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Re: 12 Easy Pieces (or not?)
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27 Mar 2017, 00:32
Bunuel wrote: 9. Julie is putting M marbles in a row in a repeating pattern: blue, white, red, green, black, yellow, pink. If the row begins with blue marble and ends with red marble, then which of the following could be the value of M? A. 22 B. 30 C. 38 D. 46 E. 54
There are total of 7 different color marbles in a pattern. Now, as the row begins with blue marble and ends with red marble (so ends with 3rd marble in a pattern) then M=7k+3. The only answer choice which is multiple of 7 plus 3 is 38=35+3.
Answer: C. Hi Bunuel, I tried to refer other posts for this answer but I did not understand. Can you please elaborate this? how M=7k+3..



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Re: 12 Easy Pieces (or not?)
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27 Mar 2017, 00:48
RMD007 wrote: Bunuel wrote: 9. Julie is putting M marbles in a row in a repeating pattern: blue, white, red, green, black, yellow, pink. If the row begins with blue marble and ends with red marble, then which of the following could be the value of M? A. 22 B. 30 C. 38 D. 46 E. 54
There are total of 7 different color marbles in a pattern. Now, as the row begins with blue marble and ends with red marble (so ends with 3rd marble in a pattern) then M=7k+3. The only answer choice which is multiple of 7 plus 3 is 38=35+3.
Answer: C. Hi Bunuel, I tried to refer other posts for this answer but I did not understand. Can you please elaborate this? how M=7k+3.. The row begins with blue marble and ends with red marble. What cases can we have? {blue, white, red} = 7*0 + 3 {blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*1 + 3 {blue, white, red, green, black, yellow, pink} {blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*2 + 3 {blue, white, red, green, black, yellow, pink} {blue, white, red, green, black, yellow, pink} {blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*3 + 3 ... As you can see in any case the number of marbles must be a multiple of 7 plus 3. Hope it's clear.
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Re: 12 Easy Pieces (or not?)
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29 Mar 2017, 13:59
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).
Answer: D. Is it not the other way around where 1/100 is the least number and 1/18 is the max negative number. I would have picked A as the answer.



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Re: 12 Easy Pieces (or not?)
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29 Mar 2017, 21:38
vs224 wrote: 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).
Answer: D. Is it not the other way around where 1/100 is the least number and 1/18 is the max negative number. I would have picked A as the answer. No. 1/18 = ~0.06 and 1/100 = 0.01 0.06 < 0.01 As you can see 1/100 is to the right of 1/18. Attachment:
MSP3701ihg50f15748di7300006532e8b04fda4c6c.gif [ 1.41 KiB  Viewed 998 times ]
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Re: 12 Easy Pieces (or not?)
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28 May 2017, 22:08
Bunuel wrote: 11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p? A. 97 B. 151 C. 209 D. 211 E. 219
What is the largest possible sum of these three numbers that we can form? Maximize the first digit: 76+73+71=220=even, so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: 67+73+71=211=prime.
Answer: D. Was it implicit that we could only use available digits only once ? I wasted time using same digits multiple times to form different no . Though I ended up with ( 77,73,61 ) which too gave 211 but I arrived at it by brute force. Was there something in the language that I should be careful to identify such scenarios so I don't make this mistake again ?



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Re: 12 Easy Pieces (or not?)
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28 May 2017, 23:34
booksknight wrote: Bunuel wrote: 11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p? A. 97 B. 151 C. 209 D. 211 E. 219
What is the largest possible sum of these three numbers that we can form? Maximize the first digit: 76+73+71=220=even, so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: 67+73+71=211=prime.
Answer: D. Was it implicit that we could only use available digits only once ? I wasted time using same digits multiple times to form different no . Though I ended up with ( 77,73,61 ) which too gave 211 but I arrived at it by brute force. Was there something in the language that I should be careful to identify such scenarios so I don't make this mistake again ? We are given a data set which has three 7's in it. If we could use each number in the set multiple times, then why 7's were written three times?
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Re: 12 Easy Pieces (or not?)
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11 Jul 2017, 04:44
Bunuel wrote: 2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x? A. 5 B. 6 C. 7 D. 18 E. 20
Also tricky. Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: 9, 8, 7, 6, 4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=9\) then \(x_{max}x_{min}=9(9)=18\).
Answer: D. That is tricky. I took 3<x<10 So minimum value of x= 4 Maximum value of x= 9 Bunuel When a question involves ^2 or any even power, then we have to consider the negative value? Will that be the case in all situations?



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Re: 12 Easy Pieces (or not?)
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11 Jul 2017, 04:50
Shiv2016 wrote: Bunuel wrote: 2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x? A. 5 B. 6 C. 7 D. 18 E. 20
Also tricky. Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: 9, 8, 7, 6, 4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=9\) then \(x_{max}x_{min}=9(9)=18\).
Answer: D. That is tricky. I took 3<x<10 So minimum value of x= 4 Maximum value of x= 9 Bunuel When a question involves ^2 or any even power, then we have to consider the negative value? Will that be the case in all situations? Yes, x^2 = (positive integer) has two solutions for x, positive and negative.
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Re: 12 Easy Pieces (or not?)
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12 Jul 2017, 17:28
Bunuel wrote: 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 colours. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480
Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.
Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.
Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.
Answer: B. Hi Bunuel, Is this option possible: *R*R_R*R*R*? Or "in a row" means we can not have any space in between? Thank you



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Re: 12 Easy Pieces (or not?)
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12 Jul 2017, 21:26
Soul777 wrote: Bunuel wrote: 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 colours. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480
Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.
Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.
Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.
Answer: B. Hi Bunuel, Is this option possible: *R*R_R*R*R*? Or "in a row" means we can not have any space in between? Thank you In R*R*R*R*R* each * is a place for 2 blue, 2 green and 1 yellow marbles. So, * cannot accommodate say 2 marbles.
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Re: 12 Easy Pieces (or not?)
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18 Aug 2017, 14:21
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.
Answer: E. I couldn't understand (25+65)*1.5=135 miles apart i above solution. Why are we multiple combined rate with 1.5.



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Re: 12 Easy Pieces (or not?)
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19 Aug 2017, 03:39
ammuseeru wrote: 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.
Answer: E. I couldn't understand (25+65)*1.5=135 miles apart i above solution. Why are we multiple combined rate with 1.5. Because 1.5 hours before they meet, the distance left to cover would be (combined rate)*(time) = (25+65)*1.5 = 135 miles.
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Re: 12 Easy Pieces (or not?)
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17 Dec 2017, 08:09
The worst case scenario will be when 5 white chips are selected in a row. So the answer should be 6 (Option 'C').



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Re: 12 Easy Pieces (or not?)
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17 Dec 2017, 09:08




Re: 12 Easy Pieces (or not?) &nbs
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