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Given a description in words of the costs of some items in terms of an unknown cost, write down an expression for the total cost of a selection of items. Then simplify the expression, and finally evaluate it at a given point.

The word problem is about the costs of sweets in a sweet shop.

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

{pname} eats a lot of sweets. You are trying to work out the cost of the sweets that {pname} ate last week.

{pname} ate $\\var{a1}$ packets of lollipops, $\\var{b1}$ packets of toffee and $\\simplify{{c1}}$ packets of jelly sweets.

You know that a packet of toffee costs $£1$ more than a packet of lollipops, and a packet of jelly sweets costs half as much as a packet of toffees.

", "advice": "

a)

We are told that the price of a packet of lollipops is represented by the letter $x$.

A packet of toffee costs $£1$ more than a packet of lollipops, i.e. $x+1$.

A packet of jelly sweets costs half as much as a packet of toffee, so $\\frac{1}{2}(x+1)$.

b)

To find the total cost, multiply the expressions above for the cost of each kind of sweet by the number of packets eaten, and add them together.

Without simplifying, we obtain:

\\begin{align}
\\text{Cost} &= \\simplify[]{{a1}x+{b1}(x+1) + {c1}*(1/2)*(x+1)} \\\\
&= \\simplify[]{{a1}x+{b1}(x+1) + {c1/2}*(x+1)}
\\text{.}
\\end{align}

c)

The first step in simplifying this expression is to expand both sets of brackets:

\\begin{align}
\\simplify[]{ {a1}x + {b1}(x+1) + {c1/2}*(x+1)} &= \\simplify[]{ {a1}x + {b1}x + {b1}*1 + {c1/2}x + {c1/2}*1} \\\\
&= \\simplify[] { {a1}x + {b1}x + {b1} + {c1/2}x + {c1/2} } \\text{.}
\\end{align}

Finally, collect like terms:

\\begin{align}
\\simplify[] { {a1}x + {b1}x + {b1} + {c1/2}x + {c1/2} } &= \\simplify[]{ {a1+b1+c1/2}x + {b1+c1/2} } \\text{.}
\\end{align}

d)

Once we know that the price of a packet of lollipops is $£2$, we can substitute this for $x$ in the equation above.

\\begin{align}
\\text{Cost}&=\\simplify{ {a1+b1+c1/2}x+{b1+c1/2} }\\\\
&=\\var{a1+b1+c1/2} \\times 2+\\var{b1+c1/2} \\\\
&=\\var{(a1+b1+c1/2)*2+b1+c1/2} \\text{.}
\\end{align}

So {pname} spent $£\\var{total}$ on sweets last week.

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Number of packets of toffee eaten

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Number of packets of jelly sweets eaten.

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Number of packets of lollipops eaten

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The total spent.

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Let the cost of a packet of lollipops be $£x$.

Write an expression in terms of $x$ for the cost of each kind of sweet:

Lollipops: £[[0]]

Toffees: £[[1]]

Jelly sweets: £[[2]]

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Write an algebraic expression for the overall cost of the sweets {pname} ate, in terms of $x$.

£[[0]]

Now simplify your expression for the total cost.

£[[0]]

You find out that a packet of lollipops costs $£2$.

Calculate {pname}'s total expenditure on sweets last week.

£[[0]]

Don't use brackets

"}, "valuegenerators": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Using Surds, Rationalising the Denominator", "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": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "tags": ["Fractions", "fractions", "rationalise the denominator", "surds", "Surds", "taxonomy"], "metadata": {"description": "

Manipulate surds and rationalise the denominator of a fraction when it is a surd.

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

To include a square root sign in your answer use sqrt(). For example, to write $\\sqrt{3}$, type sqrt(3) into the answer box. If you are entering a number multiplied by the square root of some other number, for example $3\\sqrt{5}$, type 3*sqrt(5) into the answer box.

", "advice": "

a)

Surds can be manipulated using the rule

\\[\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}.\\]

We are asked to state which of $\\sqrt{\\var{p}}$, $\\sqrt{\\simplify{{a}*{n}^2}}$, and $\\sqrt{\\var{a}}$ can be simplified further. Commonly, surds can be simplified if the number inside of the square root has a square number as a factor.

Here, $\\var{p}$ is a prime number which means that its only divisors are $\\var{p}$ and $1$.

Therefore, $\\sqrt{\\var{p}}$ cannot be simplified any further.

Similarly, $\\var{a}$ is also a prime number, so $\\sqrt{\\var{a}}$ also cannot be simplified any further.

On the other hand, $\\simplify{{a}*{n}^2}$ is not a prime number and we can use the previous rule to simplify $\\sqrt{\\simplify{{a}*{n}^2}}$ as

\\[
\\begin{align}
\\sqrt{\\simplify{{a}*{n}^2}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{a}}\\\\
&= \\simplify{{n}*sqrt({a})}.
\\end{align}
\\]

b)

Using the same rule of manipulation as in part a), we can simplify $\\sqrt{\\simplify{{n}^2*{p}}}$ as

\\[
\\begin{align}
\\sqrt{\\simplify{{n}^2*{p}}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{p}}\\\\
&= \\simplify{{n}*sqrt({p})}.
\\end{align}
\\]

c)

Here, we can use both of the rules for manipulating surds:

\\[\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b} \\text{.} \\]

\\[ \\sqrt{\\frac{a}{b}} = \\frac{\\sqrt{a}}{\\sqrt{b}} \\text{.} \\]

We can simplify $\\displaystyle\\frac{ \\sqrt{\\simplify{{a}*{v}}} }{ \\sqrt{\\var{a}} }$ as follows.

\\[
\\begin{align}
\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} &= \\frac{\\sqrt{\\var{a}} \\times \\sqrt{\\var{v}}}{\\sqrt{\\var{a}}} \\\\[0.5em]
&= \\frac{\\sqrt{\\var{a}}}{\\sqrt{\\var{a}}} \\times \\sqrt{\\var{v}} \\\\[0.5em]
&= \\simplify{{sqrt(a)/sqrt(a)}} \\times \\sqrt{\\var{v}} \\\\[0.5em]
&= \\sqrt{\\var{v}} \\text{.}
\\end{align}
\\]

Or,

\\[
\\begin{align}
\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} &= \\sqrt{\\frac{\\simplify{{a}*{v}}}{\\var{a}}} \\\\[0.5em]
&= \\sqrt{\\var{v}} \\text{.}
\\end{align}
\\]

d)

We can simplify the fraction as

\\[
\\begin{align}
\\frac{\\sqrt{\\simplify{({b}{m})^2*{s}}}}{\\var{m}} &= \\frac{\\sqrt{\\simplify{({b*m})^2}} \\times \\sqrt{\\var{s}}}{\\var{m}} \\\\[0.5em]
&= \\frac{\\simplify{{b*m}} \\times \\sqrt{\\var{s}}}{\\var{m}} \\\\[0.5em]
&= \\simplify{{b}*sqrt({s})} \\text{.}
\\end{align}
\\]

e)

\\[
\\begin{align}
\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2{a})+{n}sqrt({b}^2*{a})} &= \\var{d}\\sqrt{\\var{a}} - \\var{b}(\\sqrt{\\simplify{{v}^2}} \\times \\sqrt{\\var{a}})+\\var{n}(\\sqrt{\\simplify{{b}^2}} \\times \\sqrt{\\var{a}}) \\\\
&= \\var{d}\\sqrt{\\var{a}} -\\var{b}(\\simplify{{v}*sqrt({a})})+\\var{n}(\\simplify{{b}*sqrt({a})}) \\\\
&= \\simplify{{d}sqrt({a})}-\\simplify{{b}*{v}sqrt({a})}+\\simplify{{n}*{b}sqrt({a})} \\\\
&= \\simplify{({d}-{b}*{v}+{n}*{b})sqrt({a})} \\text{.}
\\end{align}
\\]

f)

We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{\\sqrt{a}}$, by multiplying the top and bottom by $\\sqrt{a}$.

Therefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$, we multiply top and bottom by $\\sqrt{\\var{a}}$.

\\[
\\begin{align}
\\frac{1}{\\sqrt{\\var{a}}} &= \\frac{1}{\\sqrt{\\var{a}}} \\times \\frac{\\sqrt{\\var{a}}}{\\sqrt{\\var{a}}} \\\\[0.5em]
&= \\frac{\\sqrt{\\var{a}}}{\\var{a}} \\text{.}
\\end{align}
\\]

g)

We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$ by multiplying the top and bottom by $a-\\sqrt{b}$.

Therefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$, we multiply the top and bottom by $\\var{n} - \\sqrt{\\var{a}}$.

\\[
\\begin{align}
\\frac{1}{\\var{n}+\\sqrt{\\var{a}}} &= \\frac{1}{\\var{n}+\\sqrt{\\var{a}}} \\times \\frac{\\var{n}-\\sqrt{\\var{a}}}{\\var{n}-\\sqrt{\\var{a}}} \\\\[0.5em]
&=\\frac{\\var{n}-\\sqrt{\\var{a}}}{(\\var{n}+\\sqrt{\\var{a}})(\\var{n}-\\sqrt{\\var{a}})} \\\\[0.5em]
&=\\frac{\\var{n}-\\sqrt{\\var{a}}}{\\simplify{{n}^2}-\\var{a}} \\\\[0.5em]
&=\\frac{\\var{n}-\\sqrt{\\var{a}}}{\\simplify{{n}^2-{a}}} \\text{.}
\\end{align}
\\]

h)

We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{a-\\sqrt{b}}$ by multiplying the top and bottom by $a+\\sqrt{b}$.

Therefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}}$, we multiply the top and bottom by $\\var{d+p}+\\sqrt{\\var{p}}$.

\\[
\\begin{align}
\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} &= \\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} \\times \\frac{\\var{d+p}+\\sqrt{\\var{p}}}{\\var{d+p}+\\sqrt{\\var{p}}} \\\\[0.5em]
&=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{(\\var{d+p}-\\sqrt{\\var{p}})(\\var{d+p}+\\sqrt{\\var{p}})} \\\\[0.5em]
&=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{\\simplify{{d+p}^2}-\\var{p}} \\\\[0.5em]
&=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{\\simplify{{d+p}^2-{p}}} \\\\[0.5em]
&=\\simplify{{t}/{(d+p)^2-p}}(\\var{d+p}+\\sqrt{\\var{p}}) \\\\[0.5em]
&= \\simplify[all,!noleadingMinus]{({t*(d+p)}+{t}*sqrt({p}))/({(d+p)^2-p})} \\text{.}
\\end{align}
\\]

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Parts c and e

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Fraction in answer for part d.

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parts b and d

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Parts a, d,e and h

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parts d and e

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shorter list of primes for parts a,c,e,f and g

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Which of the following can be simplified further?

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Can be simplified further

", "

Cannot be simplified further

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$\\sqrt{\\var{p}}$

", "

$\\sqrt{\\simplify{{a}*{n}^2}}$

", "

$\\sqrt{\\var{a}}$

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Simplify $\\sqrt{\\simplify{{n}^2*{p}}}$.

$\\sqrt{\\simplify{{n}^2*{p}}} =$ [[0]]$\\sqrt{\\var{p}}$.

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Recall the first rule of surds

$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.

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Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} =$ [[0]].

You could use either of the following rules:

$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.

$\\displaystyle\\sqrt{\\frac{a}{b}} = \\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}}$.

You must simplify your answer further.

"}, "notallowed": {"strings": ["/"], "showStrings": false, "partialCredit": 0, "message": "

You must simplify your answer further.

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Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{({b}{m})^2*{s}}}}{\\var{m}}$.

$\\displaystyle\\frac{\\sqrt{\\simplify{({b}*{m})^2*{s}}}}{\\var{m}} =$ [[0]]$\\sqrt{\\var{s}}$.

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Simplify $\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2*{a})+{n}sqrt({b}^2*{a})}$.

$\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2*{a})+{n}sqrt({b}^2*{a})} =$ [[0]].

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Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{1}{\\sqrt{\\var{a}}} =$ [[0]] [[1]] .

To rationalise the denominator of fractions in the form $\\frac{1}{\\sqrt{a}}$, multiply the top and bottom by $\\sqrt{a}$.

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Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}} =$ [[0]] [[1]] .

To rationalise the denominator of fractions in the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$, multiply the top and bottom by $a-\\sqrt{b}$.

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Rationalise the denominator of the fraction $\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}}$.

$\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} =$ [[0]] [[1]] .

To rationalise the denominator of fractions in the form, $\\displaystyle\\frac{1}{a-\\sqrt{b}}$, multiply the top and bottom by ${a+\\sqrt{b}}$.

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