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Algebra Level 4

Carlos ran a race at a competition. The course was measured at 1000 m {1000\text{ m}} correct to two significant figures and his time was given as 156 s {156\text{ s}} correct to the nearest second. If the difference between his maximum and minimum possible average speed can be expressed as a b m/s \frac{a}{b}\text{ m/s} , where a a and b b are positive integers , find a + b a+b .


The answer is 107583.

Armain Labeeb by Armain Labeeb , 20, Bangladesh

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1 solution

Armain Labeeb
Jul 19, 2016

Basic Upper and Lower bound question.

Using a table I am going to find out the upper and lower bound for the distance and time.

The lower bound for the distance is 1000 − 5 = 995 m 1000-5=\boxed{995m}

The lower bound for the time is 156 − 0.5 = 155.5 s 156-0.5=\boxed{155.5s}

The upper bound for the distance is 1000 + 5 = 1005 m 1000+5=\boxed{1005m}

The upper bound for the time is 156 + 0.5 = 156.5 s 156+0.5=\boxed{156.5s}

Lower Bound Upper Bound
Distance 995 m 995m 1005 m 1005m
Time 155.5 s 155.5s 156.5 s 156.5s

As time is inversely proportional to speed, we have,

S m a x = D m a x T m i n = 1005 155.5 m / s = 2010 311 m / s S m i n = D m i n T m a x = 995 156.5 m / s = 1990 313 m / s a b = 2010 311 − 1990 313 = 2010 ( 313 ) − 1990 ( 311 ) 311 × 313 = 629130 − 618890 97343 = 10240 97343 (10240 and 97343 are co prime) ∴ a + b = 10240 + 97343 = 107583 \large\begin{aligned} { S }_{ max } & =\frac { { D }_{ max } }{ { T }_{ min } } & \\ & =\frac { 1005 }{ 155.5 } { m }/{ s } & =\boxed { \frac { 2010 }{ 311 } m/s } \\ { S }_{ min } & =\frac { { D }_{ min } }{ { T }_{ max } } & \\ & =\frac { 995 }{ 156.5 } { m }/{ s } & =\boxed { \frac { 1990 }{ 313 } m/s } \\ \frac { a }{ b } & =\frac { 2010 }{ 311 } -\frac { 1990 }{ 313 } & \\ & =\frac { 2010(313)-1990(311) }{ 311\times 313 } & \\ & =\frac { 629130-618890 }{ 97343 } & \\ & =\frac { 10240 }{ 97343 } & \text{(10240 and 97343 are co prime)} \\ \therefore \quad \quad \quad a+b & =10240+97343 & \\ & =\boxed{107583} & \end{aligned}

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why is the upper bound 1005m???

Silkin Mong - 4 years, 10 months ago

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Ahh yes.. I was waiting for someone to ask that! ' The upper bound is actually 1004. 9 ˙ 1004.\dot { 9 } which is infinitely close to 1005 1005 so we can consider to be 1005 1005 . Actually it can be written as 1004. 9 ˙ = 1005 1004.\dot { 9 }=1005

Armain Labeeb - 4 years, 10 months ago

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