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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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02 Feb 2012, 03:15
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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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16 Apr 2016, 09:25
bikographer 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.

Considering (2), and rearranging it,

2x - 3y < -1 .......... (i)

substituting x=-2 and y=-1 in (i), we get

-4 + 3 = -1 which is not <-1

This means that the condition doesn't hold good, and hence x < y doesn't hold good.

Could you please point out where exactly I am making a mistake in my process?

$$3x + 4 < 2y + 3$$ means that $$3x<2y-1$$, not 2x - 3y < -1.
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The Discreet Charm of the DS [#permalink]

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16 Apr 2016, 18:14
Bunuel wrote:
bikographer 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.

Considering (2), and rearranging it,

2x - 3y < -1 .......... (i)

substituting x=-2 and y=-1 in (i), we get

-4 + 3 = -1 which is not <-1

This means that the condition doesn't hold good, and hence x < y doesn't hold good.

Could you please point out where exactly I am making a mistake in my process?

$$3x + 4 < 2y + 3$$ means that $$3x<2y-1$$, not 2x - 3y < -1.

Bunuel, I was talking only about the second statement, and in your reply you have talked about the first statement. My doubt, which I have highlighted by taking x = -2 and y = -1 and applying these values to statement (2), still remains.
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Re: The Discreet Charm of the DS [#permalink]

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17 Apr 2016, 08:05
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bikographer 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.

Considering (2), and rearranging it,

2x - 3y < -1 .......... (i)

substituting x=-2 and y=-1 in (i), we get

-4 + 3 = -1 which is not <-1

This means that the condition doesn't hold good, and hence x < y doesn't hold good.

Could you please point out where exactly I am making a mistake in my process?

The statements in DS questions are true. So, if you pick numbers you should pick such that they satisfy the statement. x and y cannot be -2 and -1 respectively because they do not satisfy 2x - 3 < 3y - 4.
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The Discreet Charm of the DS [#permalink]

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06 Nov 2016, 04:18
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

I think the option should be E
Option 1 -x * -y will result in positive, where x is negative.

Can you explain further..??
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Re: The Discreet Charm of the DS [#permalink]

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06 Nov 2016, 05:06
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hardik0491 wrote:
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

I think the option should be E
Option 1 -x * -y will result in positive, where x is negative.

Can you explain further..??

If x is negative, then x*|y| = negative*non-negative = non-positive, which cannot be prime, since only positive numbers are primes.
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Re: The Discreet Charm of the DS [#permalink]

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15 Dec 2016, 10:16
Hello Bunuel,

For question 9, option B -- What if we assign the -100 for x and -99 for y? x<y, but the equation (rearranged) 3y-2x>1 does not hold true. Please help!
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Re: The Discreet Charm of the DS [#permalink]

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15 Dec 2016, 10:23
Shru91 wrote:
Hello Bunuel,

For question 9, option B -- What if we assign the -100 for x and -99 for y? x<y, but the equation (rearranged) 3y-2x>1 does not hold true. Please help!

(2) says that 2x - 3 < 3y - 4, so when plugging numbers for x and y they must satisfy this inequality. After that you should check whether you get x < y or not. Not vise-versa.
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Re: The Discreet Charm of the DS [#permalink]

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10 Jun 2017, 10:30
Bunuel wrote:
4. How many numbers of 5 consecutive positive integers is divisible by 4?

(1) The median of these numbers is odd --> the median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) 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.

Bunuel Can you pls help in understanding this : "Out of 2 consecutive even integers only one is a multiple of 4. Sufficient." from statement 1.Thanks
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Re: The Discreet Charm of the DS [#permalink]

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10 Jun 2017, 10:50
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.

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

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10 Jun 2017, 11:47
KARISHMA315 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.

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?

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

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26 Jul 2017, 04:52
Is this valid?

Statement 1
2x²+9<9x
2x²<9x-9
2x²<9(x-1)
2x²÷(x-1)<9
plugin 1
2(1)²÷(1-1)=2÷0=undefined

plugin 2
2(2)²÷(2-1)=8÷1<9 valid
any no.greater than 2 will give no greater than 9
if we take a negative no than the sign will change and inequality will not be valid.

Statement 2
|x+10|=2x+8

|x+10|=2(x+4)

|x+10|w÷2=X+4

Only X=2 satisfies the equation

|2+10|÷2=12÷2=6

2+4=6

[qote="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.

Check this for more on solving inequalities like the one in the first statement:
http://gmatclub.com/forum/x2-4x-94661.html#p731476
http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/everything-is ... me#p868863
http://gmatclub.com/forum/xy-plane-7149 ... ic#p841486

Hope it helps.[/quote]

Posted from my mobile device
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Re: The Discreet Charm of the DS [#permalink]

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05 Sep 2017, 23:43
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

Hi Bunuel,

I understood that x*|y|=positive; but if y<=0 then |y|=-y and in this case x can be negative.
Am I missing something?

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

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05 Sep 2017, 23:49
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Yashodhan123 wrote:
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

Hi Bunuel,

I understood that x*|y|=positive; but if y<=0 then |y|=-y and in this case x can be negative.
Am I missing something?

Thanks.

Yes.

An absolute value of a number cannot be negative: |a| is positive or 0, no matter whether a itself is negative or not.

If y <= 0, then |y| = -y, yes, but even in this case -y = -negative = positive.
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Re: The Discreet Charm of the DS [#permalink]

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18 Oct 2017, 08:10
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 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.

Please explain this part-since it's not odd/2 then x and y are not equal..

how can u say its not odd..pls help!!

Thanks
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The Discreet Charm of the DS [#permalink]

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18 Oct 2017, 08:30
Bunuel wrote:
3. If a, b and c are integers, is abc an even integer?

In order the product of the integers to be even at leas on of them must be even

(1) b is halfway between a and c --> on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at leas on of them is be even? Not necessarily, consider $$a=1$$, $$b=3$$ and $$c=5$$. Of course it's also possible that $$b=even$$, for example if $$a=1$$ and $$b=7$$. Not sufficient.

(2) a = b - c --> $$a+c=b$$. Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient.

Why are we assuming that a and c could be even..? or could be odd?

Its also possible that a is even and c is odd..in that case b can be odd

Pls explain thanks
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18 Oct 2017, 09:58
zanaik89 wrote:
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 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.

Please explain this part-since it's not odd/2 then x and y are not equal..

how can u say its not odd..pls help!!

Thanks

Please re-read the highlighted part. We got that if $$x=y$$ then the total time would be odd/2. (2) says that it's NOT odd/2, therefore x does not equal to y.
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Re: The Discreet Charm of the DS [#permalink]

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18 Oct 2017, 10:00
zanaik89 wrote:
Bunuel wrote:
3. If a, b and c are integers, is abc an even integer?

In order the product of the integers to be even at leas on of them must be even

(1) b is halfway between a and c --> on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at leas on of them is be even? Not necessarily, consider $$a=1$$, $$b=3$$ and $$c=5$$. Of course it's also possible that $$b=even$$, for example if $$a=1$$ and $$b=7$$. Not sufficient.

(2) a = b - c --> $$a+c=b$$. Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient.

Why are we assuming that a and c could be even..? or could be odd?

Its also possible that a is even and c is odd..in that case b can be odd

Pls explain thanks

It's not clear what are you trying to say here. If a is even, then abc is even.
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Re: The Discreet Charm of the DS [#permalink]

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18 Oct 2017, 17:50
Bunuel wrote:
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.

Hi Bunuel,

I have two questions..

1) Cant we do 6a=3b simplify further and plug it back into equation and solve?

6a=3b can be written as 2a=b?

2) In your solution u have got x^2=2x how are thee two values of x? how did u get x=0?

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

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18 Oct 2017, 20:04
zanaik89 wrote:
Bunuel wrote:
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.

Hi Bunuel,

I have two questions..

1) Cant we do 6a=3b simplify further and plug it back into equation and solve?

6a=3b can be written as 2a=b?

2) In your solution u have got x^2=2x how are thee two values of x? how did u get x=0?

Thanks

1. Not clear what you mean at all. Please show your work.

2. x^2=2x;

x^2 - 2x = 0

x(x - 2) = 0

x = 0 or x = 2.

P.S. You should know such basic staff before attempting questions, especially hard ones as in this set.
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Re: The Discreet Charm of the DS [#permalink]

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07 Nov 2017, 06:24
Thanks for the compilation guys! It was super helpful

Out of curiosity. What difficulty level are these questions?
Re: The Discreet Charm of the DS   [#permalink] 07 Nov 2017, 06:24

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