3 Mark Question

Question 13 Marks

Find x, if
$\Big( \frac{3}{5}\Big)^{-3}\times\Big(\frac{3}{2}\Big)^{5}=\Big(\frac{3}{2}\Big)^{2\text{x}+1}$

Answer

$\Big( \frac{3}{5}\Big)^{-3}\times\Big(\frac{3}{2}\Big)^{5}=\Big(\frac{3}{2}\Big)^{2\text{x}+1}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-3+5}=\Big(\frac{3}{2}\Big)^{2\text{x}+1}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{2}=\Big(\frac{3}{2}\Big)^{2\text{x}+1}$
Comparing, we get:
$2\text{x}+1=2$
$\Rightarrow2\text{x}=2-1=1$
$\Rightarrow\text{x}=\frac{1}{2}$

View full question & answer→

Question 23 Marks

By what number should $\Big(\frac{1}{2}\Big)^{-1}$ be multiplied so that the product may be equal to $\Big(\frac{-4}{7}\Big)^{-1}?$

Answer

Let x be multiplied, the
$\text{x}\times\Big (\frac{1}{2}\Big)^{-1}=\Big(\frac{-4}{7}\Big )^{-1}$
$\text{x}=\Big(\frac{-4}{7}\Big )^{-1}\div\Big(\frac{1}{2}\Big)^{-1}$
$=\Big(\frac{7}{-4}\Big) ^1\div\Big(\frac{2}{1}\Big)^1=\frac{7}{-4}\div2$
$=\frac{7}{-4}\times\frac{1}{2}=\frac{7}{-8}=\frac{-7}{8}$
$\therefore$ Required number $=\frac{-7}{8}$

View full question & answer→

Question 33 Marks

Find x, if
$\Big( \frac{2}{5}\Big)^{-3}\times\Big(\frac{3}{2}\Big)^{15}=\Big(\frac{2}{5}\Big)^{2\text{x}+1}$

Answer

$\Big( \frac{2}{5}\Big)^{-3}\times\Big(\frac{3}{2}\Big)^{15}=\Big(\frac{2}{5}\Big)^{2\text{x}+1}$
$\Rightarrow\Big(\frac{2}{5}\Big)^{-3+15}=\Big(\frac{2}{5}\Big)^{2+3\text{x}}$
$\Rightarrow\Big(\frac{2}{5}\Big)^{12}=\Big(\frac{2}{5}\Big)^{2+3\text{x}}$
Comparing, we get:
$2+3\text{x}=12$
$3\text{x}=12-2=10$
$\text{x}=\frac{10}{3}$

View full question & answer→

Question 43 Marks

By what number should $\Big(\frac{5}{3}\Big)^{-2}$ be multiplied so that the product may be $\Big(\frac{7}{3}\Big)^{-1}?$

Answer

Let x be multiplied, then
$\text{x}\times\Big(\frac{5}{3}\Big)^{-2}=\Big(\frac{7}{3}\Big)^{-1}\Rightarrow\text{x}=\Big(\frac{7}{3}\Big)^{-1}\div\Big(\frac{5}{3}\Big)^{-2}$
$\Rightarrow\text{x}=\Big(\frac{3}{7}\Big)^1\div\Big(\frac{3}{5}\Big)^2=\frac{3}{7}\div\frac{9}{25}=\frac{3}{7}\times\frac{25}{9}$
$=\frac{25}{21}$
$\therefore$ Required number $=\frac{25}{21}$

View full question & answer→

Question 53 Marks

Find x, if
$\Big( \frac{1}{2}\Big)^{-19}\div\Big(\frac{-1}{2}\Big)^{8}=\Big(\frac{-1}{2}\Big)^{-2\text{x}+1}$

Answer

$\Big( \frac{1}{2}\Big)^{-19}\div\Big(\frac{-1}{2}\Big)^{8}=\Big(\frac{-1}{2}\Big)^{-2\text{x}+1}$
$\Rightarrow\Big(\frac{-1}{2}\Big)^{-19-8}=\Big(\frac{-1}{2}\Big)^{-4\text{x}+1}$
$\Rightarrow\Big(\frac{1}{4}\Big)^{-27}=\Big(\frac{1}{4}\Big)^{-2\text{x}+1}$
Comparing, we get:
$-2\text{x}+1=-27$
$-2\text{x}=-27-1=-28$
$\text{x}=\frac{-28}{-2}=14$
$\therefore\text{x}=14$

View full question & answer→

Question 63 Marks

Simplify:
$\bigg\{\Big(\frac{2}{3}\Big)\bigg\}\times\Big(\frac{1}{3}\Big)^{-4}\times3^{-1}\times6^{-1}$

Answer

$\bigg\{\Big(\frac{2}{3}\Big)\bigg\}\times\Big(\frac{1}{3}\Big)^{-4}\times3^{-1}\times6^{-1}$
$=\Big(\frac{2}{3}\Big)^{2\times3}\times\Big(\frac{3}{1}\Big)^4\times\frac{1}{3^1}\times\frac{1}{6^1}$
$=\Big(\frac{2}{3}\Big)^6\times(3)^4\times\frac{1}{3}\times\frac{1}{6}$
$=\frac{2\times2\times2\times2\times2\times2}{3\times3\times3\times3\times3\times3}\times3\times3\times 3\times3\times\frac{1}{3}\times\frac{1}{6}$
$=\frac{64}{3\times3\times3\times6}=\frac{64}{81\times2}=\frac{32}{81}$

View full question & answer→

Question 73 Marks

If x $=\Big(\frac{4}{5}\Big)^{-2}\div\Big (\frac{1}{4}\Big)^{2},$ find the value of $\text{x}^{1}$

Answer

$=\Big(\frac{4}{5}\Big)^{-2}\div\Big (\frac{1}{4}\Big)^{2}$
$\Big(\frac{4}{5}\Big)^{-2}\div\Big(\frac{1}{4}\Big)^{2}$
$=\Big(\frac{5}{4}\div\frac{1}{4}\Big)^{2}$
$=\Big(\frac{5}{4}\times\frac{4}{1}\Big)^{2}=(5)^2=25$
$\therefore\text{x}^1= (25)^{-1}=\frac{1}{25}$

View full question & answer→

Question 83 Marks

By what number should $(-15)^{-1}$ be divided so that the quotient may be equal to $(-5)^{-1}?$

Answer

Let the number should be divided by = x
their $(-15)^{-1}\div \text{x}=(-5)^{-1}\Rightarrow \frac{1}{-15}\div \text{x}$
$=\frac{1}{(-5)^{1}}$
$\Rightarrow\frac{-1}{15}\times\frac{1}{\text{x}}=\frac{1}{-5}$
$\Rightarrow\frac{-1}{15\text{x}}=\frac{1}{-5}\Rightarrow-15\text{x}=-5$
$\Rightarrow\text{x}=\frac{-5}{-15}=\frac{1}{3}$
$\Rightarrow$ Required number $=\frac{1}{3}$

View full question & answer→

Question 93 Marks

If x $=\Big(\frac{3}{2}\Big)^2\times\Big (\frac{2}{3}\Big)^{-4},$ find the value of $\text{x}^{-2}$

Answer

$\text{x}=\Big(\frac{3}{2}\Big)^2\times\Big (\frac{2}{3}\Big)^{-4}$
$\Big(\frac{3}{2}\Big)^2\times\Big(\frac{2}{3}\Big)^{-4}$
$=\Big(\frac{3}{2}\Big)^{2+4}=\Big(\frac{3}{2}\Big)^6$
$\text{x}^2=\bigg[\Big(\frac{3}{2}\Big)^6\bigg]=\Big(\frac{3}{2}\Big)^{-12}$
$=\Big(\frac{2}{3}\Big)^{12}$

View full question & answer→

Question 103 Marks

By what number should $(-15)^{-1}$ be ddivded so that the quotient may be equal to $(-15)^{-1}?$

Answer

Let $(-15)^{-1}$ be divided by x
$\therefore(-15)^{-1}\div\text{x}=(-5)^{-1}$
$\Rightarrow\Big(\frac{1}{-15}\Big)\div\text{x}=\Big(\frac{1}{-5}\Big)^1$
$\Rightarrow\frac{1}{-15}\times\frac{1}{\text{x}}=\frac{1}{-5}$
$\Rightarrow\frac{1}{-15\text{x}}=\frac{1}{-5}$
$\Rightarrow\text{x}=\frac{-5}{-15}=\frac{1}{3}$
$\therefore$ Required number $=\frac{1}{3}$

View full question & answer→

Question 113 Marks

Find the value of x for which $5^{2\text{x}}\div5^{3}=5^5.$

Answer

$5^{2\text{x}}\div5^{3}=5^5$
$\Rightarrow5^{2\text{x}}\div5^5=5^{-3}$
$\Rightarrow5^{2\text{x}}=5^{5-3}=5^{2}$
Comparing, we get:
$2\text{x}=2\Rightarrow\text{x}=\frac{2}{2}=1$
$\therefore\text{x}=1$

View full question & answer→

Question 123 Marks

Find x, if
$\Big(\frac{8}{3}\Big)^{2\text{x}+3}\times\Big(\frac{8}{3}\Big)^{5}=\Big(\frac{8}{3}\Big)^{\text{x}+2}$

Answer

$\Big(\frac{8}{3}\Big)^{2\text{x}+3}\times\Big(\frac{8}{3}\Big)^{5}=\Big(\frac{8}{3}\Big)^{\text{x}+2}$
$\Rightarrow\Big(\frac{8}{3}\Big)^{2\text{x}+1+5}=\Big(\frac{2}{5}\Big)^{2+3\text{x}}$
Comaring, we get:
$2\text{x}+1+5=\text{x+2}$
$\Rightarrow2\text{x}-\text{x}=2-1-5\Rightarrow\text{x}=-4$
$\therefore\text{x}=-4$

View full question & answer→

Maths 3 Mark Question

Take a timed test