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The Discreet Charm of the DS

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The Discreet Charm of the DS [#permalink]

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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

Solution: the-discreet-charm-of-the-ds-126962-20.html#p1039633

2. Is xy<=1/2?
(1) x^2+y^2=1
(2) x^2-y^2=0

Solution: the-discreet-charm-of-the-ds-126962-20.html#p1039634

3. If a, b and c are integers, is abc an even integer?
(1) b is halfway between a and c
(2) a = b - c

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039637

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

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039645

5. What is the value of integer x?
(1) 2x^2+9<9x
(2) |x+10|=2x+8

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039650

6. If a and b are integers and ab=2, is a=2?
(1) b+3 is not a prime number
(2) a>b

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039651

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

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039655

8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?
(1) a+b>14
(2) a-c>6

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039662

9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3
(2) 2x - 3 < 3y - 4

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039665

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.

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039671

11. If x and y are integers, is x a positive integer?
(1) x*|y| is a prime number.
(2) x*|y| is non-negative integer.

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039678

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

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039680
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Re: The Discreet Charm of the DS [#permalink]

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New post 27 May 2012, 03:21
Here is my approach :
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

Ans: x & y are odd integer.
statement1: x=1,y=1 or x=3,y=1 not sufficient
statement2: time of completion is 10.30am-9.45am =45min =3/4Hr; i.e rate and time consumed by both is same.
hence, statement B is sufficient.

Ans : B

2. Is xy<=1/2?
(1) x^2+y^2=1
(2) x^2-y^2=0

Ans: statement 1: x^2+y^2=1,i choose the no: x=y=sqrt(1/2); hence sufficient
x=0,y=1 ;hence sufficient
x=1,y=0 ;hence sufficient

i didnt find any number which doesnot comply to statement 1.
so, sufficient.
statement2: x^2-y^2=0 ==> mod(x) = mod(y)
==> x=y
& x=-y not sufficient

Ans: A

3. If a, b and c are integers, is abc an even integer?
(1) b is halfway between a and c
(2) a = b - c

Ans: a,b,c are integers,not in sequence.
statement 1: b is half way between a & c.
a=2,b=4,c=6 abc=48 even
a=2,b=3,c=6 abc=36 even
a=3,b=5,c=7 abc=105 odd

statement 1 Not sufficient
Statement 2: a=b-c ==> b=a+c ; we cant say that abc will be even or odd because we dont know whether a,b,c is odd or even.
Not sufficient
on combining both statement also, we cant say anything about abc.

Ans: E.

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

Ans: E ( No explanation)

5. What is the value of integer x?
(1) 2x^2+9<9x
(2) |x+10|=2x+8
Ans:
statement 1: 2x^2+9<9x ==>2x^2-9x+9<0
==>(2x-1)(x-3)<0
so, 1/2<x<3 or x>3&x<1/2
Not sufficient

Statement 2: |x+10| = 2x+8
if x>10;
x+10=2x+8 ==>x=2 but (x>10)
if x<10;
-x-10=2x+8 ==>x= -6 and (x<10)
so, x=-6
Sufficient

Ans B

6. If a and b are integers and ab=2, is a=2?
(1) b+3 is not a prime number
(2) a>b
Ans: ab=2 ==> a=2/b

statement 1: b+3 is not a prime number i.e
b+3=1,4,6,8 so, b could be = -2,1,3,5
Not sufficient
Statement 2: a>b
and ab=2 and a&b are integers..only possible value is
a=2 & b=1

Sufficient
Ans B


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
Ans:
total oranges =76
No of customer =19
how many bought only 1 oranges?
statement1:
if none bought more than 4, then,max no of oranges bought is 19x4 =76 oranges.
in short, each customer has bought 4 oranges.
sufficient
statement 2: customer can buy any no of oranges totaling 76. 4-4=0 even, 5-3=2 even,and many more.
not sufficient

Ans : A

8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?
(1) a+b>14
(2) a-c>6
Ans: x=0.abcd
7/9=0.777777

statement 1: a+b>14
(a,b) :(7,8) ,(7,9),(8,9),(8,8),(9,9)
x=0.abcd ; replacing the value of a&b
x=0.78cd
x=0.79cd
x=0.89cd
x=0.88cd
x=0.99cd
all are greater then 0.77777 hence
Sufficient

statment 2: a-c>6
(a,c): (9,2) (7,0) and many more
x=0.92cd is >0.7777 ok
x=0.70cd is <0.7777 not ok

Not sufficient

Ans A


9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3
(2) 2x - 3 < 3y - 4
Ans: x,y <0
statement 1: 3x+4<2y+3 ==>3x-2y+1<0 not sufficient

statement 2: 2x-3<3y-4 ==> 2x-3y +1<0 not sufficient

on combining both statement and solving for x& y
x< -1/5 & y< 1/5
so, y>x for interval (-1/5 to 1/5) since both are -ve so interval should be (-1/5 to 0)
and y=x for (-infinity to -1/5)
Not sufficient

Ans E


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.
Ans: f(10,x)=11 ==> (10+x)/GCF(10,x) =11 ==>x = GCF(10,x)-10

Statement 1: x could be =1,4,9,16,25..
GCF of (10,1) , (10,4),(10,9) will be different.
Not sufficient

Statement 2:
x= 2 , no of factor 2 (1&2) ok
x= 4 , no of factor 3 (1,2,4) ok
x= 10 , no of factor 4 (1,2,5,10) not ok
not sufficient

on combining I & II
we can get value like 1,4,25 which satisfy both the statement
but no unique value of x can be found.

Ans E

11. If x and y are integers, is x a positive integer?
(1) x*|y| is a prime number.
(2) x*|y| is non-negative integer.
Ans:
statement 1: x*|y| is prime no
no information about +ve or -ve no.
Not sufficient

Satement 2: for x*|y| has to non-ve integer both x& y has to -ve or +ve simultaneously
any value inside mode is always positive. mode(y) = positive
to make x*|y| +ve, X has to be positive.

hence sufficient.
Ans B


12. If 6a=3b=7c, what is the value of a+b+c?
(1) ac=6b
(2) 5b=8a+4c
Ans: 6a=3b=7c= k
a=k/6
b=k/3
c=k/7
a+b+c = (k/6)+(k/3)+k/7) if we can find the value of K, we wil have our answer.

Statement 1: ac=6b ==>(k/6)(k/7) = 6.k/3 ==>k=84
Sufficient

statement 2: 5b=8a+4c
==> 5.k/3 = (8k/6)+(4.k/7)
no value of k can be found.

Hence
Ans A.

please check my approach ans suggest if anything is missing or wrong.
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Re: The Discreet Charm of the DS [#permalink]

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New post 04 Jun 2012, 21:39
Bunuel wrote:
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.



\((x-\frac{3}{2})(x-3)<0\)

I agree one solution of this inequality is

\((x-\frac{3}{2})\) > 0 , (x-3)<0 => \((\frac{3}{2})\) < x < 3

However, Don't u think this can also resort to

\((x-\frac{3}{2})\) < 0 , (x-3)>0 => x < \((\frac{3}{2})\) , x > 3

and in that case x can have infinite values.

and if that is the case Stmt-1 alone would not be sufficient.
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Re: The Discreet Charm of the DS [#permalink]

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New post 05 Jun 2012, 00:51
Expert's post
anordinaryguy wrote:
Bunuel wrote:
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.



\((x-\frac{3}{2})(x-3)<0\)

I agree one solution of this inequality is

\((x-\frac{3}{2})\) > 0 , (x-3)<0 => \((\frac{3}{2})\) < x < 3

However, Don't u think this can also resort to

\((x-\frac{3}{2})\) < 0 , (x-3)>0 => x < \((\frac{3}{2})\) , x > 3


and in that case x can have infinite values.

and if that is the case Stmt-1 alone would not be sufficient.


The second case of \((x-\frac{3}{2})<0\) and \((x-3)>0\) is not possible. This condition leads to \(x<\frac{3}{2}\) and \(x>3\), but \(x\) can not be simultaneously less than \(\frac{3}{2}\) (to make \(x-\frac{3}{2}\) negative) and more than 3 (to make \(x-3\) positive).

For more on how to solve such kind of inequalities check:
x2-4x-94661.html#p731476 (Check this first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
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Re: The Discreet Charm of the DS [#permalink]

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New post 20 Jun 2012, 07:51
Hi Bunuel,
Where can I find the AO and explainations. I can see them anywhere in the thread ?

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Re: The Discreet Charm of the DS [#permalink]

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New post 20 Jun 2012, 08:05
Expert's post
vaibhavalw wrote:
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.

Hope it helps.
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Re: The Discreet Charm of the DS [#permalink]

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New post 20 Jun 2012, 10:20
Thanks a lot :)
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Re: The Discreet Charm of the DS [#permalink]

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New post 21 Jun 2012, 05:38
Hi Bunuel,

Are you still giving kudos for posting detailed answers to these questions? Can I post my answers?

Thanks..
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Re: The Discreet Charm of the DS [#permalink]

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New post 21 Jun 2012, 05:42
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Re: The Discreet Charm of the DS [#permalink]

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New post 22 Jun 2012, 15:46
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
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Re: The Discreet Charm of the DS [#permalink]

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New post 27 Jun 2012, 08:03
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 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.

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

Answer: A.

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Re: The Discreet Charm of the DS [#permalink]

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New post 27 Jun 2012, 08:17
Expert's post
imhimanshu wrote:
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 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.

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

Answer: A.


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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Re: The Discreet Charm of the DS [#permalink]

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New post 08 Jul 2012, 10:47
Bunuel wrote:
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.


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 \(x<y\) :( :(
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New post 28 Jul 2012, 07:36
Bunuel wrote:
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.


Why we recalled \((x-y)^2\geq{0}\) ?
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New post 29 Jul 2012, 01:55
Expert's post
AnanJammal wrote:
Bunuel wrote:
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.


Why we recalled \((x-y)^2\geq{0}\) ?


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 \(xy\leq{\frac{1}{2}}\).
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New post 27 Aug 2012, 02:40
Bunuel wrote:
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 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.....

What do you think?

Cheers,
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New post 28 Aug 2012, 02:57
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.

B is the answer.
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New post 28 Aug 2012, 04:22
Expert's post
OldFritz wrote:
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.....

What do you think?

Cheers,
Der alte Fritz


Typo edited. Thank you.
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New post 04 Sep 2012, 20:34
piyushksharma wrote:
Bunuel wrote:
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.

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
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New post 15 Nov 2012, 23:10
gmatDeep wrote:
Bunuel wrote:
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.


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.
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New post 06 Dec 2012, 23:19
Expert's post
Bunuel wrote:
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)
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Re: The Discreet Charm of the DS   [#permalink] 06 Dec 2012, 23:19

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