Some problems describe numbers indirectly: the digits of a number, the parts of a fraction or people’s ages at different times. Let letters stand for the basic unknowns, write each described quantity as an expression, then form and solve two equations. **Quantity** | **Expression** two-digit number with tens digit x and ones digit y | 10x+y the same number with its digits reversed | 10y+x fraction with numerator x and denominator y | xy add 3 to the numerator, subtract 1 from the denominator | x+3y-1 age now x years; age in n years; age n years ago | x, x+n, x-n Why 10x+y: a two-digit number is worth ten times its tens digit plus its ones digit; for example, 47=4 x 10+7. Writing xy would mean x multiplied by y, which is a different number. Method - Choose letters for the basic unknowns (the digits, the numerator and denominator, or the present ages). - Write each quantity in the question as an expression in those letters, using the table. - Form one equation for each fact. For an equation between fractions, cross-multiply to remove the fractions. - Solve, then answer the question that was asked (the number, the fraction or the ages), and check it against the words. Example A fraction becomes 1 when 4 is added to its numerator, and becomes 12 when 9 is added to its denominator. Find the fraction. Let the fraction be xy. Then x+4y=1, so y=x+4; and xy+9= 12, so 2x=y+9. 2x=(x+4)+9 x=13 Then y=17 and the fraction is 1317. Check: 1717=1 and 1326= 12. Ages example: a father is four times as old as his son. In 8 years’ time he will be three times as old as his son. Let the son be s years old and the father f years old. Then f=4s and f+8=3(s+8). 4s+8=3s+24 s=16 So the son is 16 and the father is 64. Check: in 8 years they are 24 and 72, and 72=3 x 24. Check: digits must be whole numbers from 0 to 9, and a tens digit cannot be 0. Ages must be sensible: a person cannot have a negative age n years ago.

Numbers, digits, fractions and ages

Secondary 2

Some problems describe numbers indirectly: the digits of a number, the parts of a fraction or people’s ages at different times. Let letters stand for the basic unknowns, write each described quantity as an expression, then form and solve two equations.

Quantity	Expression
two-digit number with tens digit $x$ and ones digit $y$	$10x+y$
the same number with its digits reversed	$10y+x$
fraction with numerator $x$ and denominator $y$	$\dfrac{x}{y}$
add $3$ to the numerator, subtract $1$ from the denominator	$\dfrac{x+3}{y-1}$
age now $x$ years; age in $n$ years; age $n$ years ago	$x$, $x+n$, $x-n$

Why $10x+y$: a two-digit number is worth ten times its tens digit plus its ones digit; for example, $47=4\times10+7$. Writing $xy$ would mean $x$ multiplied by $y$, which is a different number.

Method

Choose letters for the basic unknowns (the digits, the numerator and denominator, or the present ages).
Write each quantity in the question as an expression in those letters, using the table.
Form one equation for each fact. For an equation between fractions, cross-multiply to remove the fractions.
Solve, then answer the question that was asked (the number, the fraction or the ages), and check it against the words.

Example

A fraction becomes $1$ when $4$ is added to its numerator, and becomes $\dfrac{1}{2}$ when $9$ is added to its denominator. Find the fraction.

Let the fraction be $\dfrac{x}{y}$. Then $\dfrac{x+4}{y}=1$, so $y=x+4$; and $\dfrac{x}{y+9}=\dfrac{1}{2}$, so $2x=y+9$.

$\begin{aligned} 2x&=(x+4)+9 \\ x&=13 \end{aligned}$

Then $y=17$ and the fraction is $\dfrac{13}{17}$. Check: $\dfrac{17}{17}=1$ and $\dfrac{13}{26}=\dfrac{1}{2}$.

Ages example: a father is four times as old as his son. In $8$ years’ time he will be three times as old as his son. Let the son be $s$ years old and the father $f$ years old. Then $f=4s$ and $f+8=3(s+8)$.

$\begin{aligned} 4s+8&=3s+24 \\ s&=16 \end{aligned}$

So the son is $16$ and the father is $64$. Check: in $8$ years they are $24$ and $72$, and $72=3\times24$.

Check: digits must be whole numbers from $0$ to $9$, and a tens digit cannot be $0$. Ages must be sensible: a person cannot have a negative age $n$ years ago.

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Quantity	Expression
two-digit number with tens digit \(x\) and ones digit \(y\)	\(10x+y\)
the same number with its digits reversed	\(10y+x\)
fraction with numerator \(x\) and denominator \(y\)	\(\dfrac{x}{y}\)
add \(3\) to the numerator, subtract \(1\) from the denominator	\(\dfrac{x+3}{y-1}\)
age now \(x\) years; age in \(n\) years; age \(n\) years ago	\(x\), \(x+n\), \(x-n\)