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Arithmetic I

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Proportion & Variation

Proportion And Variation

MODULES

Basics of Proportion
Continued Proportion
Componedo & Dividendo
Sum Rule
Other Proportions
Basics of Variation
Combined Variation
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

PRACTICE

Proportion & Variation : Level 1
Proportion & Variation : Level 2
Proportion & Variation : Level 3
ALL MODULES

CAT 2025 Lesson : Proportion & Variation - Sum Rule

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1.3.2 Sum Rule

Property: If ab=cd=ef=\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} =ba​=dc​=fe​= .... =k= k=k, then k=a+c+e+....b+d+f+....k= \dfrac{a + c + e + ....}{b + d + f + ....}k=b+d+f+....a+c+e+....​

Example 6

If ab=cd=ef=5\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = 5ba​=dc​=fe​=5, then a3+c3+e3b3+d3+f3=\dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} =b3+d3+f3a3+c3+e3​= ?
(1) 111           (2) 555           (3) 252525           (4) 125125125          

Solution

ab=cd=ef=5\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = 5ba​=dc​=fe​=5

⇒a3b3=c3d3=e3f3=125⇒\dfrac{a^{3}}{b^{3}} = \dfrac{c^{3}}{d^{3}} = \dfrac{e^{3}}{f^{3}} = 125⇒b3a3​=d3c3​=f3e3​=125

Applying sum rule, a3b3=c3d3=e3f3=a3+c3+e3b3+d3+f3=125\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}} = 125b3a3​=d3c3​=f3e3​=b3+d3+f3a3+c3+e3​=125

Answer: (4) 125125125


1.3.3 Extension to sum rule

Property: If ab=cd=ef=...=k\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = ... = kba​=dc​=fe​=...=k, then k=papb=qcqd=rerf=...=pa+qc+re+....pb+qd+rf+...k = \dfrac{pa}{pb} = \dfrac{qc}{qd} = \dfrac{re}{rf} = ... = \dfrac{pa + qc + re + ....}{pb + qd + rf + ...}k=pbpa​=qdqc​=rfre​=...=pb+qd+rf+...pa+qc+re+....​

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

Example 7

If ab=cd\dfrac{a }{b} = \dfrac{c}{d}ba​=dc​, then 5a−4b5c−4d=\dfrac{5a - 4b}{5c - 4d} =5c−4d5a−4b​= ?
(1) ab\dfrac{a}{b}ba​           (2) 5a4b\dfrac{5a}{4b}4b5a​           (3) ac\dfrac{a}{c}ca​           (4) 5a4c\dfrac{5a}{4c}4c5a​          

Solution

Note that the numerator contains aaa and bbb, while the denominator contains ccc and ddd.

∴ Applying alternendo, ab=cd⇒ac=bd\dfrac{a}{b} = \dfrac{c}{d} ⇒ \dfrac{a}{c} = \dfrac{b}{d}ba​=dc​⇒ca​=db​

Applying the extension to sum rule,

ac=bd=5a5c=−4b−4d=5a−4b5c−4d\dfrac{a}{c} = \dfrac{b}{d} = \dfrac{5a}{5c} = \dfrac{-4b}{-4d} = \dfrac{5a - 4b}{5c - 4d}ca​=db​=5c5a​=−4d−4b​=5c−4d5a−4b​

Answer: (3) ac\dfrac{a}{c}ca​

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


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