// Numbas version: exam_results_page_options {"metadata": {"description": "

Questions on powers, the laws of indices, and exponential growth.

", "licence": "Creative Commons Attribution 4.0 International"}, "timing": {"allowPause": true, "timedwarning": {"action": "none", "message": ""}, "timeout": {"action": "none", "message": ""}}, "showstudentname": true, "name": "Powers and indices [L0 Randomised]", "feedback": {"advicethreshold": 0, "showanswerstate": true, "intro": "", "feedbackmessages": [], "showactualmark": true, "allowrevealanswer": true, "showtotalmark": true}, "showQuestionGroupNames": false, "navigation": {"preventleave": true, "showresultspage": "oncompletion", "showfrontpage": true, "browse": true, "reverse": true, "allowregen": true, "onleave": {"action": "none", "message": ""}}, "percentPass": 0, "question_groups": [{"pickQuestions": 1, "pickingStrategy": "all-ordered", "name": "Group", "questions": [{"name": "Calculate powers of ten", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "advice": "

When $n$ is positive, we multiply $10$ by itself $n$ times,

\n

\\[\\text{e.g. } 10^3 = 10 \\times 10 \\times 10 = 1000 \\text{ .}\\]

\n

When $n$ is negative, we can think of $10^{-n}$ as $\\frac{1}{10^{n}}$,

\n

\\[\\text{e.g. } 10^{-3} = \\frac{1}{10^3} = \\frac{1}{1000} = 0.001\\text{ .}\\]

\n

When $n = 0$:

\n

\\[10^{0} = 1 \\text{ .}\\]

\n

Generally, we can think of $10^n$ as a number in standard form $1 \\times 10^n$. Then $n$ always tells us the number of decimal places to move the decimal point in $1.0$, for example

\n

\\[10^{-3} = 1.0 \\times 10^{-3} \\text{ and since } n = - 3 \\text{, we go } 3 \\text{ places back as follows: } 1.0 ⇒ 0.1 ⇒ 0.01 ⇒ 0.001 \\text{ .}\\]

\n

A complete table of powers of ten for $n$ from $-6$ to $6$ is: 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$10^n$
$-6$$0.000001$
$-5$$0.00001$
$-4$$0.0001$
$-3$$0.001$
$-2$$0.01$
$-1$$0.1$
$0$$1$
$1$$10$
$2$$100$
$3$$1000$
$4$$10000$
$5$$100000$
$6$$1000000$
\n

", "statement": "

Powers of ten can be useful while working with standard index numbers. Fill in the following table of powers of ten:

", "variables": {"n": {"name": "n", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(2..5)"}}, "tags": ["powers", "powers of 10", "standard index numbers", "taxonomy"], "ungrouped_variables": ["n"], "functions": {}, "preamble": {"js": "", "css": ""}, "type": "question", "variable_groups": [], "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": "1000"}, "metadata": {"description": "

Fill in a table of powers of 10.

", "licence": "Creative Commons Attribution 4.0 International"}, "parts": [{"scripts": {}, "showCorrectAnswer": true, "gaps": [{"notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "variableReplacements": [], "mustBeReducedPC": 0, "minValue": "{-n-1}", "correctAnswerFraction": false, "allowFractions": false, "correctAnswerStyle": "plain", "variableReplacementStrategy": "originalfirst", "scripts": {}, "maxValue": "{-n-1}", "showCorrectAnswer": true, "type": "numberentry", "marks": 1, "showFeedbackIcon": true}, {"notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "variableReplacements": [], "mustBeReducedPC": 0, "minValue": "10^{-n+1}", "correctAnswerFraction": false, "allowFractions": false, "correctAnswerStyle": "plain", "variableReplacementStrategy": "originalfirst", "scripts": {}, "maxValue": "10^{-n+1}", "showCorrectAnswer": true, "type": "numberentry", "marks": 1, "showFeedbackIcon": true}, {"notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "variableReplacements": [], "mustBeReducedPC": 0, "minValue": "10^{n}", "correctAnswerFraction": false, "allowFractions": false, "correctAnswerStyle": "plain", "variableReplacementStrategy": "originalfirst", "scripts": {}, "maxValue": "10^{n}", "showCorrectAnswer": true, "type": "numberentry", "marks": 1, "showFeedbackIcon": true}, {"notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "variableReplacements": [], "mustBeReducedPC": 0, "minValue": "{n+2}", "correctAnswerFraction": false, "allowFractions": false, "correctAnswerStyle": "plain", "variableReplacementStrategy": "originalfirst", "scripts": {}, "maxValue": "{n+2}", "showCorrectAnswer": true, "type": "numberentry", "marks": 1, "showFeedbackIcon": true}], "type": "gapfill", "prompt": "

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$10^n$
[[0]]$\\var{10^(-n-1)}$
$\\var{-n+1}$[[1]]
$0$$1$
$1$$10$
$\\var{n}$[[2]]
[[3]]$\\var{10^(n+2)}$
\n

", "marks": 0, "variableReplacements": [], "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst"}]}, {"name": "Square and cube numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "variable_groups": [], "variables": {"x": {"templateType": "anything", "definition": "sort([x1,x2,x3,x4,x5])", "description": "

Sorted list of integers from 1 to 12.

", "name": "x", "group": "Ungrouped variables"}, "x1": {"templateType": "anything", "definition": "random(1..3)", "description": "", "name": "x1", "group": "Ungrouped variables"}, "y2": {"templateType": "anything", "definition": "y^2", "description": "

Squared number in part c).

", "name": "y2", "group": "Ungrouped variables"}, "x5": {"templateType": "anything", "definition": "random(10..12 except x1 except x2 except x3 except x4)", "description": "", "name": "x5", "group": "Ungrouped variables"}, "x4": {"templateType": "anything", "definition": "random(7..10 except x1 except x2 except x3)", "description": "", "name": "x4", "group": "Ungrouped variables"}, "x3": {"templateType": "anything", "definition": "random(5..7 except x1 except x2)", "description": "", "name": "x3", "group": "Ungrouped variables"}, "ly": {"templateType": "anything", "definition": "y2 - 2y + random(2..4)", "description": "

Lower bound in part c).

", "name": "ly", "group": "Ungrouped variables"}, "x2": {"templateType": "anything", "definition": "random(3..5 except x1)", "description": "", "name": "x2", "group": "Ungrouped variables"}, "uy": {"templateType": "anything", "definition": "y2 + 2y - random(2..5)", "description": "

Upper bound in part c).

", "name": "uy", "group": "Ungrouped variables"}, "y": {"templateType": "anything", "definition": "random(3..12 except x)", "description": "

Answer to part c).

", "name": "y", "group": "Ungrouped variables"}}, "type": "question", "parts": [{"showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[0]*x[0]", "maxValue": "x[0]*x[0]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[1]*x[1]", "maxValue": "x[1]*x[1]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[2]*x[2]", "maxValue": "x[2]*x[2]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[3]*x[3]", "maxValue": "x[3]*x[3]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[4]*x[4]", "maxValue": "x[4]*x[4]", "marks": 1, "variableReplacements": []}], "prompt": "

Find the following:

\n

$\\var{x[0]}^2 =$  [[0]]

\n

$\\var{x[1]}^2 =$ [[1]]

\n

$\\var{x[2]}^2 =$ [[2]]

\n

$\\var{x[3]}^2 =$ [[3]]

\n

$\\var{x[4]}^2 =$ [[4]]

\n

"}, {"showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[0]*x[0]*x[0]", "maxValue": "x[0]*x[0]*x[0]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[1]*x[1]*x[1]", "maxValue": "x[1]*x[1]*x[1]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[2]*x[2]*x[2]", "maxValue": "x[2]*x[2]*x[2]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[3]*x[3]*x[3]", "maxValue": "x[3]*x[3]*x[3]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[4]*x[4]*x[4]", "maxValue": "x[4]*x[4]*x[4]", "marks": 1, "variableReplacements": []}], "prompt": "

Find the following:

\n

$\\var{x[0]}^3 =$  [[0]]

\n

$\\var{x[1]}^3 =$ [[1]]

\n

$\\var{x[2]}^3 =$ [[2]]

\n

$\\var{x[3]}^3 =$ [[3]]

\n

$\\var{x[4]}^3 =$ [[4]]

"}, {"showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "y^2", "maxValue": "y^2", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "y", "maxValue": "y", "marks": "1", "variableReplacements": []}], "prompt": "

Find a square number $y^2$ between $\\var{ly}$ and $\\var{uy}$ and its integer root $y$.

\n

$y^2 = $  [[0]]

\n

$y = $  [[1]]

"}], "advice": "

a)

\n

Squared integers are called square numbers. It may be useful to remember the first few square numbers to be able to use them without a calculator.

\n

Here:

\n

\\[ \\begin{align} \\var{x[0]}^2 &= \\var{x[0]} \\times \\var{x[0]} \\\\&= \\var{x[0]^2} \\text{.}\\end{align}\\]

\n

\\[ \\begin{align} \\var{x[1]}^2 &= \\var{x[1]} \\times \\var{x[1]} \\\\&= \\var{x[1]^2} \\text{.}\\end{align}\\]

\n

\\[ \\begin{align} \\var{x[2]}^2 &= \\var{x[2]} \\times \\var{x[2]} \\\\&= \\var{x[2]^2} \\text{.}\\end{align}\\]

\n

\\[ \\begin{align} \\var{x[3]}^2 &= \\var{x[3]} \\times \\var{x[3]} \\\\&= \\var{x[3]^2} \\text{.}\\end{align}\\]

\n

\\[ \\begin{align} \\var{x[4]}^2 &= \\var{x[4]} \\times \\var{x[4]} \\\\&= \\var{x[4]^2} \\text{.}\\end{align}\\]

\n

\n

b)

\n

Cubed integers are called cubed numbers. To obtain these, we would typically always use a calculator.

\n

We can either cube the number $x$:

\n

\\[ \\begin{align} \\var{x[0]}^3 &= \\var{x[0]} \\times \\var{x[0]} \\times \\var{x[0]} \\\\&= \\var{x[0]^3} \\text{,} \\end{align}\\]

\n

or we can multiply the square number $(x_n)^2$ from part a) by the appropriate $x_n$:

\n

\\[ \\begin{align} \\var{x[0]}^3 &= \\var{x[0]}^2 \\times \\var{x[0]} \\\\&= \\var{x[0]^2} \\times \\var{x[0]}\\\\&= \\var{x[0]^3} \\text{.} \\end{align}\\]

\n

\\[ \\begin{align} \\var{x[1]}^3 &= \\var{x[1]} \\times \\var{x[1]} \\times \\var{x[1]} \\\\ &= \\var{x[1]^3} \\text{.}\\end{align}\\]

\n

\\[ \\begin{align} \\var{x[2]}^3 &= \\var{x[2]} \\times \\var{x[2]} \\times \\var{x[2]} \\\\ &= \\var{x[2]^3} \\text{.}\\end{align}\\]

\n

\\[ \\begin{align} \\var{x[3]}^3 &= \\var{x[3]} \\times \\var{x[3]} \\times \\var{x[3]} \\\\ &= \\var{x[3]^3} \\text{.}\\end{align}\\]

\n

\\[ \\begin{align} \\var{x[3]}^4 &= \\var{x[4]} \\times \\var{x[4]} \\times \\var{x[4]} \\\\ &= \\var{x[4]^3} \\text{.}\\end{align}\\]

\n

\n

c)

\n

Here is a table of square numbers for integers from $1$ to $15$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$y$$1$$2$$3$$4$$5$$6$$7$$8$$9$$10$$11$$12$$13$$14$$15$
$y^2$$1$$4$$9$$16$$25$$36$$49$$64$$81$$100$$121$$144$$169$$196$$225$
\n

The only square number between $\\var{ly}$ and $\\var{uy}$ is $\\var{y2}$.

\n

To calculate $y$ we must calculate the square root of $y^2$,

\n

\\[ \\sqrt{\\var{y2}} = \\var{y} \\text{.}\\]

\n

This is our integer $y$.

\n

\n

", "tags": ["cube", "indices", "multiplication", "powers", "roots", "square", "taxonomy"], "preamble": {"js": "", "css": ""}, "rulesets": {}, "functions": {}, "ungrouped_variables": ["x1", "x2", "x3", "x4", "x5", "x", "y2", "y", "ly", "uy"], "statement": "

Try the following questions on square and cube numbers.

", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Find the squares, and cubes, of some numbers.

\n

Finally, find a square number between two given limits.

"}, "variablesTest": {"condition": "", "maxRuns": "1000"}}, {"name": "Always, sometimes or never: square and cube numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "type": "question", "statement": "

For each of the following statements, select one option from \"Always\", \"Sometimes\" or \"Never\". 

\n

Select:

\n", "variablesTest": {"condition": "", "maxRuns": "1000"}, "variables": {"a": {"group": "Ungrouped variables", "name": "a", "description": "

This variable was created solely for the purpose of being able to publish this question.

", "templateType": "anything", "definition": "1"}}, "functions": {}, "tags": ["always", "cubic numbers", "never", "sometimes", "square numbers", "taxonomy"], "variable_groups": [], "parts": [{"layout": {"expression": "", "type": "all"}, "scripts": {}, "minMarks": 0, "type": "m_n_x", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "displayType": "radiogroup", "choices": ["

i)      $x^3$ is greater than $x^2$

", "

ii)      $x^2$ is greater than $x$

", "

iii)      If $x$ is negative, $x^2$ is negative

", "

iv)      If $x$ is negative, $x^3$ is negative

", "

v)      $x^2 = x$

", "

vi)      $x^2 = - x$

", "

vii)      $(x+1)^2 \\gt x$

", "

viii)      $(x+1)^3 \\gt x$

", "

ix)      $x^3 \\times x = x^2 \\times x^2$

", "

x)      $x^2$ has the opposite sign to $x$

", "

xi)      $x^3$ has the opposite sign to $x$

"], "showFeedbackIcon": true, "answers": ["

Always

", "

Sometimes

", "

Never

"], "warningType": "none", "matrix": [["0", "1", 0], [0, "1", "0"], ["0", 0, "1"], ["1", "0", 0], ["0", "1", 0], ["0", "1", 0], ["1", "0", 0], [0, "1", 0], ["1", 0, "0"], [0, "1", 0], [0, 0, "1"]], "variableReplacements": [], "marks": 0, "maxMarks": 0, "shuffleChoices": false, "maxAnswers": 0, "minAnswers": 0, "shuffleAnswers": false}], "ungrouped_variables": ["a"], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Decide whether statements about square and cube numbers are always true, sometimes true or never true.

"}, "preamble": {"css": "", "js": ""}, "advice": "

i)

\n

Suppose $x$ is negative, for example $x = -5$.

\n

\\[ \\begin{align} x^2 &= (-5)^2 \\\\&= 25 \\\\ x^3 &= (-5)^3 \\\\&= -125 \\\\x^3 &\\lt x^2\\text{.} \\end{align} \\]

\n

Now let $x$ be positive, for example $x = 5$.

\n

\\[ \\begin{align} x^2 &= 5^2 \\\\&= 25 \\\\ x^3 &= 5^3 \\\\&= 125 \\\\x^3 &\\gt x^2\\text{.}\\end{align} \\]

\n

Therefore, $x^3$ is sometimes greater than $x^2$.

\n

\n

\n

ii)

\n

This is true for either $x \\gt 1$ or $x \\lt 0$ but false for $0 \\leq x \\leq 1$. For example, let $x=0.5$. Then

\n

\\[ \\begin{align} \\text{When } x &= 0.5\\text{,} \\\\x^2 &= 0.25\\text{, so} \\\\ x^2 &\\lt x \\text{.} \\end{align} \\]

\n

Therefore, $x^2$ is sometimes greater than $x$.

\n

\n

\n

iii)

\n

Multiplying two negative numbers gives a positive answer and multiplying two postive numbers gives a positive answer. Therefore, $x^2$ is never negative.

\n

\n

\n

iv)

\n

Multiplying a negative number by itself an odd number of times always gives a negative answer. For example, let $x = -1$. Then

\n

\\[\\begin{align} x^3 &= (-1)^3 \\\\&= -1\\times-1\\times-1 \\\\&=1\\times-1 \\\\&= - 1 \\text{.} \\end{align}\\]

\n

Therefore, if $x$ is negative, $x^3$ is always negative.

\n

\n

\n

v)

\n

This is true when $x = 1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $x$.

\n

\n

\n

vi)

\n

This is true when $x = -1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $-x$.

\n

\n

\n

vii)

\n

When $x$ is positive, for example $x = 5$:

\n

\\[ \\begin{align} (x+1)^2 &= (5 + 1)^2 \\\\&= 6^2 \\\\&= 36 \\gt x = 5 \\end{align} \\]

\n

When $x = 0$:

\n

\\[ \\begin{align} (x+1)^2 &= (0 + 1)^2 \\\\&= 1^2 \\\\&= 1 \\gt x = 0 \\end{align} \\]

\n

When $x$ is negative, such as $x = -4$:

\n

\\[ \\begin{align} (x+1)^2 &= (-4 + 1)^2 \\\\&= (-3)^2 \\\\&= 9 \\gt x = -4 \\end{align} \\]

\n

To see the behaviour of $(x+1)^2$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$x$$-3$$-2$$-1$$-0.5$$0$$0.5$$1$$2$\n

$3$

\n
$(x+1)^2$$4$$1$$0$$0.25$$1$$2.25$$4$$9$$16$
\n

Therefore, $(x+1)^2$ is always greater than $x$.

\n

\n

\n

viii)

\n

When $x$ is positive, for example $x = 5$:

\n

\\[ \\begin{align} (x+1)^3 &= (5 + 1)^3 \\\\&= 6^3 \\\\&= 216 \\gt x = 5 \\end{align} \\]

\n

When $x = 0$:

\n

\\[ \\begin{align} (x+1)^3 &= (0 + 1)^3 \\\\&= 1^3 \\\\&= 1 \\gt x = 0 \\end{align} \\]

\n

When $x$ is negative, such as $x = -4$:

\n

\\[ \\begin{align} (x+1)^2 &= (-4 + 1)^3 \\\\&= (-3)^3 \\\\&= -27 \\lt x = -4 \\end{align} \\]

\n

To see the behaviour of $(x+1)^3$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$x$$-3$$-2$$-1$$-0.5$$0$$0.5$$1$$2$\n

$3$

\n
$(x+1)^3$$-8$$-1$$0$$0.125$$1$$3.375$$8$$27$$64$
\n

Therefore, $(x+1)^3$ is sometimes greater than $x$.

\n

\n

\n

ix)

\n

We can write 

\n

\\[x^3\\times x = x \\times x \\times x \\times x = x^2 \\times x^2\\text{.}\\]

\n

Therefore, $x^3 \\times x$ always equals $x^2 \\times x^2$.

\n

\n

\n

x)

\n

Since $x^2$ is always positive, $x^2$ only has the opposite sign to $x$ when $x$ is negative. 

\n

Therefore, $x^2$ sometimes has the opposite sign to $x$.

\n

\n

\n

xi)

\n

As seen in part iv), multiplying a negative number by itself an odd number of times always gives a negative answer. It is also true that multiplying a positive number by itself an odd number of times will give a positive answer. So, $x^3$ always has the same sign as $x$, since we have an odd power.

\n

Therefore, $x^3$ never has the opposite sign to $x$.

\n

\n

"}, {"name": "Laws of Indices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "tags": ["indices", "laws of indices", "powers", "taxonomy"], "metadata": {"description": "

This question aims to test understanding and ability to use the laws of indices.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Using the laws of indices, simplify each expression down to its simplest form. Recall that $a^{0} = 1$ for any number $a$.

", "advice": "

a)

\n

Here we are using the rule of indices: $a^m \\times a^n = a^{m+n}$.

\n

Using this rule, 

\n

\\[
\\begin{align}
a^\\var{x} \\times a^\\var{y}\\ &= a^\\simplify[all, !collectNumbers]{{x}+{y}}\\\\
&= a^\\var{x+y}.
\\end{align}
\\]

\n

b)

\n

We are asked to find $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$.

\n

Notice there is a constant in front of each of the terms.

\n

To do this, write the product out explicitly, as

\n

\\[\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} = \\var{c} \\times \\var{d} \\times a^\\var{p} \\times a^\\var{q}.\\]

\n

We know that $\\var{c} \\times \\var{d} = \\var{c*d}$, and using the rule of indices: $a^\\var{p} \\times a^\\var{q} = a^\\var{p+q}$.

\n

Therefore:

\n

\\begin{align}
\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}&= \\var{c*d} \\times a^\\var{p+q} \\\\
&= \\simplify{{c*d}*a^{p+q}}.
\\end{align}

\n

c)

\n

Here we are using: $a^m \\div a^n = a^{m-n}$.

\n

We are asked to simplify the expression, $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$.

\n

To do this, we just have to use the previously mentioned rule of indices. We write this out explicity as

\n

\\[\\simplify{{b}*a^{x}/({g}*a^{y})} = \\simplify{{b}/{g}} \\times \\simplify{a^{x}/(a^{y})}.\\]

\n

Using rules of indices,

\n

\\begin{align}                                                                                                                                                                                                                                                                                           \\frac{a^\\var{x}}{a^\\var{y}} &= a^\\var{x} \\div a^\\var{y}\\\\
&= a^\\simplify[all, !collectNumbers]{{x}-{y}}\\\\
&= a^\\var{x-y}.
\\end{align}

\n

Therefore,

\n

\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\simplify{a^{{x}-{y}}}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}

\n

Alternatively, 

\n

Using the rule of indices: $a^{-m}  = \\displaystyle\\frac{1}{a^{m}}$, we can rewrite the question as:

\n

\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\frac{a^\\var{x}}{a^\\var{y}}\\\\
&= \\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}}.
\\end{align}

\n

And then using the rule: $a^m \\times a^n = a^{m+n}$, this becomes:

\n

\\begin{align}
\\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}} &= \\simplify{{b}/{g}} \\times a^\\simplify[all,!collectNumbers]{{x}+(-{y})}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}

\n

d)

\n

The question asks us to simplify $(\\simplify{{c}*a^{p}})^{\\var{q}}$.

\n

To do this we use the rules:

\n

\\[(a^{m})^{n} = a^{mn},\\]

\n

\\[(ab)^m = a^mb^m.\\]

\n

We can then expand the equation as

\n

\\[(\\simplify{{c}*a^{p}})^{\\var{q}}= \\var{c}^{\\var{q}} \\times (a^{\\var{p}})^{\\var{q}}.\\]

\n

Then using the rule of indices mentioned previously,

\n

\\[
\\begin{align}
(\\simplify{{c}*a^{p}})^{\\var{q}}&= \\simplify{{c}^{q}} \\times a^\\var{p*q}\\\\
&= \\simplify{{c}^{q}*a^{p*q}}.
\\end{align}
\\]

\n

e)

\n

The question asks us to simplify $\\sqrt[\\var{d}]{\\var{x}^\\var{d}a}$.

\n

To do this we use the rules:

\n

\\[a^\\frac{1}{m} = \\sqrt[m]{a},\\]

\n

\\[(ab)^m = a^mb^m.\\]

\n

We can expand the expression as follows:

\n

\\[
\\begin{align}
\\sqrt[\\var{d}]{a} &= (\\simplify{a})^\\frac{1}{\\var{d}}\\\\
&= a^\\frac{1}{\\var{d}}.
\\end{align}
\\]

\n

f)

\n

The question requires us to simplify $\\sqrt[\\var{c}]{a^\\var{q}}$.

\n

Here, we use the rule of indices: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$, allowing us to expand the expression as follows:

\n

\\[
\\begin{align}
\\sqrt[\\var{c}]{\\simplify{a^{q}}} &= \\simplify[fractionnumbers,all]{(a^{q})^{{1}/{{c}}}}\\\\
&= \\simplify[fractionnumbers,all]{a^{{q}/{c}}}.
\\end{align}
\\]

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Used in part c

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Used in parts b,d and f

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Used in parts a,c and e

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Used in parts b and e

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Used in parts b and d

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Used in parts b,d and f

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Used in part c

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\n

Used in parts a,c and f

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Write $a^{\\var{x}} \\times a^{\\var{y}}$ as a single power of $a$.

\n

\n

$a^{\\var{x}} \\times a^{\\var{y}} =$ [[0]].

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Use the rule: $a^m \\times a^n = a^{m+n}$.

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Write $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$ as an integer multiplied by a single power of $a$.

\n

$\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} =$ [[0]].

\n

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Write $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$ as a number multiplied by a single power of $a$.

\n

$\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})} =$ [[0]].

\n

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You could use one of the following rules:

\n

$a^m \\div a^n = a^{m-n}$.

\n

$a^{-m} = \\displaystyle\\frac{1}{a^m}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "answer": "{b}/{g}a^{x-y}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "mustmatchpattern": {"pattern": "$n/$n`? * a^?`?", "partialCredit": 0, "message": "You haven't simplified: your answer is not in the form $\\frac{?}{?} \\cdot a^?$.", "nameToCompare": ""}, "valuegenerators": [{"name": "a", "value": ""}]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

Write $(\\simplify{{c}*a^{p}})^{\\var{q}}$ as an integer multiplied by a single power of $a$.

\n

$(\\simplify{{c}*a^{p}})^{\\var{q}} =$ [[0]].

\n

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Use the rules:

\n

$(ab)^m = a^mb^m$.

\n

$(a^m)^n = a^{mn}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "answer": "{c^{q}}*a^{p*q}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "mustmatchpattern": {"pattern": "$n`?*a^?`?", "partialCredit": 0, "message": "You must write your answer as an integer multiplied by a power of $a$.", "nameToCompare": ""}, "valuegenerators": [{"name": "a", "value": ""}]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

Write $\\sqrt[\\var{d}]{a}$ as a single power of $a$. 

\n

$\\sqrt[\\var{d}]{a} =$ [[0]].

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Use the rule: $a^\\frac{1}{m} = \\sqrt[m]{a}$.

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You must input your answer as a single power of a.

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Write $\\sqrt[\\var{q}]{a^\\var{c}}$ as a single power of $a$.

\n

$\\sqrt[\\var{q}]{a^\\var{c}} =$ [[0]].

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Use the rule: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$.

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You must input your answer as a single power of a.

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Working with numbers that are very large or very small can be tricky.

\n

Standard form allows us to simplify these numbers, using powers of 10.

\n
The standard index form can be defined as
\n
\\[A \\times 10^n,\\]
\n
where $1 ≤ A < 10$ and $n$ is an integer, e.g. $2.26 \\times 10^5$ is a standard form of a number 226000.
\n
\n

 

\n

Write the following in standard index form (for example, for $2.01\\times 10^5$ we would write 2.01*10^5 in the gap).

\n
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$\\var{A[2]*10^2} = $  [[0]]

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$\\var{A3dp*10^(-1)} = $  [[0]]

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$\\var{precround(A5dp*10^7,0)} =$  [[0]]

", "type": "gapfill"}, {"scripts": {}, "variableReplacements": [], "marks": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "gaps": [{"checkingtype": "absdiff", "scripts": {}, "showpreview": true, "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "answersimplification": "!collectnumbers", "checkingaccuracy": 0.001, "expectedvariablenames": [], "marks": "1", "notallowed": {"message": "", "partialCredit": 0, "showStrings": false, "strings": ["10^5"]}, "vsetrange": [0, 1], "musthave": {"message": "", "partialCredit": 0, "showStrings": false, "strings": ["*10^", "-5"]}, "variableReplacements": [], "answer": "{{small5}*10^5}*10^(-5)", "checkvariablenames": false, "type": "jme"}], "showCorrectAnswer": true, "prompt": "

$\\var{small5} = $  [[0]]

", "type": "gapfill"}], "ungrouped_variables": ["A3dp", "A5dp", "small5", "A", "ran", "B", "int"], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Convert a variety of numbers from decimal to standard index form.

"}, "preamble": {"css": "", "js": ""}, "advice": "

Converting from decimal to a standard form, we are looking for $A \\times 10^n$.

\n

We need make the first number ($A$) between 1 and 10, so we put the decimal place after the first non-zero digit.

\n

 

\n

a)

\n

In $\\var{A[2]*10^2}$, the first non-zero digit is $\\var{siground(A[2] - 0.5, 1)}$ so we get $A = \\var{A[2]}$.

\n

If we moved the decimal place in $\\var{A[2]}$ so it matches our original number $\\var{A[2]*10^2}$, we would go 2 places to the right, so $n = 2$.

\n

 

\n

b)

\n

In $\\var{A3dp*10^(-1)}$, the first non-zero digit is $\\var{siground(A3dp - 0.5, 1)}$ so we get $A = \\var{A3dp}$.

\n

If we moved the decimal place in $\\var{A3dp}$ so it matches our original number $\\var{A3dp*10^(-1)}$, we would go 1 place to the left, so $n = -1$.

\n

 

\n

c)

\n

In $\\var{precround(A5dp*10^7,0)}$ the first non-zero digit is $\\var{siground(A5dp - 0.5, 1)}$ so we get $A = \\var{A5dp}$.

\n

If we moved the decimal place in $\\var{A5dp}$ so it matches our original number $\\var{precround(A5dp*10^7,0)}$, we would go 7 places to the right, so $n = 7$.

\n

 

\n

d)

\n

In $\\var{small5}$ the first non-zero digit is {siground({{small5}*10^5} - 0.5, 1)} so we get $A = \\var{small5*10^5}$.

\n

If we moved the decimal place in $\\var{small5*10^5}$ so it matches our original number $\\var{small5}$, we would go 5 places to the left, so $n = -5$.

\n

"}, {"name": "Cumulative percent decrease", "extensions": ["random_person"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "type": "question", "tags": ["decrease", "percentages", "taxonomy"], "variablesTest": {"condition": "", "maxRuns": "1000"}, "variables": {"test": {"definition": "precround(precround(price*((100-perc)/100)^5, 2)*((100-perc)/100)^(n-5), 2)", "description": "

Calculated value of price2 to ensure we mention rounding errors in advice only when needed.

", "templateType": "anything", "name": "test", "group": "Part b)"}, "pricee3": {"definition": "precround(price*((100 - perc)/100)^(testn-2),2)", "description": "", "templateType": "anything", "name": "pricee3", "group": "Part b)"}, "person": {"definition": "random_person()", "description": "", "templateType": "anything", "name": "person", "group": "Part b)"}, "testn": {"definition": "random(6..9)", "description": "

Number of months in total.

", "templateType": "anything", "name": "testn", "group": "Part b)"}, "pricee1": {"definition": "precround(price*((100 - perc)/100)^(testn),2)", "description": "

The resulting price after the total of testn months.

", "templateType": "anything", "name": "pricee1", "group": "Part b)"}, "n": {"definition": "if(pricee2 < threshold, testn-1, testn)", "description": "", "templateType": "anything", "name": "n", "group": "Part b)"}, "threshold": {"definition": "siground(pricee1+5,2)", "description": "", "templateType": "anything", "name": "threshold", "group": "Part b)"}, "price": {"definition": "random(300..800) + 0.99", "description": "

The original price.

", "templateType": "anything", "name": "price", "group": "Part a)"}, "price2": {"definition": "if(pricee2 < threshold, pricee2, pricee1)", "description": "", "templateType": "anything", "name": "price2", "group": "Part b)"}, "perc": {"definition": "random(2..4 #0.5)", "description": "

Percentage decrease per month.

", "templateType": "anything", "name": "perc", "group": "Part a)"}, "pricee2": {"definition": "precround(price*((100 - perc)/100)^(testn-1),2)", "description": "", "templateType": "anything", "name": "pricee2", "group": "Part b)"}}, "statement": "

A smartphone's value decreases by $\\var{perc}$% every month. The original price when it is released is $£\\var{price}$.

", "variable_groups": [{"name": "Part a)", "variables": ["price", "perc"]}, {"name": "Part b)", "variables": ["threshold", "pricee1", "pricee2", "pricee3", "testn", "test", "price2", "n", "person"]}], "parts": [{"showCorrectAnswer": true, "scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "stepsPenalty": 0, "gaps": [{"correctAnswerFraction": false, "precisionMessage": "

Round your answer to $2$ decimal places.

", "precisionPartialCredit": 0, "scripts": {}, "maxValue": "precround(price*((100-perc)/100)^5, 2)", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "precision": "2", "precisionType": "dp", "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "mustBeReduced": false, "minValue": "precround(price*((100-perc)/100)^5, 2)", "showPrecisionHint": true, "marks": "2", "variableReplacements": [], "strictPrecision": false, "showCorrectAnswer": true, "type": "numberentry"}], "showFeedbackIcon": true, "prompt": "

How much will the smartphone be worth after $5$ months?

\n

£ [[0]]

", "steps": [{"scripts": {}, "variableReplacements": [], "type": "information", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "showFeedbackIcon": true, "prompt": "

The original price of the phone is $£\\var{price}$ and we are told that the price decreases by $\\var{perc}$% every month.

\n

", "marks": 0}, {"correctAnswerFraction": false, "scripts": {}, "maxValue": "1-{perc}/100", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "prompt": "

What is the decimal multiplier for the decrease in the smartphones each month?

", "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "mustBeReduced": false, "minValue": "1-{perc}/100", "variableReplacements": [], "marks": "0.5", "showCorrectAnswer": true, "type": "numberentry"}, {"precisionMessage": "

Round your answer to $2$ decimal places.

", "precisionPartialCredit": 0, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "prompt": "

Multiply the original price by the decimal multiplier to obtain the price after 1 month.

", "mustBeReducedPC": 0, "mustBeReduced": false, "minValue": "{price}*(1-{perc}/100)", "variableReplacements": [], "marks": "0.5", "strictPrecision": false, "showCorrectAnswer": true, "correctAnswerFraction": false, "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "precision": "2", "maxValue": "{price}*(1-{perc}/100)", "precisionType": "dp", "correctAnswerStyle": "plain", "showPrecisionHint": true}, {"precisionMessage": "

Round your answer to $2$ decimal places.

", "precisionPartialCredit": 0, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "prompt": "

Multiply your answer above by the decimal multiplier to obtain the price after 2 months. 

\n

Note that this is the same as multiplying the original price by $d^2$, where $d$ is the decimal multiplier.

", "mustBeReducedPC": 0, "mustBeReduced": false, "minValue": "{price}*(1-{perc}/100)^2", "variableReplacements": [], "marks": "0.5", "strictPrecision": false, "showCorrectAnswer": true, "correctAnswerFraction": false, "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "precision": "2", "maxValue": "{price}*(1-{perc}/100)^2", "precisionType": "dp", "correctAnswerStyle": "plain", "showPrecisionHint": true}], "marks": 0}, {"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "scripts": {}, "maxValue": "n-5", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "mustBeReduced": false, "minValue": "n-5", "variableReplacements": [], "marks": "2", "showCorrectAnswer": true, "type": "numberentry"}], "showFeedbackIcon": true, "prompt": "

{person['name']} has $£\\var{threshold}$ to spend on a smartphone. After how many more full months will {person['pronouns']['they']} be able to afford the smartphone?

\n

[[0]] months

", "marks": 0}], "ungrouped_variables": [], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Given the original price of a smartphone and the rate at which it decreases, calculate its price after a given number of months. In the second part, calculate the time remaining until the price goes below a certain point.

"}, "preamble": {"css": "", "js": ""}, "functions": {}, "advice": "

a)

\n

We can use the multiplier method to calculate the new price. If the price decreases by {perc}%, this its value is {100-perc}% of the original value after 1 month. The decimal multiplier for {100-perc}% is

\n

\\[\\frac{\\var{100-perc}}{100} = \\var{(100-perc)/100} \\text{.}\\]

\n

Each month our smartphone's value can be found by multiplying the previous month's value by the decimal multiplier. For example, after the first month, the value is

\n

\\[ \\var{(100-perc)/100} \\times\\mathrm{£}\\var{price} = \\mathrm{£}\\var{dpformat(price*(100-perc)/100,2)}\\text{.} \\]

\n

To calculate the price after 5 months, we multiply the original price of the smartphone by our multiplier 5 times:

\n

\\[ \\begin{align} \\text{Final worth} &= \\var{price} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\\\&=  \\var{price} \\times \\var{(100-perc)/100}^{5} \\\\&= £\\var{precround(price*((100-perc)/100)^5, 2)} {.} \\end{align}\\]

\n

b)

\n

From part a), the value after 5 months is  £$\\var{precround(price*((100-perc)/100)^5, 2)}$. Continuing to multiply the price by the decimal multiplier,

\n

\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100} = £\\var{precround(precround(price*((100-perc)/100)^5, 2)*(100-perc)/100, 2)}\\]

\n

\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100}^2 = £\\var{precround(precround(price*((100-perc)/100)^6, 2)*(100-perc)/100, 2)}\\]

\n

\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100}^3 = £\\var{precround(precround(price*((100-perc)/100)^7, 2)*(100-perc)/100, 2)}\\]

\n

\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100}^4 = £\\var{precround(precround(price*((100-perc)/100)^8, 2)*(100-perc)/100, 2)}\\]

\n

\n

The smartphone's value will be below $£\\var{threshold}$ after {n-5} more months ({n} months in total since its release).

"}]}], "duration": 0, "type": "exam", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Matthew James Sykes", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2582/"}], "extensions": ["random_person"], "custom_part_types": [], "resources": []}