Bunuel
Yesterday, on an audio streaming platform, the number of songs streamed was thrice the number of podcasts streamed. Yesterday, was the platform’s revenue from song streams greater than its revenue from podcast streams?
(1) The average (arithmetic mean) revenue earned per podcast stream on the platform yesterday was 12 cents more than the average revenue earned per song stream on the platform yesterday.
(2) Yesterday, the average revenue per stream, calculated across both songs and podcasts, exceeded 20 cents.
Let number of songs streamed = s
Let number of podcasts streamed = p
Let average revenue from one song streamed = Rs
Let average revenue from one podcast streamed = Rp
Also given that s = 3p
To answer whether s * Rs > p * Rp
=> 3p * Rs > p * Rp
=> 3Rs > Rp --------- (a)
We have to answer whether 3Rs > Rp?
(1) The average (arithmetic mean) revenue earned per podcast stream on the platform yesterday was 12 cents more than the average revenue earned per song stream on the platform yesterday.
Rp = Rs + 12
Using this in equation (a)
3Rs > Rp
=> 3Rs > Rs + 12
=> 2Rs > 12
=> Rs > 6
Hence this statement satisfies the condition only when Rs > 6 else it will fail
Insufficient(2) Yesterday, the average revenue per stream, calculated across both songs and podcasts, exceeded 20 cents.
=> \(\frac{(s * Rs + p * Rp) }{ (s + p)}\) > 20
=> \(\frac{(3p * Rs + p * Rp) }{ (3p + p)}\) > 20
=> \(\frac{(3p * Rs + p * Rp) }{ (4p)}\) > 20
=> 3Rs + Rp > 80
With this we cannot say if 3Rs > Rp
3Rs could be smaller than Rs or vice versa
Insufficient Using both statements together we get
Rp = Rs + 12 ------------ (b)
and
3Rs + Rp > 80
=> 3Rs + Rs + 12 > 80
=> 4Rs > 68
=> Rs > 17
Let's test some values
Let Rs = 18, then Rp = 18+12= 30
3Rs = 54
=> 3Rs > Rp
With a linear relationship between Rs and Rp, we can conclude that 3Rs > Rp
Going further will only increase the difference between 3Rs and Rp, i.e 3Rs >> Rp
Both statements together are SufficientIMHO Option C