Fractions and Rational Numbers: MCQ test, Set 2

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

$\displaystyle \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} =$
A $\displaystyle \frac{15}{4}$ B $\displaystyle \frac{4}{60}$ C $\displaystyle \frac{1}{60}$ D $\displaystyle \frac{4}{15}$

Answer and solution

Answer: (D) $\displaystyle \frac{4}{15}$

$4$ parts of $\displaystyle \frac{1}{15}$ are added; the denominator does not change.

$\displaystyle = \frac{4}{15}$
$\displaystyle 1 + \frac{11}{15}$ written as an improper fraction is
A $\displaystyle \frac{12}{15}$ B $\displaystyle \frac{12}{16}$ C $\displaystyle \frac{26}{15}$ D $\displaystyle \frac{26}{11}$

Answer and solution

Answer: (C) $\displaystyle \frac{26}{15}$

$\displaystyle 1 = \frac{15}{15}$

$\displaystyle \frac{15}{15} + \frac{11}{15} = \frac{26}{15}$
Among $\displaystyle \frac{9}{9}, \frac{5}{13}, \frac{5}{14}, \frac{3}{7}$ , the improper fraction is
A $\displaystyle \frac{9}{9}$ B $\displaystyle \frac{5}{13}$ C $\displaystyle \frac{5}{14}$ D $\displaystyle \frac{3}{7}$

Answer and solution

Answer: (A) $\displaystyle \frac{9}{9}$

In an improper fraction the numerator is greater than or equal to the denominator.

$9 = 9$ , so $\displaystyle \frac{9}{9}$ is improper.
$\displaystyle \frac{143}{9}$ as a mixed fraction is
A $\displaystyle 15\frac{1}{9}$ B $\displaystyle 15\frac{8}{9}$ C $\displaystyle 14\frac{8}{9}$ D $\displaystyle 16\frac{8}{9}$

Answer and solution

Answer: (B) $\displaystyle 15\frac{8}{9}$

$143 \div 9$ : quotient $15$ , remainder $8$

$\displaystyle \frac{143}{9} = 15 + \frac{8}{9} = 15\frac{8}{9}$
$\displaystyle 6\frac{1}{3}$ as an improper fraction is
A $\displaystyle \frac{9}{3}$ B $\displaystyle \frac{19}{3}$ C $\displaystyle \frac{7}{3}$ D $\displaystyle \frac{17}{3}$

Answer and solution

Answer: (B) $\displaystyle \frac{19}{3}$

$\displaystyle 6\frac{1}{3} = \frac{(6 \times 3) + 1}{3}$

$\displaystyle = \frac{18 + 1}{3} = \frac{19}{3}$
A like fraction of $\displaystyle \frac{14}{15}$ is
A $\displaystyle \frac{28}{16}$ B $\displaystyle \frac{15}{14}$ C $\displaystyle \frac{14}{3}$ D $\displaystyle \frac{28}{15}$

Answer and solution

Answer: (D) $\displaystyle \frac{28}{15}$

Like fractions have the same denominator.

$\displaystyle \frac{28}{15}$ has denominator $15$ .
$24$ in the form $\displaystyle \frac{p}{q}$ is
A $\displaystyle \frac{1}{24}$ B $\displaystyle \frac{24}{24}$ C $\displaystyle \frac{-24}{1}$ D $\displaystyle \frac{24}{1}$

Answer and solution

Answer: (D) $\displaystyle \frac{24}{1}$

Every integer $\displaystyle a = \frac{a}{1}$ .

$\displaystyle 24 = \frac{24}{1}$
$\displaystyle \frac{7}{11} = \frac{\square}{66}$ . The number in the box is
A $6$ B $42$ C $62$ D $13$

Answer and solution

Answer: (B) $42$

$66 \div 11 = 6$

$\displaystyle \frac{7}{11} = \frac{7 \times 6}{11 \times 6} = \frac{42}{66}$
The simplest form of $\displaystyle \frac{-26}{-28}$ is
A $\displaystyle \frac{-13}{-14}$ B $\displaystyle \frac{26}{28}$ C $\displaystyle \frac{13}{14}$ D $\displaystyle -\frac{13}{14}$

Answer and solution

Answer: (C) $\displaystyle \frac{13}{14}$

$\displaystyle \frac{-26}{-28} = \frac{26}{28}$ , since $(-)/(-) = (+)$

H.C.F of $26, 28 = 2$ , so $\displaystyle = \frac{13}{14}$
Among $\displaystyle \frac{28}{36}, \frac{-9}{-11}, -\frac{7}{6}, \frac{8}{-5}$ , the one in standard form is
A $\displaystyle \frac{8}{-5}$ B $\displaystyle \frac{28}{36}$ C $\displaystyle \frac{-9}{-11}$ D $\displaystyle -\frac{7}{6}$

Answer and solution

Answer: (D) $\displaystyle -\frac{7}{6}$

Standard form: numerator and denominator have no common divisor other than 1, and the denominator is positive.

$\displaystyle \frac{8}{-5}$ and $\displaystyle \frac{-9}{-11}$ have negative denominators; $\displaystyle \frac{28}{36}$ has the common factor $4$ .

So $\displaystyle -\frac{7}{6}$ is in standard form.
The descending order of $-28, 5, -33, 36, -42$ is
A $36, 5, -28, -33, -42$ B $-42, -33, -28, 5, 36$ C $-42, 36, -33, -28, 5$ D $36, 5, -42, -33, -28$

Answer and solution

Answer: (A) $36, 5, -28, -33, -42$

Negative numbers come before $0$ and the positive numbers; $-42 < -33 < -28$ .

Descending order: $36, 5, -28, -33, -42$
The smallest among $\displaystyle -\frac{2}{3}, \frac{15}{8}, \frac{3}{2}, \frac{1}{2}$ is
A $\displaystyle -\frac{2}{3}$ B $\displaystyle \frac{15}{8}$ C $\displaystyle \frac{3}{2}$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (A) $\displaystyle -\frac{2}{3}$

L.C.M of $3, 8, 2 = 2 \times 3 \times 4 = 24$

$\displaystyle -\frac{2}{3} = -\frac{16}{24}, \frac{15}{8} = \frac{45}{24}, \frac{3}{2} = \frac{36}{24}, \frac{1}{2} = \frac{12}{24}$

The smallest numerator is $-16$ , so the smallest is $\displaystyle -\frac{2}{3}$ .
To compare $\displaystyle \frac{3}{10}$ and $\displaystyle \frac{4}{11}$ by cross multiplication, $3 \times 11$ is compared with
A $10 \times 11$ B $3 \times 4$ C $10 \times 4$ D $3 \times 10$

Answer and solution

Answer: (C) $10 \times 4$

Rule: $\displaystyle \frac{p}{q} > \frac{r}{s}$ if $ps > qr$ .

Here $p = 3, q = 10, r = 4, s = 11$ , so $3 \times 11$ is compared with $10 \times 4$ .
$\displaystyle \frac{21}{27} + \frac{5}{27} - \frac{40}{27} =$
A $\displaystyle -\frac{14}{81}$ B $\displaystyle \frac{66}{27}$ C $\displaystyle -\frac{14}{27}$ D $\displaystyle \frac{14}{27}$

Answer and solution

Answer: (C) $\displaystyle -\frac{14}{27}$

Like fractions: keep the denominator, add and subtract the numerators.

$\displaystyle = \frac{21 + 5 - 40}{27} = -\frac{14}{27}$
$\displaystyle \frac{10}{9} - \frac{13}{12} =$
A $\displaystyle -\frac{1}{36}$ B $\displaystyle \frac{1}{36}$ C $\displaystyle -\frac{3}{36}$ D $\displaystyle -\frac{3}{21}$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{36}$

L.C.M of $9, 12 = 3 \times 3 \times 4 = 36$

$\displaystyle = \frac{40 - 39}{36}$

$\displaystyle = \frac{1}{36}$
$\displaystyle \frac{11}{9} - \frac{7}{12} + \frac{4}{3} =$
A $\displaystyle -\frac{71}{36}$ B $\displaystyle \frac{8}{36}$ C $\displaystyle \frac{71}{36}$ D $\displaystyle \frac{8}{24}$

Answer and solution

Answer: (C) $\displaystyle \frac{71}{36}$

L.C.M of $9, 12, 3 = 3 \times 3 \times 4 = 36$

$\displaystyle = \frac{44 - 21 + 48}{36}$

$\displaystyle = \frac{92 - 21}{36}$

$\displaystyle = \frac{71}{36}$
$\displaystyle 9 - \frac{16}{9} =$
A $\displaystyle \frac{65}{9}$ B $\displaystyle -\frac{7}{9}$ C $\displaystyle \frac{97}{9}$ D $\displaystyle -\frac{65}{9}$

Answer and solution

Answer: (A) $\displaystyle \frac{65}{9}$

Write $\displaystyle 9 = \frac{9}{1}$ .

$\displaystyle 9 - \frac{16}{9} = \frac{81 - 16}{9} = \frac{65}{9}$
$\displaystyle \frac{7}{17} \times \frac{17}{9} \times \frac{9}{13} =$
A $\displaystyle \frac{13}{7}$ B $\displaystyle \frac{7}{13}$ C $\displaystyle \frac{7}{1989}$ D $1$

Answer and solution

Answer: (B) $\displaystyle \frac{7}{13}$

$\displaystyle = \frac{7 \times 17 \times 9}{17 \times 9 \times 13}$

Cancelling $17$ and $9$ : $\displaystyle = \frac{7}{13}$
$\displaystyle -\frac{17}{13} \times \left(-\frac{13}{14}\right) =$
A $\displaystyle \frac{10}{9}$ B $\displaystyle -\frac{17}{14}$ C $\displaystyle \frac{17}{14}$ D $\displaystyle \frac{238}{169}$

Answer and solution

Answer: (C) $\displaystyle \frac{17}{14}$

Sign: $(-) \times (-) = (+)$

$\displaystyle = \frac{17 \times 13}{13 \times 14}$

$\displaystyle = \frac{17 \times 13}{13 \times (2 \times 7)}$

Cancelling the common factor $13$ : $\displaystyle = \frac{17}{2 \times 7} = \frac{17}{14}$
$\displaystyle 8\frac{1}{8} \times 6 =$
A $\displaystyle \frac{195}{4}$ B $\displaystyle \frac{189}{4}$ C $\displaystyle \frac{65}{48}$ D $\displaystyle \frac{385}{8}$

Answer and solution

Answer: (A) $\displaystyle \frac{195}{4}$

$\displaystyle 8\frac{1}{8} = \frac{(8 \times 8) + 1}{8} = \frac{65}{8}$

$\displaystyle \frac{65}{8} \times \frac{6}{1} = \frac{195}{4}$ (cancelling common factors)
The reciprocal of $\displaystyle -\frac{3}{13}$ is
A $\displaystyle -\frac{13}{3}$ B $\displaystyle \frac{13}{3}$ C $\displaystyle -\frac{3}{13}$ D $\displaystyle \frac{3}{13}$

Answer and solution

Answer: (A) $\displaystyle -\frac{13}{3}$

Reciprocal of $\displaystyle \frac{c}{d}$ is $\displaystyle \frac{d}{c}$ ; the sign stays.

Reciprocal of $\displaystyle -\frac{3}{13}$ is $\displaystyle -\frac{13}{3}$ .
$\displaystyle 8\frac{1}{6} \div 8 =$
A $\displaystyle \frac{7}{6}$ B $\displaystyle \frac{49}{48}$ C $\displaystyle \frac{48}{49}$ D $\displaystyle \frac{196}{3}$

Answer and solution

Answer: (B) $\displaystyle \frac{49}{48}$

$\displaystyle 8\frac{1}{6} = \frac{(8 \times 6) + 1}{6} = \frac{49}{6}$

$\displaystyle \frac{49}{6} \times \frac{1}{8} = \frac{49}{48}$
The smallest among $\displaystyle \frac{1}{30}, \frac{1}{24}, \frac{1}{7}, \frac{1}{14}$ is
A $\displaystyle \frac{1}{14}$ B $\displaystyle \frac{1}{24}$ C $\displaystyle \frac{1}{7}$ D $\displaystyle \frac{1}{30}$

Answer and solution

Answer: (D) $\displaystyle \frac{1}{30}$

$7 < 14 < 24 < 30$ (all have the same sign)

$\displaystyle \Rightarrow \frac{1}{7} > \frac{1}{14} > \frac{1}{24} > \frac{1}{30}$

So the smallest is $\displaystyle \frac{1}{30}$ .
If $-15 < 2$ , then multiplying both sides by $3$ gives
A $-45 = 6$ B $-45 > 6$ C $45 < -6$ D $-45 < 6$

Answer and solution

Answer: (D) $-45 < 6$

Rule: $a < b \Rightarrow ma < mb$ if $m > 0$ (the sign stays).

$3 \times \left(-15\right) = -45$ , $3 \times 2 = 6$ , so $-45 < 6$
The part of the number line from $-1$ to $0$ is divided into $11$ equal parts. The 8th point to the left of $0$ is
A $\displaystyle \frac{8}{11}$ B $\displaystyle -\frac{8}{11}$ C $\displaystyle -\frac{3}{11}$ D $\displaystyle -\frac{11}{8}$

Answer and solution

Answer: (B) $\displaystyle -\frac{8}{11}$

Each part $\displaystyle = \frac{1}{11}$ ; to the left of $0$ the numbers are negative.

$8$ parts to the left of $0$ : $\displaystyle -\frac{8}{11}$

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MCQ test, Set 2