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

\frac{7}{15}

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\frac{9}{15}

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\frac{6}{15}

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\frac{8}{15}

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Solution

The correct option is D $\frac{8}{15}$
In the right side of the given triangle, the sum of the fractions is:

$\frac{1}{15} + \frac{5}{15} + \frac{5}{15} = \frac{11}{15}$

And the sum of the fractions of bottom side of the triangle is:

$\frac{2}{15} + \frac{4}{15} + \frac{5}{15} = \frac{11}{15}$

It is given that the sum of the fractions on each side of the triangle is same, therefore, the sum of the left side of the triangle will also be same that is $\frac{11}{15}$ .

Now, let the empty space on the left side be $x$ , then we have

$\frac{1}{15} + x + \frac{2}{15} = \frac{11}{15} \Rightarrow x + \frac{3}{15} = \frac{11}{15} \Rightarrow x = \frac{11}{15} - \frac{3}{15} \Rightarrow x = \frac{8}{15}$

Hence, the empty space fills with the fraction $\frac{8}{15}$ .

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Name the property of multiplication illustrated by each of the following statements :

(i) $\frac{- 15}{8} \times \frac{- 12}{7} = \frac{- 12}{7} \times \frac{- 15}{8}$

(ii) $(\frac{- 2}{3} \times \frac{7}{9}) \times \frac{- 9}{5} = \frac{- 2}{3} \times (\frac{7}{9} \times \frac{- 9}{5})$

(iii) $\frac{- 3}{4} \times (\frac{- 3}{4} \times \frac{- 5}{6}) + (\frac{- 3}{4} \times \frac{7}{8})$

(iv) $\frac{- 16}{9} \times 1 = 1 \times \frac{- 16}{9} = \frac{- 16}{9}$

(v) $\frac{- 11}{15} \times \frac{15}{- 11} = \frac{15}{- 11} \times \frac{- 11}{15} = 1$

(vi) $\frac{- 7}{5} \times 0 = 0$

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