Fractions and Rational Numbers: MCQ test, Set 9

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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.

A bar is divided into 16 equal parts and 7 parts are shaded. The part that is not shaded is
A $\displaystyle \frac{16}{9}$ B $\displaystyle \frac{7}{16}$ C $\displaystyle \frac{9}{16}$ D $\displaystyle \frac{9}{7}$

Answer and solution

Answer: (C) $\displaystyle \frac{9}{16}$

Unshaded parts $= 16 - 7 = 9$

Unshaded region $\displaystyle = \frac{9}{16}$
Two rectangles are each divided into 8 equal parts. All parts of the first and 5 parts of the second are shaded. The total shaded region is
A $\displaystyle \frac{8}{13}$ B $\displaystyle \frac{13}{8}$ C $\displaystyle \frac{5}{8}$ D $\displaystyle \frac{13}{16}$

Answer and solution

Answer: (B) $\displaystyle \frac{13}{8}$

First figure $= 1$ , second $\displaystyle = \frac{5}{8}$

$\displaystyle 1 + \frac{5}{8} = \frac{8}{8} + \frac{5}{8} = \frac{13}{8}$
In the fraction $\displaystyle \frac{15}{22}$ , the denominator is
A $22$ B $15$ C $\displaystyle \frac{15}{22}$ D $\displaystyle \frac{22}{15}$

Answer and solution

Answer: (A) $22$

In a fraction, the upper number is the numerator and the lower number is the denominator.

So the denominator of $\displaystyle \frac{15}{22}$ is $22$ .
$\displaystyle \frac{28}{11}$ as a mixed fraction is
A $\displaystyle 6\frac{2}{11}$ B $\displaystyle 2\frac{6}{11}$ C $\displaystyle 2\frac{5}{11}$ D $\displaystyle 3\frac{6}{11}$

Answer and solution

Answer: (B) $\displaystyle 2\frac{6}{11}$

$28 \div 11$ : quotient $2$ , remainder $6$

$\displaystyle \frac{28}{11} = 2 + \frac{6}{11} = 2\frac{6}{11}$
$\displaystyle 4\frac{1}{11}$ as an improper fraction is
A $\displaystyle \frac{5}{11}$ B $\displaystyle \frac{43}{11}$ C $\displaystyle \frac{15}{11}$ D $\displaystyle \frac{45}{11}$

Answer and solution

Answer: (D) $\displaystyle \frac{45}{11}$

$\displaystyle 4\frac{1}{11} = \frac{(4 \times 11) + 1}{11}$

$\displaystyle = \frac{44 + 1}{11} = \frac{45}{11}$
From $\displaystyle \frac{8}{9}, \frac{8}{5}, \frac{10}{11}, \frac{20}{11}, \frac{8}{13}, \frac{17}{11}$ , the like fractions are
A $\displaystyle \frac{8}{5}, \frac{10}{11}, \frac{8}{13}$ B $\displaystyle \frac{17}{11}, \frac{20}{11}, \frac{10}{11}$ C $\displaystyle \frac{17}{11}, \frac{8}{9}, \frac{20}{11}$ D $\displaystyle \frac{8}{9}, \frac{8}{5}, \frac{8}{13}$

Answer and solution

Answer: (B) $\displaystyle \frac{17}{11}, \frac{20}{11}, \frac{10}{11}$

Like fractions have the same denominator.

$\displaystyle \frac{17}{11}, \frac{20}{11}, \frac{10}{11}$ all have denominator $11$ .
Among $0$ , $\displaystyle \frac{-7}{11}$ , $\displaystyle \frac{15}{0}$ , $-4$ , the one that is not a rational number is
A $\displaystyle \frac{-7}{11}$ B $0$ C $\displaystyle \frac{15}{0}$ D $-4$

Answer and solution

Answer: (C) $\displaystyle \frac{15}{0}$

A rational number is of the form $\displaystyle \frac{p}{q}$ , $p, q$ integers and $q \ne 0$ .

$\displaystyle 0 = \frac{0}{1}$ and $\displaystyle -4 = \frac{-4}{1}$ are rational numbers.

In $\displaystyle \frac{15}{0}$ the denominator is $0$ , so it is not a rational number.
Which of the following is NOT equivalent to $\displaystyle \frac{11}{12}$ ?
A $\displaystyle \frac{33}{36}$ B $\displaystyle \frac{77}{84}$ C $\displaystyle \frac{66}{72}$ D $\displaystyle \frac{12}{13}$

Answer and solution

Answer: (D) $\displaystyle \frac{12}{13}$

$\displaystyle \frac{66}{72}$ , $\displaystyle \frac{33}{36}$ , $\displaystyle \frac{77}{84}$ are got by multiplying $11$ and $12$ by the same number.

$\displaystyle \frac{12}{13}$ is got by adding $1$ ; adding does not give an equivalent fraction.
The simplest form of $\displaystyle \frac{117}{63}$ is
A $\displaystyle \frac{13}{7}$ B $\displaystyle \frac{39}{7}$ C $\displaystyle \frac{39}{21}$ D $\displaystyle \frac{13}{21}$

Answer and solution

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

H.C.F of $117, 63 = 9$

$\displaystyle \frac{117}{63} = \frac{117 \div 9}{63 \div 9} = \frac{13}{7}$
The standard form of $\displaystyle \frac{104}{-120}$ is
A $\displaystyle -\frac{13}{15}$ B $\displaystyle \frac{13}{15}$ C $\displaystyle \frac{13}{-15}$ D $\displaystyle -\frac{52}{60}$

Answer and solution

Answer: (A) $\displaystyle -\frac{13}{15}$

$\displaystyle \frac{104}{-120} = -\frac{104}{120}$ , since $(+)/(-) = (-)$

H.C.F of $104, 120 = 8$ , so $\displaystyle = -\frac{13}{15}$ (the denominator must be positive)
The ascending order of $30, -34, -26, -16, 38$ is
A $38, 30, -16, -26, -34$ B $-16, -26, -34, 30, 38$ C $-34, -26, -16, 30, 38$ D $-16, -26, 30, -34, 38$

Answer and solution

Answer: (C) $-34, -26, -16, 30, 38$

Negative numbers come before $0$ and the positive numbers; $-34 < -26 < -16$ .

Ascending order: $-34, -26, -16, 30, 38$
The greatest among $\displaystyle \frac{1}{10}, \frac{11}{10}, \frac{4}{3}, -\frac{19}{12}$ is
A $\displaystyle \frac{4}{3}$ B $\displaystyle -\frac{19}{12}$ C $\displaystyle \frac{1}{10}$ D $\displaystyle \frac{11}{10}$

Answer and solution

Answer: (A) $\displaystyle \frac{4}{3}$

L.C.M of $10, 3, 12 = 2 \times 3 \times 5 \times 2 = 60$

$\displaystyle \frac{1}{10} = \frac{6}{60}, \frac{11}{10} = \frac{66}{60}, \frac{4}{3} = \frac{80}{60}, -\frac{19}{12} = -\frac{95}{60}$

The largest numerator is $80$ , so the greatest is $\displaystyle \frac{4}{3}$ .
$\displaystyle \frac{9}{8} \;\square\; \frac{12}{11}$ . The sign that goes in the box is
A cannot be compared B $=$ C $<$ D $>$

Answer and solution

Answer: (D) $>$

$9 \times 11 = 99$ , $8 \times 12 = 96$

$\displaystyle 99 > 96 \Rightarrow \frac{9}{8} > \frac{12}{11}$
$\displaystyle \frac{49}{40} + \frac{57}{40} - \frac{37}{40} =$
A $\displaystyle -\frac{69}{40}$ B $\displaystyle \frac{143}{40}$ C $\displaystyle \frac{69}{120}$ D $\displaystyle \frac{69}{40}$

Answer and solution

Answer: (D) $\displaystyle \frac{69}{40}$

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

$\displaystyle = \frac{49 + 57 - 37}{40} = \frac{69}{40}$
$\displaystyle \frac{1}{10} - \frac{4}{5} =$
A $\displaystyle \frac{7}{10}$ B $\displaystyle -\frac{3}{15}$ C $\displaystyle -\frac{7}{10}$ D $\displaystyle -\frac{3}{10}$

Answer and solution

Answer: (C) $\displaystyle -\frac{7}{10}$

L.C.M of $10, 5 = 5 \times 2 = 10$

$\displaystyle = \frac{1 - 8}{10}$

$\displaystyle = -\frac{7}{10}$
$\displaystyle \frac{3}{4} + \frac{13}{12} - \frac{5}{8} =$
A $\displaystyle -\frac{29}{24}$ B $\displaystyle \frac{59}{24}$ C $\displaystyle \frac{29}{24}$ D $\displaystyle \frac{11}{24}$

Answer and solution

Answer: (C) $\displaystyle \frac{29}{24}$

L.C.M of $4, 12, 8 = 2 \times 2 \times 3 \times 2 = 24$

$\displaystyle = \frac{18 + 26 - 15}{24}$

$\displaystyle = \frac{44 - 15}{24}$

$\displaystyle = \frac{29}{24}$
$\displaystyle \frac{36}{13} - 2 =$
A $\displaystyle \frac{62}{13}$ B $\displaystyle \frac{10}{13}$ C $\displaystyle -\frac{10}{13}$ D $\displaystyle \frac{34}{13}$

Answer and solution

Answer: (B) $\displaystyle \frac{10}{13}$

Write $\displaystyle 2 = \frac{2}{1}$ .

$\displaystyle \frac{36}{13} - 2 = \frac{36 - 26}{13} = \frac{10}{13}$
$\displaystyle -19 \times \frac{36}{19} =$
A $\displaystyle -\frac{684}{361}$ B $\displaystyle -\frac{36}{361}$ C $36$ D $-36$

Answer and solution

Answer: (D) $-36$

$\displaystyle -19 = \frac{-19}{1}$

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

$\displaystyle = -\frac{19 \times 36}{1 \times 19}$

$\displaystyle = -\frac{19 \times (2 \times 2 \times 3 \times 3)}{1 \times 19}$

Cancelling the common factor $19$ : $\displaystyle = -\frac{2 \times 2 \times 3 \times 3}{1} = -36$
$\displaystyle \frac{13}{9} \times \frac{9}{11} \times \frac{11}{18} =$
A $1$ B $\displaystyle \frac{18}{13}$ C $\displaystyle \frac{13}{18}$ D $\displaystyle \frac{13}{1782}$

Answer and solution

Answer: (C) $\displaystyle \frac{13}{18}$

$\displaystyle = \frac{13 \times 9 \times 11}{9 \times 11 \times 18}$

Cancelling $9$ and $11$ : $\displaystyle = \frac{13}{18}$
$\displaystyle 4\frac{4}{7} \times 4\frac{1}{2} =$
A $\displaystyle \frac{127}{14}$ B $\displaystyle \frac{144}{7}$ C $\displaystyle \frac{20}{7}$ D $\displaystyle \frac{114}{7}$

Answer and solution

Answer: (B) $\displaystyle \frac{144}{7}$

Convert to improper fractions first: $\displaystyle 4\frac{4}{7} = \frac{32}{7}, \; 4\frac{1}{2} = \frac{9}{2}$

$\displaystyle \frac{32}{7} \times \frac{9}{2} = \frac{144}{7}$
$\displaystyle \frac{15}{14} \div \left(-\frac{5}{16}\right) =$
A $\displaystyle -\frac{24}{7}$ B $\displaystyle -\frac{75}{224}$ C $\displaystyle -\frac{7}{24}$ D $\displaystyle \frac{24}{7}$

Answer and solution

Answer: (A) $\displaystyle -\frac{24}{7}$

$\displaystyle \frac{15}{14} \div \left(-\frac{5}{16}\right) = \frac{15}{14} \times \left(-\frac{16}{5}\right)$ (multiply by the reciprocal)

$\displaystyle = -\frac{24}{7}$
$\displaystyle \frac{7}{9} \div 7 =$
A $9$ B $\displaystyle \frac{1}{9}$ C $\displaystyle \frac{7}{16}$ D $\displaystyle \frac{49}{9}$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{9}$

$\displaystyle \frac{7}{9} \div 7 = \frac{7}{9} \div \frac{7}{1} = \frac{7}{9} \times \frac{1}{7}$

$\displaystyle = \frac{1}{9}$
Since $-29 < -6$ , which is true?
A $\displaystyle \frac{1}{-29} < \frac{1}{-6}$ B $\displaystyle \frac{1}{-29} = \frac{1}{-6}$ C $\displaystyle \frac{1}{-29} > \frac{1}{-6}$ D $\displaystyle \frac{1}{-6} > \frac{1}{-29}$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{-29} > \frac{1}{-6}$

Rule: $\displaystyle a < b \Leftrightarrow \frac{1}{a} > \frac{1}{b}$ if $a, b$ have the same sign.

$-29 < -6$ (both negative) $\displaystyle \Rightarrow \frac{1}{-29} > \frac{1}{-6}$
If $-10 < -8$ , then multiplying both sides by $-6$ gives
A $60 = 48$ B $-60 > -48$ C $60 < 48$ D $60 > 48$

Answer and solution

Answer: (D) $60 > 48$

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

$-6 \times \left(-10\right) = 60$ , $-6 \times \left(-8\right) = 48$ , so $60 > 48$
$\displaystyle \frac{22}{9}$ lies between
A $2$ and $3$ B $-3$ and $-2$ C $3$ and $4$ D $1$ and $2$

Answer and solution

Answer: (A) $2$ and $3$

$\displaystyle \frac{22}{9} = 2 + \frac{4}{9}$

So it lies between $2$ and $3$ on the number line.

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MCQ test, Set 9