Fractions MCQ Quiz - Objective Question with Answer for Fractions - Download Free PDF

Calculation:

\(\rm \frac{x}{y}=\frac{6}{5}\)   ---(1)

Now we have,

\(\rm \frac{6}{7}-\frac{5x-y}{5x+y}\)

\(\rm \frac{6}{7}-\frac{y(5\times\frac{x}{y}-1)}{y(5\times \frac{x}{y}+1)}\)

⇒ \(\rm \frac{6}{7}-\frac{5\times\frac{6}{5}-1}{5\times\frac{6}{5}+1}\)       [using (1)]

⇒ \(\rm \frac{6}{7}-\frac{6-1}{6+1}\)

⇒ \(\frac{6}{7}-\frac{5}{7} = \frac{1}{7}\)

⇒ \(\rm \frac{6}{7}-\frac{5x-y}{5x+y}=\frac{1}{7}\)

∴ The correct answer is \(\frac{1}{7}\)

Given:

Expression: [(32 ÷ 8) × {(15 ÷ 5) + (30 ÷ 5) × (7 - 2)}]

Calculations:

Step 1: Simplify the terms inside the parentheses:

32 ÷ 8 = 4

15 ÷ 5 = 3

30 ÷ 5 = 6

7 - 2 = 5

Step 2: Simplify the multiplication:

(6 × 5) = 30

Step 3: Add the results:

3 + 30 = 33

Step 4: Multiply the results:

4 × 33 = 132

∴ The value of the expression is 132.

Calculation:

The given expression,

⇒\(\frac{0.0203 \times 2.92}{0.7 \times 0.0365 \times 2.9}\)

Remove the decimal from numerator and denominator:

⇒\(\frac{203 \times 292\times 10^6}{7 \times 365 \times 29\times 10^6}\)

Cancel out the terms from numerator and denominator:

⇒\(\frac{292}{365}\)

⇒0.8

Option 1 is the correct answer.

Given:

The expression is \(\dfrac{7}{10} ÷ 1 \dfrac{2}{5} \text { of } \dfrac{3}{4}-1 \dfrac{1}{4} \text { of } \dfrac{2}{3} ÷ 4 \dfrac{1}{6}+\dfrac{1}{15}\)

Formula used:

Follow the BODMAS rule according to the table given below:

Calculation:

\(\dfrac{7}{10} ÷ 1 \dfrac{2}{5} \text { of } \dfrac{3}{4}-1 \dfrac{1}{4} \text { of } \dfrac{2}{3} ÷ 4 \dfrac{1}{6}+\dfrac{1}{15}\)

⇒ \(\dfrac{7}{10}\) ÷ \(\dfrac{7}{5}\) × \(\dfrac{3}{4}\) - \(\dfrac{5}{4}\) × \(\dfrac{2}{3}\) ÷ \(\dfrac{25}{6}\) + \(\dfrac{1}{15}\)

⇒ \(\dfrac{7}{10}\) ÷ \(\dfrac{21}{20}\) - \(\dfrac{5}{6}\) ÷ \(\dfrac{25}{6}\) + \(\dfrac{1}{15}\)

⇒ \(\dfrac{7}{10}\) × \(\dfrac{20}{21}\) - \(\dfrac{5}{6}\)​ × \(\dfrac{6}{25}\) + \(\dfrac{1}{15}\)

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∴ \(\dfrac{7}{10} ÷ 1 \dfrac{2}{5} \text { of } \dfrac{3}{4}-1 \dfrac{1}{4} \text { of } \dfrac{2}{3} ÷ 4 \dfrac{1}{6}+\dfrac{1}{15}\) is \(\dfrac{8}{15}\).37" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-14-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-51-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-575-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-28-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-22-Frame" role="presentation" style="position: relative;" tabindex="0">

Concept Used:

Calculation:

⇒ \(\rm 864÷ \left\{\frac{3}{4}\left[\frac{16}{15}\right]-\frac{2}{3}\right\} \)

⇒ 864 ÷ {3/4 × 16/15 - 2/3}

⇒ 864 ÷ {4/5 - 2/3}

⇒ 864 ÷ {\(\frac{4 × 3\ - \ 2 × 5}{5 × 3}\)}

⇒ 864 ÷ {\(\frac{12 \ - \ 10}{15}\)}

⇒ 864 ÷ 2/15

⇒ 864 × 15/2

⇒ 6480

∴ The simplified value is 6480

Solution:

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 25/2 + 37/3 + 73/6

= (75 + 74 + 73)/6

= 222/6

= 37

Shortcut Trick

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 12 + 12 + 12 + (1/2 + 1/3 + 1/6)

= 36 + 1 = 37

Let that fraction be x.

⇒ x + 5/8 = 1

⇒ x = 1 – 5/8

(7/13) = 0.538

(2/3) = 0.666

(4/11) = 0.3636

(5/9) = 0.5555

Out of 2/3, 7/13, 4/11, 5/9

2/3 is the largest number followed by 5/9 then 7/13 and the smallest is 4/11.

\(\Rightarrow 4\frac{1}{5}-\left[ 3\frac{1}{3}-\left\{ 2\frac{1}{2}-\left( \frac{1}{3}+\frac{1}{6}-\frac{1}{12} \right) \right\} \right]\)

\(\Rightarrow \frac{{21}}{5} - \left[ {\frac{{10}}{3}-\left\{ {\frac{5}{2} - \left( {\frac{{4{\rm{\;}} + {\rm{\;}}2{\rm{\;}}-{\rm{\;}}1}}{{12}}} \right)} \right\}} \right]\)

\(\Rightarrow \frac{{21}}{5} - \left[ {\frac{{10}}{3} - \left\{ {\frac{5}{2} - \frac{5}{{12}}} \right\}} \right]\)

\(\Rightarrow \frac{{21}}{5} - \left[ {\frac{{10}}{3} - \frac{{25}}{{12}}} \right]\)

\(\Rightarrow \frac{{21}}{5} - \frac{{15}}{{12}}\)

\(\Rightarrow \frac{{21}}{5} - \frac{5}{4}\)

\(\Rightarrow \frac{{84 - 25{\rm{\;}}}}{{20}}\)

⇒ 59/20

Let the number of people be x and number of the chair be y.

Number of available chair = y × (9/13) = 9y/13

Number of empty chair = y - (9y/13) = 4y/13

Given, Number of empty chairs = 28

According to the question

4y/13 = 28

y = 28 × (13/4) = 91

Total number of chairs = 91

Number of chairs in which people are sitting = 91 - 28 = 63

Number of people who sit = x × (7/9) = 7x/9

According to the question

7x/9 = 63

x = 63 × (9/7) = 81

Total number of people are = 81

Formula used:

a² - b² = (a + b)(a - b)

Calculation

⇒ \(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2}+\frac{q^2-(p-r)^2}{(p+q)^2-r^2}+\frac{r^2-(p-q)^2}{(q+r)^2-p^2}\)

⇒ [(p + q - r)(p - q + r)]/[(p + q + r)(p - q + r)] + [(p + q - r)(q - p + r)]/[(p + q + r)(p + q - r)] + [(p - q + r)(q  -p + r)]/[(p + q + r)(q - p + r)]

⇒ [(p + q - r)]/[(p + q + r)] + [q - p + r)]/[(p + q + r)] + [(p - q + r)]/[(p + q + r)]

⇒ [(p + q - r)]/[(p + q + r)] + [q - p + r)]/[(p + q + r)] + [(p - q + r)]/[(p + q + r)]

⇒ (p + q + r)/(p + q + r)

⇒ 1.

The value is 1.

Shortcut Trick

Let put p = q = r = 1

So,

\(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2}+\frac{q^2-(p-r)^2}{(p+q)^2-r^2}+\frac{r^2-(p-q)^2}{(q+r)^2-p^2}\)

⇒ \(\rm\frac{1^2-(1-1)^2}{(1+1)^2-1^2}+\frac{1^2-(1-1)^2}{(1+1)^2-1^2}+\frac{1^2-(1-1)^2}{(1+1)^2-1^2}\)

⇒ \(\rm\frac{1-0}{(4-1)}+\frac{1-0}{(4-1)}+\frac{1-0}{(4-1)}\)

⇒ 1/3 + 1/3 + 1/3 = 1

Hence, The value is 1.

Given:

(5x - 2y) ∶ (x - 2y) = 9 ∶ 17

Calculation:

The given ratio can be written as:

(5x - 2y)/(x - 2y) = 9/17

17 × (5x - 2y) = 9 × (x - 2y)

85x - 34y = 9x - 18y

76x = 16y

x/y = 16/76

x/y = 4/19

9 × (4/19)/13 = 36/247

So, 9x/13y = 36/247.

Given:

The fractions are 13/19, 25/ 31, 28/31, 70/79.

Calculation:

The values are-

13/19 = 0.68

25/31 = 0.80

28/31 = 0.90

70/79 = 0.88

∴ Option C is correct.

Given:

\(\frac{{5{\rm{x}}}}{{1{\rm{\;}} + {\rm{\;}}\frac{1}{{1{\rm{\;}} + {\rm{\;}}\frac{{\rm{x}}}{{1{\rm{\;}} - {\rm{\;x}}}}}}}}{\rm{\;}} = {\rm{\;}}1\)

Calculation:

\(\Rightarrow {\rm{\;}}\frac{{5{\rm{x}}}}{{1{\rm{\;}} + {\rm{\;}}\frac{{1 - {\rm{x}}}}{{1 - {\rm{x\;}} + {\rm{\;x}}}}}}{\rm{\;}} = {\rm{\;}}1 \)

\(\Rightarrow {\rm{\;}}\frac{{5{\rm{x}}}}{{1{\rm{\;}} + {\rm{\;}}1 - {\rm{x}}}}{\rm{\;}} = {\rm{\;}}1\)

⇒ 5x/(2 – x) = 1

⇒ 5x = 2 – x

⇒ 6x = 2

⇒ x = 2/6

∴ The required value of x is 1/3.

Concept Used:

Follow BODMAS rule to solve this question, as per the order given below,

Calculations:

Given expression is,

\(\Rightarrow 5\frac{1}{6} - 3\frac{4}{9} + ? = \frac{7}{3} \times 4\frac{1}{6}\)

\(\Rightarrow {\rm{\;}}\frac{{31}}{6} - \frac{{31}}{9} + ? = \frac{7}{3} \times \frac{{25}}{6}\)

\(\Rightarrow 31\left( {\frac{1}{6} - \frac{1}{9}} \right) + \;? = \left( {\frac{{175}}{{18}}} \right)\)

\(\Rightarrow {\rm{\;}}31(\frac{3}{{54}}) + ? = \frac{{175}}{{18}}\)

\(\Rightarrow {\rm{\;}}\frac{{31}}{{18}} + ? = \frac{{175}}{{18}}\)

⇒ ? = 144/18

⇒ ? = 8