From the question given below, I am trying to, firstly, find the proportion of iPhone users from the information provided in the Business Scenario.

Secondly, I am trying to find the share of Android logins given the updated information.

For these questions, I didn't see the importance of the total average logins of 15 or 18 (for part 2) and so I just used the information of the iPhone logins and Android logins. I wasn't sure how/whether I needed to use the information from the overall average logins?

Therefore, for the first part, my calculation was $$\text{proportion} = \frac{9}{9+19} * 1000000 = 321.429 $$

For part 2 I got 56% using the following calculation $$\frac{19}{15+19} $$

Are these answers correct?



Check calculations

Initial Scenario : 1 M Users with 15 login/month average.

This means : 15 M logins in a month.

Average number of login per month : 9 for iPhone Users and 19 for Android Users.

The equation will be : $19 \times \text A + 9 \times (1 - \text A) = 15$, where $\text A$ is the number of Android Users and $\text {iP} = 1 - \text A$ is the number of iPhone Users.

Solving, we get : $10 \times \text A =6$, and thus $\text A=0,6$.

In conclusion, the Initial Scenario is : 600 K Android Users and 400 K iPhone Users.

Final Scenario : in the final scenario the number of logins is increased by 20%, and thus amounts to 18 M logins ($15 \times 1.2$).

iPhone logins increase by $\dfrac 2 3$ , i.e. from 9 to 15.

The new equation will be : $0,6 \times y + 15 \times 0,4 = 18$, where $y$ is the new average number of logins of Android Users.

Solving for $y$, we get $y=20$, that is the new number of logins per month of Android Users.


Initial Scenario : average number of login per month : 9 for iPhone Users and 19 for Android Users.

Final Scenario : average number of login per month : 15 for iPhone Users and 20 for Android Users.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.