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At company X, 44% of the employees that own laptops do not [#permalink]

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07 Jul 2012, 05:36

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At company X, 44% of the employees that own laptops do not own cellphone, and 30% of the employees that own cellphone do not own laptops. If the number of employees without laptops is 40% less than the employees with laptops, the what percent of the company's employees neither own a laptop nor a cellphone ??

A. 17.5 B. 22.5 C. 24 D. 26 E. 28

In this tricky problem I'm stuck on 30% of the employees. Is this question properly formulated , it seems a bit confused ....

At company X, 44% of the employees that own laptops do not own cellphone, and 30% of the employees that own cellphone do not own laptops. If the number of employees without laptops is 40% less than the employees with laptops, the what percent of the company's employees neither own a laptop nor a cellphone ??

A. 17.5 B. 22.5 C. 24 D. 26 E. 28

In this tricky problem I'm stuck on 30% of the employees. Is this question properly formulated , it seems a bit confused ....

Say x% of the employees own laptops. Since "the number of employees without laptops is 40% less than the employees with laptops", then \(100-x=0.6*x\) --> \(x=62.5\), so 62.5% of the employees own laptops.

Next: "44% of the employees that own laptops do not own cellphone", so 56% of the employees that own laptops own cellphone, so own both laptops and cellphone --> 0.56*62.5={Both}.

"30% of the employees that own cellphone do not own laptops", so 70% of the employees that own cellphone own laptops, so own both laptops and cellphone --> 0.7*y={Both}, where y is percentage of the employees who own cellphone.

From above: 0.56*62.5=0.7*y --> y=50 and {Both}=0.7*y=35, so 50% of the employees own cellphone and 35% of the employees own both laptops and cellphone.

Re: At company X, 44% of the employees that own laptops do not [#permalink]

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09 Jul 2012, 13:40

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Hello Bunuel, Can you tell me what am I doing wrong...

I assume that let the total number of ppl be x, then let the intersection be y and remaining be z...then .44x is the ppl with laptops but not cellphones; .3x is the number of ppl with cellphone but not laptops

Thus as per the equation: .3x+z <ppl who do not have laptops> = 0.6(.44x+y). ...so they are basically asking z/x*100 ....But this is not yielding the right answer

Re: At company X, 44% of the employees that own laptops do not [#permalink]

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09 Jul 2012, 15:16

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Considering the total of people with laptops are 100 and without are 60 so in total are 160

Then you have laptop and NO laptop in the tree. Then under laptop you have cell and No cell and the same under NO laptop

So, under laptop you have 0.56 % cellphone and 0.44 % under NO cell phone.

Under the second branch of tree you have people with cellphone that is X (you do not know nothing about this value) and HERE IS THE TRICKY part of this difficult problem.

When the statement says: 30% of people with cellphone do not have laptop you have to consider 30 % of 0.56 (people with cellphone under LAPTOP) + people with celphone under NO laptop ----->> y= 0.3 (y + 0.56x ).

Even though the amazing explanation of bunuel I fought with this part really tricky (at least for me) an entire evening.

Re: At company X, 44% of the employees that own laptops do not [#permalink]

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10 Jul 2012, 03:43

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I spent some time writing this difficult problem and made sure the wording was not ambiguous and close to the types I have seen on the GMAT. I usually approach these overlapping set problems with a tree format, although many students prefer drawing a table. Nevertheless, it is a difficult problem that trips me up as well.

Here is a link to the problem that I had posted on my blog along with a video explanation:

Re: At company X, 44% of the employees that own laptops do not [#permalink]

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10 Jul 2012, 03:50

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dabral wrote:

I spent some time writing this difficult problem and made sure the wording was not ambiguous and close to the types I have seen on the GMAT. I usually approach these overlapping set problems with a tree format, although many students prefer drawing a table. Nevertheless, it is a difficult problem that trips me up as well.

Here is a link to the problem that I had posted on my blog along with a video explanation:

Dear Dabral , infact I read this question on your very well done website and I'm confortable too with tree diagram, infact using table I didn't spot the connection between cellphone . Compliments

At first look seemed wording (my fault). indeed a really good question. Thanks for your work.
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Re: At company X, 44% of the employees that own laptops do not [#permalink]

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13 Aug 2012, 04:36

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I solved it using table, although it took around 4 minutes but it was neat and accurate. I initially took X for total laptop owners and Y for Total Cellphone owners. Then solved for total people without laptops =.6X (laptop owners[X]-without laptop[WL]=40%of laptop owners[X]) Then the net total becomes X+.6X=1.6X. Solve for Y under cellphone column i.e Y=.3Y+.56X resulting in Y=.8X and thats it. In the end i got the value under "neither cellphone nor laptop column" = .36X. and (.36X/1.6X)*100=22.5%

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Re: At company X, 44% of the employees that own laptops do not [#permalink]

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15 Aug 2013, 07:15

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It took 5 min. complex problem but not difficult.
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Re: At company X, 44% of the employees that own laptops do not [#permalink]

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15 Aug 2013, 08:05

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carcass wrote:

At company X, 44% of the employees that own laptops do not own cellphone, and 30% of the employees that own cellphone do not own laptops. If the number of employees without laptops is 40% less than the employees with laptops, the what percent of the company's employees neither own a laptop nor a cellphone ??

A. 17.5 B. 22.5 C. 24 D. 26 E. 28

In this tricky problem I'm stuck on 30% of the employees. Is this question properly formulated , it seems a bit confused ....

Fact I "44% of the employees that own laptops do not own cellphone, and 30% of the employees that own cellphone do not own laptops."

=> 56% of the employees that own laptops own cellphone as well, and 70% of the employees that own cellphone own laptops as well --> employees having both laptops and cellphones

=> 56% of laptop owners = 70% of cell phone owners => ratio of owners: Laptop(L):Cellphone(C) = 70/56= 5:4

Fact II

"the number of employees without laptops is 40% less than the employees with laptops"

People with Laptop = L% People without Laptop = 0.6L%

L + 0.6L = 100 => L=62.5 of total employees

If 62.5% employees own Laptops, then from the ratio (4/5)*62.5 = 50% employees own Cellphones.

and 56% of Laptop owners own Cellphones => 62.5*(56/100) = 35% total employees own both Laptops and Cellphones

Now, Laptop Owners = 62.5% Cellphone Owners = 50% Both Laptops and Cellphones owners = 35%

Owners of Either Laptop or Cellphones or Both = 62.5 + 50 - 35 = 77.5% Employees with neither Laptop nor Cellphones = 100-77.5 = 22.5%

Ans B.
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Re: At company X, 44% of the employees that own laptops do not [#permalink]

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23 Oct 2013, 07:41

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Let number of employees who own a laptop be 100 => L=100 Therefore, the number of employees who do not own a laptop = L - 0.40L = 100 - 0.4*100 = 60 The number of employees who own a laptop but not a cell phone = 0.44*L = 44 Therefore, the number of employees who own both a laptop and a cellphone = 100-44 = 56 Let the number of employees who own a cellphone be c. Given that 30% of c do not own a laptop. Therefore 70% of c own a laptop. 0.70*c = 56 => c=80 Therefore, number of employees who own neither a laptop nor a cellphone = 80-44 = 36. Therefore, percentage of such employees = (36/160)*100 = 22.5

Re: At company X, 44% of the employees that own laptops do not [#permalink]

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05 Oct 2014, 04:18

carcass wrote:

At company X, 44% of the employees that own laptops do not own cellphone, and 30% of the employees that own cellphone do not own laptops. If the number of employees without laptops is 40% less than the employees with laptops, the what percent of the company's employees neither own a laptop nor a cellphone ??

A. 17.5 B. 22.5 C. 24 D. 26 E. 28

In this tricky problem I'm stuck on 30% of the employees. Is this question properly formulated , it seems a bit confused ....

I am wondering where am i going wrong with my approach. Could you please help me pinning down the flaw

Y - total # of employees Only Laptop = (44/ 100) Y Only Cellphone = (30/ 100) Y Laptop and Cellphone- W

So Laptop - ( 44/100)Y + W

Given condition:

{{[( 44/100)Y + W ] - [(30 / 100) Y] } / ( 44/100)Y + W ] } = (4/10)

On solving W - (3/50) Y

So adding (44/ 100) Y +(30/ 100) Y+(3/50) Y = 80/100 Y

Y - 80/100 Y = 20/100 Y So I am getting 20%
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At company X, 44% of the employees that own laptops do not [#permalink]

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04 Jan 2015, 07:00

carcass wrote:

At company X, 44% of the employees that own laptops do not own cellphone, and 30% of the employees that own cellphone do not own laptops. If the number of employees without laptops is 40% less than the employees with laptops, the what percent of the company's employees neither own a laptop nor a cellphone ??

A. 17.5 B. 22.5 C. 24 D. 26 E. 28

In this tricky problem I'm stuck on 30% of the employees. Is this question properly formulated , it seems a bit confused ....

--------------------------LAPTOP---------------NO LAPTOP--------TOTAL CELL PHONE-----------0.56X=0.7Y---------0.3Y-----------------Y----- NO CELL PHONE-------0.44X---------------- ------------------- --- TOTAL-------------------X-----------------------100-X--------------100-- GIVEN : ( X-100+X)/X = 0.4 ; X=62.5 ; WE CAN SOLVE FOR Y AND REST OF THE CELLS . ANSWERS IS 22.5
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Re: At company X, 44% of the employees that own laptops do not [#permalink]

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