The flowchart represents a mathematical algorithm that takes two

Question

https://gmatclub.com/forum/download/file.php?id=89609 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 S = S + x. In this particular process, the current values of the variables S and x are added together and the sum then becomes the value of S. For example, if the value of S is 4 and the value of x is 9 before the process S = S + x is completed, then the value of S is 13 and the value of x is 9 after the process is completed. Use the drop-down menus to fill in the blanks in the following statements based on the algorithm represented by the flowchart. If 12 and 20 are entered as the values for x and y, respectively, then immediately after the process x = 3x is completed for the second time, the value of S is GRAPH_1_PLACEHOLDER . If 15 and 17 are entered as the values for x and y, respectively, then the process S = S + x is completed GRAPH_2_PLACEHOLDER times before the algorithm stops. GMAT-Club-Forum-ivkh393i.png

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=89609

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 S = S + x. In this particular process, the current values of the variables S and x are added together and the sum then becomes the value of S. For example, if the value of S is 4 and the value of x is 9 before the process S = S + x is completed, then the value of S is 13 and the value of x is 9 after the process is completed.

Use the drop-down menus to fill in the blanks in the following statements based on the algorithm represented by the flowchart.

If 12 and 20 are entered as the values for x and y, respectively, then immediately after the process x = 3x is completed for the second time, the value of S is  GRAPH_1_PLACEHOLDER .

If 15 and 17 are entered as the values for x and y, respectively, then the process S = S + x is completed  GRAPH_2_PLACEHOLDER  times before the algorithm stops.

Official Solution:  
 
Drop-down 1:

Start with:

S = 0, x = 12, y = 20

Since 20 is not divisible by 3:

S = 0 + 12 = 12
y = 20 - 1 = 19

Since 19 is not divisible by 3:

S = 12 + 12 = 24
y = 19 - 1 = 18

Since 18 is divisible by 3:

x = 3 * 12 = 36
y = 18/3 = 6

This is the first time the process x = 3x is completed.

Since 6 is also divisible by 3:

x = 3 * 36 = 108

This is the second time the process x = 3x is completed. The value of S has not changed since it became 24.

Therefore, the value of S is  24 .
 
Drop-down 2:

Start with x = 15 and y = 17. We only need to count how many times the process S = S + x is completed.

17 is not divisible by 3, so S = S + x is completed for the first time, and y becomes 16.
16 is not divisible by 3, so S = S + x is completed for the second time, and y becomes 15.
15 is divisible by 3, so x becomes 45 and y becomes 5.
5 is not divisible by 3, so S = S + x is completed for the third time, and y becomes 4.
4 is not divisible by 3, so S = S + x is completed for the fourth time, and y becomes 3.
3 is divisible by 3, so x becomes 135 and y becomes 1.
1 is not divisible by 3, so S = S + x is completed for the fifth time, and y becomes 0.

The algorithm then stops.

Therefore, the process S = S + x is completed  5  times.
 
Correct answer:  
Drop-down 1: "24"
Drop-down 2: "5"

dhart · Answer

24 &5

abhikc3004 · Answer

Making a table and simulating the logic is helpful:

Case 1:

S 
(=S+x) 
  
  x 
(=3x)  
  y  
(=   y/3   
or  y-1)    
 
  0 
 12 
 20 
 
  24 
 12 
 18 
 
  24 
 36 
 6 
 
  24 
 108 
 2

Hence,  S = 24

Case 2:

Condition
Counter 
  S 
(=S+x)    
  x
(=3x)   
  y  
(=   y/3   
or  y-1)

0 
 15 
 17 
 
  1 
 15 
 15 
 16 
 
  2 
 30 
 15 
 15 
 
  - 
 75 
 45 
 5 
 
  3 
 75 
 45 
 4 
 
  4 
 120 
 45 
 3 
 
  - 
 120 
 135 
 1 
 
  5 
 255 
 135 
 0 (end)

The algorithm stops at  y = 0 , hence no. of iterations of  S = S+x  is  5 .

Bunuel  - this took me around 4 minutes to compute, which is feel is a lot. Is there a shortcut/smarter way to tackling such questions?

The flowchart represents a mathematical algorithm that takes two

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\(S \\ (=S+x)\)	\(x \\ (=3x)\)	\(y \) (= \( y/3 \) or \(y-1) \)
0	12	20
24	12	18
24	36	6
24	108	2

Condition Counter	\(S \\ (=S+x) \)	\(x\\ (=3x)\)	\(y \) (= \( y/3 \) or \(y-1) \)
	0	15	17
1	15	15	16
2	30	15	15
-	75	45	5
3	75	45	4
4	120	45	3
-	120	135	1
5	255	135	0 (end)

GMAT Graphs and Tables (G&T) Questions

The flowchart represents a mathematical algorithm that takes two

Prep Toolkit

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My Rewards