2 Marks Questions

Question 512 Marks

Taking $\text{x}=\frac{-4}{9},\text{y}=\frac{5}{12}$ and $\text{z}=\frac{7}{18},$ find $x + (y + z)$

Answer

Given, $\text{x}=\frac{-4}{9},\text{y}=\frac{5}{12}$ and $\text{z}=\frac{7}{18}$
Here, $x + (y + z)$
$=\frac{-4}{9}+\Big(\frac{5}{12}+\frac{7}{18}\Big)$
$=\frac{-4}{9}+\Big(\frac{5\times3+7\times2}{36}\Big)$
$=\frac{-4}{9}+\frac{29}{36}=\frac{-4\times4+29}{36}=\frac{13}{36}$

View full question & answer→

Question 522 Marks

Reduce the following rational numbers in its lowest form: $\frac{91}{-364}$

Answer

$\frac{91}{-364}$ can be written as $=\frac{91+91}{-364+91}$
$[$dividing numerator and denominator by $HCF$ of $91$ and $364$ i.e. $91]$
$=\frac{-91\times\frac{1}{91}}{-364\times\frac{1}{91}}$
$\Big[\because\text{Reciprocal of }91=\frac{1}{91}\Big]$
$=\frac{1}{4},$ which is the lowest form.

View full question & answer→

Question 532 Marks

Write the following as rational numbers in their standard forms. $240 ÷ (-840)$

Answer

Here, $ 240 + (-840) =\frac{240}{-840}$
$\therefore HCF$ of $240$ and $840 = 120$
On dividing numerator and denominator by their $HCF,$
we get $\frac{240\div120}{-840\div120}=\frac{2}{-7}$
$=\frac{2\times(-1)}{-7\times(-1)}$
$[$multipying numerator denominator by $(-1)$ for positive denominator$]$
$=\frac{-2}{7}$

View full question & answer→

Question 542 Marks

Taking $\text{x}=\frac{-4}{9},\text{y}=\frac{5}{12}$ and $\text{z}=\frac{7}{18},$ findThe rational number when added to $z$ gives us $x.$

Answer

Given, $\text{x}=\frac{-4}{9},\text{y}=\frac{5}{12}$ and $\text{z}=\frac{7}{18}$
Let $A$ be added to $z$ to give $x.$
$\therefore\text{A}+\text{z}=\text{x}$
$\Rightarrow\text{A}+\frac{7}{18}=\frac{-4}{9}$
$\Rightarrow\text{A}=\frac{-4}{9}-\frac{7}{18}=\frac{-4\times2-7\times1}{18}$
$=\frac{-8-7}{18}=\frac{-15}{18}=\frac{-5}{6}$

View full question & answer→

Question 552 Marks

Express the following rational numbers in its standard form: $\frac{299}{-161}$

Answer

Given rational number is $\frac{299}{-161}.$
For standard form of given rational number, $=\frac{299\div23}{-161\div23}$
$[\therefore HCF$ of $299$ and $61 = 23]$
$=\frac{-13}{7}=\frac{13\div(-1)}{13\div(-1)}$
$[$dividing by $(-1)$ in both numerator and denominator$]$
$=\frac{-13}{7}$
Hence, the standard form of $\frac{299}{-161}$ is $\frac{-13}{7}.$

View full question & answer→

MATHS 2 Marks Questions