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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

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PRACTICE

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

CAT 2025 Lesson : Averages - Weighted Average

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3. Weighted Average

Weighted average method is applied where we need to find the average of groups of different sizes. Where
x1,x2,...,xnx_{1}, x_{2}, ... , x_{n}x1​,x2​,...,xn​ are the values and w1,w2,...,wnw_{1}, w_{2}, ... , w_{n}w1​,w2​,...,wn​ are the weights, the weighted average is

x‾=w1x1+w2x2+w3x3+...+wnxnw1+w2+w3+...+wn=∑i=1nwixi∑i=1nwi\overline{x} = \dfrac{w_{1} x_{1} + w_{2} x_{2} + w_{3} x_{3} + ... + w_{n} x_{n}}{w_{1} + w_{2} + w_{3} + ... + w_{n}} = \dfrac{\displaystyle\sum_{i=1}^n w_{i} x_{i}}{\displaystyle\sum_{i=1}^n w_{i}}x=w1​+w2​+w3​+...+wn​w1​x1​+w2​x2​+w3​x3​+...+wn​xn​​=i=1∑n​wi​i=1∑n​wi​xi​​

Average is to be found for the values, i.e., the
x\bm{x}x -terms. Weights, i.e., the w\bm{w}w -terms are the number of occurrences of these xxx-terms.

(Note: To find the weighted average of two
xxx-values, we can use the alligation method detailed in the Mixtures & Alligations lesson.)

Example 6

The average of the percentage marks scored by Standard X students in four schools - R, S, T and U are 75,80,6875, 80, 6875,80,68 and 909090 respectively. If number of students in R, S, T and U are 48,60,8448, 60, 8448,60,84 and 969696 respectively, what is the average of the percentage marks scored by students in the 4 schools combined?

Solution

In this question, the averages of the respective schools are provided. However, the number of students (the weights) in each school is different.

Average =48×75+60×80+84×68+96×9048+60+84+96= \dfrac{48 \times 75 + 60 \times 80 + 84 \times 68 + 96 \times 90}{48 + 60 + 84 + 96}=48+60+84+9648×75+60×80+84×68+96×90​

=3600+4800+5712+8640288= \dfrac{3600 + 4800 + 5712 + 8640}{288}=2883600+4800+5712+8640​ =22752288=79= \dfrac{22752}{288} = 79=28822752​=79

Alternatively (Recommended Method)

The impact of weights is always relative to one another.

∴ Weights of
48,60,8448, 60, 8448,60,84 and 969696 can be simplified as a ratio and then applied.

48:60:84:96=4:5:7:848 : 60 : 84 : 96 = 4 : 5 : 7 : 848:60:84:96=4:5:7:8

Average
=4×75+5×80+7×68+8×904+5+7+8=300+400+476+72024=189624=3164=79= \dfrac{4 \times 75 + 5 \times 80 + 7 \times 68 + 8 \times 90}{4 + 5 + 7 + 8} = \dfrac{300 + 400 + 476 + 720}{24} = \dfrac{1896}{24} = \dfrac{316}{4} = 79=4+5+7+84×75+5×80+7×68+8×90​=24300+400+476+720​=241896​=4316​=79

Answer:
797979

Example 7

In a class, 30%30\%30% of the students play 111 sport, 30%30\%30% play 222 sports and 40%40\%40% play 333 sports. What is the average number of sports played by a student?

Solution

We need to find the average for the number of sports, which are the xxx-values. The weights here are the number of students, which are given as percentages.

Average
=0.3×1+0.3×2+0.4×30.3+0.3+0.4=0.3+0.6+1.21=2.1= \dfrac{0.3 \times 1 + 0.3 \times 2 + 0.4 \times 3}{0.3 + 0.3 + 0.4} = \dfrac{0.3 + 0.6 + 1.2}{1} = 2.1=0.3+0.3+0.40.3×1+0.3×2+0.4×3​=10.3+0.6+1.2​=2.1

Note:
100%=1100 \% = 1100%=1. As the sum of percentages equals 100%100 \%100%, going forward, we needn't separately add them up in the denominator.

Answer:
2.12.12.1

Example 8

The average age of 909090 officers in a division is 363636 in 200020002000. In 2004,452004, 452004,45 new officers, whose average age was 313131, joined the division. Besides this, if no other officer left or joined the division between 200020002000 and 201020102010, what was the average age of the officers in the division in 201020102010?

Solution

As each of these officers get 444 years older in 200420042004, their average age will also increase by 444.
∴ Average age of the
909090 officers in 2004=36+4=402004 = 36 + 4 = 402004=36+4=40

Average age of the
454545 officers joining in 2004=312004 = 312004=31

Ratio of weights
=90:45=2:1= 90 : 45 = 2 : 1=90:45=2:1

Average age of all the officers in
2004=2×40+1×312+1=372004 = \dfrac{2 \times 40 + 1 \times 31}{2 + 1} = 372004=2+12×40+1×31​=37

666 years later, the age of each of these officers would have increased by 666.
∴ Average of all the officers in
2010=37+6=432010 = 37 + 6 = 432010=37+6=43

Alternatively

This is not the recommended method. This involves applying simple average, i.e. sum of values by the number of items.

Sum of ages of
909090 officers in 2004=40×90=36002004 = 40 \times 90 = 36002004=40×90=3600

Sum of ages of
404040 officers in 2004=31×45=13952004 = 31 \times 45 = 13952004=31×45=1395

Average age in
2004=3600+139590+45=372004 = \dfrac{3600 + 1395}{90 + 45} = 372004=90+453600+1395​=37

Average age in 2010
=37+6=43= 37 + 6 = 43=37+6=43

Answer:
434343

3.1 Weighted Average using Assumed Mean

This is similar to the assumed mean in the previous section. The difference here is the application of weights.

Where
aaa is the Assumed Mean, xix_{i}xi​ is a value and wiw_{i}wi​ is a weight,

Weighted Average
=a+∑i=1n(xi−a)×wi∑i=1nwi=a+∑i=1ndi×wi∑i=1nwi= a + \dfrac{\displaystyle\sum_{i=1}^n (x_{i} - a) \times w_{i}}{\displaystyle\sum_{i=1}^n w_{i}} = a + \dfrac{\displaystyle\sum_{i=1}^n d_{i} \times w_{i}}{\displaystyle\sum_{i=1}^n w_{i}}=a+i=1∑n​wi​i=1∑n​(xi​−a)×wi​​=a+i=1∑n​wi​i=1∑n​di​×wi​​

Example 9

A group of students were asked by their teacher to count the number of leaves in a tree. 20%,15%,25%,30%20 \%, 15 \%, 25 \%, 30 \%20%,15%,25%,30% and 10%10 \%10% of the students responded 3345,3346,3348,33493345, 3346, 3348, 33493345,3346,3348,3349 and 335033503350 respectively. What was the average number reported by the students?

Solution

Let the assumed average, a=3348a = 3348a=3348

Leaves (xi)\bm{(x_{i})}(xi​) wi\bm{w_{i}}wi​ a di=(xi−a)\bm{d_{i} = (x_{i} - a)}di​=(xi​−a) di×wi\bm{d_{i} \times w_{i}}di​×wi​
334533453345 0.200.200.20 334833483348 −3-3−3 −0.60-0.60−0.60
334633463346 0.150.150.15 334833483348 −2-2−2 −0.30-0.30−0.30
334833483348 0.250.250.25 334833483348 000 000
334933493349 0.300.300.30 334833483348 111 0.300.300.30
335033503350 0.100.100.10 334833483348 222 0.200.200.20
Total −0.40-0.40−0.40


Average
=a+∑i=1ndi×wi∑i=1nwi=3348+−0.41=3347.6= a + \dfrac{\displaystyle\sum_{i=1}^n d_{i} \times w_{i}}{\displaystyle\sum_{i=1}^n w_{i}} = 3348 + \dfrac{-0.4}{1} = 3347.6=a+i=1∑n​wi​i=1∑n​di​×wi​​=3348+1−0.4​=3347.6

Answer:
3347.63347.63347.6
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