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03 Apr 2020, 04:19
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
69% (02:30) correct
31%
(02:57)
wrong
based on 136
sessions
History
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A website is attracting 448,000 users per day. Suppose that users only visit the site between 8am and 10pm, and do so evenly between these time points. Before noon, two percent of users click on ads displayed on the website. After noon, only one percent of users click on ads. How many users click on ads between 11:30am and 12:30pm?
A. 960
B. 320
C. 160
D. 480
E. 640
A. 960
B. 320
C. 160
D. 480
E. 640
03 Apr 2020, 20:44
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So, we have 448000 users distributed evenly, so time frame is 14 hrs.
Let us check how many come in an hour...=\(\frac{448000}{14}=32000\)...
Before noon, 2% click on the ads —— 32000*2/100=640
After noon, 1% click on the ads —— 32000*1/100=320
Now 11:30am to 12:30 pm consists of half hour before noon and half hour after noon.
So the answer is \(\frac{1}{2}(640)+\frac{1}{2}(320)=320+160=480\)
D
08 Apr 2020, 02:01
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Just a thought:
Just because users visit a website evenly doesn't mean that they will click on ads evenly too. Lets says per hour 100 users visit the site and in the first hour 50 users click on ads and in the second hour 90 users click on ads. In that case we won't be able to solve the question. In order to solve the question, it should have been mentioned that users click on ads evenly.
chetan2u can this question be solved, if users do NOT click on ads evenly?
