Linda left home and drove for $2$ hours. She stopped for lunch then drove for another $3$ hours at a rate that is $10$ mph higher than the rate before she had lunch. If the total distance Linda travelled is $230$ miles, what was the rate before lunch?

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Solution

If $x$ is the rate at which Linda drove before lunch the rate after lunch is equal $x + 10$ . The total distance $D$ travelled by Linda is given by
$D = 2 x + 3 (x + 10)$
and is equal to $230$ miles
Hence, $2 x + 3 (x + 10) = 230$
$\Rightarrow 2 x + 3 x + 30 = 230$
$\Rightarrow 5 x = 230 - 30$
$\Rightarrow 5 x = 200$
$\Rightarrow x = 40$ miles/hour

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Linda left home and drove for 2 hours. She stopped for lunch then drove for another 3 hours at a rate that is 10mph higher than the rate before she had lunch. If the total distance Linda travelled is 230 miles, what was the rate before lunch?

If x is the rate at which Linda drove before lunch the rate after lunch is equal x+10. The total distance D travelled by Linda is given byD=2x+3(x+10)and is equal to 230 miles Hence, 2x+3(x+10)=230⇒2x+3x+30=230⇒5x=230−30⇒5x=200⇒x=40 miles/hour

Linda left home and drove for $2$ hours. She stopped for lunch then drove for another $3$ hours at a rate that is $10$ mph higher than the rate before she had lunch. If the total distance Linda travelled is $230$ miles, what was the rate before lunch?