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Manager
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Re: 12 Easy Pieces (or not?)
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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



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Re: 12 Easy Pieces (or not?)
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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 yx? A. 4<yx<4 B. 2<yx<4 C. 12<yx<4 D. 12<yx<12 E. 4<yx<12
To get max value of yx take max value of y and min value of x: 9(3)=12; To get min value of yx take min value of y and max value of x: 7(5)=12;
Hence, the range of all possible values of yx is 12<yx<12.
Answer: D. Why is it not A? Can’t we simply subtract x from y? Then the answer will be A? Posted from my mobile device



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Re: 12 Easy Pieces (or not?)
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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 yx? A. 4<yx<4 B. 2<yx<4 C. 12<yx<4 D. 12<yx<12 E. 4<yx<12
To get max value of yx take max value of y and min value of x: 9(3)=12; To get min value of yx take min value of y and max value of x: 7(5)=12;
Hence, the range of all possible values of yx is 12<yx<12.
Answer: D. 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). 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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Re: 12 Easy Pieces (or not?)
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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 yx? A. 4<yx<4 B. 2<yx<4 C. 12<yx<4 D. 12<yx<12 E. 4<yx<12
To get max value of yx take max value of y and min value of x: 9(3)=12; To get min value of yx take min value of y and max value of x: 7(5)=12;
Hence, the range of all possible values of yx is 12<yx<12.
Answer: D. 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). 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!



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Re: 12 Easy Pieces (or not?)
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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 yx? A. 4<yx<4 B. 2<yx<4 C. 12<yx<4 D. 12<yx<12 E. 4<yx<12
To get max value of yx take max value of y and min value of x: 9(3)=12; To get min value of yx take min value of y and max value of x: 7(5)=12;
Hence, the range of all possible values of yx is 12<yx<12.
Answer: D. Why is it not A? Can’t we simply subtract x from y? Then the answer will be A? Posted from my mobile deviceHere's how to solve using your approach: A: 7<y<9 B: 3<x<5 this means B': 5<x<3 A+B':: 12< yx < 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 subtractingThat’s a great insight! I’ll have this fitted in my brains! Thanks a ton



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Re: 12 Easy Pieces (or not?)
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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).
Answer: B. 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. Would you please explain? Thanks



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Re: 12 Easy Pieces (or not?)
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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).
Answer: B. 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. Would you please explain? 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.




Re: 12 Easy Pieces (or not?)
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