A new computer can perform x calculations in y seconds and an older computer can perform r calculations in s minutes. If these two computers work simultaneously, how many calculations can be performed in t minutes ?

Options

A. `t((x)/(60y)+(r )/(s))`
B. `t((60x)/(y)+(r )/(s))`
C. `t((x)/(y)+(r )/(s))`
D. `t((x)/(y)+(60r)/(s))`

Answer

B

Explanation

Take the computers one at a time. Put everything in terms of seconds : t minutes is 60t seconds and s minutes is 60s seconds.
The new computer performs `(x)/(y)` calculations per second, which is `(60tx)/(y)` calculations in t minutes. The old computer performs `(r )/(60s)` calculations per second, which is `(60tr)/(60s)=(tr)/(s )` calculations in t minutes. The number of calculations they perform together is
`(60tx)/(y)+(tr)/(s)=((60x)/(y)+(r )/(s))`