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By the correct number, replace $\square $ in each of the following:
$
  {\text{(a) }}\dfrac{2}{7} = \dfrac{8}{\square } \\
  {\text{(b) }}\dfrac{5}{8} = \dfrac{{10}}{\square } \\
  {\text{(c) }}\dfrac{{45}}{{60}} = \dfrac{{15}}{\square } \\
  {\text{(d) }}\dfrac{{18}}{{24}} = \dfrac{\square }{4} \\
 $

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Hint- Here, we will be obtaining the values of the missing numbers by cross multiplication.
Here let us suppose the missing number be \[x = {\text{ }}\square \]. Now, we will cross multiply the equation and find the value of \[x\].
\[{\text{(a)}}\] Let \[\dfrac{2}{7} = \dfrac{8}{x}\]
Now cross multiplying the above equation, we get
\[ \Rightarrow 2x = 8 \times 7 \Rightarrow 2x = 56 \Rightarrow x = 28\].
Therefore, the correct equation is \[\dfrac{2}{7} = \dfrac{8}{{28}}\].
\[{\text{(b)}}\] Let \[\dfrac{5}{8} = \dfrac{{10}}{x}\]
Now cross multiplying the above equation, we get
\[ \Rightarrow 5x = 8 \times 10 \Rightarrow 5x = 80 \Rightarrow x = 16\].
Therefore, the correct equation is \[\dfrac{5}{8} = \dfrac{{10}}{{16}}\].
\[{\text{(c)}}\] Let \[\dfrac{{45}}{{60}} = \dfrac{{15}}{x}\]
Now cross multiplying the above equation, we get
\[ \Rightarrow 45x = 60 \times 15 \Rightarrow x = \dfrac{{60 \times 15}}{{45}} \Rightarrow x = 20\]
Therefore, the correct equation is \[\dfrac{{45}}{{60}} = \dfrac{{15}}{{20}}\].
\[{\text{(d)}}\] Let \[\dfrac{{18}}{{24}} = \dfrac{x}{4}\]
Now cross multiplying the above equation, we get
\[ \Rightarrow 4 \times 18 = 24x \Rightarrow x = \dfrac{{4 \times 18}}{{24}} \Rightarrow x = 3\]
Therefore, the correct equation is\[\dfrac{{18}}{{24}} = \dfrac{3}{4}\].

Note- These types of problems are solved by simply cross multiplying the given equation with one unknown and then finally solving for that unknown.

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