Among the computers at a store, 55 have a hard drive with at least 1

Question

Among the computers at a store, 55 have a hard drive with at least 1 terabyte capacity, and 40 have at least a 3.2-gigahertz CPU. If the total number of computers exceeds 100, and 20 of the computers have neither a hard drive with 1-terabyte or greater capacity nor a CPU of 3.2 gigahertz or greater, which of the following could be the number of computers that have both a 1-terabyte or greater hard drive and a 3.2-gigahertz or greater CPU?

I. 10
II. 15
III. 20

A. I only
B. II only
C. III only
D. I and II
E. II and III

Show Answer

Official Answer

A

I. 10 
II. 15 
III. 20 
 
A. I only 
B. II only 
C. III only 
D. I and II 
E. II and III

sahibr · Accepted Answer

In such questions, I personally always find it easier to map the data in a tabular form. (I don't know if there is a name for this method) Anywho, first and foremost, using the data given in the question, make a table as given below: untitled.png I have marked the unknown ie. computers >1TB and >3.2 G Hz as x I have marked the other unknown ie. Total no. of computers in the store as T Now using the table, we can form the following equations: untitled (2).png Now lets take the bottommost horizontal row, (Even the rightmost vertical column can be considered

) We know 55 + 60 - x = T 115-x = T <------------- (1) We know that T has to be greater than 100. Plugging in the values into (1), we get I) x = 10, T = 105 ----> Hence, possible. II) x= 15, T = 100 ----> Not possible as T>100. III) x = 20, T = 95 -----> Again, not possible. Hence only (I) is possible. Option A is our answer.

Loser94 · Answer

Total no of computers with 1tb HD=55
with 3.2 Tb CPU=40
computers with don't have anything =20
Now depending upon total no of computers , number of computers having both will change
Total comp are more than 100 .
So assuming comp will be around 101 will give us computers have HD and CPU or both will be 81
and total of COmp with HD and CPU= 55+40=95
so comp with both will be 95-81=14.

So answer will be less than or equal to 14

So Option A is 5
Answer will be A.

CrackverbalGMAT · Answer

Let A = At least 1 TB Hard Drive = 55

Let B = At least 3.2GHZ CPU = 40

Neither = (A U B)' = 20

The Universal set U = Total = A U B + (AUB)'

A U B = A + B - (A  \cap  B)

Therefore U = A + B - (A  \cap  B) + (A U B)'

(A  \cap  B) = A + B + (A U B)' - U

If the total computers, U = 101, then

(A  \cap  B) = 55 + 40 + 20 - 101 = 14

If the total computers, U = 102, then

(A  \cap B) = 55 + 40 + 20 - 102 = 13

So as the total number of computers increase, (A  \cap  B) decreases. Therefore (A  \cap  B)  \leq  14

Only I satisfies this.

Option A

Arun Kumar

MathRevolution · Answer

Total computers > 100. Let them be 101.

101 - 20 = 55 + 40 - Both

=> Both < 95 - 81 = 14

Only option A satisifes it.

Answer A

taha1234 · Answer

OFFICIAL EXPLANATION

Analyze the question

This Roman numeral format question states that 55 objects have a big hard drive, that 40 have fast processors, and that 20 have neither characteristic. Also, the total number of these objects is greater than 100.

State the task

Use the overlapping sets formula to determine which of the Roman numeral choices could be the number of objects that have both characteristics and select the choice that represents the correct Roman numeral(s).

Approach strategically

The formula for overlapping sets is Group 1 + Group 2 − Both + Neither = Total. In this case, substitute a greater than sign for the equal sign. So 55 + 40 − B + 20 > 100, where B is the number of machines that are both big and fast. Solve for B: 115 − B > 100; 15 > B. Thus, only values less than 15 will work, and that means only Statement I works. (A) is correct.

Confirm your answer

Plug 15, which is Statement II, into the formula: Total = 55 + 40 − 15 + 20 = 100, which is not greater than 100. Moreover, if subtracting 15 results in a total that is too small, than subtracting 20 will as well, and Statement I only is confirmed as the correct choice.

TAKEAWAY: Pay special attention to inequalities, noting that if a value is less/greater than some number (e.g., y > 100), the value cannot be that number (y ≠ 100).

kakakakaak · Answer

but why we are fixin the other two variables .. even they can also vary... as least possible is 55 and 40. so can anyone explain what is wron in my approach

bumpbot · Answer

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Among the computers at a store, 55 have a hard drive with at least 1

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Among the computers at a store, 55 have a hard drive with at least 1

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