A computer is rendering two scenes. The computer has already rendered 540 thousand pixels ( kpx ) of a kitchen scene and renders another 90kpx each minute. The computer has rendered 960kpx of a garden scene and renders 60kpx more pixels each minute. After how many more minutes will the scenes have the same amounts rendered? minutes

Question

A computer is rendering two scenes. The computer has already rendered 540 thousand pixels (  kpx ) of a kitchen scene and renders another  90kpx each minute. The computer has rendered  960kpx of a garden scene and renders  60kpx more pixels each minute. After how many more minutes will the scenes have the same amounts rendered? minutes

Bytelearn · Accepted Answer

Denoting the number of minutes: Let's denote the number of minutes after which both scenes will have the same amount rendered as 't'. We know that the kitchen scene has already 540kpx rendered and it renders at a rate of 90kpx per minute. So, the total rendered for the kitchen scene after 't' minutes will be: 540kpx + 90kpx * tCalculating the total rendered for the kitchen scene: Similarly, the garden scene has already 960kpx rendered and it renders at a rate of 60kpx per minute. So, the total rendered for the garden scene after 't' minutes will be: 960kpx + 60kpx * tCalculating the total rendered for the garden scene: We want to find the value of 't' when both scenes have the same amount rendered. Therefore, we set the two expressions equal to each other: 540kpx + 90kpx * t = 960kpx + 60kpx * tSetting the expressions equal to each other: Now, we solve for 't'. First, we'll subtract 60kpx * t from both sides to get the terms involving 't' on one side: 90kpx * t - 60kpx * t = 960kpx - 540kpxSolving for 't': Simplifying both sides gives us: 30kpx * t = 420kpxDividing both sides to solve for 't': Now, we divide both sides by 30kpx to solve for 't': t = 420kpx / 30kpxPerforming the division to find 't': Performing the division gives us: t = 14

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