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GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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23 Jul 2012, 16:53
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The flowchart represents a mathematical algorithm that takes two positive integers as the input and returns a positive integer as the output. Processes are indicated in the rectangular symbols in the flowchart. Each process is symbolized by an equation, such as T = T + a. In this particular process, the current values of the variables T and a are added together and the sum then becomes the value of T. For example, if the value of T is 3 and the value of a is 7 before the process T = T + a is completed, then the value of T is 10 and the value of a is 7 after the process is completed. Use the dropdown menus to fill in the blanks in the following statements based on the algorithm represented by the flowchart. If 24 and 35 are entered as the values for a and b, respectively, then the first nonzero value of T is: 24, 48, 96, 192, 384. If 35 and 27 are entered as the values for a and b, respectively, then after the process b=\(\frac{b}{2}\) is completed for the second time, the value of b is: 3, 6, 12, 13, 26.
Last edited by bmwguy on 23 Jul 2012, 17:27, edited 1 time in total.



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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23 Jul 2012, 17:00
bmwguy wrote: Attachment: preppic5.jpg The flowchart represents a mathematical algorithm that takes two positive integers as the input and returns a positive integer as the output. Processes are indicated in the rectangular symbols in the flowchart. Each process is symbolized by an equation, such as T = T + a. In this particular process, the current values of the variables T and a are added together and the sum then becomes the value of T. For example, if the value of T is 3 and the value of a is 7 before the process T = T + a is completed, then the value of T is 10 and the value of a is 7 after the process is completed. Use the dropdown menus to fill in the blanks in the following statements based on the algorithm represented by the flowchart. If 24 and 35 are entered as the values for a and b, respectively, then the first nonzero value of T is: 24, 48, 96, 192, 384. If 35 and 27 are entered as the values for a and b, respectively, then after the process is completed for the second time, the value of b is: 3, 6, 12, 13, 26. I think your second question was transcribed incorrectly"the process" doesn't tell us WHICH process should be repeated twice!
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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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23 Jul 2012, 17:29
KapTeacherEli wrote: I think your second question was transcribed incorrectly"the process" doesn't tell us WHICH process should be repeated twice! I fixed it. Thank you!



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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14 Aug 2012, 07:39
I got this question correct by simply putting in the values and following the schematic. Was there supposed to be any cool way of figuring out without just plugging the numbers in?



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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14 Aug 2012, 09:42
Injuin wrote: I got this question correct by simply putting in the values and following the schematic. Was there supposed to be any cool way of figuring out without just plugging the numbers in? Not that I can see. Near as I can tell, this is similar to "symbol" questions on the quantitative sectionsimple substitution obscured by a difficult way of expressing the math.
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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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07 Jun 2013, 19:22
Can anyone explain the first question? The value for T is not output until b=1 so we would have to iterate then loop until b=1 so a T is output. Maybe I'm just missing something.



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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09 Jun 2013, 21:33
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the 1st question start>T=0>ENTER 35 and 27> IS 27 ODD? YES >T=0+24=24 ANSW IS 24
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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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10 Jun 2013, 08:51
I understand the first value greater than zero for T is 24 as this is the first iteration but we don't output a value for T from the flow chart until b=1 so until we can get b=1 from 37 (b1 then b/2) we never actually output a value for T. From a computer programming perspective I think this question if flawed if the answer to the first part is 24.



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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11 Jun 2013, 16:33
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Nwsmith11 wrote: I understand the first value greater than zero for T is 24 as this is the first iteration but we don't output a value for T from the flow chart until b=1 so until we can get b=1 from 37 (b1 then b/2) we never actually output a value for T. From a computer programming perspective I think this question if flawed if the answer to the first part is 24. Hey, Whether or not a value is actually an 'output' is irrelevant; usually, in a program, this may be implemented as showing the user a value, or perhaps passing the value to another process. If they ask about the 'value of T', all that matters is what value it takes on in memory; if you were stepping through the program in a debugger, and you checked for what the variable 'T' contains in memory at each step, you would see all of those values changing regardless of the value actually being 'outputted'. Hope that helps.
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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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11 Jun 2013, 16:34
Quote: Hey,
Whether or not a value is actually an 'output' is irrelevant; usually, in a program, this may be implemented as showing the user a value, or perhaps passing the value to another process.
If they ask about the 'value of T', all that matters is what value it takes on in memory; if you were stepping through the program in a debugger, and you checked for what the variable 'T' contains in memory at each step, you would see all of those values changing regardless of the value actually being 'outputted'.
Hope that helps. Awesome thanks! Guess I was reading too far into the question.



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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11 Jun 2013, 17:15
KapTeacherEli wrote: Injuin wrote: I got this question correct by simply putting in the values and following the schematic. Was there supposed to be any cool way of figuring out without just plugging the numbers in? Not that I can see. Near as I can tell, this is similar to "symbol" questions on the quantitative sectionsimple substitution obscured by a difficult way of expressing the math. I agree that given the particular questions that they asked about this pattern, solving it this way is probably the easiest. If they asked questions that required looping through the block diagram many times, however, I feel that pattern recognition would be very important for maximizing time efficiency. For example, suppose that the first question asked "What is the final value of T?" In that case, iterating through the loop would be too mechanical, and I think it would be more useful to recognize the pattern and calculate the final answer that way. So essentially, we see that with B = 35 as an initial value, the highest power of 2 contained within 35 is 5, so we know that A will be doubled 5 times, making the final value for A equivalent to 24 * 2^5 = 24*32 = 768. From there, we recognize that A is added to T for every time B takes on an odd value: starting with 35, we can quickly see that 35 is odd, 17 is odd, and 1 is odd, so T will be composed of the first, second, and last values that are taken on by A; i.e, 24, 24*2, 24*32 = 24*35 = 840. In fact, this algorithm just implements T = A * B, which would be crucial to recognize if we were given very large numbers. I suppose a formal proof for this is probably going too far, but here's a somewhat intuitive explanation: Suppose we have some arbitrary value for A, and B = 61. The highest power of 2 that is less than 61 is 5 (32), so we know that A will be doubled 5 times; i.e, multiplied by 32. So to get T = A*61, we need to show that T is composed of A*32 + A*(6132 = 29). Since we know that only A*(powers of 2) are added to T, we can break down T as follows: 61 = 32 + 16 + 8 + 4 + 1 = the sum of the following powers of 2: (0, 2, 3, 4, 5). We can see this relationship by breaking down B = 61 as follows: 61  1 = 60  Odd 60/2 = 30  First Halving; 2^1 30/2 = 15, 151 = 14  2^2  Odd 14/2 = 7, 71 = 6  2^3  Odd 6/2 = 3, 31 = 2  2^4  Odd 2/2 = 1  2^5  Odd * Sorry for the poor formatting * We can see that the number is halved 5 times, corresponding to the 5 times that A is doubled. Beside each layer is marked the corresponding 'doubling' of A. Now, we recognize that an odd number is encountered at the 2^0, 2^2, 2^3, 2^4, and 2^5 layers, meaning we would be adding A*(1 + 4 + 8 + 16 + 32) to T, equating to 61*A. I know that was kind of longwinded but hopefully it's useful/interesting to someone. I found it pretty neat.
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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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30 Jun 2013, 02:31
mattce wrote: KapTeacherEli wrote: Injuin wrote: I got this question correct by simply putting in the values and following the schematic. Was there supposed to be any cool way of figuring out without just plugging the numbers in? Not that I can see. Near as I can tell, this is similar to "symbol" questions on the quantitative sectionsimple substitution obscured by a difficult way of expressing the math. I agree that given the particular questions that they asked about this pattern, solving it this way is probably the easiest. If they asked questions that required looping through the block diagram many times, however, I feel that pattern recognition would be very important for maximizing time efficiency. For example, suppose that the first question asked "What is the final value of T?" In that case, iterating through the loop would be too mechanical, and I think it would be more useful to recognize the pattern and calculate the final answer that way. So essentially, we see that with B = 35 as an initial value, the highest power of 2 contained within 35 is 5, so we know that A will be doubled 5 times, making the final value for A equivalent to 24 * 2^5 = 24*32 = 768. From there, we recognize that A is added to T for every time B takes on an odd value: starting with 35, we can quickly see that 35 is odd, 17 is odd, and 1 is odd, so T will be composed of the first, second, and last values that are taken on by A; i.e, 24, 24*2, 24*32 = 24*35 = 840. In fact, this algorithm just implements T = A * B, which would be crucial to recognize if we were given very large numbers. I suppose a formal proof for this is probably going too far, but here's a somewhat intuitive explanation: Suppose we have some arbitrary value for A, and B = 61. The highest power of 2 that is less than 61 is 5 (32), so we know that A will be doubled 5 times; i.e, multiplied by 32. So to get T = A*61, we need to show that T is composed of A*32 + A*(6132 = 29). Since we know that only A*(powers of 2) are added to T, we can break down T as follows: 61 = 32 + 16 + 8 + 4 + 1 = the sum of the following powers of 2: (0, 2, 3, 4, 5). We can see this relationship by breaking down B = 61 as follows: 61  1 = 60  Odd 60/2 = 30  First Halving; 2^1 30/2 = 15, 151 = 14  2^2  Odd 14/2 = 7, 71 = 6  2^3  Odd 6/2 = 3, 31 = 2  2^4  Odd 2/2 = 1  2^5  Odd * Sorry for the poor formatting * We can see that the number is halved 5 times, corresponding to the 5 times that A is doubled. Beside each layer is marked the corresponding 'doubling' of A. Now, we recognize that an odd number is encountered at the 2^0, 2^2, 2^3, 2^4, and 2^5 layers, meaning we would be adding A*(1 + 4 + 8 + 16 + 32) to T, equating to 61*A. I know that was kind of longwinded but hopefully it's useful/interesting to someone. I found it pretty neat. What a brilliant post. Took me a while to figure this out, but this is great. Thank you.



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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27 Apr 2014, 09:28
If 35 and 27 are entered as the values for a and b, respectively, then after the process b=\(\frac{b}{2}\) is completed for the second time, the value of b is: 3, 6, 12, 13, 26. [/quote] Could someone help me solve the second question ?Thanks



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Re: GMATPrep IR: Flowchart representing mathematical algorithm [#permalink]
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19 May 2014, 11:21
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gmatgambler wrote: If 35 and 27 are entered as the values for a and b, respectively, then after the process b=\(\frac{b}{2}\) is completed for the second time, the value of b is: 3, 6, 12, 13, 26. Could someone help me solve the second question ?Thanks[/quote] To solve the second question you take the new values for A and B (35 & 27) and you reinput them into the flowchart. Keep in mind it asks you for the value of B after the process of b = b/2 is completed for the SECOND time. So treating the flow chart as new, you would simply follow the process twice (until you reach B=B/2 for the second time!), Starting from the beginning I'll go through the flow the first time around, Start> 'T = 0'> 'Enter A and B' A = 35, B = 27 > 'Is B Odd' Yes, its 27 > 'T=T+A' 0 + 35 = 35, t = 35 (no relevance to this question but you continue in this direction) > 'B = B  1' B = 27 1, B = 26 > 'A = 2A' A = 2(35), A = 70 (again, no relevance to this question but you continue in this direction) > 'B = B/2' B = 26/2, B = 13 > 'Is B =1?' Because at this point B = 13, the answer is NO, and because the answer is NO you follow the flow chart back to the question 'Is B Odd'From there you continue through the flow chart like before until you get to 'B = B 1', B = 131, B = 12 Continuing on you will arrive back to 'B = B/2', which gives us our answer of B = 12/2, B = 6. The first time around at B=B/2 leaves us with the answer 13, but because the question specifically asks for the value of B the SECOND TIME AROUND the value of B = 6




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