Statistics and Sets Problems

[swerve]

A new cell phone plan is offering to price based on average monthly use. Brandon and Jodie are comparing their average use to determine the best plan for them. Brandon's average monthly usage in 2001 was q minutes. Was this less than, greater than, or equal to Jodie's 2001 average monthly usage, in minutes?

1) From January to August 2001, Jodie's average monthly usage was 1.5q minutes.
2) From April to December 2001, Jodie's average monthly usage was 1.5q minutes.

The OA is B

Source: Manhattan Prep

[deloitte247]

Brandon's average monthly use in 2001 =q min
Let Jodie's average monthly use in 2001=x minutes
Jodie's 2001 average monthly usage in minutes
i.e To check if q.x or q<x or q=x

Statement 1
From January to August = 8 months
$$1.5q=\frac{Total\ \min}{8}$$
and total months for 8 months =12q

If Jodie did not make any more calls from September to December or
$$x=\frac{12q}{12}=q\min$$
If Jodie made a call of at least 1 min from September to December then,
$$x=\frac{12q+4}{12}=\frac{12q}{12}+\frac{4}{12}=\left(q+\frac{1}{3}\right)\min$$
With the information in Statement 1
x is either equal to q or greater than q (with 1/3) but since we cannot decipher the logical position or value of x.

$$Statement\ 1\ is\ INSUFFICIENT.$$

Statement 2
From April to December, Jolie's average Monthly usage is 1.5q min
April to December = 9 months
$$Average=\ \frac{Total\ \min s\ spent\ on\ calls}{no\ of\ months\ }$$
$$1.5q=\ \frac{Total\ \min utes\ }{9}$$
Total minutes for 9 months =13.5q
If Jodie did not make any call from January to march then,
$$x=\frac{13.5q}{12}=1.125q\min
$$ $$If\ Jodie\ made\ at\ least\ one\ \min\ of\ call\ from\ Jan\ to\ Mar\ then$$
$$x=\frac{13.5q+3}{12}=\left(1.125q+\frac{1}{4}\right)\min$$
$$both\ cases\ x>q,\ Hence\ statement\ 2\ alone\ is\ SUFFICIENT.
$$ $$answer\ \ is\ Option\ B$$