Re: In a company of 100 employees, 75 have at Laptop, 80 have at least one
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08 Dec 2019, 04:19
The answer is C: 45.
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.