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Question

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(a) \[\dfrac{3}{8}\]

(b) \[\dfrac{1}{3}\]

(c) \[\dfrac{13}{36}\]

(d) \[\dfrac{25}{72}\]

Answer

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Hint: Make the fraction \[\dfrac{a}{b}\text{ and }\dfrac{c}{d}\] minimum by taking 2 smallest and 2 largest numbers from the set. We have to select a,b,c and d in such a format that numerator terms like a and c should be smaller and denominator terms like b and d should be bigger to find the minimum value.

__Complete step-by-step answer:__

We are given the set of numbers {1, 2, 3, ….., 9}. If a, b, c, d are four distinct numbers chosen from this set, then we have to find the minimum value of

\[L=\dfrac{a}{b}+\dfrac{c}{d}\]

To find the minimum value of \[\dfrac{a}{b}+\dfrac{c}{d}\], we have to choose 4 numbers such that \[\left( \dfrac{a}{b} \right)\text{ and }\left( \dfrac{c}{d} \right)\] have minimum values individually and hence \[\dfrac{a}{b}+\dfrac{c}{d}\] would also have minimum value.

Now, we know that if we take any fraction say, \[\dfrac{N}{D}\] where N is numerator and D is denominator and want to make it minimum, then we have to select the smallest possible number as N and biggest possible number as D.

Hence, to get minimum values of fractions \[\dfrac{a}{b}\text{ and }\dfrac{c}{d}\], we will select two largest numbers from the set {1, 2, 3…..9} and two smallest numbers from set {1, 2, 3…..9}

So, the two largest numbers are 8 and 9 and two smallest numbers are 1 and 2 from the set.

Since, we know that for \[\left( \dfrac{a}{b} \right)\text{ and }\left( \dfrac{c}{d} \right)\] to be minimum, a and c must be taken as numbers 1 and 2, while b and d must be taken as 8 and 9.

Now, let us put a = 1 and c = 2. Therefore, we get

\[L=\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{1}{b}+\dfrac{2}{d}\]

Now if b = 9 and d = 8, we get,

\[\begin{align}

& L=\dfrac{1}{9}+\dfrac{2}{8} \\

& =\dfrac{8+18}{72} \\

& =\dfrac{26}{72} \\

& =\dfrac{13}{36} \\

& =0.36111 \\

\end{align}\]

Therefore, we get L = 0.36111 (Approx)

Now, if b= 8 and d = 9. We get,

\[L=\dfrac{1}{8}+\dfrac{2}{9}\]

\[\begin{align}

& =\dfrac{9+16}{72} \\

& =\dfrac{25}{72} \\

& =0.34722 \\

\end{align}\]

Therefore, in this case we get L = 0.34722 (Approx)

As we can see that,

\[0.36111>0.34722\]

Or, \[\dfrac{13}{36}>\dfrac{25}{72}\]

Therefore we get minimum values of \[\dfrac{a}{b}+\dfrac{c}{d}\] as \[\dfrac{25}{72}\].

Hence, option (d) is correct.

Note: Here, some students take \[\dfrac{a}{b}\text{ as }\dfrac{1}{9}\] and \[\dfrac{c}{d}\text{ as }\dfrac{2}{8}\] and get the wrong answer \[\dfrac{13}{36}\] which is option (c). But they must keep in mind that we not only have to make \[\dfrac{a}{b}\text{ and }\dfrac{c}{d}\] minimum but we also need to make \[\left( \dfrac{a}{b}+\dfrac{c}{d} \right)\] minimum. Therefore, we take \[\dfrac{a}{b}=\dfrac{2}{9}\text{ and }\dfrac{c}{d}=\dfrac{1}{8}\] which makes \[\dfrac{a}{b}+\dfrac{c}{d}\] minimum.

We are given the set of numbers {1, 2, 3, ….., 9}. If a, b, c, d are four distinct numbers chosen from this set, then we have to find the minimum value of

\[L=\dfrac{a}{b}+\dfrac{c}{d}\]

To find the minimum value of \[\dfrac{a}{b}+\dfrac{c}{d}\], we have to choose 4 numbers such that \[\left( \dfrac{a}{b} \right)\text{ and }\left( \dfrac{c}{d} \right)\] have minimum values individually and hence \[\dfrac{a}{b}+\dfrac{c}{d}\] would also have minimum value.

Now, we know that if we take any fraction say, \[\dfrac{N}{D}\] where N is numerator and D is denominator and want to make it minimum, then we have to select the smallest possible number as N and biggest possible number as D.

Hence, to get minimum values of fractions \[\dfrac{a}{b}\text{ and }\dfrac{c}{d}\], we will select two largest numbers from the set {1, 2, 3…..9} and two smallest numbers from set {1, 2, 3…..9}

So, the two largest numbers are 8 and 9 and two smallest numbers are 1 and 2 from the set.

Since, we know that for \[\left( \dfrac{a}{b} \right)\text{ and }\left( \dfrac{c}{d} \right)\] to be minimum, a and c must be taken as numbers 1 and 2, while b and d must be taken as 8 and 9.

Now, let us put a = 1 and c = 2. Therefore, we get

\[L=\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{1}{b}+\dfrac{2}{d}\]

Now if b = 9 and d = 8, we get,

\[\begin{align}

& L=\dfrac{1}{9}+\dfrac{2}{8} \\

& =\dfrac{8+18}{72} \\

& =\dfrac{26}{72} \\

& =\dfrac{13}{36} \\

& =0.36111 \\

\end{align}\]

Therefore, we get L = 0.36111 (Approx)

Now, if b= 8 and d = 9. We get,

\[L=\dfrac{1}{8}+\dfrac{2}{9}\]

\[\begin{align}

& =\dfrac{9+16}{72} \\

& =\dfrac{25}{72} \\

& =0.34722 \\

\end{align}\]

Therefore, in this case we get L = 0.34722 (Approx)

As we can see that,

\[0.36111>0.34722\]

Or, \[\dfrac{13}{36}>\dfrac{25}{72}\]

Therefore we get minimum values of \[\dfrac{a}{b}+\dfrac{c}{d}\] as \[\dfrac{25}{72}\].

Hence, option (d) is correct.

Note: Here, some students take \[\dfrac{a}{b}\text{ as }\dfrac{1}{9}\] and \[\dfrac{c}{d}\text{ as }\dfrac{2}{8}\] and get the wrong answer \[\dfrac{13}{36}\] which is option (c). But they must keep in mind that we not only have to make \[\dfrac{a}{b}\text{ and }\dfrac{c}{d}\] minimum but we also need to make \[\left( \dfrac{a}{b}+\dfrac{c}{d} \right)\] minimum. Therefore, we take \[\dfrac{a}{b}=\dfrac{2}{9}\text{ and }\dfrac{c}{d}=\dfrac{1}{8}\] which makes \[\dfrac{a}{b}+\dfrac{c}{d}\] minimum.