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Math Expert V
Joined: 02 Sep 2009
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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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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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1
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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GMAT 1: 710 Q49 V38 Re: The Discreet Charm of the DS  [#permalink]

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

Let 6a=3b=7c=K (where K is a constant)
Thus, a=K/6
b= K/3
c=K/7

Therefore, we need to find the value of K/6 + K/3 + K/7 = 9K/14. OR if we find the value of just K it will give us the answer.

Stmt 1: ac=6b
$$K/6*K/7 = 6*K/3;$$

$$K^2/42-2K=0$$

$$K(K/42-2)=0$$

So, K = 0 or K= 84. NOT Sufficient.

Stmt 2:
5b=8a+4c
5K/3=8K/6 +4K/7
40K/21-5K/3=0
5K/3=0
K=0 SUFFICIENT

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

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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.

Dear sir,
Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0?
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BelalHossain046 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.

Dear sir,
Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0?

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.

Hope it's clear.
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Re: The Discreet Charm of the DS  [#permalink]

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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.

Shouldn't we consider zero as positive ?
2 is not sufficient as |y| can be zero so x can be negative or positive
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1
teaserbae 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.

Shouldn't we consider zero as positive ?
2 is not sufficient as |y| can be zero so x can be negative or positive

ZERO.

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.
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Re: The Discreet Charm of the DS  [#permalink]

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Bunuel wrote:
5. What is the value of integer x?

(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, until now whenever I saw modulus, I was told to begin by removing the mod and considering both signs.
Therefore, I approached it as
x+10=2x+8 and -(x+10)=2x+8. Just wanted to understand when to use this method and when to use the take it as an absolute value method?
Thanks!
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Re: The Discreet Charm of the DS  [#permalink]

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With respect to question no 8 above, (If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?), why would the answer not just be A? Since we can determine the least possible value of both a and b, and thereby deduce that x would be greater than 0.77.
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tejasvis7 wrote:
With respect to question no 8 above, (If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?), why would the answer not just be A? Since we can determine the least possible value of both a and b, and thereby deduce that x would be greater than 0.77.

Solution is here: https://gmatclub.com/forum/the-discreet ... l#p1039662

Here is what it says about (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.
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The Discreet Charm of the DS  [#permalink]

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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
Could you please change 4*14x (as highlighted) to 4*6x because c=6x. (small typo)
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dhritidutta 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
Could you please change 4*14x (as highlighted) to 4*6x because c=6x. (small typo)

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Edited. Thank you.
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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.

What if the time taken by both would have been 3/2??

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

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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.

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.

When you solve the second statement we get x=2 or x=-6??

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

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ssshyam1995 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.

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.

When you solve the second statement we get x=2 or x=-6??

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Doesn't the solution give only one value of x which satisfy the equation? Does x = -6 satisfy |x+10|=2x+8?
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Re: The Discreet Charm of the DS  [#permalink]

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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.

In statement 2 if y was more negative then wouldn't x be greater than y??

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ssshyam1995 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.

In statement 2 if y was more negative then wouldn't x be greater than y??

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[quote="Bunuel"][b]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?[/b]

Notice that the greatest common factor of 10 and x, GCF(10,x), naturally must be a factor of 10: 1, 2, 5, and 10. Thus from f(10,x)=11 we can get four different values of x:

GCF(10,x)=1 --> [m]f(10,x)=11=[fraction]10+x/1[/fraction][/m] --> [m]x=1[/m];
GCF(10,x)=2 --> [m]f(10,x)=11=[fraction]10+x/2[/fraction][/m] --> [m]x=12[/m];
GCF(10,x)=5 --> [m]f(10,x)=11=[fraction]10+x/5[/fraction][/m] --> [m]x=45[/m];
GCF(10,x)=10 --> [m]f(10,x)=11=[fraction]10+x/10[/fraction][/m] --> [m]x=100[/m].

(1) x is a square of an integer --> [m]x[/m] can be 1 or 100. Not sufficient.

(2) The sum of the distinct prime factors of x is a prime number ---> distinct primes of 12 are 2 and 3: [m]2+3=5=prime[/m], distinct primes of 45 are 3 and 5: [m]3+5=8\neq{prime}[/m] and distinct primes of 100 are 2 and 5: [m]2+5=7=prime[/m]. [m]x[/m] can be 12 or 100. Not sufficient.

(1)+(2) [m]x[/m] can only be 100. Sufficient.

Notice that the greatest common factor of 10 and x, GCF(10,x), naturally must be a factor of 10: 1, 2, 5, and 10. Is this like a rule?? What if the question had given us LCM(10,x)??

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Re: The Discreet Charm of the DS  [#permalink]

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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.

For statement 1 could we have taken (x+y)square instead of (x-y)square??

Posted from my mobile device Re: The Discreet Charm of the DS   [#permalink] 25 Nov 2019, 03:53

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