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Arithmetic II

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Averages

Averages

MODULES

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Basics & Assumed Mean
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Weighted Average
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GM & HM
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Common Types
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Median, Mode & Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

PRACTICE

Averages : Level 1
Averages : Level 2
Averages : Level 3
ALL MODULES

CAT 2025 Lesson : Averages - GM & HM

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4. Geometric Mean

Geometric mean is another type of average.

Where xi,x2,x3,...,xnx_{i}, x_{2}, x_{3}, ..., x_{n}xi​,x2​,x3​,...,xn​ are nnn elements,

Geometric Mean (GM) =(x1×x2×x3×...×xn)1n= {(x_{1} \times x_{2} \times x_{3} \times ... \times x_{n})}^{\frac{1}{n}}=(x1​×x2​×x3​×...×xn​)n1​

A related concept is CAGR, or Compound Annual Growth Rate. This is covered in Interest and Growth.

Example 10

The population of Karnavati grew by 20%,25%20 \%, 25 \%20%,25% and 10%10 \%10% in 2016,20172016, 20172016,2017 and 201820182018 respectively. What is the compound annual growth rate during this period?

(111) 161616%            (222) 181818%            (333) 202020%           (444) 222222%           

Solution

Let the population of Karnavati in 201520152015 be ppp.

Population of Karnavati in 2018=p×1.2×1.25×1.1=1.65p2018 = p \times 1.2 \times 1.25 \times 1.1 = 1.65 p2018=p×1.2×1.25×1.1=1.65p

CAGR is the constant compound rate at which the same growth of 65%65 \%65% could have been achieved over 333 years. This is the geometric mean of the three multiplication factors. Let the CAGR be rrr.

p×(1+r100)3=1.65pp \times {\left(1 + \dfrac{r}{100} \right)}^{3} = 1.65 pp×(1+100r​)3=1.65p

⇒ 1+r100=(1.65)131 + \dfrac{r}{100} = {(1.65)}^{\frac{1}{3}}1+100r​=(1.65)31​

Going through the options, we note that (1.18)3=1.18×1.18×1.18≈1.643(1.18)^{3} = 1.18 \times 1.18 \times 1.18 \approx 1.643(1.18)3=1.18×1.18×1.18≈1.643

(Note: In most CAT exams, the onscreen calculators only have the basic arithmetic functions. As the exponential function is not present, we need to work with the options)

∴ 1+r100=1.181 + \dfrac{r}{100} = 1.181+100r​=1.18

⇒ r=18%r = 18 \%r=18%

Answer: (222) 18%18 \%18%


4.1 GM of 2 terms

GM of 222 terms, say aaa and bbb, is (ab)12=ab{(ab)}^{\frac{1}{2}} = \sqrt{ab}(ab)21​=ab​

Example 11

The arithmetic mean of 222 positive numbers is 6 while the geometric mean is 333 \sqrt{3}33​. What is the difference between the two numbers?

Solution

Let the two numbers be aaa and bbb

a+b2=6\dfrac{a + b}{2} = 6 2a+b​=6 ⇒ a+b=12 a + b = 12 a+b=12 ⇒ b=12−a⟶(1) b = 12 - a \longrightarrow (1)b=12−a⟶(1)

ab=33\sqrt{ab} = 3 \sqrt{3} ab​=33​ ⇒ ab=27⟶(2) ab = 27 \longrightarrow (2)ab=27⟶(2)

Substituting (1)(1)(1) in (2)(2)(2),
⇒ a(12−a)=27 a(12 - a) = 27a(12−a)=27

⇒ a2−12a+27=0a^{2} - 12a + 27 = 0a2−12a+27=0

a=9a = 9a=9 or 333 , so b=3b = 3b=3 or 999

Difference between aaa and b=9−3=6b = 9 - 3 = 6b=9−3=6

Answer: 666


5. Harmonic Mean

Where xi,x2,x3,....,xnx_{i}, x_{2}, x_{3}, ...., x_{n}xi​,x2​,x3​,....,xn​ are nnn elements,

Harmonic Mean (HM) =n1x1+1x2+1x3+...+1xn= \dfrac{n}{\dfrac{1}{x_{1}} + \dfrac{1}{x_{2}} + \dfrac{1}{x_{3}} + ... + \dfrac{1}{x_{n}}}=x1​1​+x2​1​+x3​1​+...+xn​1​n​

For 222 positive numbers aaa and bbb, upon simplification, HM =21a+1b=2aba+b= \dfrac{2}{\dfrac{1}{a} + \dfrac{1}{b}} = \dfrac{2ab}{a + b}=a1​+b1​2​=a+b2ab​

5.1 Applicability

Let's say for 222 terms, aaa and bbb, a×b=Pa \times b = Pa×b=P

If we need to find the average of a when different scenarios are provided where PPP is equal and aaa is different, we apply Harmonic Mean.
For instance, Speed ×\times× Time === Distance and Price per unit ×\times× Quantity === Revenue

Where Distances are equal, Average Speed is the harmonic mean of the individual speeds. [covered in Time & Speed lesson]

Where Revenues are equal, Average Price per unit is the harmonic mean of the individual price per units.

Example 12

A trader earned Rs. 500500500 by selling at Rs. 555 per unit, Rs. 500500500 by selling at Rs. 101010 per unit and Rs. 500500500 by selling at Rs. 202020 per unit. What was his average price per unit (rounded to 222 decimals)?

Solution

As the revenues are equal here, the average price per unit is the harmonic mean of individual price per units.

Average Price per unit =315+110+120=34+2+120=607=8.57= \dfrac{3}{\dfrac{1}{5} + \dfrac{1}{10} + \dfrac{1}{20}} = \dfrac{3}{\dfrac{4 + 2 + 1}{20}} = \dfrac{60}{7} = 8.57=51​+101​+201​3​=204+2+1​3​=760​=8.57

Note: The revenue of Rs. 500500500 does not matter while calculating the average. Only the fact that the revenues earned were equal matters.

Alternatively

When in doubt you can always apply the general average principle.

Average Price per unit =Total RevenueTotal units sold= \dfrac{\text{Total Revenue}}{\text{Total units sold}}=Total units soldTotal Revenue​

Number of Units sold =5005+50010+50020=100+50+25=175= \dfrac{500}{5} + \dfrac{500}{10} + \dfrac{500}{20} = 100 + 50 + 25 = 175=5500​+10500​+20500​=100+50+25=175

Average Price per unit =500+500+500175=607=8.57= \dfrac{500 + 500 + 500}{175} = \dfrac{60}{7} = 8.57=175500+500+500​=760​=8.57

Answer: 8.578.578.57


5.2 AM, GM and HM

For any set of positive real numbers, AM ≥\ge≥ GM ≥\ge≥ HM. [Remember the terms in alphabetical order]

AM = GM = HM, is only when all the terms are equal. For instance, for the terms in the set {3,3,3,3,33, 3, 3, 3, 33,3,3,3,3}, AM === GM === HM =3= 3=3.

Even if one of the terms is different, like {3,3,3,3,23, 3, 3, 3, 23,3,3,3,2}, AM > GM > HM
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