Answer:
Let the average speed of the truck be \[x\,\,km/hr\]. Then, new average speed of truck \[=(x+20)\,\,km/hr\]. Time taken by truck to cover\[150\,km=\frac{150}{x}hrs\]. and time taken by truck to cover \[200km=\frac{200}{x+20}\,hrs\]. \[\therefore \frac{150}{x}+\frac{200}{x+20}=5\] \[\frac{150(x+20)+200x}{x(x+20)}=5\] \[150x+3000+200x=5x(x+20)\] \[350x+3000=5{{x}^{2}}+100x\] \[5{{x}^{2}}-250x-3000=0\] \[{{x}^{2}}-50x-600=0\] \[{{x}^{2}}-60x+10x-600=0\] \[x(x-60)+10(x-60)=0\] \[(x-10)(x-60)=0\] \[x=-10\] or \[60\] Since speed cannot be negative. So, \[x=60\] \[\therefore \] First speed of truck \[=60\text{ }km/hr\].
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