Question

If a, b, c, d are four distinct numbers chosen from the set {1, 2, 3, ….., 9}, then the minimum value of $\dfrac{a}{b}+\dfrac{c}{d}$ is:
(a) $\dfrac{3}{8}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{13}{36}$
(d) $\dfrac{25}{72}$

Answer 1

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,

& L=\dfrac{1}{9}+\dfrac{2}{8} \\\ & =\dfrac{8+18}{72} \\\ & =\dfrac{26}{72} \\\ & =\dfrac{13}{36} \\\ & =0.36111 \\\ \end{aligned}$$ Therefore, we get L = 0.36111 (Approx) Now, if b= 8 and d = 9. We get, $$L=\dfrac{1}{8}+\dfrac{2}{9}$$ $$\begin{aligned} & =\dfrac{9+16}{72} \\\ & =\dfrac{25}{72} \\\ & =0.34722 \\\ \end{aligned}$$ Therefore, in this case we get L = 0.34722 (Approx) As we can see that, $$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.