1.2.1 Directed Numbers, PT3 Practice

Posted on June 3, 2017 by user

1.2.1 Directed Numbers, PT3 Practice 1

Question 1:

Complete the following calculation.

$\begin{array}{l} - 4.6 \times (\frac{3}{5} - 1 \frac{3}{4}) = - 4.6 \times (0.6 - x) \\ = - 4.6 \times x \\ = x \end{array}$

Solution:

$\begin{array}{l} - 4.6 \times (\frac{3}{5} - 1 \frac{3}{4}) = - 4.6 \times (0.6 - 1.75) \\ = - 4.6 \times - 1.15 \\ = 5.29 \end{array}$

Question 2:

Complete the following calculation.

$\begin{array}{l} - 43 + 35 \div 1 \frac{1}{6} = - 43 + 35 \div \frac{7}{6} \\ = - 43 + 35 \times x \\ = - 43 + x \\ = - 13 \end{array}$

Solution:

$\begin{array}{l} - 43 + 35 \div 1 \frac{1}{6} = - 43 + 35 \div \frac{7}{6} \\ = - 43 + 35 \times \frac{6}{7} \\ = - 43 + 30 \\ = - 13 \end{array}$

Question 3:
Calculate each of the following.

(a) (–9) × 14

(b) 6.7 – 3.2 × (–0.5)

Solution:

(a) (–9) × 14 = –126

(b) 6.7 – 3.2 × (–0.5) = 6.7 – (–1.6)

= 6.7 + 1.6

= 8.3

= 36

Question 4:

Calculate the value of

$0.6 + (- 1 \frac{2}{3}) \div (\frac{4}{15}) - 2$

Solution:

$\begin{array}{l} 0.6 + (- 1 \frac{2}{3}) \div (\frac{4}{15}) - 2 \\ = 0.6 + [- \frac{5}{3} \times \frac{15^{5}}{4}] - 2 \\ = 0.6 + [- \frac{25}{4}] - 2 \\ = 0.6 - 6 \frac{1}{4} - 2 \\ = 0.6 - 6.25 - 2 \\ = - 7.65 \end{array}$

Question 5:

Calculate the value of

$(- 0.6 \times 1 \frac{1}{4}) - [- 1.6 \times (- 1.5)] \div 0.4$

Solution:

$\begin{array}{l} (- 0.6 \times 1 \frac{1}{4}) - [- 1.6 \times (- 1.5)] \div 0.4 \\ = ({- 0.6}^{- 0.3} \times \frac{5}{4^{2}}) - [\frac{{- 1.6}^{- 4} \times (- 1.5)}{{0.4}^{1}}] \\ = - \frac{1.5}{2} - 6 \\ = - 0.75 - 6 \\ = - 6.75 \end{array}$