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Proportion & Variation

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CAT 2025 Lesson : Proportion & Variation - Sum Rule

Property: If

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

....

= k

, then

k= \dfrac{a + c + e + ....}{b + d + f + ....}

\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = 5

, then

\dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} =

?
(1)

1

(2)

5

(3)

25

(4)

125

\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = 5

⇒\dfrac{a^{3}}{b^{3}} = \dfrac{c^{3}}{d^{3}} = \dfrac{e^{3}}{f^{3}} = 125

Applying sum rule,

\dfrac{a^{3}}{b^{3}} = \dfrac{c^{3}}{d^{3}} = \dfrac{e^{3}}{f^{3}} = \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = 125

Answer: (4)

125

Property: If

\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = ... = k

, then

k = \dfrac{pa}{pb} = \dfrac{qc}{qd} = \dfrac{re}{rf} = ... = \dfrac{pa + qc + re + ....}{pb + qd + rf + ...}

Note that multiplying and dividing a ratio by the same term does not change its value.

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

, then

\dfrac{5a - 4b}{5c - 4d} =

?
(1)

\dfrac{a}{b}

(2)

\dfrac{5a}{4b}

(3)

\dfrac{a}{c}

(4)

\dfrac{5a}{4c}

Note that the numerator contains

a

and

b

, while the denominator contains

c

and

d

.

∴ Applying alternendo,

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

Applying the extension to sum rule,

\dfrac{a}{c} = \dfrac{b}{d} = \dfrac{5a}{5c} = \dfrac{-4b}{-4d} = \dfrac{5a - 4b}{5c - 4d}

Answer: (3)

\dfrac{a}{c}

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

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