Given: At a lab, bacteria P multiplies itself in every 18 days, while bacteria Q multiplies itself in every 15 days.
Asked: Approximately by what percent is the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself in a 3-year period?
3-year = 3*365 = 1095 days
The number of times bacteria P multiplies itself in 3-year period = 1095/18 = 61 times
The number of times bacteria Q multiplies itself in 3-year period = 1095/15 = 73 times
The percent by which the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself in a 3-year period = ((1095/15)/(1095/18) - 1)*100% = 300/15 % = 20%
IMO C
Asked: Approximately by what percent is the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself in a 3-year period?
3-year = 3*365 = 1095 days
The number of times bacteria P multiplies itself in 3-year period = 1095/18 = 61 times
The number of times bacteria Q multiplies itself in 3-year period = 1095/15 = 73 times
The percent by which the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself in a 3-year period = ((1095/15)/(1095/18) - 1)*100% = 300/15 % = 20%
IMO C
Statistics : Posted by Kinshook • on 24 Sep 2018, 01:31 • Replies 6 • Views 4945
