No questions found
(a) These are the masses, in kg, of ten items.
38 6 21 50 32 7 15 9 27 25
The maximum total mass that each bin can hold is 60 kg.
Amara uses Method 1 to put these ten items into bins.
The table shows how she puts the first 6 items into bins.
\begin{table}[h]
\centering
\begin{tabular}{|c|c|c|}
\hline
\textbf{Bin} & \textbf{Mass of items in bin} & \textbf{Unused mass in bin} \\
\hline
1 & 38, 6, 7 & 22, 16, 9 \\
\hline
2 & 21, 32 & 39, 7 \\
\hline
3 & 50 & 10 \\
\hline
4 & & \\
\hline
5 & & \\
\hline
\end{tabular}
\end{table}
Complete Amara’s table to show that she needs 5 bins. [4]
(b) These are the masses, in kg, of six items.
8 16 13 10 5 3
The maximum total mass that each bin can hold is 20 kg.
\begin{table}[h]
\centering
\begin{tabular}{|c|c|c|}
\hline
\textbf{Bin} & \textbf{Mass of items in bin} & \textbf{Unused mass in bin} \\
\hline
1 & 8 & 12 \\
\hline
2 & & \\
\hline
3 & & \\
\hline
4 & & \\
\hline
5 & & \\
\hline
\end{tabular}
\end{table}
Use Method 1 to complete the table for all six items.
The first item has been put in for you.
You may not need all the bins. [2]
536
550
570
Amara wants to see if she can use fewer bins.
She puts her items in order of mass before she puts them in bins.
She uses this method.
Method 2 Put the masses in order, largest first.
Then use Method 1.
These are the masses, in kg, of the ten items from Question 1(a).
38 6 21 50 32 7 15 9 27 25
(a) Write these ten masses in order, largest first.
......... , ......... , ......... , ......... , ......... , ......... , ......... , ......... , ......... , .........
(b) The maximum total mass that each bin can hold is 60 kg.
Complete the table using Method 2.
[Table: Bin | Mass of items in bin | Unused mass in bin]
1 | |
2 | |
3 | |
4 | |
5 | |
(c) Find and compare the percentage of unused storage for these ten items when using Method 1 and Method 2.
Use your answers from Question 1(a) and Question 2(b).
(d) A best solution uses the smallest possible number of bins.
A set of items with a total mass of 270 kg is put into 4 bins.
The maximum total mass that each bin can hold is 80 kg.
Show that this is a best solution.
(e) Amara tries another way to improve Method 1.
Method 3 Look for items that combine to make as many full bins as possible and place these first.
For the remaining items, use Method 2.
These are the masses, in kg, of eight items.
21 10 30 19 13 7 28 4
The maximum total mass that each bin can hold is 40 kg.
Does Method 3 give a best solution for these items?
Show how you decide.
[Table: Bin | Mass of items in bin | Unused mass in bin]
1 | |
2 | |
3 | |
4 | |
5 | |
535
500
564
Amara has six items.
The mass, in kg, of each item is an integer.
These are the masses of her six items.
9 4 7 15 x y
The maximum total mass that each bin can hold is \textbf{20 kg}.
Amara finds that Method 1 needs fewer bins than Method 2.
Find the value of \(x\) and the value of \(y\).
\(x = \) ...........................................
\(y = \) ........................................... [4]
510
531
595
Amara now also uses the dimensions of each bin and of each item to find the maximum number of items that will go into a bin.
Each item is a cuboid and is the same height as the bin. Amara looks at the rectangular top of each bin to work out the maximum number of items that will go into it.
A bin is now full either because it contains the maximum total mass or because it contains the maximum number of items that will go into it.
She uses this method.
Method 4: Work out the maximum number of items for each bin using dimensions. Then use Method 2.
(a) The rectangular top of Amara’s bin measures 90 cm by 90 cm. The rectangular top of each item measures 60 cm by 30 cm.
Find the maximum number of items that will go into a bin using these dimensions. [2]
(b) These are the masses, in kg, of eleven items with tops measuring 60 cm by 30 cm.
41, 33, 22, 18, 16, 14, 8, 7, 6, 5, 4
The maximum total mass that each bin can hold is 90 kg. The top of each bin measures 90 cm by 90 cm. Each bin costs $13.50.
Use Method 4 to put these items into bins. Show that this is a best solution and find the cost of this solution.
[Table_1]
$ ..................................................... [5]
591
520
547
Some chemicals decay naturally. This means that their mass reduces. This task is about the decreasing mass of these chemicals. The half-life is the time it takes for the mass of a chemical to become half of its mass. Lee uses different models to look at some of these chemicals.
(a) The half-life of chemical A is 100 minutes. Lee has 400 grams of this chemical. Lee uses this model for the mass, in grams, of chemical A which remains at time t minutes. $$A(t) = 400 \left( \frac{1}{2} \right)^\frac{t}{100}$$ At the end of 1 half-life, $t = 100$ and half of the mass of chemical A, 200 g, remains. At the end of 2 half-lives, $t = 200$ and 100 g of chemical A remains.
(i) Find the mass, in grams, of chemical A which remains at the end of 4 half-lives. [2]
(ii) On the axes below, sketch the graph of $y = A(t)$ for $0 \leq t \leq 600$. [2]
(iii) Find the decrease in mass from $t = 50$ to $t = 100$. [2]
(iv) Find the number of half-lives when the mass remaining is 10 grams. Give your answer correct to 2 decimal places. [3]
(b) The half-life of chemical B is 30 minutes. Lee has 200 grams of this chemical.
(i) Change the model from part (a) to find the mass, in grams, of chemical B which remains at time t minutes. $$B(t) = \ldots \times \left( \frac{1}{2} \right)^\ldots$$ [1]
(ii) Find the fraction of the mass of chemical B which remains at the end of 300 minutes. [2]
(c) The quarter-life is the time it takes for the mass of a chemical to become a quarter of its mass. At the end of 1 quarter-life, one quarter of the mass remains.
The quarter-life of chemical C is 48 minutes. Lee has 240 grams of this chemical. He changes the model from part (a) so that it finds the mass, in grams, of chemical C which remains at time t minutes. His model is valid for 6 quarter-lives.
Complete Lee’s model. $$C(t) = \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$$ for $\ldots \ldots \ldots \ldots \leq t \leq \ldots \ldots \ldots \ldots$. [3]
571
565
573
Lee now uses this model in his calculations for the mass of a chemical which remains at time $t$ minutes.
$$N(t) = N_0 \times 3^{kt}$$ where $k$ is a constant
(a)
(i) Explain why $N_0$ is the mass of the chemical Lee starts with.
.............................................................................................................................................. [1]
(ii) Write an expression in terms of $N_0$ for the mass remaining at the end of 1 half-life.
.............................................................. [1]
(iii) Lee thinks $3^{kH} = \frac{1}{2}$ , for any starting mass $N_0$ and any half-life $H$ minutes.
Show that Lee is correct.
[1]
(b) The half-life of chemical $D$ is 10 minutes.
(i) Use logarithms to show that the value of $k$ for chemical $D$ is $-0.063$, correct to 3 decimal places.
[2]
(ii) Lee has 40 grams of chemical $D$.
Use $N(t)$ to find the time it takes for the mass of chemical $D$ to become 10% of its mass.
.............................................................. [4]
582
593
551
Lee has 60 grams of chemical $E$.
This decreases by 11 grams in 1.6 minutes.
(a) Use $N(t) = N_0 \times 3^{kt}$, where $k$ is a constant, to find the half-life of chemical $E$ in minutes. ....................................................... [5]
(b) Change the model from Question 5(a) to find the mass, in grams, of chemical $E$ which remains at time $t$ minutes and use it to check your answer to Question 7(a). [1]
590
587
534
