CAT 2025 Lesson : Averages - Weighted Average

The average of the percentage marks scored by Standard X students in four schools - R, S, T and U are $75, 80, 68$ and $90$ respectively. If number of students in R, S, T and U are $48, 60, 84$ and $96$ 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 $= \dfrac{48 \times 75 + 60 \times 80 + 84 \times 68 + 96 \times 90}{48 + 60 + 84 + 96}$

$= \dfrac{3600 + 4800 + 5712 + 8640}{288}$ $= \dfrac{22752}{288} = 79$

Alternatively (Recommended Method)

The impact of weights is always relative to one another.

∴ Weights of $48, 60, 84$ and $96$ can be simplified as a ratio and then applied.

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

Average $= \dfrac{4 \times 75 + 5 \times 80 + 7 \times 68 + 8 \times 90}{4 + 5 + 7 + 8} = \dfrac{300 + 400 + 476 + 720}{24} = \dfrac{1896}{24} = \dfrac{316}{4} = 79$

Answer: $79$

In a class, $30\%$ of the students play $1$ sport, $30\%$ play $2$ sports and $40\%$ play $3$ 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 $x$-values. The weights here are the number of students, which are given as percentages.

Average $= \dfrac{0.3 \times 1 + 0.3 \times 2 + 0.4 \times 3}{0.3 + 0.3 + 0.4} = \dfrac{0.3 + 0.6 + 1.2}{1} = 2.1$

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

Answer: $2.1$

The average age of $90$ officers in a division is $36$ in $2000$. In $2004, 45$ new officers, whose average age was $31$, joined the division. Besides this, if no other officer left or joined the division between $2000$ and $2010$, what was the average age of the officers in the division in $2010$?

Solution

As each of these officers get $4$ years older in $2004$, their average age will also increase by $4$.
∴ Average age of the $90$ officers in $2004 = 36 + 4 = 40$

Average age of the $45$ officers joining in $2004 = 31$

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

Average age of all the officers in $2004 = \dfrac{2 \times 40 + 1 \times 31}{2 + 1} = 37$

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

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 $90$ officers in $2004 = 40 \times 90 = 3600$

Sum of ages of $40$ officers in $2004 = 31 \times 45 = 1395$

Average age in $2004 = \dfrac{3600 + 1395}{90 + 45} = 37$

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

Answer: $43$

A group of students were asked by their teacher to count the number of leaves in a tree. $20 \%, 15 \%, 25 \%, 30 \%$ and $10 \%$ of the students responded $3345, 3346, 3348, 3349$ and $3350$ respectively. What was the average number reported by the students?

Solution

Let the assumed average, $a = 3348$

Leaves $\bm{(x_{i})}$	$\bm{w_{i}}$	a	$\bm{d_{i} = (x_{i} - a)}$	$\bm{d_{i} \times w_{i}}$
$3345$	$0.20$	$3348$	$-3$	$-0.60$
$3346$	$0.15$	$3348$	$-2$	$-0.30$
$3348$	$0.25$	$3348$	$0$	$0$
$3349$	$0.30$	$3348$	$1$	$0.30$
$3350$	$0.10$	$3348$	$2$	$0.20$
Total				$-0.40$

Average $= a + \dfrac{\displaystyle\sum_{i=1}^n d_{i} \times w_{i}}{\displaystyle\sum_{i=1}^n w_{i}} = 3348 + \dfrac{-0.4}{1} = 3347.6$

Answer: $3347.6$