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Arithmetic I

Proportion & Variation

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CAT 2025 Lesson : Proportion & Variation - Other Proportions

1.3.4 Circular Proportion

Property: If

\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{a}

, then

a = b = c

Example 8

\dfrac{3x + 6}{2y + 6} = \dfrac{y + 3}{6} = \dfrac{4}{x + 2}

, what is the value of

x + y

Solution

\dfrac{3x + 6}{2y + 6} = \dfrac{y + 3}{6} = \dfrac{4}{x + 2}

⇒\dfrac{3x + 6}{2y + 6} = \dfrac{2y + 6}{12} = \dfrac{12}{3x + 6}

∴

3x + 6 = 2y + 6 = 12

⇒x = 2

and

y = 3

∴

x + y = 5

Answer:

5

1.4 Common Approach to solving

Two common ways to solve questions are substitution and equating to a constant,

\bm{k}

1.4.1 Substitution

This method is useful when we do not have an idea on how to proceed. We substitute values that satisfy the equations.

Example 9

\dfrac{a}{2b + c} = \dfrac{b}{2c + a} = \dfrac{c}{2a + b}

, then the value of

\dfrac{4b + 5c}{3a}

could be
(1)

1

(2)

3

(3)

\dfrac{1}{3}

(4)

0.3

Solution

The given identity of

\dfrac{a}{2b + c} = \dfrac{b}{2c + a} = \dfrac{c}{2a + b}

is true when

a = b = c.

∴ Let

a = b = c = 1

\dfrac{4b + 5c}{3a} = \dfrac{9}{3} = 3

Answer:(2)

3

1.4.2 Equating to $\bm{k}$

When the identities are multiplied or squared, then we can equate the identities to a constant

\bm{k}

and solve.

Example 10

\dfrac{a}{b} = \dfrac{c}{d}

, then

\dfrac{3ac - 2bd}{3ac + 2bd} =

?
(1)

\dfrac{ac}{bd}

(2)

\dfrac{3ac}{2bd}

(3)

\dfrac{3a - 2b}{3a + 2b}

(4)

\dfrac{3c^{2} - 2d^{2}}{3c^{2} + 2d^{2}}

Solution

Let

k = \dfrac{a}{b} = \dfrac{c}{d}

Then

k^{2} = \dfrac{ac}{bd} = \dfrac{a^{2}}{b^{2}} = \dfrac{c^{2}}{d^{2}}

Value will remain the same if

ac

and

bd

can be replaced by

a^{2}

and

b^{2}

c^{2}

and

d^{2}

respectively.

∴

\dfrac{3ac - 2bd}{3ac + 2bd} = \dfrac{3c^{2} - 2d^{2}}{3c^{2} + 2d^{2}}

Answer: (4)

\dfrac{3c^{2} - 2d^{2}}{3c^{2} + 2d^{2}}

Note: Alternatively, you can substitute values. However, the stated approach saves time.

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CAT 2025 Lesson : Proportion & Variation - Other Proportions

1.3.4 Circular Proportion

Example 8

Solution

1.4 Common Approach to solving

1.4.1 Substitution

Example 9

Solution

1.4.2 Equating to k\bm{k}k

Example 10

Solution

1.4.2 Equating to $\bm{k}$