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Sunny goes mountain biking every Saturday. The trail is $96$ km long, and is uphill all the way to the top.       
His uphill speed is $16$ km per hour slower than his normal speed.    
His downhill speed is $8$ km faster than his normal speed.    
If a round trip on the trail takes $6$ hours, what is his normal speed (in km/h)?   

 Feb 9, 2021
 #1
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Let  the  normal speed  =  S

 

And  distance / rate  = time....so

 

Time Up  +  Time Down =   6...so....

 

96 / ( S  -16)    + 96 / (S + 8)   =    6                 divide  through by 6

 

16/ (S - 16)  + 16/ ( S + 8)   =  1

 

[16 ( S + 8)  +  16 ( S -16)  ]  /  [ (S -16) ( S + 8)  ] =   1

 

[ 32S   - 128  ]   =   (S  -16) ( S + 8)

 

32S  -  128  =   S^2  - 8S   -  128

 

32S  =  S^2 - 8S

 

40S  = S^2

 

S^2  - 40S  =   0

 

S ( S -40)  =  0

 

S - 40  =  0 

 

S = 40  km/h

 

 

cool cool cool

 Feb 9, 2021

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