Question

A Web music store offers two versions of a popular song. The size of the standard version is 2.6 megabytes (MB). The size of the high-quality version is 4.7 MB.Yesterday, there were 1110 downloads of the song, for a total download size of 4629 MB. How many downloads of the high-quality version were there?

Answer 1

According to the statement:

0. the size of the standard version is ,s1 = 2.6MB,,

,

1. the size of the high-quality version is, s2 = 4.7MB,,

,

2. there were ,N = 1100, downloads of the song,

,

3. the total download size is ,S = 4629,, in MB.

Now, we can write the following equations:

(E1) The total number of downloads N is the sum of the number of downloads of each version:

$N=n_1+n_2,$

where n1 is the number of downloads of the standard version and n2 is the number of downloads of the high-quality version.

(E2) The total downloaded size S is the sum of the product of the size of each version by the number of downloads of that version:

$S=s_1\cdot n_1+s_2\cdot n_2\text{.}$

We want to find n2, the number of downloads of the high-quality version.

Now, replacing the values of the statement in the equations above we have:

$\begin{gathered} (E3)\rightarrow1100=n_1+n_{2,} \\ (E4)\rightarrow4629=2.6_{}\cdot n_1+4.7_{}\cdot n_2\text{.} \end{gathered}$

Using equation (E3) we express n1 in terms of n2:

$(E5)\rightarrow n_1=1100-n_2\text{.}$

Replacing the last equation in equation (E4) we have:

$(E4)\rightarrow4629=2.6_{}\cdot(1100-n_2)+4.7_{}\cdot n_2\text{.}$

Solving the last equation for n2 we get:

$\begin{gathered} 4629=2.6\cdot1100-2.6\cdot n_2+4.7\cdot n_2,_{} \\ 4629=2860+2.1\cdot n_2, \\ 4629-2860=2.1\cdot n_2, \\ 2.1\cdot n_2=1769, \\ n_2=\frac{1769}{2.1}, \\ n_2\cong842.38, \\ n_2\cong842. \end{gathered}$

Answer

Round to the integer, there were 842 downloads of the high-quality version.