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# 12 Easy Pieces (or not?)

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Manager
Joined: 13 Apr 2019
Posts: 179
Location: India
Concentration: Marketing, Operations
GMAT 1: 690 Q49 V35
GPA: 3.5
WE: General Management (Retail)
Re: 12 Easy Pieces (or not?)  [#permalink]

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04 Aug 2019, 03:49
6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

_ _ _ _ _ _ _ _ _ _

Since no adjacent colors can be same, reds can only be positioned in 2 ways (either position numbers 1,3,5,7,9 or numbers 2,4,6,8,10)

given this, whatever order we now place remaining 5, they will always meet the Anna's condition. So, number of ways will 2*5!/(2!*2!*1!) =60
Manager
Joined: 01 Jan 2019
Posts: 90
Concentration: Finance, Entrepreneurship
GPA: 3.24
Re: 12 Easy Pieces (or not?)  [#permalink]

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04 Aug 2019, 13:27
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Why is it not A? Can’t we simply subtract x from y? Then the answer will be A?

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Math Expert
Joined: 02 Sep 2009
Posts: 60460
Re: 12 Easy Pieces (or not?)  [#permalink]

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04 Aug 2019, 13:58
Shef08 wrote:
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Why is it not A? Can’t we simply subtract x from y? Then the answer will be A?

Posted from my mobile device

You can only apply subtraction when the signs of inequalities are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

For more check Manipulating Inequalities.
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Manager
Joined: 01 Jan 2019
Posts: 90
Concentration: Finance, Entrepreneurship
GPA: 3.24
Re: 12 Easy Pieces (or not?)  [#permalink]

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04 Aug 2019, 20:30
Bunuel wrote:
Shef08 wrote:
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Why is it not A? Can’t we simply subtract x from y? Then the answer will be A?

Posted from my mobile device

You can only apply subtraction when the signs of inequalities are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

For more check Manipulating Inequalities.

Thank you, Bunuel!
Manager
Joined: 01 Jan 2019
Posts: 90
Concentration: Finance, Entrepreneurship
GPA: 3.24
Re: 12 Easy Pieces (or not?)  [#permalink]

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04 Aug 2019, 20:57
azhrhasan wrote:
Shef08 wrote:
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Why is it not A? Can’t we simply subtract x from y? Then the answer will be A?

Posted from my mobile device

Here's how to solve using your approach:

A: -7<y<9
B: -3<x<5 this means B': -5<-x<3

A+B':: -12< y-x < 12

Never subtract or multiply or divide inequalities until you are sure of the sign. On the other hand, Addition can be done always irrespective of the sign.
Always use the addition approach on inequalities and never try to shortcut it by multiplying or dividing or subtracting

That’s a great insight! I’ll have this fitted in my brains! Thanks a ton
Intern
Joined: 22 Dec 2018
Posts: 21
WE: Medicine and Health (Health Care)
Re: 12 Easy Pieces (or not?)  [#permalink]

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28 Aug 2019, 06:37
Bunuel wrote:
10. If $$n$$ is an integer and $$\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}$$, then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

Also no need for algebraic manipulation. 1/10^(n+1) is 10 times less than 1/10^n, and both when expressed as decimals are of a type 0.001 (some number of zeros before 1) --> so the given expression to hold true we should have: 0.001<0.00737<0.01, which means that n=2 (1/10^n=0.01 --> n=2).

Hi Bunuel,

I am unable to understand how did we get 0.001 for 1/10^(n+1).
From what i understand 1/10^(n+1) is of form 1/10^n x 10^1.

Thanks
Intern
Joined: 22 Dec 2018
Posts: 21
WE: Medicine and Health (Health Care)
Re: 12 Easy Pieces (or not?)  [#permalink]

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28 Aug 2019, 07:06
Bunuel wrote:
swatjazz wrote:
Bunuel wrote:
10. If $$n$$ is an integer and $$\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}$$, then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

Also no need for algebraic manipulation. 1/10^(n+1) is 10 times less than 1/10^n, and both when expressed as decimals are of a type 0.001 (some number of zeros before 1) --> so the given expression to hold true we should have: 0.001<0.00737<0.01, which means that n=2 (1/10^n=0.01 --> n=2).

Hi Bunuel,

I am unable to understand how did we get 0.001 for 1/10^(n+1).
From what i understand 1/10^(n+1) is of form 1/10^n x 10^1.

Thanks

1/10^1 = 0.1
1/10^2 = 0.01
1/10^3 = 0.001
...

Thus, both 1/10^(n+1) and 1/10^n when expressed as decimals are of a type 0.001 (some number of zeros before 1). So, not that both are equal to 0.001 but both will be 0. followed by some number of zeros before 1.

Thanks so much! It's crystal clear now. Much appreciated for quick response.
Manager
Joined: 27 Oct 2019
Posts: 54
12 Easy Pieces (or not?)  [#permalink]

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21 Nov 2019, 09:51
hmasand wrote:
Is this true tho?

The question specifically asks 1.5 hours BEFORE they meet. This would mean that (assuming they met in 4 hours. [25 x 4 = 100 + 65 x 4 = 260 TOTAL 360] The question asks about the 2.5th hour (4 hours - 1.5 hours), they would have been 100 miles apart at this juncture (25 x 2.5 = 62.50 + 65 x 2.5 = 162.5)

Bunuel wrote:
3. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet?
A. 25 miles
B. 65 miles
C. 70 miles
D. 90 miles
E. 135 miles

Make it simple! The question is: how far apart will they be exactly 1.5 hours before they meet? As Fanny and Alexander's combined rate is 25+65 mph then 1.5 hours before they meet they'll be (25+65)*1.5=135 miles apart.

This is the same doubt I have..
Bunuel can you please clarify this doubt..

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Manager
Joined: 27 Oct 2019
Posts: 54
Re: 12 Easy Pieces (or not?)  [#permalink]

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21 Nov 2019, 10:00
Bunuel wrote:
12. If $${-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}$$ and $${-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}$$, what is the least value of $$x^2*y$$ possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6

To get the least value of $$x^2*y$$, which obviously will be negative, try to maximize absolute value of $$x^2*y$$, as more is the absolute value of a negative number "more" negative it is (the smallest it is).

To maximize $$|x^2*y|$$ pick largest absolute values possible for $$x$$ and $$y$$: $$(-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}$$. Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).

Is there any other way to do this questions??

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Math Expert
Joined: 02 Sep 2009
Posts: 60460
Re: 12 Easy Pieces (or not?)  [#permalink]

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22 Nov 2019, 01:09
ssshyam1995 wrote:
hmasand wrote:
Is this true tho?

The question specifically asks 1.5 hours BEFORE they meet. This would mean that (assuming they met in 4 hours. [25 x 4 = 100 + 65 x 4 = 260 TOTAL 360] The question asks about the 2.5th hour (4 hours - 1.5 hours), they would have been 100 miles apart at this juncture (25 x 2.5 = 62.50 + 65 x 2.5 = 162.5)

Bunuel wrote:
3. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet?
A. 25 miles
B. 65 miles
C. 70 miles
D. 90 miles
E. 135 miles

Make it simple! The question is: how far apart will they be exactly 1.5 hours before they meet? As Fanny and Alexander's combined rate is 25+65 mph then 1.5 hours before they meet they'll be (25+65)*1.5=135 miles apart.

This is the same doubt I have..
Bunuel can you please clarify this doubt..

Posted from my mobile device

I don't understand what is done there. If you want to solve this question long way, then:
1. The would meet in 360/(25 + 65) = 4 hours.
2. In 2.5 hours they together would cover 2.5*(25 + 65) = 225 miles, so 360 - 225 = 135 miles would be left to cover in the remaining 1.5 hours.
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12 Easy Pieces (or not?)  [#permalink]

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12 Dec 2019, 08:38
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

I have a question Bunuel,

So, lowest / minimum value of x is not -3. its -2 and the highest of x is 4 and the same applies for Y. It is not inclusive, according to the question. But why does our range have to be with those values ?
-11<y-x<11 ( not inclusive of 11. Makes no difference to the answer, but I want to know )
What am I doing wrong ?
12 Easy Pieces (or not?)   [#permalink] 12 Dec 2019, 08:38

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