If the algorithm above is applied, assigning 100 as the value of x and

Question

Step 1: Assign a value for x and a value for y
Step 2: If x < y, then go to Step 5; otherwise, go to step 3
Step 3: Divide by 2 and set the result to be the new value of x
Step 4: Go to step 2
Step 5: Write the value of x
Step 6: End
 
If the algorithm above is applied, assigning 100 as the value of x and 3 as the value of y, how many times is step 2 performed before step 5 is reached?

A. Three
B. Four
C. Five
D. Six
E. Seven

gmatophobia · Accepted Answer

The steps above can be represented in a flow diagram as shown below - Screenshot 2024-04-11 004004.png Iteration 1 : x = 100 , y = 3 x < y ⇒ No x = \frac{x}{2} = 50 Iteration 2 : x = 50 , y = 3 x < y ⇒ No x = \frac{x}{2} = 25 Iteration 3 : x = 25 , y = 3 x < y ⇒ No x = \frac{x}{2} = 12.50 Iteration 4 : x = 12.50 , y = 3 x < y ⇒ No x = \frac{x}{2} = 6.25 Iteration 5 : x = 6.25 , y = 3 x < y ⇒ No x = \frac{x}{2} = 3.125 Iteration 6 : x = 3.125 , y = 3 x < y ⇒ No x = \frac{x}{2} = 1.XX Iteration 7 : x = 1.XX , y = 3 x < y ⇒ Yes Go to Step 5 Hence step 2 is performed seven times before step 5 is reached. Option E

GMATinsight · Answer

LOOP-1
Step 1: Assign a value for x=100 and a value for y=3
Step 2: If x < y, then go to Step 5; otherwise, go to step 3 (Since x > y so go to step-3)
Step 3: Divide by 2 (100/2=50) and set the result to be the new value of x
Step 4: Go to step 2

LOOP-2
Step 1: Assign a value for x=50 and a value for y=3
Step 2: If x < y, then go to Step 5; otherwise, go to step 3 (Since x > y so go to step-3)
Step 3: Divide by 2 (50/2=25) and set the result to be the new value of x
Step 4: Go to step 2

LOOP-3
Step 1: Assign a value for x=25 and a value for y=3
Step 2: If x < y, then go to Step 5; otherwise, go to step 3 (Since x > y so go to step-3)
Step 3: Divide by 2 (25/2=12.5) and set the result to be the new value of x
Step 4: Go to step 2

LOOP-4
Step 1: Assign a value for x=12.5 and a value for y=3
Step 2: If x < y, then go to Step 5; otherwise, go to step 3 (Since x > y so go to step-3)
Step 3: Divide by 2 (12.5/2=6.25) and set the result to be the new value of x
Step 4: Go to step 2

LOOP-5
Step 1: Assign a value for x=6.25 and a value for y=3
Step 2: If x < y, then go to Step 5; otherwise, go to step 3 (Since x > y so go to step-3)
Step 3: Divide by 2 (6.25/2=3.125) and set the result to be the new value of x
Step 4: Go to step 2

LOOP-6
Step 1: Assign a value for x=3.125 and a value for y=3
Step 2: If x < y, then go to Step 5; otherwise, go to step 3 (Since x > y so go to step-3)
Step 3: Divide by 2 (3.125/2=1.5625) and set the result to be the new value of x
Step 4: Go to step 2

LOOP-7
Step 1: Assign a value for x=1.5625 and a value for y=3
Step 2: If x < y, then go to Step 5; otherwise, go to step 3 (Since x < y so go to step-5)
Step 3: Divide by 2 and set the result to be the new value of x
Step 4: Go to step 2
Step 5: Write the value of x=1.5625
Step 6: End

i.e. 7 time step-2 is performed before step-5 is reached

Answer: Option E

Swagatalaxmi

Step 1: Assign a value for x and a value for y
Step 2: If x < y, then go to Step 5; otherwise, go to step 3
Step 3: Divide by 2 and set the result to be the new value of x
Step 4: Go to step 2
Step 5: Write the value of x
Step 6: End

If the algorithm above is applied, assigning 100 as the value of x and 3 as the value of y, how many times is step 2 performed before step 5 is reached?

A. Three
B. Four
C. Five
D. Six
E. Seven

KarishmaB · Answer

It is an iterative algorithm in which the value of x is written after it is made smaller than y. The important thing to evaluate is how many times we encounter step 2, not just how many times division by 2 is done.

We start with x = 100 in step 1 and go to step 2 for the first time with x = 100. Since x < y, we will go to step 3 and make x = 50. 
Next we will got o step 2 again with 50.

Now count on your fingers: 100, 50, 25, 12.5, 6.75, 3.something (still greater than y so will be divided again), 1.5something (Now it is less than y so will be printed)

Hence 7 times we will arrive at step 2.

Answer (E)

daynesj24 · Answer

This one isn't too challenging, however, how would I know the first 100 and 3 counts as one iteration?

HarshavardhanR · Answer

That first 100 and 3 should count, logically.

Step 1: x = 100; y = 3.
Step 2: is x < y? No (100 is not < 3). So move to Step 3.

We are going through Step 2 for the first time right here. This would count as the first time step 2 is executed.

Overall ->

(1) x = 100;  100 not < 3 ; so, new x = 100/2 = 50
(2) x = 50;  50 not < 3 ; so, new x = 50/2 = 25
(3) x = 25;  25 not < 3 ; so, new x = 25/2 = 12.5
(4) x = 12.5;  12.5 not < 3 ; so, new x = 12.5/2 = 6.25
(5) x = 6.25;  6.25 not < 3 ; so, new x = 6.25/2 = 3.125
(6) x = 3.125;  3.125 not < 3 ; so, new x = 3.125/2 = 1.something
(7) x = 1.something;  it is < 3 ; so, finally, move to step 5 (value of x) and end the algo.

How many times was step 2 performed (from the start of the algo) before step 5 was reached? 7.

___
Harsha
Passionate about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

wicoci2835 · Answer

I did it like this,
here, 100 is divided by 2 until it reaches a value less than 3, which can be represented as
   <3
n has to be equal to six.
So, in the sixth iteration, 100 becomes less than 3 in step 3. 
However, until it goes to step 2 again, it doesn't go to step 5, so in iteration 7, 100<3 is verified.
Answer: Seven.

froppy · Answer

Why would the starting value be a step..................

sarthak1701 · Answer

The key to note here is that ASSESSING whether x is larger than y is the 2nd step and NOT dividing the value of x by 2.

bumpbot · Answer

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If the algorithm above is applied, assigning 100 as the value of x and

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