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04 Feb 2025, 12:07
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D
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Difficulty:
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62% (01:26) correct
38%
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In a room of 100 people, 99% are left-handed. How many left-handed people must leave the room to reduce the percentage to 98%?
A. 1
B. 2
C. 49
D. 50
E. 98
A. 1
B. 2
C. 49
D. 50
E. 98
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13 Feb 2025, 01:45
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Official Solution:
In a room of 100 people, 99% are left-handed. How many left-handed people must leave the room to reduce the percentage to 98%?
A. 1
B. 2
C. 49
D. 50
E. 98
There are 99 left-handed people and 1 right-handed person in the room. After some number of left-handed people leave, the right-handed person will still remain and must represent 2% of the remaining people. Since 2% is equivalent to \(\frac{1}{50}\) of the total, the number of remaining people must be 50 (with the 1 right-handed person making up 2% of this new total). This means that 50 left-handed people must leave the room to bring the percentage of left-handed individuals down to 98%.
Answer: D
In a room of 100 people, 99% are left-handed. How many left-handed people must leave the room to reduce the percentage to 98%?
A. 1
B. 2
C. 49
D. 50
E. 98
There are 99 left-handed people and 1 right-handed person in the room. After some number of left-handed people leave, the right-handed person will still remain and must represent 2% of the remaining people. Since 2% is equivalent to \(\frac{1}{50}\) of the total, the number of remaining people must be 50 (with the 1 right-handed person making up 2% of this new total). This means that 50 left-handed people must leave the room to bring the percentage of left-handed individuals down to 98%.
Answer: D
General Discussion
04 Feb 2025, 12:14
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If there are 100 people, 99% left-handed are 99 people.
Let x be the number of people who leave
(99-x)/(100-x) = 0.98
99-x = 98-0.98x
1=0.02x
x =50
Let x be the number of people who leave
(99-x)/(100-x) = 0.98
99-x = 98-0.98x
1=0.02x
x =50
Bunuel
