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$

{\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} \\

$

Answer

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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.

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.