Umbrella Corporation monitors a network of 6,000 computers.

Question

Umbrella Corporation monitors a network of 6,000 computers. At the start of the virus outbreak, 4 of these computers get infected. The infection weakens with each round of spreading:

In the first iteration, each of the 4 initially infected machines infects 8 others. 
 In the second iteration, each machine infected in the first iteration infects 4 others. 
 In the third iteration, each machine infected in the second iteration infects 2 others. 
 In the fourth iteration, each machine infected in the third iteration infects 1 other.

Select for  2 iterations  the number of computers that remain uninfected after two iterations are completed, and select for  4 iterations  the number of computers that remain uninfected after four iterations are completed. Make only two selections, one in each column.

Bunuel · Accepted Answer

The network contains 6,000 computers in total, and the outbreak begins with 4 of them infected.

Iteration 1 : Each of the 4 infects 8 others, producing 32 new infections. Adding these to the initial 4 gives 36 infected computers.
 
Iteration 2 : The 32 from the first round each infect 4 others, producing 128 new infections. Adding this to the 36 infected so far gives a cumulative total of 164 after two iterations. The number of unaffected computers is 6,000 - 164 =  5,836 .
 
Iteration 3 : The 128 from the second round each infect 2 others, producing 256 new infections. Cumulative total = 164 + 256 = 420.
 
Iteration 4 : The 256 from the third round each infect 1 other, producing 256 new infections. Cumulative total = 420 + 256 = 676 after four iterations. The number of unaffected computers is 6,000 - 676 =  5,324 .

I29-166

Dereno · Answer

Total computers within the network = 6000

At the START, the number of infections = 4

First iteration :

Each of the 4 initially infected machine infects 8 others.

So, the number of infections = 4*8 = 32.

Second Iteration:

The number of infected computers at the beginning of second iteration = 32.

The number of Infections at second iteration = 32*4 = 128.

The number of infected machines TILL the Second iteration = 4+32+128 =  164 .

The number of UNINFECTED machine = 6000 - 164 =  5836 .

Third Iteration:

The number of infected machines at the beginning of third iteration = 128.

The number of infections at the third iteration = 128*2 = 256.

The number of infected machines TILL the Third iteration = 4+32+128+256=  420 .

The number of UNINFECTED machine = 6000 - 420 =  5580 .

Fourth Iteration:

The number of infected machines at the beginning of the fourth iteration = 256.

The number of infections at the fourth iteration = 256*1 = 256.

The number of infected machines TILL the end of Fourth Iteration = 4+32+128+256+256 = 676.

The number of UNINFECTED machine = 6000 - 676 =  5324 .

The number of UNINFECTED machines after 2nd Iteration = 5836.

The number of UNINFECTED machines after 4th Iteration = 5324.

findingmyself · Answer

The key is to understabd "WHAT infects OTHERS means"

Say here: In the first iteration, each of the 4 initially infected machines infects 8 others.= The total infected after first iteration is not 8*4=32, Instead it is 4(initially infected) which then infects 8 others each which is 8*4=32, totalling to 4+32=36 infected monitors

In the second iteration, each machine infected in the first iteration infects 4 others: Here we have 4+32+32*4=164

In the third iteration, each machine infected in the second iteration infects 2 others: Here we have 4+32+32*4+32*4*2

In the fourth iteration, each machine infected in the third iteration infects 1 other: 4+32+(32*4)+(32*4*2)+(32*4*2)=676

Each time we are adding the previous + newly infected

Thus, our answer is 6000-164=5836 and 6000-676= 5324

AbhishekP220108 · Answer

The key to solving this problem is how efficiently we process the data. Let's dive in
The total number of computers is 6000, and initially, 4 were infected.

1st iteration- each initial 4 will infect 8 others so total infected- 32 in 1st iteration, after 1st iteration total infected are 4+32
2nd iteration- each 32 infected in 1st will infect 4 others, so infected now- 32*4=128, total infected after 2nd iteration are 4+32+128=164
3rd iteration- each 128 infected in 2nd iteration will infect 2 others, so infected now -128*2=256, total infected after 3rd iteration are 4+32+128+256=420
4th iteration-each 256 infected in 3rd iteration will infect 1 other, so infected now-256, total infected after 4th iteration are 4+32+128+256+256=676

we have to select the nos that remain uninfected after 2nd and 4th iteration

after 2nd uninfected=6000-164=5836
after 4th uninfected=6000-676=5324

hope it helps

Umbrella Corporation monitors a network of 6,000 computers.

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2 iterations	4 iterations
		5,260
		5,324
		5,452
		5,488
		5,836
		5,872

GMAT Two-part Analysis (TPA) Questions

Umbrella Corporation monitors a network of 6,000 computers.

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

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My Rewards