Mathematics
08.02.2020 10:46
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There are 21 wheels at the bike shop. The wheels will be used to build tricycles

There are 21 wheels at the bike shop. The wheels will be used to build tricycles and bicycles. Some day, the owner would like to build unicycles , too. There will behalf as many tricycles as bicycles. How many of each type of bike will be built?
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paulusl19
paulusl19
5,0(85 marks)

There are 3 tricycles and 9 bicycles.

Step-by-step explanation:

Since we have given that

Number of wheels at the bike shop = 21

Number of wheels bicycle has = 2

Number of wheels tricycle has = 3

Let the number of tricycle be 'x'.

Let the number of bicycle be '2x'.

According to question, we get that

3x+2\times 2x=21\\\\3x+4x=21\\\\7x=21\\\\x=\dfrac{21}{7}\\\\x=3

So, Number of bicycle would be 3x=3\times 3=9

Hence, there are 3 tricycles and 9 bicycles.

freakymnt
freakymnt
4,7(72 marks)

\large\underline{\sf{Solution-}}

Given series is

\rm \longmapsto\:\displaystyle\sum_{k=1}^{\infty}\rm \displaystyle\sum_{n=1}^{\infty}\rm  \frac{k}{ {2}^{n + k} }

can be further rewritten as

\rm \:  =  \: \:\displaystyle\sum_{k=1}^{\infty}\rm \displaystyle\sum_{n=1}^{\infty}\rm  \frac{k}{ {2}^{k} . {2}^{n} }

can be further rewritten as

\rm \:  =  \: \displaystyle\sum_{k=1}^{\infty}\rm \bigg[\dfrac{k}{ {2}^{k}}\displaystyle\sum_{n=1}^{\infty}\rm  \dfrac{1}{ {2}^{n} } \bigg]

\rm \:  =  \: \displaystyle\sum_{k=1}^{\infty}\rm \dfrac{k}{ {2}^{k} }\bigg(\dfrac{1}{2}  + \dfrac{1}{ {2}^{2} }  + \dfrac{1}{ {2}^{3} }  -  -  -  \infty \bigg)

So, its an infinite GP series with

\purple{\rm \longmapsto\:a = \dfrac{1}{2}}

\purple{\rm \longmapsto\:r = \dfrac{1}{2}}

We know,

Sum of infinite GP series with common ratio r ( - 1 < r < 1 ) and first term a is given by

\boxed{\tt{  \:  \: S_ \infty  =  \frac{a}{1 - r} , \:  \: provided \: that \:  |r|  < 1}}

So, 1using this, we get

\rm \:  =  \: \displaystyle\sum_{k=1}^{\infty}\rm  \frac{k}{ {2}^{k} }\bigg( \frac{ \frac{1}{2} }{1 -  \frac{1}{2} }\bigg)

\rm \:  =  \: \displaystyle\sum_{k=1}^{\infty}\rm  \frac{k}{ {2}^{k} }\bigg( \frac{ \frac{1}{2} }{ \frac{2 - 1}{2} }\bigg)

\rm \:  =  \: \displaystyle\sum_{k=1}^{\infty}\rm  \frac{k}{ {2}^{k} }\bigg( \frac{ \frac{1}{2} }{ \frac{1}{2} }\bigg)

\rm \:  =  \: \displaystyle\sum_{k=1}^{\infty}\rm  \frac{k}{ {2}^{k} }

\rm \:  =  \: \dfrac{1}{2}  + \dfrac{2}{ {2}^{2} }  + \dfrac{3}{ {2}^{3} }  + \dfrac{4}{ {2}^{4} }  +  -  -  -  \infty

Let assume that

\rm \longmapsto\:S_ \infty   =  \: \dfrac{1}{2}  + \dfrac{2}{ {2}^{2} }  + \dfrac{3}{ {2}^{3} }  + \dfrac{4}{ {2}^{4} }  +  -  -  -  \infty

Now, its an infinite Arithmetico Geometrico Series, So multiply by 1/2 we get

\rm \longmapsto\:\dfrac{1}{2} S_ \infty   =  \: \dfrac{1}{ {2}^{2} }  + \dfrac{2}{ {2}^{3} }  + \dfrac{3}{ {2}^{4} }  + \dfrac{4}{ {2}^{5} }  +  -  -  -  \infty

On Subtracting above two equations, we get

\rm \longmapsto\:\dfrac{1}{2} S_ \infty   =  \: \dfrac{1}{2}  + \dfrac{1}{ {2}^{2} }  + \dfrac{1}{ {2}^{3} }  + \dfrac{1}{ {2}^{4} }  +  -  -  -  \infty

\rm \longmapsto\:\dfrac{1}{2} S_ \infty   =  \:  \dfrac{ \dfrac{1}{2} }{1 -  \dfrac{1}{2} }

\rm \longmapsto\:\dfrac{1}{2} S_ \infty   =  \:  \dfrac{ \dfrac{1}{2} }{ \dfrac{1}{2} }

\rm \longmapsto\:\dfrac{1}{2} S_ \infty   =  \: 1

\bf\implies \:S_ \infty  = 2

Hence,

\\ \rm \longmapsto\:\boxed{\tt{  \:  \:  \:  \: \displaystyle\sum_{k=1}^{\infty}\rm \displaystyle\sum_{n=1}^{\infty}\rm  \frac{k}{ {2}^{n + k} }  = 2 \: \:   \:  \: }} \\

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