CBSE Class 9 Mathematics Ratio and Proportion Notes Set 09

Ratio as a Percentage

A ratio \( a:b \) becomes a percentage by writing \( \dfrac ab\times100 \), so \( 3:40 \) is \( 7\dfrac12\% \). Percentages are convenient for comparing ratios because they all have the common denominator 100. When one mixture is \( \dfrac38 \) alcohol and another \( \dfrac25 \), compare \( 37.5\% \) with \( 40\% \): the second is stronger by \( 2.5 \) percentage points. If a known part forms a given percentage of the whole, form an equation: if 22% of the boys are 44 in number, the whole is \( \dfrac{44\times100}{22}=200 \).

Changing a Ratio by Adding or Subtracting

Adding the same positive number to both terms of a ratio pushes it towards 1: a ratio with the larger first term gets smaller, a ratio with the smaller first term gets larger. For example \( 4:3 \) becomes \( 5:4 \), which is smaller, while \( 2:3 \) becomes \( 3:4 \), which is larger. To find the number which must be added or subtracted, write the new ratio as a fraction, equate it to the required ratio and cross-multiply. For \( (38-x):(31-x)=4:3 \) this gives \( x=10 \).

Homogeneous Equations and Ratios

An equation in which every term has the same total degree, such as \( 2x^2-3xy+y^2=0 \), fixes only the ratio \( x:y \). Divide throughout by \( y^2 \) and put \( w=\dfrac xy \) to get a quadratic in \( w \): here \( (2w-1)(w-1)=0 \), so \( x:y=1:2 \) or \( 1:1 \). Two linear homogeneous equations in three unknowns give the ratios \( x:z \) and \( y:z \) by eliminating one variable at a time.

Equal Ratios and the Common Multiplier

If \( \dfrac{x+y}{2}=\dfrac{y+z}{3}=\dfrac{z+x}{4}=k \), add the three equations to get \( x+y+z=\dfrac{9k}{2} \) and subtract each pair sum to find \( x,y,z \) in terms of \( k \); here \( x:y:z=3:1:5 \). When a ratio is given together with a second condition, as in \( x:2=y:3 \) with \( 6x+7y=11 \), write \( x=2k,\ y=3k \) and solve for \( k \).

Componendo and Dividendo in Equations

If \( \dfrac ab=\dfrac cd \) then \( \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d} \). This removes sums and differences of radicals. In \( \dfrac{\sqrt{2+x}+\sqrt{2-x}}{\sqrt{2+x}-\sqrt{2-x}}=2 \) it gives \( \dfrac{\sqrt{2+x}}{\sqrt{2-x}}=3 \), so \( 2+x=9(2-x) \) and \( x=\dfrac85 \).

Variation Problems

To solve a variation problem, write the relation with a constant \( k \), find \( k \) from the given pair of values, and substitute for the unknown. Joint variation multiplies the factors, as in \( V=kr^2h \) for a cylinder. If \( z\propto y^2 \) and \( y\propto\dfrac1x \), then \( z\propto\dfrac1{x^2} \). To double the surface area of a sphere, since \( S\propto R^2 \), the radius must be multiplied by \( \sqrt2 \).

Worked Examples

Example 1. Two numbers are in the ratio \( 3:4 \) and, when 12 is added to each, the ratio is \( 5:6 \). Find the numbers.
Let the numbers be \( 3x \) and \( 4x \). Then \( 6(3x+12)=5(4x+12) \), so \( 2x=12 \) and \( x=6 \). The numbers are 18 and 24.

Example 2. If \( x:5=y:3 \), find \( \dfrac{9x+7y}{9x-7y} \).
Put \( x=5k,\ y=3k \): the value is \( \dfrac{66k}{24k}=\dfrac{11}4 \).

Example 3. If \( y \) varies inversely as \( x \) and \( y=5 \) when \( x=3 \), find \( y \) when \( x=9 \).
\( y=\dfrac kx \) with \( k=15 \), so \( y=\dfrac{15}{9}=\dfrac53 \).