5 Mark Question

Question 15 Marks

Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
$\text{x}=-2,\text{y}=\frac{3}{5},\text{z}=\frac{-4}{3}$

Answer

$\text{x}=-2,\text{y}=\frac{3}{5},\text{z}=\frac{-4}{3}$
So, $(\text{x + y})+\text{z}=\Big(-2+\frac{3}{5}\Big)+\frac{-4}{3}$
$=\Big(\frac{-10}{5}+\frac{3}{5}\Big)+\frac{-4}{3}=\frac{-7}{5}+\frac{-4}{3}$
$=\frac{-21}{15}+\frac{-20}{15}=\frac{-21-20}{15}=\frac{-41}{15}$
$\text{x (y + z)}=-2+\Big(\frac{3}{5}+\frac{-4}{3}\Big)$
$=\frac{-2}{1}+\Big(\frac{9}{15}+\frac{-20}{15}\Big)=\frac{-2}{1}+\frac{-11}{15}=\frac{-30}{15}+\frac{-11}{15}$
$=\frac{-30-11}{15}=\frac{-41}{15}$
$\therefore\Big(-2+\frac{3}{5}\Big)+\frac{-4}{3}=-2+\Big(\frac{3}{5}+\frac{-4}{3}\Big)$

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Question 25 Marks

Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
$\text{x}=\frac{-7}{11},\text{y}=\frac{2}{-5},\text{z}=\frac{-3}{22}$

Answer

$\text{x}=\frac{-7}{11},\text{y}=\frac{2}{-5},\text{z}=\frac{-3}{22}$
$\Rightarrow\text{x}=\frac{-7}{11},\text{y}=\frac{2\times(-1)}{-5\times(-1)},\text{z}=\frac{-3}{22}$
$\Rightarrow\text{x}=\frac{-7}{11},\text{y}=\frac{2}{-5},\text{z}=\frac{-3}{22}$
Now $(\text{x}+\text{y})+\text{z}\Big(\frac{-7}{11}+\frac{-2}{5}\Big)+\frac{-3}{22}$
$=\frac{-35-22}{55}+\frac{-3}{22}$
$=\frac{-57}{55}+\frac{-3}{22}$ (LCM of 22, 55 = 110)
$=\frac{-114-15}{110}=\frac{-129}{110}$
and $\text{x}+(\text{y}+\text{z})=\frac{-7}{11}+\Big(\frac{-2}{5}+\frac{-3}{22}\Big)$
$=\frac{-7}{11}+\frac{-44-15}{110}$
$\frac{-7}{11}+\frac{-59}{110}$
$=\frac{-70-59}{110}=\frac{-129}{110}$
$\therefore(\text{x}+\text{y})+\text{z}=\text{x}+(\text{y}+\text{z})$

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Question 35 Marks

Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
$\text{x}=\frac{-2}{5},\text{y}=\frac{4}{3},\text{z}=\frac{-7}{10}$

Answer

$\text{x}=\frac{-2}{5},\text{y}=\frac{4}{3},\text{z}=\frac{-7}{10}$
$\therefore(\text{x}+\text{y})+\text{z}=\Big(\frac{-2}{5}+\frac{4}{3}\Big)+\frac{-7}{10}$
$=\frac{-6+20}{15}+\frac{-7}{10}$
$=\frac{14}{15}+\frac{-7}{10}$
LCM of 15, 10 = 30
$=\frac{28-21}{30}=\frac{7}{30}$
and $\text{x}+(\text{y}+\text{z})=\frac{-2}{5}+\Big(\frac{4}{3}+\frac{-7}{10}\Big)$
$=\frac{-2}{5}+\frac{40-21}{30}$
$=\frac{-2}{5}+\frac{19}{30}$
$=\frac{-12+19}{30}=\frac{7}{30}$
$\therefore(\text{x}+\text{y})+\text{z}=\text{x}+(\text{y}+\text{z})$

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Question 45 Marks

Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
$\text{x}=\frac{1}{2},\text{y}=\frac{2}{3},\text{z}=-\frac{1}{5}$

Answer

$\text{x}=\frac{1}{2},\text{y}=\frac{2}{3},\text{z}=-\frac{1}{5}$
$(\text{x}+\text{y})+\text{z}=\Big(\frac{1}{2}+\frac{2}{3}\Big)+\frac{-1}{5}$
$=\frac{3+4}{6}+\frac{-1}{5}=\frac{7}{6}+\frac{-1}{5}$
$=\frac{7}{6}-\frac{1}{5}$
$=\frac{35-6}{30}=\frac{29}{30}$
and $\text{x}+(\text{y}+\text{z})=\frac{1}{2}+\bigg(\frac{2}{3}+\frac{-1}{5}\bigg)$
$=\frac{1}{2}+\frac{10-3}{15}$
$=\frac{1}{2}+\frac{7}{15}$
$=\frac{15+14}{30}=\frac{29}{30}$
$\therefore(\text{x}+\text{y})+\text{z}=\text{x}+(\text{y}+\text{z})$

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