Fractions Test

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Fractions Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

The standard form of $$\displaystyle\frac{192}{-168}$$ is _________.

A
$$\displaystyle\frac{-2}{3}$$

B
$$\displaystyle\frac{-8}{7}$$

C
$$\displaystyle\frac{-1}{7}$$

D
$$\displaystyle\frac{-6}{7}$$

Solution

the standard form of $$ - \dfrac{192}{168}$$
$$=-\dfrac{2 \times 3\times 4\times 8}{2\times 3\times 4\times 7}$$
$$-\dfrac{8}{7}$$
Question 2
1 / -0

Add the fractions:

A
$$\dfrac{29}{24}$$

B
$$\dfrac{12}{20}$$

C
$$\dfrac{20}{24}$$

D
$$\dfrac{16}{24}$$

Solution

Out of $$12$$ squares, $$7$$ are filled, the fraction of squares filled will be =$$\dfrac{7}{12}$$

Out of $$8$$ squares, $$5$$ are filled ,the fraction of squares filled will be =$$\dfrac{5}{8}$$

Adding them, we get $$=\dfrac{7}{12}+\dfrac{5}{8}=\dfrac{14+15}{24}=\dfrac{29}{24}$$
Question 3
1 / -0

Which number should come in place of $$\square$$
$$\dfrac{4}{9}$$+$$\dfrac{7}{9}$$+$$\dfrac\square{9}$$ = 2$$\dfrac{1}{9}$$

A
$$7$$

B
$$1$$

C
$$8$$

D
$$9$$

Solution

We have,
$$\dfrac{4}{9}$$+$$\dfrac{7}{9}$$+$$\dfrac\square{9}$$ = 2$$\dfrac{1}{9}$$

$$\dfrac{4}{9}$$+$$\dfrac{7}{9}$$+$$\dfrac\square{9}$$ = $$\dfrac{19}{9}$$

$$\dfrac\square{9}$$ = $$\dfrac{19}{9}$$-$$\dfrac{4}{9}$$-$$\dfrac{7}{9}$$

$$\dfrac{\square}{9}$$ = $$\dfrac{8}{9}$$
Thus $$\square$$ = 8
Question 4
1 / -0

What should be placed in the empty space '?' so that the sum of fractions on each side of the triangle is same?

A
$$\dfrac{7}{15}$$

B
$$\dfrac{9}{15}$$

C
$$\dfrac{6}{15}$$

D
$$\dfrac{8}{15}$$

Solution

We have, $$\dfrac{2}{15}$$ + $$\dfrac{4}{15}$$ + $$\dfrac{5}{15}$$ = $$\dfrac{11}{15}$$,
$$\dfrac{1}{15}$$ + $$\dfrac{5}{15}$$ + $$\dfrac{5}{15}$$ = $$\dfrac{11}{15}$$
And $$\dfrac{1}{15}$$ + ? + $$\dfrac{2}{15}$$ = $$\dfrac{11}{15}$$ or, $$\dfrac{1}{15}$$ + $$\dfrac{8}{15}$$ + $$\dfrac{2}{15}$$ = $$\dfrac{11}{15}$$
$$\therefore$$ ? = $$\dfrac{8}{15}$$
Question 5
1 / -0

In the given number line, point M represents which of the following fractional numbers?

A
$$\displaystyle\frac{2}{28}$$

B
$$\displaystyle\frac{6}{5}$$

C
$$\displaystyle\frac{2}{3}$$

D
$$\displaystyle\frac{7}{10}$$

Solution

Point m is at $$1.2$$ on the number line.
And $$\dfrac{6}{5}$$ corresponds to $$1.2$$
Hence the correct answer is option B
Question 6
1 / -0

Using appropriate properties of addition, find the following:
(i) $$\cfrac{4}{5}+\cfrac{11}{7}+\cfrac{-7}{5}+\cfrac{-2}{7}$$

A
$$\dfrac{24}{35}$$

B
$$\dfrac{14}{35}$$

C
$$\dfrac{34}{35}$$

D
$$\dfrac{44}{35}$$

Solution

(i) $$\cfrac{4}{5}+\cfrac{11}{7}+\cfrac{-7}{5}+\cfrac{-2}{7}$$

$$\Rightarrow \cfrac{4}{5}+\cfrac{11}{7}-\cfrac{7}{5}-\cfrac{2}{7}$$

$$\Rightarrow$$ L.C.M. $$(5,7)=35$$

$$\Rightarrow \cfrac{4\times 7+11\times 5-7\times 7-2\times 5}{35}$$

$$\Rightarrow \cfrac{28+55-49-10}{35}$$$$ = \cfrac {24}{35}$$
Question 7
1 / -0

Subtract the fractions:

A
$$\dfrac{2}{3}$$

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

C
$$\dfrac{5}{12}$$

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

Solution

Out of three rectangles, two of them are filled ,the fraction filled will be $$\dfrac{2}{3}$$
Out of four rectangles, one of them is filled ,the fraction filled will be $$\dfrac{1}{4}$$
Subtracting them we get $$\dfrac{2}{3}-\dfrac{1}{4}=\dfrac{8-3}{12}=\dfrac{5}{12}$$
Question 8
1 / -0

Evaluate:
$$\dfrac{1}{\frac{7}{9}} + \dfrac{1}{\frac{5}{3}}$$

A
$$\dfrac{66}{27}$$

B
$$\dfrac{66}{35}$$

C
$$\dfrac{27}{66}$$

D
$$\dfrac{35}{66}$$

Solution

Given $$\dfrac{1}{\frac{7}{9}}+\dfrac{1}{\frac{5}{3}}$$

$$=\dfrac{9}{7}+\dfrac{3}{5}$$ --------[since $$\dfrac{1}{\frac{1}{x}}=x$$]

$$=\dfrac{45+21}{35}=\dfrac{66}{35}$$

Option B.
Question 9
1 / -0

Draw number line and locate the points on them. Which of the following are equidistant from the preceding points

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

B
$$\dfrac{1}{8} ,\dfrac{2}{8}, \dfrac{3}{8}, \dfrac{7}{8}$$

C
$$\dfrac{2}{5} ,\dfrac{3}{5}, \dfrac{8}{5}, \dfrac{4}{5}$$

D
none of these

Solution

R.E.F image
Difference between 2 adjacent point $$ = 1/4 $$
Question 10
1 / -0

Evaluate: $$\dfrac{3}{7} + \dfrac{3}{5} + \left( {\dfrac{{ - 2}}{3}} \right)$$

A
$$\dfrac {38}{105}$$

B
$$\dfrac {19}{105}$$

C
$$\dfrac {38}{108}$$

D
$$\dfrac {19}{108}$$

Solution

$$\dfrac{3}{7} +\dfrac{3}{5} +\dfrac{-2}{3} = \dfrac{15+21}{35} +\dfrac{-2}{3} = \dfrac{36}{35} - \dfrac{2}{3} = \dfrac{108 - 70}{105} = \dfrac{38}{105}$$