Rational Numbers Test

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Rational Numbers Test - 29

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Weekly Quiz Competition

Question 1
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Four rational number equivalent to $$ \displaystyle \frac{5}{8} $$ are:

A
$$ \displaystyle \frac{10}{6} $$,$$ \displaystyle \frac{12}{24} $$,$$ \displaystyle \frac{20}{32} $$,$$ \displaystyle \frac{25}{40} $$

B
$$ \displaystyle \frac{10}{16} $$,$$ \displaystyle \frac{15}{24} $$,$$ \displaystyle \frac{20}{32} $$,$$ \displaystyle \frac{25}{40} $$

C
$$ \displaystyle \frac{10}{16} $$,$$ \displaystyle \frac{15}{24} $$,$$ \displaystyle \frac{8}{32} $$,$$ \displaystyle \frac{25}{40} $$

D
$$ \displaystyle \frac{8}{16} $$,$$ \displaystyle \frac{15}{24} $$,$$ \displaystyle \frac{20}{32} $$,$$ \displaystyle \frac{25}{40} $$

Solution

$$ \displaystyle \frac{5}{8} $$= $$ \displaystyle \frac{5\times 2}{8\times 2} $$= $$ \displaystyle \frac{5\times 3}{8\times 3} $$= $$ \displaystyle \frac{5\times 4}{8\times 4} $$= $$ \displaystyle \frac{5\times 5}{8\times 5} $$

$$ \displaystyle \therefore \frac{5}{8} $$= $$ \displaystyle \frac{10}{16} $$= $$ \displaystyle \frac{15}{24} $$= $$ \displaystyle \frac{20}{32} $$= $$ \displaystyle \frac{25}{40} $$

Thus four rational number equivalent to $$ \displaystyle \frac{5}{8} $$ are :
$$ \displaystyle \frac{10}{16} $$, $$ \displaystyle \frac{15}{24} $$, $$ \displaystyle \frac{20}{32} $$, $$ \displaystyle \frac{25}{40} $$
Question 2
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Fill in the blank: $$\left [\dfrac {6}{5}\times \left (\dfrac {-2}{12}\right )\right ] + \left [\dfrac {6}{5} \times \dfrac {4}{12}\right ] = \dfrac {6}{5}\times \left [..... + \dfrac {4}{12}\right ]$$

A
$$\left (\dfrac {-2}{12}\right )$$

B
$$\left (\dfrac {-1}{12}\right )$$

C
$$-12$$

D
$$\dfrac {4}{12}$$

Solution

Distributive property over addition for rational numbers states that
$$(a\times b) + (a\times c) = a\times (b + c)$$
$$\therefore \left [\dfrac {6}{5}\times \left (\dfrac {-2}{12}\right )\right ] + \left [\dfrac {6}{5} \times \dfrac {4}{12}\right ]$$
$$= \dfrac {6}{5}\left [\left (\dfrac {-2}{12}\right ) + \dfrac {4}{12}\right ]$$.
Question 3
1 / -0

$$ \displaystyle \frac{5}{13}+\_\_=\frac{5}{13} $$

A
$$0$$

B
$$1$$

C
$$ \displaystyle \frac{5}{13} $$

D
$$ \displaystyle \frac{2}{13} $$

Solution

If we add $$0$$ to any number, we get the same number
$$\therefore \dfrac{5}{13}+0=\dfrac{5}{13}$$
Hence, the answer is $$0$$.
Question 4
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The value of $$\left [\dfrac {3}{4}\times \dfrac {4}{12}\right ] + \left [\dfrac {3}{4} \times \dfrac {-3}{9}\right ]$$ is

A
$$\dfrac {3}{4}$$

B
$$0$$

C
$$1$$

D
$$\dfrac {1}{2}$$

Solution

$$\left [\dfrac {3}{4}\times \dfrac {4}{12}\right ] + \left [\dfrac {3}{4} \times \dfrac {-3}{9}\right ]$$
$$= \dfrac {3}{4}\times \left [\dfrac {4}{12} + \dfrac {-3}{9}\right ]$$ ($$\because$$ distributive property)
$$= \dfrac {3}{4} \times 0$$
$$= 0$$.
Question 5
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If $$(81\div 9)\div 3=a$$ and $$81\div(9\div 3)=b$$, then which of the following is correct?

A
$$a=b$$

B
$$a>b$$

C
$$b>a$$

D
All of the above

Solution

Associative property is not satisfied by division.
$$(81\div 9)\div 3=a$$ $$81\div(9\div3)=b$$
$$\Rightarrow 9\div 3=3=a$$ $$81\div3 =27 = b$$

$$\therefore a\neq b$$ and $$b > a$$

So, option $$C$$ is correct.
Question 6
1 / -0

The value of $$\dfrac {1}{2} \times \left (\dfrac {1}{3} + \dfrac {4}{9}\right )$$ is

A
$$\dfrac {7}{18}$$

B
$$\dfrac {1}{2}$$

C
$$\dfrac {7}{12}$$

D
$$\dfrac {7}{24}$$

Solution

$$\dfrac {1}{2} \times \left (\dfrac {1}{3} + \dfrac {4}{9}\right ) = \dfrac {1}{2} \times \dfrac {1}{3} + \dfrac {1}{2} \times \dfrac {4}{9}$$ ($$\because$$ distributive property)
$$= \dfrac {1}{6} + \dfrac {2}{9}$$

$$= \dfrac {3 + 4}{18}$$

$$= \dfrac {7}{18}$$.
Question 7
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Find five rational numbers between $$1$$ and $$2.$$

A
$$\dfrac {1}{10}, \dfrac {2}{10}, \dfrac {3}{10}, \dfrac {4}{10}, \dfrac {5}{10}$$

B
$$\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}, \dfrac {5}{5}$$

C
$$\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}, \dfrac {1}{5}, \dfrac {1}{6}$$

D
$$\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$$

Solution

$$\textbf{Step -1: Find required rational numbers.}$$
$$\text{We need five rational numbers between }1\text{ and }2.$$
$$\text{So, multiply and divide the numbers with a natural number greater or equal to } 5+1=6.$$
$$\text{Lets take }7.$$
$$1=1\times\dfrac{7}{7}=\dfrac{7}{7}$$
$$\text{and }2=2\times\dfrac{7}{7}=\dfrac{14}{7}$$
$$\text{Thus, }1\text{ and }2\text{ becomes }\dfrac{7}{7}\text{ and }\dfrac{14}{7}\text{ respectively.}$$
$$\text{Since, }7<8<9<10<11<12<13<14$$
$$\therefore 1=\dfrac{7}{7}<\dfrac{8}{7}<\dfrac{9}{7}<\dfrac{10}{7}<\dfrac{11}{7}<\dfrac{12}{7}<\dfrac{13}{7}<\dfrac{14}{7}=2$$
$$\text{Hence, five rational numbers between }1\text{ and }2\text{ are }\dfrac{8}{7},\dfrac{9}{7},\dfrac{10}{7},\dfrac{11}{7},\text{ and }\dfrac{12}{7}.$$
$$\textbf{Hence , the correct option is D.}$$
Question 8
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Find the value of $$x$$ in $$\dfrac {4}{3} \times \left [x + \dfrac {1}{13} \right ] = \dfrac {4}{3}\times \dfrac {8}{11} + \dfrac {4}{3}\times \dfrac {1}{13}.$$

A
$$\dfrac {8}{11}$$

B
$$\dfrac {11}{8}$$

C
$$\dfrac {4}{3}$$

D
$$\dfrac {1}{13}$$

Solution

$$\dfrac {4}{3} \times \left (x + \dfrac {1}{13} \right ) = \dfrac {4}{3}\times x + \dfrac {4}{3} \times \dfrac {1}{13}$$

The distributive property over addition for rational numbers states that for any three rational numbers $$a, b$$ and $$c :$$
$$a (b + c) = (a\times b) + (a\times c)$$

$$\dfrac {4}{3}\times \dfrac {8}{11} + \dfrac {4}{3} \times \dfrac {1}{13} = \dfrac {4}{3} \times \left (\dfrac {8}{11} \times \dfrac {1}{13}\right )$$
Question 9
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The value of $$\left (\dfrac {5}{9}\times \dfrac {6}{11}\right ) + \left (\dfrac {1}{11}\times \dfrac {3}{9}\right )$$ is

A
$$\dfrac {1}{5}$$

B
$$\dfrac {1}{9}$$

C
$$\dfrac {1}{11}$$

D
$$\dfrac {1}{3}$$

Solution

$$\dfrac {5}{9}\times \dfrac {6}{11} + \dfrac {1}{11}\times \dfrac {3}{9}$$

$$= \dfrac {1}{11}\left [\dfrac {30}{9} + \dfrac {3}{9}\right ]$$ ($$\because$$ distributive property)
[Also, commutative property was used to get the terms in line]

$$= \dfrac {1}{11} \left [\dfrac {33}{9}\right ]$$

$$= \dfrac {1}{3}$$
Question 10
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The value of $$\dfrac {6}{11} \times \left [\left (\dfrac {-7}{6}\right ) - \left (\dfrac {11}{7}\right )\right ]$$ is

A
$$\dfrac {17}{77}$$

B
$$\dfrac {-17}{77}$$

C
$$\dfrac {-115}{77}$$

D
$$\dfrac {115}{77}$$

Solution

$$\dfrac {6}{11} \times \left [\left (\dfrac {-7}{6}\right ) - \left (\dfrac {11}{7}\right )\right ]$$

$$= \dfrac {6}{11} \times \dfrac {-7}{6} - \dfrac {6}{11}\times \dfrac {11}{7}$$ ($$\because$$ distributive property)

$$= \dfrac {-7}{11} - \dfrac {6}{7}$$

$$= \dfrac {-49 - 66}{77}$$

$$= \dfrac {-115}{77}$$.

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Rational Numbers Test - 29

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Rational Numbers Test - 29

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Question 1

$$ \displaystyle \frac{10}{6} $$,$$ \displaystyle \frac{12}{24} $$,$$ \displaystyle \frac{20}{32} $$,$$ \displaystyle \frac{25}{40} $$

$$ \displaystyle \frac{10}{16} $$,$$ \displaystyle \frac{15}{24} $$,$$ \displaystyle \frac{20}{32} $$,$$ \displaystyle \frac{25}{40} $$

$$ \displaystyle \frac{10}{16} $$,$$ \displaystyle \frac{15}{24} $$,$$ \displaystyle \frac{8}{32} $$,$$ \displaystyle \frac{25}{40} $$

$$ \displaystyle \frac{8}{16} $$,$$ \displaystyle \frac{15}{24} $$,$$ \displaystyle \frac{20}{32} $$,$$ \displaystyle \frac{25}{40} $$

Solution

Question 2

$$\left (\dfrac {-2}{12}\right )$$

$$\left (\dfrac {-1}{12}\right )$$

$$-12$$

$$\dfrac {4}{12}$$

Solution

Question 3

$$0$$

$$1$$

$$ \displaystyle \frac{5}{13} $$

$$ \displaystyle \frac{2}{13} $$

Solution

Question 4

$$\dfrac {3}{4}$$

$$0$$

$$1$$

$$\dfrac {1}{2}$$

Solution

Question 5

$$a=b$$

$$a>b$$

$$b>a$$

All of the above

Solution

Question 6

$$\dfrac {7}{18}$$

$$\dfrac {1}{2}$$

$$\dfrac {7}{12}$$

$$\dfrac {7}{24}$$

Solution

Question 7

$$\dfrac {1}{10}, \dfrac {2}{10}, \dfrac {3}{10}, \dfrac {4}{10}, \dfrac {5}{10}$$

$$\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}, \dfrac {5}{5}$$

$$\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}, \dfrac {1}{5}, \dfrac {1}{6}$$

$$\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$$

Solution

Question 8

$$\dfrac {8}{11}$$

$$\dfrac {11}{8}$$

$$\dfrac {4}{3}$$

$$\dfrac {1}{13}$$

Solution

Question 9

$$\dfrac {1}{5}$$

$$\dfrac {1}{9}$$

$$\dfrac {1}{11}$$

$$\dfrac {1}{3}$$

Solution

Question 10

$$\dfrac {17}{77}$$

$$\dfrac {-17}{77}$$

$$\dfrac {-115}{77}$$

$$\dfrac {115}{77}$$

Solution

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