Fractions and Rational Numbers: MCQ test, Set 6

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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 strip is divided into 11 equal parts and 4 parts are shaded. The part that is not shaded is
A $\displaystyle \frac{11}{7}$ B $\displaystyle \frac{4}{11}$ C $\displaystyle \frac{7}{4}$ D $\displaystyle \frac{7}{11}$

Answer and solution

Answer: (D) $\displaystyle \frac{7}{11}$

Unshaded parts $= 11 - 4 = 7$

Unshaded region $\displaystyle = \frac{7}{11}$
Two bars are each divided into 9 equal parts. All parts of the first and 8 parts of the second are shaded. The total shaded region is
A $\displaystyle \frac{8}{9}$ B $\displaystyle \frac{17}{9}$ C $\displaystyle \frac{17}{18}$ D $\displaystyle \frac{9}{17}$

Answer and solution

Answer: (B) $\displaystyle \frac{17}{9}$

First figure $= 1$ , second $\displaystyle = \frac{8}{9}$

$\displaystyle 1 + \frac{8}{9} = \frac{9}{9} + \frac{8}{9} = \frac{17}{9}$
In the fraction $\displaystyle \frac{22}{7}$ , the numerator is
A $\displaystyle \frac{22}{7}$ B $\displaystyle \frac{7}{22}$ C $22$ D $7$

Answer and solution

Answer: (C) $22$

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

So the numerator of $\displaystyle \frac{22}{7}$ is $22$ .
$\displaystyle \frac{31}{10}$ as a mixed fraction is
A $\displaystyle 4\frac{1}{10}$ B $\displaystyle 1\frac{3}{10}$ C $\displaystyle 3\frac{9}{10}$ D $\displaystyle 3\frac{1}{10}$

Answer and solution

Answer: (D) $\displaystyle 3\frac{1}{10}$

$31 \div 10$ : quotient $3$ , remainder $1$

$\displaystyle \frac{31}{10} = 3 + \frac{1}{10} = 3\frac{1}{10}$
$\displaystyle 23\frac{1}{8}$ as an improper fraction is
A $\displaystyle \frac{24}{8}$ B $\displaystyle \frac{183}{8}$ C $\displaystyle \frac{31}{8}$ D $\displaystyle \frac{185}{8}$

Answer and solution

Answer: (D) $\displaystyle \frac{185}{8}$

$\displaystyle 23\frac{1}{8} = \frac{(23 \times 8) + 1}{8}$

$\displaystyle = \frac{184 + 1}{8} = \frac{185}{8}$
A like fraction of $\displaystyle \frac{13}{8}$ is
A $\displaystyle \frac{8}{13}$ B $\displaystyle \frac{1}{8}$ C $\displaystyle \frac{1}{9}$ D $\displaystyle \frac{13}{4}$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{8}$

Like fractions have the same denominator.

$\displaystyle \frac{1}{8}$ has denominator $8$ .
Among $0$ , $\displaystyle \frac{6}{0}$ , $\displaystyle \frac{-11}{13}$ , $-7$ , the one that is not a rational number is
A $\displaystyle \frac{6}{0}$ B $0$ C $\displaystyle \frac{-11}{13}$ D $-7$

Answer and solution

Answer: (A) $\displaystyle \frac{6}{0}$

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

$\displaystyle 0 = \frac{0}{1}$ and $\displaystyle -7 = \frac{-7}{1}$ are rational numbers.

In $\displaystyle \frac{6}{0}$ the denominator is $0$ , so it is not a rational number.
Which of the following is NOT equivalent to $\displaystyle \frac{10}{9}$ ?
A $\displaystyle \frac{30}{27}$ B $\displaystyle \frac{11}{10}$ C $\displaystyle \frac{60}{54}$ D $\displaystyle \frac{70}{63}$

Answer and solution

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

$\displaystyle \frac{60}{54}$ , $\displaystyle \frac{30}{27}$ , $\displaystyle \frac{70}{63}$ are got by multiplying $10$ and $9$ by the same number.

$\displaystyle \frac{11}{10}$ is got by adding $1$ ; adding does not give an equivalent fraction.
The simplest form of $\displaystyle \frac{-15}{-9}$ is
A $\displaystyle -\frac{5}{3}$ B $\displaystyle \frac{-5}{-3}$ C $\displaystyle \frac{15}{9}$ D $\displaystyle \frac{5}{3}$

Answer and solution

Answer: (D) $\displaystyle \frac{5}{3}$

$\displaystyle \frac{-15}{-9} = \frac{15}{9}$ , since $(-)/(-) = (+)$

H.C.F of $15, 9 = 3$ , so $\displaystyle = \frac{5}{3}$
Among $\displaystyle \frac{-13}{-14}, \frac{6}{20}, -\frac{1}{12}, \frac{15}{-13}$ , the one in standard form is
A $\displaystyle -\frac{1}{12}$ B $\displaystyle \frac{-13}{-14}$ C $\displaystyle \frac{6}{20}$ D $\displaystyle \frac{15}{-13}$

Answer and solution

Answer: (A) $\displaystyle -\frac{1}{12}$

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

$\displaystyle \frac{15}{-13}$ and $\displaystyle \frac{-13}{-14}$ have negative denominators; $\displaystyle \frac{6}{20}$ has the common factor $2$ .

So $\displaystyle -\frac{1}{12}$ is in standard form.
The descending order of $30, 4, -2, -31, -36$ is
A $-36, -31, -2, 4, 30$ B $30, 4, -36, -31, -2$ C $30, 4, -2, -31, -36$ D $-36, -31, 30, 4, -2$

Answer and solution

Answer: (C) $30, 4, -2, -31, -36$

Negative numbers come before $0$ and the positive numbers; $-36 < -31 < -2$ .

Descending order: $30, 4, -2, -31, -36$
The smallest among $\displaystyle \frac{1}{3}, -\frac{7}{10}, -\frac{1}{12}, \frac{19}{12}$ is
A $\displaystyle \frac{19}{12}$ B $\displaystyle -\frac{1}{12}$ C $\displaystyle \frac{1}{3}$ D $\displaystyle -\frac{7}{10}$

Answer and solution

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

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

$\displaystyle \frac{1}{3} = \frac{20}{60}, -\frac{7}{10} = -\frac{42}{60}, -\frac{1}{12} = -\frac{5}{60}, \frac{19}{12} = \frac{95}{60}$

The smallest numerator is $-42$ , so the smallest is $\displaystyle -\frac{7}{10}$ .
To compare $\displaystyle \frac{4}{3}$ and $\displaystyle \frac{16}{11}$ by cross multiplication, $4 \times 11$ is compared with
A $4 \times 16$ B $3 \times 16$ C $3 \times 11$ D $4 \times 3$

Answer and solution

Answer: (B) $3 \times 16$

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

Here $p = 4, q = 3, r = 16, s = 11$ , so $4 \times 11$ is compared with $3 \times 16$ .
$\displaystyle \frac{71}{56} + \frac{76}{56} - \frac{38}{56} =$
A $\displaystyle \frac{109}{56}$ B $\displaystyle \frac{185}{56}$ C $\displaystyle -\frac{109}{56}$ D $\displaystyle \frac{109}{168}$

Answer and solution

Answer: (A) $\displaystyle \frac{109}{56}$

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

$\displaystyle = \frac{71 + 76 - 38}{56} = \frac{109}{56}$
$\displaystyle \frac{5}{6} - \frac{13}{9} =$
A $\displaystyle \frac{11}{18}$ B $\displaystyle -\frac{8}{15}$ C $\displaystyle -\frac{8}{18}$ D $\displaystyle -\frac{11}{18}$

Answer and solution

Answer: (D) $\displaystyle -\frac{11}{18}$

L.C.M of $6, 9 = 3 \times 2 \times 3 = 18$

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

$\displaystyle = -\frac{11}{18}$
$\displaystyle \frac{11}{16} - \frac{5}{4} - \frac{3}{8} =$
A $\displaystyle -\frac{15}{16}$ B $\displaystyle \frac{3}{16}$ C $\displaystyle \frac{15}{16}$ D $\displaystyle \frac{3}{28}$

Answer and solution

Answer: (A) $\displaystyle -\frac{15}{16}$

L.C.M of $16, 4, 8 = 2 \times 2 \times 2 \times 2 = 16$

$\displaystyle = \frac{11 - 20 - 6}{16}$

$\displaystyle = \frac{11 - 26}{16}$

$\displaystyle = -\frac{15}{16}$
$\displaystyle 3 - \frac{7}{9} =$
A $\displaystyle -\frac{4}{9}$ B $\displaystyle \frac{34}{9}$ C $\displaystyle \frac{20}{9}$ D $\displaystyle -\frac{20}{9}$

Answer and solution

Answer: (C) $\displaystyle \frac{20}{9}$

Write $\displaystyle 3 = \frac{3}{1}$ .

$\displaystyle 3 - \frac{7}{9} = \frac{27 - 7}{9} = \frac{20}{9}$
$\displaystyle 18 \times \frac{1}{30} =$
A $\displaystyle \frac{1}{540}$ B $\displaystyle \frac{3}{5}$ C $\displaystyle \frac{19}{30}$ D $\displaystyle \frac{18}{540}$

Answer and solution

Answer: (B) $\displaystyle \frac{3}{5}$

$\displaystyle 18 = \frac{18}{1}$

$\displaystyle = \frac{18 \times 1}{1 \times 30}$

$\displaystyle = \frac{(2 \times 3 \times 3) \times 1}{1 \times (2 \times 3 \times 5)}$

Cancelling the common factors $2, 3$ : $\displaystyle = \frac{3}{5}$
$\displaystyle \frac{15}{7} \times \frac{7}{5} \times \frac{5}{16} =$
A $\displaystyle \frac{15}{16}$ B $\displaystyle \frac{16}{15}$ C $1$ D $\displaystyle \frac{15}{560}$

Answer and solution

Answer: (A) $\displaystyle \frac{15}{16}$

$\displaystyle = \frac{15 \times 7 \times 5}{7 \times 5 \times 16}$

Cancelling $7$ and $5$ : $\displaystyle = \frac{15}{16}$
$\displaystyle 4\frac{5}{6} \times 4 =$
A $\displaystyle \frac{101}{6}$ B $\displaystyle \frac{38}{3}$ C $\displaystyle \frac{58}{3}$ D $\displaystyle \frac{29}{24}$

Answer and solution

Answer: (C) $\displaystyle \frac{58}{3}$

$\displaystyle 4\frac{5}{6} = \frac{(4 \times 6) + 5}{6} = \frac{29}{6}$

$\displaystyle \frac{29}{6} \times \frac{4}{1} = \frac{58}{3}$ (cancelling common factors)
The reciprocal of $\displaystyle -\frac{41}{23}$ is
A $\displaystyle -\frac{23}{41}$ B $\displaystyle -\frac{41}{23}$ C $\displaystyle \frac{23}{41}$ D $\displaystyle \frac{41}{23}$

Answer and solution

Answer: (A) $\displaystyle -\frac{23}{41}$

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

Reciprocal of $\displaystyle -\frac{41}{23}$ is $\displaystyle -\frac{23}{41}$ .
$\displaystyle 4\frac{1}{4} \div 7 =$
A $\displaystyle \frac{23}{28}$ B $\displaystyle \frac{17}{28}$ C $\displaystyle \frac{28}{17}$ D $\displaystyle \frac{119}{4}$

Answer and solution

Answer: (B) $\displaystyle \frac{17}{28}$

$\displaystyle 4\frac{1}{4} = \frac{(4 \times 4) + 1}{4} = \frac{17}{4}$

$\displaystyle \frac{17}{4} \times \frac{1}{7} = \frac{17}{28}$
The greatest among $\displaystyle \frac{1}{15}, \frac{1}{30}, \frac{1}{28}, \frac{1}{18}$ is
A $\displaystyle \frac{1}{28}$ B $\displaystyle \frac{1}{18}$ C $\displaystyle \frac{1}{15}$ D $\displaystyle \frac{1}{30}$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{15}$

$15 < 18 < 28 < 30$ (all have the same sign)

$\displaystyle \Rightarrow \frac{1}{15} > \frac{1}{18} > \frac{1}{28} > \frac{1}{30}$

So the greatest is $\displaystyle \frac{1}{15}$ .
If $0 < 13$ , then multiplying both sides by $-9$ gives
A $0 > -117$ B $0 > 117$ C $0 = -117$ D $0 < -117$

Answer and solution

Answer: (A) $0 > -117$

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

$-9 \times 0 = 0$ , $-9 \times 13 = -117$ , so $0 > -117$
$\displaystyle -\frac{22}{7}$ lies between
A $-5$ and $-4$ B $-3$ and $-2$ C $-4$ and $-3$ D $3$ and $4$

Answer and solution

Answer: (C) $-4$ and $-3$

$\displaystyle -\frac{22}{7} = -3 - \frac{1}{7}$

So it lies between $-4$ and $-3$ on the number line.

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MCQ test, Set 6