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# In a company of 100 employees, 75 have at Laptop, 80 have at least one

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Joined: 27 Oct 2017
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In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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07 Dec 2019, 18:13
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GMATBusters’ Quant Quiz Question -7

In a company of 100 employees, 75 have at Laptop, 80 have at least one cell phone, and 55 have at least one Alexa. If x and y are respectively the greatest and lowest possible number of employees that have all three of these devices, what is the value of x – y ?
A. 65
B. 55
C. 45
D. 35
E. 25

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Posts: 12
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Schools: HBS
Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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07 Dec 2019, 21:36

The maximum can be the lowest of the at least possibility which is 55. The leas can be >=3. i just picked the lowest available in answer choice
Director
Joined: 25 Jul 2018
Posts: 712
Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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07 Dec 2019, 22:06
In total, there are 100 employees
75 laptop —> 25 no laptop
80 cell phone —> 20 no phone
55 Alexa —> 45 no Alexa.

All three devices:
—> the greatest possible number —55 = x

The lowest possible number — 55–45= 10 =y

—> x —y = 55–10 = 45

Posted from my mobile device
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 00:23
If we find the maximum number of employees who has all the three- laptop, Mobile phone and Alexa- the count is 75

If we find the minimum number of employees who has all the three- laptop, Mobile phone and Alexa- the count is 10

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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 00:43
we have
100= 75+80+55-2*( all three)
x= 55 i.e max
and min over lap is 10 each
∆ ; 55-10 ; 45
IMO C

In a company of 100 employees, 75 have at Laptop, 80 have at least one cell phone, and 55 have at least one Alexa. If x and y are respectively the greatest and lowest possible number of employees that have all three of these devices, what is the value of x – y ?
A. 65
B. 55
C. 45
D. 35
E. 25
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 00:57
1
In order to find maximum number of employees who have all three devices, We need to find out maximum common number which comes to 55 which would be value of x
Now to find out minimum we are assuming out of 80 employee having laptops, 25 are different employees having at least one cell phone, so here 55 employees are common and out of 55 employees having at least one Alexa 25 employees are different having at least one Alexa, so here 30 employees are common, so minimum common number comes to 30 which is y
So x-y comes to 55-30 which is 25, hence Option E
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Joined: 15 Apr 2017
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 01:43
1
Ans:C
Maximum value of intersection of all the three sets( Laptop, cell phone, Alexa)= Minimum value of(any three set)=55
Minimum Value of three set = A+B+C-2*Total Value
Hence, Minimum Value of three set = 210-2*100=10
hence x-y= 55-10=45
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Joined: 22 Feb 2018
Posts: 741
Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 04:14
1
2
In a company of 100 employees, 75 have at Laptop, 80 have at least one cell phone, and 55 have at least one Alexa. If x and y are respectively the greatest and lowest possible number of employees that have all three of these devices, what is the value of x – y ?
A. 65
B. 55
C. 45
D. 35
E. 25

The obvious maximum that have all 3 is 55, because it is limited by the SMALLEST number.

The minimum is simply the sum of the max of each people who don't have the product, so:

100-80 = 20 don't have Cell phones
100-75 = 25 don't have Laptops
and 100-55 = 45 don't have Alexa

So, a total of 20+25+45 = 90 combined who might not have some combination of the 3 products. So subtract that from 100, to give you the minimum of the people who could have all 3 and you get 100-90 = 10.
55-10 = 45

Imo. C
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 04:19

EXPLANATION:
Let's assume:
A = number of people having one device.
B = number of people having two devices.
C = number of people having three devices.
D = number of people having zero devices.

So, total number of people A + B + C + D = 100.
And number of devices: A + 2B + 3C + D = 210.

For this question, we are assuming that D, or number of people having no devices, is zero (although it is not given in the question, I would show that we are not getting the answer if we assume there are people having no devices).

So, A + B + C = 100
A + 2B + 3C = 210.

The above equations must be equal, to satisfy the condition of same people having the same devices.

To maximize people having 3 devices, we minimize people having 2 and 1 device.
Hence, we assume A + B = 0, and likewise, A + 2B = 0.

So, 100 - C = 210 - 3C.
This means 2C = 110.
This means, C = 55.

This is the maximum value, or X.

Now, to get minimum value, or Y, let's assume no people have 3 devices. So C = 0.

This means A + B = 100, and A + 2B = 210.
Solving both, we get A = -10. Which is not possible (number of people can't be negative integer).

Now, let's assume number of people having three devices as 10 (as in the answer choice, difference is of 10).
So C = 10. This means 3C = 30.

So, A + B = 90, and A + 2B = 180.
Solving, we get A = 0, and B = 90. This is a valid solution, as it shows there are 90 people having 2 devices each (so 180 devices), and zero people having one device each, and 10 people having three devices each (30 devices). So 100 people having 210 devices is satisfied.

So, Y = 10.
Hence, X - Y = 55 - 10 = 45, the answer.

Why there are no people with zero devices?

If we assume D = 20 (as it is given 80 people have atleast cell phone, so the maximum number of people who would have zero devices is 20, as the same 80 people can possess all three devices):

So equation would be:
A + B + C = 80.
A + 2B + 3C = 210.

Going by the same thought process we applied earlier, first we maximize C, to get X. (A + B, and A + 2B assumed to be zero).
So we get 80 - C = 210 - 3C.
C = 65 = X.

Now, to find minimal value, we again try to get the valid solution, on C = 10, 20, 30, and so forth.

We would find that, only on C = 50, we would get the valid equation:
A + B = 80 - 50
A + 2B = 210 - 150

We would get A = 0, B = 30, and C = 50. Means, zero people with one device, 30 people with 2 devices, and minimum 50 people with 3 devices, and 20 people with 0 devices.

So, again, Y = 50.

But, X - Y = 65 - 50 = 15, which is not given in the answer choice.

Hence, we assume, zero people have no devices.
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 05:16
Clearly, the maximum no. of employees which can have all 3 devices is 55, which is limited by the number of Alexas.
Therefore, x = 55

For minimum number of employees,
We will find the sum of all the employees which might not have any of the product.

100 - 75 = 25 do not have laptop
100 - 80 = 20 do not have cell phone
100 - 55 = 45 do not have Alexa

Therefore, 25 + 20 + 45 = 90 employees might not have all of the three products.

Therefore, y = 100 - 90 = 10
Now, x - y = 55 - 10 = 45

Posted from my mobile device
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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08 Dec 2019, 05:37
the maximum is limited by the smallest value which is 55. all values are above this number. so y= 55
if we see carefully, the difference among the items,

employee with No Laptop =100-75 = 25
employee with No Cell phone =100-80 = 20
employee with No Alexa 100-55 = 45

20+25+45 = 90 people dont have all three items.
That mean 10 may have three items as none is not specified and considered as zero.
so min x =10
the difference is 55-10= 45
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Joined: 09 Dec 2019
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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09 Dec 2019, 05:23
1
Let the employees be numbered from 1 to 100.

For greatest number of employees that have all three devices, let the distribution be as follows:
Laptops to be given to employees numbered from 1 to 75
Cell phones to be given to employees numbered from 1 to 80
Alexa to be given to employees numbered from 1 to 55

So, employees numbered 1 to 55 have all three devices.
⇒ x=55

For least number of employees that have all three devices, let the distribution be as follows:

Alexa to be given to employees numbered from 1 to 55
Laptops to be given to employees numbered from 56 to 100
Now, we are left with 30 laptops which we have to distribute among employees numbered from 1 to 55 (who already have Alexa)
Let employees numbered from 1 to 30 be given the remaining 30 laptops.
So, employees numbered from 1 to 30 have two devices each and from 31 to 100 have one device each.

Let cell phones be given to employees numbered from 31 to 100.
Now, we are left with 10 cell phones which we have to distribute among employees numbered 1 to 30 (who already have Alexa and laptop)
Let employees numbered from 1 to 10 be given cell phones.

So, employees numbered 1 to 10 have all three devices.
⇒ y=10

Therefore, x - y = 55 - 10 = 45
Senior Manager
Joined: 12 Dec 2015
Posts: 493
In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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03 Jan 2020, 08:57
In a company of 100 employees, 75 have at Laptop, 80 have at least one cell phone, and 55 have at least one Alexa. If x and y are respectively the greatest and lowest possible number of employees that have all three of these devices, what is the value of x – y ?
A. 65
B. 55
C. 45 --> correct
D. 35
E. 25

SOLUTION:
let's say each bar/dash(-) =5
----------------:80
----------------:75
----------------: 55
so maximum overlap = x = 55 (left 11 blue overlap bars/dashes)

--------------------: 80
--------------------: 55
--------------------: 75
so minimum overlap = y = 10 (middle 2 blue overlap bars/dashes)

So x-y=55-10=45
Senior Manager
Joined: 12 Dec 2015
Posts: 493
In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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03 Jan 2020, 09:02
In a company of 100 employees, 75 have at Laptop, 80 have at least one cell phone, and 55 have at least one Alexa. If x and y are respectively the greatest and lowest possible number of employees that have all three of these devices, what is the value of x – y ?
A. 65
B. 55
C. 45 --> correct
D. 35
E. 25

SOLUTION:
let's say each bar/dash(-) =5
----------------:80
----------------:75
----------------: 55
so maximum overlap = x = 55 (left 11 blue overlap bars/dashes)

--------------------: 80
--------------------: 55
--------------------: 75
so minimum overlap = y = 10 (middle 2 blue overlap bars/dashes)

So x-y=55-10=45
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Joined: 27 Jan 2017
Posts: 13
Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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29 Mar 2020, 06:03
Max for Min(75,55,80)=55 so tot 100 one 0
Min for 75+80+55=210-100=110 but none (110) cannot be more than total (100) so none=110-100=10 thus None max is 100 therefore common Min=10
Max-min common= 55-10=45
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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01 Apr 2020, 08:39
One of those 700 level question that I did mentally!
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Joined: 16 Jan 2019
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Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one  [#permalink]

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03 Apr 2020, 12:26
Can anyone please post exact methodical solution?
Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one   [#permalink] 03 Apr 2020, 12:26