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Out of 2000 employees in an office 48% preferred Coffee (C ), 54% liked (T), 64% used to smoke (S). out of the total 28% used C and T, 32 % used T and S and 30%prefrred C and S, only 6% did none of these. The number of employees preferring only coffee is
- a)100
- b)260
- c)160
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Out of 2000 employees in an office 48% preferred Coffee (C ), 54% like...
Given data:
Total number of employees = 2000
Percentage of employees who prefer Coffee (C) = 48%
Percentage of employees who like Tea (T) = 54%
Percentage of employees who smoke (S) = 64%
Percentage of employees who use C and T = 28%
Percentage of employees who use T and S = 32%
Percentage of employees who prefer C and S = 30%
Percentage of employees who do not prefer any of these = 6%
To find:
The number of employees preferring only Coffee
Solution:
Let's assume the number of employees who prefer only coffee, only tea and only smoke as x, y and z respectively.
Number of employees who prefer both Coffee and Tea = 28% of 2000 = 560
Number of employees who prefer both Tea and Smoke = 32% of 2000 = 640
Number of employees who prefer both Coffee and Smoke = 30% of 2000 = 600
Number of employees who prefer only Coffee = x
Number of employees who prefer only Tea = y
Number of employees who prefer only Smoke = z
Number of employees who prefer Coffee and Tea but not Smoke = 48% - x - 560 - y
Number of employees who prefer Tea and Smoke but not Coffee = 54% - y - 640 - (48% - x - 560 - y)
Number of employees who prefer Coffee and Smoke but not Tea = 64% - z - 600 - (48% - x - 560 - y)
Number of employees who do not prefer any of these = 6%
Total number of employees = x + y + z + (48% - x - 560 - y) + (54% - y - 640 - (48% - x - 560 - y)) + (64% - z - 600 - (48% - x - 560 - y)) + 6%
Simplifying the above equation, we get:
2000 = 6% + 46% + x + y + z
1944 = x + y + z
Number of employees who prefer only Coffee = x = (48% - 28% - 30% + 6%) of 2000 = 160.
Therefore, the number of employees who prefer only Coffee is 160.
Total number of employees = 2000
Percentage of employees who prefer Coffee (C) = 48%
Percentage of employees who like Tea (T) = 54%
Percentage of employees who smoke (S) = 64%
Percentage of employees who use C and T = 28%
Percentage of employees who use T and S = 32%
Percentage of employees who prefer C and S = 30%
Percentage of employees who do not prefer any of these = 6%
To find:
The number of employees preferring only Coffee
Solution:
Let's assume the number of employees who prefer only coffee, only tea and only smoke as x, y and z respectively.
Number of employees who prefer both Coffee and Tea = 28% of 2000 = 560
Number of employees who prefer both Tea and Smoke = 32% of 2000 = 640
Number of employees who prefer both Coffee and Smoke = 30% of 2000 = 600
Number of employees who prefer only Coffee = x
Number of employees who prefer only Tea = y
Number of employees who prefer only Smoke = z
Number of employees who prefer Coffee and Tea but not Smoke = 48% - x - 560 - y
Number of employees who prefer Tea and Smoke but not Coffee = 54% - y - 640 - (48% - x - 560 - y)
Number of employees who prefer Coffee and Smoke but not Tea = 64% - z - 600 - (48% - x - 560 - y)
Number of employees who do not prefer any of these = 6%
Total number of employees = x + y + z + (48% - x - 560 - y) + (54% - y - 640 - (48% - x - 560 - y)) + (64% - z - 600 - (48% - x - 560 - y)) + 6%
Simplifying the above equation, we get:
2000 = 6% + 46% + x + y + z
1944 = x + y + z
Number of employees who prefer only Coffee = x = (48% - 28% - 30% + 6%) of 2000 = 160.
Therefore, the number of employees who prefer only Coffee is 160.
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Community Answer
Out of 2000 employees in an office 48% preferred Coffee (C ), 54% like...
NCC) = 960
n (T) = 1080
n (s) = 1280
n(C∩T)= 560
n(T∩S) = 640
n(c∩s) = 600
n(TUSUC)' = 120
n(TUSUC) = n(U)-n(TUSUC)'
= 2000 - 120
-1880
n (TUSUC) = nCA) +n(c) + n (s) - n(c∩T)-n(T∩s)-n(c∩s)+n(T∩s∩c)
n(T∩s∩c)=360
n(c only) = n(c) - [n(c∩T)+n (c∩s) ]+n(T∩s∩c)
= 960 - [560 + 600] +360 = 160
n (T) = 1080
n (s) = 1280
n(C∩T)= 560
n(T∩S) = 640
n(c∩s) = 600
n(TUSUC)' = 120
n(TUSUC) = n(U)-n(TUSUC)'
= 2000 - 120
-1880
n (TUSUC) = nCA) +n(c) + n (s) - n(c∩T)-n(T∩s)-n(c∩s)+n(T∩s∩c)
n(T∩s∩c)=360
n(c only) = n(c) - [n(c∩T)+n (c∩s) ]+n(T∩s∩c)
= 960 - [560 + 600] +360 = 160
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Out of 2000 employees in an office 48% preferred Coffee (C ), 54% liked (T), 64% used to smoke (S). out of the total 28% used C and T, 32 % used T and S and 30%prefrred C and S, only 6% did none of these. The number of employees preferring only coffee isa)100b)260c)160d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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