \(x^2-2xy+y^2\geq{0}\), which leads to \(1-2xy\geq{0}\) (since \(x^2+y^2=1\)) and finally to \(xy\leq{\frac{1}{2}}\).
_________________

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 either an integer or integer/2: 0.5 hours, 1 hour, 1.5 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 neither an integer nor integer/2 then \(x\) and \(y\) are not equal. Sufficient.

Answer: B.

Hi Bunuel, since we know that, if x=y, then the total time taken by both working together is (x/2 or y/2), and since both x and y are odd, then the total time x/2 will be only non-integer, precisely, one-half of odd integers or mathematically stated, = Integer (both odd and even) + 0.5, eg. 0.5, 1.5, 2.5, 3.5, 4.5 etc. I don't think we can have integers at all. All that I have written is restricted to when x=y, of course.

Though everything else you wrote, including the final answer, is correct.....

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

From the question stem: 1. B and C both starts painting at 9:45. 2. x and y are odd numbers.

From stem1: x=1 or 3 and y=1 or 3 can satisfy the inequality. So, we can not attain the solution without other hints. From stem2: B and C completed the painting on the same time. As, they started and completed on the same time, the required same amount of time to complete the painting task. So, x=y.

Hi Bunuel, since we know that, if x=y, then the total time taken by both working together is (x/2 or y/2), and since both x and y are odd, then the total time x/2 will be only non-integer, precisely, one-half of odd integers or mathematically stated, = Integer (both odd and even) + 0.5, eg. 0.5, 1.5, 2.5, 3.5, 4.5 etc. I don't think we can have integers at all. All that I have written is restricted to when x=y, of course.

Though everything else you wrote, including the final answer, is correct.....

(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.

Hope it helps.

Hi bunuel, Isn't |x+10|=2x+8 be written as Either x+10=2x+8 or x+10=-(2x+8) ? and then this should be solved? Please help on this one.

Hey not sure if you already understood bunuel's solution, but I didn't..i figured it out though according to the way we do it. just like you said solve for x+10=2x+8 or x+10=-(2x+8) ....once you get answers for x, just plug them into the original equation and see if they work..only x=2 works! so just make sure to check absolute value solutions after you find them

(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.

Bunuel, while solving (1), how do you know which nos to plugin in and test? Can you suggest some approach please? I got this one wrong because for the values I plugged in, I was always getting \(x<y\)

I'm quite confused by the fundamental concept here. I sub in x = -1.1 and y = -1 (which makes x<y), and equation 2x - 3 < 3y - 4 doesn't work, but then if you switch them around, x = -1 and y = -1.1, the equation still doesn't work.

What are we trying to find with substituting in numbers here? I feel like I have tied a knot in my head. Someoone please help? Thanks.

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

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 \(x \neq y\)

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

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)

Thanks a lot for your explanation and response to my pm
_________________

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