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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 - Combined Variation

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2.3 Joint Variation

Joint variation is a form of variation involving two or more variables.
D=s×tD = s \times tD=s×t, an increase of speed or time will result in an increase in the distance covered. So, distance jointly varies with speed and time.

In the case of volume of a cylinder,
V=πr2hV = \pi r^{2} hV=πr2h. Here, V∝r2V \propto r^{2}V∝r2 and V ∝h\propto h∝h.

So,
VVV jointly varies with r2r^{2}r2 and hhh.

This is represented as
V=Kr2hV = K r^{2}hV=Kr2h. In this case the value of KKK is π\piπ.

2.4 Combined Variation

Combined variations contain combinations of Direct, Inverse and Joint variations.

For instance, if
a∝ba \propto ba∝b and a∝1ca \propto \dfrac{1}{c}a∝c1​ and a∝da \propto da∝d, then the same can be combined and expressed as

a∝bdca \propto \dfrac{bd}{c}a∝cbd​ or a=Kbdca = \dfrac{Kbd}{c}a=cKbd​

Example 15

xxx varies directly with the cube of yyy. xxx is also inversely proportional to the square root of zzz. When xxx = 323232, yyy = 444 and zzz = 646464. What is the value of xxx when yyy = 333 and zzz = 161616?

Solution

x∝y3x \propto y^{3}x∝y3 and x∝1zx \propto \dfrac{1}{\sqrt{z}}x∝z​1​

⇒x=Ky3z⇒x = \dfrac{Ky^{3}}{\sqrt{z}}⇒x=z​Ky3​

Substituting
xxx = 323232, yyy = 444 and zzz = 646464,

32=K×436432 = \dfrac{K \times 4^{3}}{\sqrt{64}}32=64​K×43​

⇒K=4⇒K = 4⇒K=4

When
yyy = 333 and zzz = 161616,

x=4×3316=27x = \dfrac{4 \times 3^{3}}{\sqrt{16}} = 27x=16​4×33​=27

Answer:
272727



Example 16

If xxx varies as the cube of yyy, and yyy varies as the fifth root of zzz, then xxx varies as the nthn^{th}nth power of zzz, where nnn is:
[FMS 2011]

(1) 115\dfrac{1}{15}151​           (2) 53\dfrac{5}{3}35​           (3) 35\dfrac{3}{5}53​           (4) 151515          

Solution

x∝y3x \propto y^{3}x∝y3 and y∝z15y \propto z^{\frac{1}{5}}y∝z51​ are given in the question

y∝z15⇒y3∝z35y \propto z^{\frac{1}{5}} ⇒ y^{3} \propto z^{\frac{3}{5}}y∝z51​⇒y3∝z53​

∴
x∝y3∝z35x \propto y^{3} \propto z^{\frac{3}{5}}x∝y3∝z53​

Answer:
35\frac{3}{5}53​



Example 17

The value of a gem is directly proportional to the square of its weight. When Anu dropped the gem, it broke into three pieces whose weights were in the ratio 1:2:31 : 2 : 31:2:3. What is the percentage reduction in the value?
[FMS 2011]

Solution

v=Kw2v = Kw^{2}v=Kw2 where vvv and www are the value and weight of a gem respectively.

If the weight of the gem is
666 gm, then the 333 smaller pieces weight 111 gm, 222 gm and 333 gm.

Value of
666 gm gem =K×62=36K= K \times 6^{2} = 36K=K×62=36K

Sum of value of
333 smaller gems =K×12+K×22+K×32=14K= K \times 1^{2} + K \times 2^{2} + K \times 3^{2} = 14K=K×12+K×22+K×32=14K

%\%% Reduction in value =36K−14K36K×100%=1118×100%=61.11%= \dfrac{36K - 14K}{36K} \times 100 \% = \dfrac{11}{18} \times 100 \% = 61.11 \%=36K36K−14K​×100%=1811​×100%=61.11%

Answer:
61.11%61.11 \%61.11%



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