Real Number & Integers - Practice 2
Get ready to tackle 7 practice questions chosen from past-year exam papers. These questions focus on the most frequently tested types in the Real Number System chapter, helping you practice effectively and build confidence.
Related Lessons:
Practice 2 - Question 1
Arrange the following numbers in ascending order:
\(0.\dot{6} \quad \frac{3}{5} \quad \frac{5}{7} \quad \frac{66}{100}
\)
Solution:
To arrange the answers in ascending order, use your calculator to convert all the items to 3 significant figures. If two values are the same, convert to 4 or 5 significant figures before performing comparision.
\(\frac{3}{5} \quad \frac{66}{100} \quad 0.\dot{6} \quad \frac{5}{7}
\)
Practice 2 - Question 2
If the temperature in the freezer is \(-21°C\) and the temperature outside the freezer is \(18°C\), find the difference between these two temperatures.
Solution:
Diff=\(18-\left(-21\right)=39°C\)
Practice 2 - Question 3
Estimate \(\frac{\left(-2.04\right)^3-\left[\sqrt{15.99}\div1.58\right]}{\left[\left(2.96\right)^2+35.52\right]}\), giving your answer correct to 1 significant figure.
Solution:
Note that in estimation question, there can more than one solution. So it is important for students to show the entire working.
\[
\begin{align*}
\frac{\left(-2.04\right)^3-\left[\sqrt{15.99}\div1.58\right]}{\left[\left(2.96\right)^2+35.52\right]}&≈\frac{\left(-2\right)^3-\left[\sqrt{16}\div2\right]}{\left[\left(3\right)^2+36\right]}\\
&=\frac{-8-2}{45}\\
&=\frac{-10}{50}\\
&=-0.2\left(1s.f.\right)
\end{align*}
\]
Practice 2 - Question 4
On a particular day, the temperature at the bottom of a mountain was \(32.5°C\) and the temperature at the top of the mountain was \(-5.2°C\).
a) What was the difference between the temperature at the top and the temperature at the bottom of the mountain?
b) Given that the mountain is 2500 m tall and assuming that the temperature drops per metre height is constant, find the height of the mountain where the temperature is \(5°C\).
Solution:
a) \(32.5-\left(-5.2\right)=37.7°C\)
b)\[
\begin{align*}
\frac{h}{2500}&=\frac{5-32.5}{-5.2-32.5}\\
&=\frac{5-32.5}{-5.2-32.5}\times2500\\
&=1823.607\\
&=1820metres\left(3s.f.\right)
\end{align*}
\]
Practice 2 - Question 5
Given that \(87\times132=11,484\), showing all working, write down the exact value of
a) \(0.087\times13200\).
b) \(0.11484\div0.0087\)
Solution:
For this type of questions, students are required to show all the workings step by step.
a) \[
\begin{align*}
0.087\times13200&=\frac{87}{1000}\times132\times100\\
&=\frac{87\times132}{10}\\
&=\frac{11484}{10}\\
&=1148.4
\end{align*}
\]
b) \[
\begin{align*}
87\times132&=11,484\\
\frac{87}{10,000}\times132\times\frac{1}{10}&=\frac{11,484}{10,000}\times\frac{1}{10}\\
0.0087\times13.2&=0.11484\\
0.11484&=0.0087\times13.2\\
0.11484\div0.0087=13.2
\end{align*}
\]
Practice 2 - Question 6
Write down the number halfway between \(\frac{5}{7}\) and \(\frac{6}{7}\) in the form \(\frac{m}{n}\) where m and n are integers.
Solution:
\(\frac{\left(\frac{5}{7}+\frac{6}{7}\right)}{2}=\frac{11}{14}\)Practice 2 - Question 7
From the list of numbers
\(-16,\ 0,\ 1,\ \sqrt2,\ \pi,\ 17\)
write down
a) all natural number (s)
b) all irrational number (s)
c) all prime number (s)
d) the smallest integer
Solution:
a) Natural numbers are integer numbers more than 0
Therefore, the answers are 1 and 17.
b) Irrational numbers are numbers that cannot be express as a fraction, and they are \(\sqrt2,\ \pi\).
c) Prime numbers are any numbers that can be divided by 1 and by itself. Therefore the only prime number is 17.
d) The smallest integer is \(-16\).
Step By Step Full Solution
If a picture says a thousand words, a video tells the entire story — from the first step to the final answer. Below is the complete step-by-step video solution for the 7 questions above, carefully walking you through the full working and reasoning to strengthen your understanding.