Maths MCQ

Consider the fractions $\frac{2}{3},\frac{3}{5}$ and $\frac{11}{15}$
$\text{LCM}$ of $3, 5$ and $15 = 15$
Firstly, convert the fractions into equivalent fractions with denominator $15$
$\Rightarrow\frac{2}{3}=\frac{2\times5}{3\times5}=\frac{10}{15}$
$\Rightarrow\frac{3}{5}=\frac{3\times3}{5\times3}=\frac{9}{15}$
Now,
$9 < 10 < 11$
$\therefore\ \frac{9}{15} < \frac{10}{15} < \frac{11}{15}$
$\frac{3}{5} < \frac{2}{3} < \frac{11}{15}$

Consider the fractions $\frac{3}{4},\frac{2}{3}$ and $\frac{12}{15}$
$\text{LCM}$ of $4, 3$ and $15 = 60$
Firstly, convert the fractions into equivalent fractions with denominator $60$
$\Rightarrow\frac{3}{4}=\frac{3\times15}{4\times15}=\frac{45}{60}$
$\Rightarrow\frac{2}{3}=\frac{2\times20}{3\times20}=\frac{40}{60}$
$\Rightarrow\frac{12}{15}=\frac{12\times4}{15\times4}=\frac{48}{60}$
Now,
$40 < 45 < 48$
$\therefore\ \frac{40}{60} < \frac{45}{60} < \frac{48}{60}$
$\frac{2}{3} < \frac{3}{4} < \frac{12}{15}$

Consider the fractions $\frac{2}{3},\frac{4}{7},\frac{8}{11}$ and $\frac{5}{9}$
$\text{LCM}$ of $3, 7, 9$ and $11 = 693$
Firstly, convert the fractions into equivalent fractions with denominator $693$
$\Rightarrow\frac{2}{3}=\frac{2\times231}{3\times231}=\frac{462}{693}$
$\Rightarrow\frac{4}{7}=\frac{4\times99}{7\times99}=\frac{396}{693}$
$\Rightarrow\frac{8}{11}=\frac{8\times63}{11\times63}=\frac{504}{693}$
$\Rightarrow\frac{5}{9}=\frac{5\times77}{9\times77}=\frac{385}{693}$
Now,
$385 < 396 < 462 < 504$
$\therefore\ \frac{385}{693} < \frac{396}{693} < \frac{462}{693} < \frac{504}{693}$
$\Rightarrow\frac{5}{9} < \frac{4}{7} < \frac{2}{3} < \frac{8}{11}$
Thus, the smallest of the given fractions is $\frac{5}{9}$

A fraction whose numerator is less than the denominator is called a proper fraction.
The numerator in each of the fractions $\frac{17}{3},\frac{12}{5},1\frac{3}{4}=\frac{7}{4}$ is more than the denominator, so these fractions are improper fractions.
The numerator of the fraction $\frac{13}{17}$ is less than the denominator, so this fraction is a proper fraction.

Consider the fractions $\frac{3}{4},\frac{5}{6},\frac{2}{3},\frac{1}{2},\frac{4}{5}$ and $\frac{9}{10}$
$\text{LCM}$ of $2, 3, 4, 5, 6$ and $10 = 60$
Firstly, convert the fractions into equivalent fractions with denominator $60$
$\Rightarrow\frac{3}{4}=\frac{3\times15}{4\times15}=\frac{45}{60}$
$\Rightarrow\frac{5}{6}=\frac{5\times10}{6\times10}=\frac{50}{60}$
$\Rightarrow\frac{2}{3}=\frac{2\times20}{3\times20}=\frac{40}{60}$
$\Rightarrow\frac{1}{2}=\frac{1\times30}{2\times30}=\frac{30}{60}$
$\Rightarrow\frac{4}{5}=\frac{4\times12}{5\times12}=\frac{48}{60}$
$\Rightarrow\frac{9}{10}=\frac{9\times6}{10\times6}=\frac{54}{60}$
Now,
$30 < 40 < 45 < 48 < 50 < 54$
$\therefore\ \frac{30}{60} < \frac{40}{60} < \frac{45}{60} < \frac{48}{60} < \frac{50}{60} < \frac{54}{60}$
$\Rightarrow\frac{1}{2} < \frac{2}{3} < \frac{3}{4} < \frac{4}{5} < \frac{5}{6} < \frac{9}{10}$
Thus, the fraction $\frac{4}{5}$ is greater than $\frac{3}{4}$ and less than $\frac{5}{6}$

$2\frac{3}{5}\div\frac{5}{7}$
$=\frac{13}{5}\div\frac{5}{7}$
$=\frac{13}{5}\times\frac{7}{5}$
$=\frac{13\times7}{5\times5}$
$=\frac{91}{25}$

Let the required number be $x.$
Now,
$1\frac{3}{4}\div\text{x}=2\frac{1}{2}$
$\Rightarrow\frac{7}{4}\times\frac{1}{\text{x}}=\frac{5}{2}$
$\Rightarrow\text{x}=\frac{7}{4}\times\frac{2}{5}$
$\Rightarrow\text{x}=\frac{7\times1}{2\times5}$
$\Rightarrow\text{x}=\frac{7}{10}$
Thus, the required number is $\frac{7}{10}$

Since the number of negative terms in the product is odd.
Therefore, their product is negative.
$9\times\Big(-\frac{1}{3}\Big)\times(-3)\times\Big(-\frac{1}{9}\Big)$
$=9\times\Big(-\frac{1}{9}\Big)\times\Big(-\frac{1}{3}\Big)\times(-3)$
$=-\Big(9\times\frac{1}{9}\times\frac{1}{3}\times3\Big)$
$=-(1\times1)$
$=-1$