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Arithmetic I

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Proportion & Variation
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CAT 2025 Lesson : Proportion & Variation - Other Proportions

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1.3.4 Circular Proportion

Property: If ab=bc=ca\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{a}, then a=b=ca = b = c

Example 8

If 3x+62y+6=y+36=4x+2\dfrac{3x + 6}{2y + 6} = \dfrac{y + 3}{6} = \dfrac{4}{x + 2}, what is the value of x+yx + y?

Solution

3x+62y+6=y+36=4x+2\dfrac{3x + 6}{2y + 6} = \dfrac{y + 3}{6} = \dfrac{4}{x + 2}

⇒3x+62y+6=2y+612=123x+6⇒\dfrac{3x + 6}{2y + 6} = \dfrac{2y + 6}{12} = \dfrac{12}{3x + 6}

∴ 3x+6=2y+6=123x + 6 = 2y + 6 = 12
⇒x=2⇒x = 2 and y=3y = 3
∴ x+y=5x + y = 5

Answer: 55


1.4 Common Approach to solving

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

If a2b+c=b2c+a=c2a+b\dfrac{a}{2b + c} = \dfrac{b}{2c + a} = \dfrac{c}{2a + b}, then the value of 4b+5c3a\dfrac{4b + 5c}{3a} could be
(1) 11           (2) 33           (3) 13\dfrac{1}{3}           (4) 0.30.3          

Solution

The given identity of a2b+c=b2c+a=c2a+b\dfrac{a}{2b + c} = \dfrac{b}{2c + a} = \dfrac{c}{2a + b} is true when a=b=c.a = b = c.

∴ Let a=b=c=1a = b = c = 1

4b+5c3a=93=3\dfrac{4b + 5c}{3a} = \dfrac{9}{3} = 3

Answer:(2) 33


1.4.2 Equating to k\bm{k}

When the identities are multiplied or squared, then we can equate the identities to a constant k\bm{k} and solve.

Example 10

If ab=cd\dfrac{a}{b} = \dfrac{c}{d}, then 3ac−2bd3ac+2bd=\dfrac{3ac - 2bd}{3ac + 2bd} = ?
(1) acbd\dfrac{ac}{bd}           (2) 3ac2bd\dfrac{3ac}{2bd}           (3) 3a−2b3a+2b\dfrac{3a - 2b}{3a + 2b}           (4) 3c2−2d23c2+2d2\dfrac{3c^{2} - 2d^{2}}{3c^{2} + 2d^{2}}          

Solution

Let k=ab=cdk = \dfrac{a}{b} = \dfrac{c}{d}

Then k2=acbd=a2b2=c2d2k^{2} = \dfrac{ac}{bd} = \dfrac{a^{2}}{b^{2}} = \dfrac{c^{2}}{d^{2}}

Value will remain the same if acac and bdbd can be replaced by a2a^{2} and b2b^{2} or c2c^{2} and d2d^{2}respectively.

∴ 3ac−2bd3ac+2bd=3c2−2d23c2+2d2\dfrac{3ac - 2bd}{3ac + 2bd} = \dfrac{3c^{2} - 2d^{2}}{3c^{2} + 2d^{2}}

Answer: (4) 3c2−2d23c2+2d2\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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