Cambridge IGCSE Mathematics - International - 0607 - Advanced Paper 6 2024 Summer Zone 1

Use the method to complete the sum of the first 60 positive integers. That is $1 + 2 + 3 + \cdots + 60 = 1830$.

$\begin{array}{cccccc} 1 & 2 & 3 & \cdots & \ldots & \\ \ldots & \ldots & \ldots & \cdots & \ldots & \\ \ldots & \ldots & \ldots & \cdots & \ldots & \\ \end{array}$

$\ldots \times \ldots = 1830$

Use the method to calculate the sum of the first 128 positive integers. That is $1 + 2 + 3 + \dots + 128$.

Complete the table.
Use Question 1, Question 2 and any patterns you notice.

[Table_1]

\begin{array}{|c|c|c|}\hline \text{Number of positive integers, starting at 1} & \text{Multiplication} & \text{Sum} \\ \hline 12 & 6 \times 13 & 78 \\ \hline 26 & 13 \times 27 & 351 \\ \hline \text{Question 1} & 60 & 1830 \\ \hline \text{Question 2} & 128 & \\ \hline 204 & & 20910 \\ \hline \end{array}

1 + 2 + 3 + \ldots + n \text{ has } n \text{ positive integers and its sum is } T.
Find a formula for } T \text{ in terms of } n.

(a) Complete the table.

[Table_1]
\begin{array}{|c|c|c|c|c|c|}\hline n & \text{Sum of first } n \text{ positive integers} & \text{Sum of first } n \text{ square numbers} & \frac{S}{T} \text{ written as} \\ & \text{Calculation} & \text{Sum (S)} & \text{a fraction with denominator 3} \\ \hline 1 & 1 & 1 & 1 & \frac{3}{3} \\ \hline 2 & 1+2 & 3 & 1^2+2^2 & 5 & \frac{5}{3} \\ \hline 3 & 1+2+3 & 6 & 1^2+2^2+3^2 & 14 & \frac{\overline{3}}{3} \\ \hline 4 & 1+2+3+4 & 10 & 1^2+2^2+3^2+4^2 & & \\ \hline 5 & 1+2+3+4+5 & 15 & 1^2+2^2+3^2+4^2+5^2 & 55 & \\ \hline \end{array}

[3]
(b) Find an expression for \frac{S}{T} in terms of n.
................................................. [3]
(c) The sum of the first 60 positive integers is 1830.
Find the sum of the first 60 square numbers.
................................................. [2]
(d) Use Question 4 and Question 5(b) to help you find a formula for S in terms of n.
Write your answer as a single fraction.
................................................. [1]

(a) Complete the table.
[Table_1]

Sum of first \( n \) positive integers | Sum of first \( n \) cube numbers
Calculation | Sum \( (T) \) | Calculation | Sum \( (C) \)
1 + 2 | 3 | 1^3 + 2^3 | 9
1 + 2 + 3 | 6 | 1^3 + 2^3 + 3^3 | 36
1 + 2 + 3 + 4 | 10 | 1^3 + 2^3 + 3^3 + 4^3 | 100
1 + 2 + 3 + 4 + 5 | 15 | 1^3 + 2^3 + 3^3 + 4^3 + 5^3 | 225
1 + 2 + 3 + 4 + 5 + 6 | \text{.....} | 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 | \text{.....}

(b) Write a formula for \( C \) in terms of \( T \).
\( \text{.................................................................................} \)

(c) \( 1 + 2 + 3 + \ldots + 60 = 1830 \)
Calculate \( 1^3 + 2^3 + 3^3 + \ldots + 60^3 \).
\( \text{.............................................................} \)

(a) Complete the table. Use \textbf{Question 6(a)} to help you.

\begin{table}[h!]\begin{center}\begin{tabular}{|c|c|c|c|c|}\hline\textbf{n} & \multicolumn{2}{|c|}{\textbf{Sum of first n cube numbers}} & \multicolumn{2}{|c|}{\textbf{Sum of first n 5th powers}} & \frac{F}{C} \text{ as a fraction with denominator 3} \\ \hline & \text{Calculation} & \text{Sum } (C) & \text{Calculation} & \text{Sum } (F) \\ \hline1 & 1^3 & 1 & 1^5 & 1 & 1 = \frac{3}{3} \\ \hline2 & 1^3 + 2^3 & 9 & 1^5 + 2^5 & 33 & \frac{33}{9} = \frac{11}{3} \\ \hline3 & 1^3 + 2^3 + 3^3 & 36 & 1^5 + 2^5 + 3^5 & 276 & \frac{276}{36} = \frac{23}{3} \\ \hline4 & 1^3 + 2^3 + 3^3 + 4^3 & 100 & 1^5 + 2^5 + 3^5 + 4^5 & 1300 & \\ \hline5 & 1^3 + 2^3 + 3^3 + 4^3 + 5^3 & 225 & 1^5 + 2^5 + 3^5 + 4^5 + 5^5 & 4425 & \frac{4425}{225} = \frac{59}{3} \\ \hline6 & 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 & & 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + 6^5 & & \\ \hline\end{tabular}\end{center}\end{table}

(b) The fraction \(\frac{F}{C}\) is written with a denominator of 3. Find an expression for the numerator in terms of \(n\). .............................................. [4]

(c) \(1 + 2 + 3 + \ldots + 60 = 1830\)
Calculate \(1^5 + 2^5 + 3^5 + \ldots + 60^5\).
Write down all the numbers on your calculator display. .............................................. [2]

The graph shows how the total income of a country is shared among the population.



The point $T$ shows that 0.3 of the population of a country earn 0.3 of the total income.
The point $U$ shows that 0.6 of the population of a country earn 0.6 of the total income.

(a) Mark a point on the line and label it $Z$.
Complete this statement for your point.
The point $Z$ shows that .......... of the population of a country earn .......... of the total income.

(b) For this graph there is perfect equality because the income is shared equally among the population of a country.
The graph shows the line of perfect equality.
Write down the equation of the line.
................................................................. [1]

In reality, the total income of a country is not shared equally, so there is income inequality. In 1905 the American economist Max Lorenz invented the Lorenz curve to show income inequality. A Lorenz curve is always on or below the line of perfect equality. The graph shows the Lorenz curve for one country. The point $V$ on this Lorenz curve shows that the poorest 0.7 of the population only earn 0.4 of the total income. (a) Use another point on the Lorenz curve to make a similar statement. ........................................................................................................................................................... ........................................................................................................................................................... [1] (b) From the statement for point $V$ we can also say that the richest 0.3 of the population earn 0.6 of the total income. (i) Write calculations to show why the values in this statement are correct. ........................................................................................................................................................... ........................................................................................................................................................... [1] (ii) Write a similar statement for the point that you chose in part (a). ........................................................................................................................................................... ........................................................................................................................................................... [1]

In 1912 the Italian statistician Corrado Gini invented the $Gini \text{ coefficient}$. The Gini coefficient measures how much income inequality there is in a country. $$ \text{[Image_1 depicting a graph with Line of perfect equality and Lorenz curve]} $$ The Gini coefficient is $\text{two times}$ the shaded area between the line of perfect equality and the Lorenz curve. This means that the greater the shaded area, the greater the income inequality.

(a) When there is perfect equality of income write down the Gini coefficient.

..................................................... [1]

(b) When the income inequality is a maximum, the shaded area will be as large as possible.

Find the Gini coefficient when there is maximum inequality of income.

..................................................... [2]

Mei uses these steps to model the Gini coefficient for the curve in Question 9.
Step 1: Plot point $V$ from Question 9 on the Lorenz curve.
Step 2: Approximate the area below the curve with a rectangle and two triangles.
Step 3: Calculate the total area of the rectangle and the two triangles. This is $T$.
Step 4: Calculate the shaded area by subtracting $T$ from the area of the large triangle with vertices (0, 0), (1, 1) and (1, 0).
Step 5: Multiply the result by 2.

[Image_1: Graph with Lorenz curve]

Steps 1 and 2 in Mei's model have been done for you on the graph.

Do Steps 3, 4 and 5 to calculate her approximation of the Gini coefficient.

Mei decides to use a point $P(x, y)$ in Step 1 of her model.

(a) Find the area in Step 3 of her model as an expression without brackets in terms of $x$ and $y$.
(b) Do steps 4 and 5 in her model to show that her approximation, $G$, for the Gini coefficient is $x-y$.
(c) Give a reason why $G$ will be smaller than the actual Gini coefficient.

(a) For Country A, the equation of the Lorenz curve is $y = x^2$.
Find Mei's most accurate estimate for the Gini coefficient by finding the maximum of $G = x - x^2$.

(b) (i) Write down the coordinates of the two points that must be on every Lorenz curve.

(ii) For Country B, the equation of the Lorenz curve is $y = a - \sqrt{1-x}$ where $a$ is a constant.
Use part (i) to find the value of $a$.

(iii) Sketch the graph of $G = x - y$ for Country B.

(c) Use the most accurate estimates in Mei’s model of the Gini coefficient to compare income inequality in Country A and Country B.