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11 May 2018, 01:09
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D
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23% (03:03) correct
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In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky Sports, and 90 watch ABC. 50 students watch both Fox News and Sky Sports, 60 students watch both Sky sports and ABC and 65 students watch both Fox News and ABC. What is the minimum number of students who watch at least one of the given channels?
A. 35
B. 50
C. 135
D. 150
E. 165
A. 35
B. 50
C. 135
D. 150
E. 165
12 May 2018, 05:25
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One formula for 3 overlapping sets is as follows:
Total = Group 1 + Group 2 + Group 3 - (at least 2 of the groups) + (all 3 groups) + none.
In the problem above:
Total = 250.
Group 1 = 120.
Group 2 = 80.
Group 3 = 90.
(at least 2 of the groups) = 50+60+65 = 175.
Let x = all 3.
Let N = none.
Plugging these values into the equation above, we get:
250 = 120 + 80 + 90 - 175 + x + N.
250 = 115 + x + N
N = 250 - 115 - x.
N = 135 - x.
To MINIMIZE the number who watch at least 1 channel, we must MAXIMIZE the value of N: the number who watch NONE of the channels.
In the equation above, the value of N will be maximized if x (all 3) is as small as possible.
Of the given overlaps -- 50, 60, 65 -- the two greatest overlaps both include A:
(at least S and A) + (at least F and A) = 60+65 = 125.
Now this value ((at least S and A) + (at least F and A)) is 35 greater than the total number who watch A (90).
Implication:
At least 35 students must watch all 3 channels.
Thus, the minimum value of x = 35.
Plugging x=35 into N = 135 - x, we get:
N = 135 - 35 = 100.
Since the maximum number who watch none of the channels = 100, the minimum number who watch at least 1 channel = 250-100 = 150.
Total = Group 1 + Group 2 + Group 3 - (at least 2 of the groups) + (all 3 groups) + none.
In the problem above:
Total = 250.
Group 1 = 120.
Group 2 = 80.
Group 3 = 90.
(at least 2 of the groups) = 50+60+65 = 175.
Let x = all 3.
Let N = none.
Plugging these values into the equation above, we get:
250 = 120 + 80 + 90 - 175 + x + N.
250 = 115 + x + N
N = 250 - 115 - x.
N = 135 - x.
To MINIMIZE the number who watch at least 1 channel, we must MAXIMIZE the value of N: the number who watch NONE of the channels.
In the equation above, the value of N will be maximized if x (all 3) is as small as possible.
Of the given overlaps -- 50, 60, 65 -- the two greatest overlaps both include A:
(at least S and A) + (at least F and A) = 60+65 = 125.
Now this value ((at least S and A) + (at least F and A)) is 35 greater than the total number who watch A (90).
Implication:
At least 35 students must watch all 3 channels.
Thus, the minimum value of x = 35.
Plugging x=35 into N = 135 - x, we get:
N = 135 - 35 = 100.
Since the maximum number who watch none of the channels = 100, the minimum number who watch at least 1 channel = 250-100 = 150.
Krotishka1
14 May 2018, 16:42
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gmatbusters
To minimize the number of students who watch at least one channel, we want to maximize the number of the students who watch none of these channels.
We can create the equation:
250 = 120 + 80 + 90 - (50 + 60 + 65) + triple + none
250 = 290 - 175 + t + n
250 = 115 + t + n
135 - t = n
Since we want to maximize the value of n, we need to minimize the value of t. Let’s let a = the number of students who watch ABC only. Recall that t is the number of students who watch all 3 channels. Thus we have 65 - t students watching ABC and Fox (but not Sky) and 60 - t students watching ABC and Sky (but not Fox) and we can create the equation for all the students who watch ABC:
(65 - t) + (60 - t) + t + a = 90
125 - t + a = 90
35 = t - a
Since we want to minimize the value of t, so a must be 0. Thus t = 35
Therefore, n = 135 - t = 135 - 35 = 100.
Since there could be a maximum 100 students who watch none of the 3 channels, there must be a minimum of 250 - 100 = 150 who watch at least one channel.
Answer: D
