JEE Mathematics Ratio Proportion and Variation Notes Set 01

Variation: Basic Ideas

Direct variation. A quantity y varies directly as x (written \( y \propto x \)) when \( y = kx \) for a constant k. Then \( x = \dfrac{1}{k}y \), so x also varies directly as y. When k is positive, y increases as x increases and decreases as x decreases. For example, the circumference of a circle varies as its diameter: \( C = \pi d \), with constant \( \pi \).
Inverse variation. A quantity y varies inversely as x (\( y \propto \dfrac{1}{x} \)) when \( y = \dfrac{k}{x} \), that is, \( xy = k \). For a fixed mass of gas at constant temperature, pressure varies inversely as volume, \( PV = \text{constant} \); doubling the pressure halves the volume.
Joint variation. z varies jointly as x and y when z varies as the product xy, that is, \( z = kxy \).
Mixed variation. z varies directly as x and inversely as y when \( \dfrac{z}{x/y} = k \), that is, \( z = k\dfrac{x}{y} \).

Useful Rules

1. Always begin by writing the relation with a constant k, then use the given pair of values to find k.
2. If \( x \propto y \) and \( y \propto z \), then \( x \propto z \).
3. If \( x \propto y \), then \( x^n \propto y^n \) for any n.
4. If \( x \propto y \) and \( u \propto v \), then \( xu \propto yv \).
5. When a quantity is the sum of two parts, one varying as one thing and the other as another, write it as a sum with a separate constant for each part (see Illustration 5).

Worked Illustrations

Illustration 1. y varies inversely as x, and y = 5 when x = 3. Find the equation connecting x and y, the value of y when x = 9, and the value of x when y = 6.
Answer:
Write \( y = \dfrac{k}{x} \). Putting x = 3, y = 5 gives \( k = 15 \), so the equation is \( y = \dfrac{15}{x} \).
When x = 9: \( y = \dfrac{15}{9} = \dfrac{5}{3} \).
When y = 6: \( x = \dfrac{15}{6} = \dfrac{5}{2} \).

Illustration 2. If a varies as (b + c) and b varies as (c + a), show that a varies as b.
Answer:
Let \( a = k_1(b + c) \) and \( b = k_2(c + a) \).
From the first, \( c = \dfrac{a}{k_1} - b \). Substitute in the second: \( b = k_2\left(\dfrac{a}{k_1} - b + a\right) \).
Collect b: \( b(1 + k_2) = k_2 a \dfrac{1 + k_1}{k_1} \).
Hence \( a = \dfrac{k_1(1 + k_2)}{k_2(1 + k_1)}\, b \). The fraction is built from constants only, so a = kb for a constant k, and a varies as b.

Illustration 3. If x varies as the square of y, and x = 2 when y = 5, find x when y = 15.
Answer:
Write \( x = ky^2 \). With x = 2, y = 5: \( 2 = 25k \), so \( k = \dfrac{2}{25} \).
When y = 15: \( x = \dfrac{2}{25} \times 225 = 18 \).

Illustration 4. If (4x − 5y) varies as (x + 3y), prove that \( x^4 + y^4 \) varies as \( 2x^2y^2 + xy^3 + 2x^3y \).
Answer:
Write \( 4x - 5y = k(x + 3y) \). Then \( x(4 - k) = y(3k + 5) \), so \( x = cy \) where \( c = \dfrac{3k + 5}{4 - k} \) is a constant.
Numerator: \( x^4 + y^4 = (c^4 + 1)y^4 \).
Denominator: \( 2x^2y^2 + xy^3 + 2x^3y = (2c^2 + c + 2c^3)y^4 \).
The ratio is \( \dfrac{c^4 + 1}{2c^2 + c + 2c^3} \), a constant. Hence \( x^4 + y^4 \) varies as \( 2x^2y^2 + xy^3 + 2x^3y \).

Illustration 5. y is the sum of two quantities, one varying directly as \( x^{3/2} \) and the other inversely as \( x^{1/2} \). If y = 1 when x = 1 and y = 38 when x = 4, find y when x = 9.
Answer:
Write \( y = ax^{3/2} + \dfrac{b}{x^{1/2}} \).
x = 1, y = 1: \( a + b = 1 \). x = 4, y = 38: \( 8a + \dfrac{b}{2} = 38 \), that is \( 16a + b = 76 \).
Subtract: \( 15a = 75 \), so a = 5 and b = −4. Hence \( y = 5x^{3/2} - \dfrac{4}{x^{1/2}} \).
At x = 9: \( y = 5 \times 27 - \dfrac{4}{3} = 135 - \dfrac{4}{3} = 133\dfrac{2}{3} \).

Illustration 6. The value of a coin varies jointly as its thickness and the square of its diameter. Two coins have diameters in the ratio 4 : 5. Find the ratio of their thicknesses if the value of the first is half that of the second.
Answer:
Let value v = k t d². For the two coins, \( \dfrac{v_1}{v_2} = \dfrac{t_1}{t_2}\left(\dfrac{d_1}{d_2}\right)^2 \).
Substitute: \( \dfrac{1}{2} = \dfrac{t_1}{t_2} \times \dfrac{16}{25} \).
So \( \dfrac{t_1}{t_2} = \dfrac{25}{32} \). The thicknesses are in the ratio 25 : 32.

Illustration 7. The surface area of a sphere varies as the square of its radius. By what ratio must the radius be increased to double the surface area?
Answer:
\( S = kR^2 \). For the new sphere, \( 2S = kR_1^2 \).
Dividing, \( 2 = \dfrac{R_1^2}{R^2} \), so \( \dfrac{R_1}{R} = \sqrt{2} \).
The radius must be multiplied by \( \sqrt{2} \), that is \( R_1 : R = \sqrt{2} : 1 \).

Illustration 8. If z varies directly as \( y^2 \) and y varies inversely as x, find the relation between z and x.
Answer:
\( z = k_1 y^2 \) and \( y = \dfrac{k_2}{x} \).
Substituting, \( z = k_1\dfrac{k_2^2}{x^2} = \dfrac{k}{x^2} \) with \( k = k_1k_2^2 \).
So z varies inversely as \( x^2 \).

Illustration 9. Inside the earth, a man's weight varies as his distance from the centre; above the surface it varies inversely as the square of that distance. Take the radius of the earth as 6400 km. Find the percentage of weight lost (i) by going 16 km down a mine and (ii) by going 16 km up in a balloon.
Answer:
(i) Below the surface: \( w = kx \). On the surface the weight is 6400k. At 16 km depth the distance is 6384 km and the weight is 6384k.
Loss = 16k, so the percentage loss is \( \dfrac{16k}{6400k} \times 100 = \dfrac{1}{4}\% \).
(ii) Above the surface: \( w = \dfrac{a}{y^2} \). On the surface the weight is \( \dfrac{a}{6400^2} \); at 16 km height the distance is 6416 km.
Percentage loss \( = \left[1 - \left(\dfrac{6400}{6416}\right)^2\right] \times 100 = \left[1 - \left(\dfrac{400}{401}\right)^2\right] \times 100 = \dfrac{801}{160801} \times 100 \approx 0.5\% \), that is, about \( \dfrac{1}{2}\% \).

Illustration 10. The volume of a cylinder varies jointly as the square of its base radius and its height. A cylinder of radius 2 m and height 7 m has volume 88 cubic metres. Find the volume of a cylinder of radius 7 m and height \( 2\dfrac{3}{11} \) m.
Answer:
\( V = kr^2h \). With r = 2, h = 7, V = 88: \( 88 = 28k \), so \( k = \dfrac{22}{7} \).
For r = 7 and \( h = \dfrac{25}{11} \): \( V = \dfrac{22}{7} \times 49 \times \dfrac{25}{11} = 2 \times 7 \times 25 = 350 \) cubic metres.