Average Questions FOR Bank Exams with Detailed Solutions

​$$\textbf{Given:}\\n=14,\ \bar{x}=72.5\\\text{Average of first }8\text{ numbers}=80\\\text{Average of last }4\text{ numbers}=62.5\\\text{Difference between 9th and 10th numbers}=5\\[6pt]\textbf{Concept Used:}\\\text{Sum}=\text{Average}\times \text{Count}\\\text{The 9th and 10th numbers are the only two not in the first 8 or last 4.}\\[6pt]\textbf{Formula Used:}\\S=n\cdot \bar{x}\\a+b=S_{\text{middle}},\quad a-b=d\\[6pt]\textbf{Solution:}\\S_{14}=14\cdot 72.5=1015\\S_{8}=8\cdot 80=640\\S_{4}=4\cdot 62.5=250\\S_{9,10}=1015-640-250=125\\\text{Let 9th}=a,\ \text{10th}=b\\a+b=125,\ a-b=5\\(a+b)+(a-b)=125+5\\2a=130\\a=65\\[6pt]\textbf{Final Answer:}\\\boxed{65}$$​​

Given: 

Average weight of group of 12 persons = 40 kg

Formula Used:

Average = $$\frac{total \ weight}{Number \ of \ people}$$​

Explanation:

Required average weight = $$\frac{(12 \times 40) + 53 - 65}{12 + 1 - 1} = \frac{468}{12} = 39 \, \text{kg}$$​​

​$$\textbf{Given:}\\A:B=6:5\\B=75\%\text{ of }C\\\frac{A+B+C}{3}=106\text{ kg}\\[6pt]\textbf{Concept Used:}\\\text{Use ratios to express weights in terms of a variable and convert average to sum.}\\[6pt]\textbf{Formula Used:}\\A:B=6:5 \Rightarrow A=6k,\ B=5k\\75\%=\frac{75}{100}=\frac{3}{4}\\\text{Average}=\frac{A+B+C}{3}\\[6pt]\textbf{Solution:}\\A:B=6:5 \Rightarrow A=6k,\ B=5k\\B=\frac{3}{4}C \Rightarrow C=\frac{4}{3}B\\C=\frac{4}{3}\cdot 5k=\frac{20k}{3}\\\frac{A+B+C}{3}=106 \Rightarrow A+B+C=318\\6k+5k+\frac{20k}{3}=318\\11k+\frac{20k}{3}=318\\\frac{33k+20k}{3}=318\\\frac{53k}{3}=318\\53k=954\\k=\frac{954}{53}=18\\A=6k=6\cdot 18=108\\C=\frac{20k}{3}=\frac{20\cdot 18}{3}=120\\C-A=120-108=12\\[6pt]\textbf{Final Answer:}\\\boxed{12\text{ kg}}$$​​

​$$\textbf{Given:}\\A=B+8\\C=D+12\\D=B+12\\D=60\text{ kg}\\[6pt]\textbf{Concept Used:}\\\text{Use linear relations to find unknown weights and then compute average.}\\[6pt]\textbf{Formula Used:}\\\text{Average}=\frac{B+C+D}{3}\\[6pt]\textbf{Solution:}\\D=B+12\\60=B+12\\B=48\\C=D+12=60+12=72\\\text{Average}=\frac{48+72+60}{3}\\=\frac{180}{3}=60\\[6pt]\textbf{Final Answer:}\\\boxed{60\text{ kg}}$$

Exam Hall Method:

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​$$\textbf{Given}\\n=7\\\text{Average of 7 students}=47\text{ kg}\\\text{Average of first 4 students}=36\text{ kg}\\\text{Average of last 4 students}=55.25\text{ kg}\\[6pt]\textbf{Concept Used}\\\text{Sum}=\text{Average}\times \text{Number},\;\text{and the 4th student is common in both groups.}\\[6pt]\textbf{Formula Used}\\S=\text{Average}\times n\\W_4=(S_{1-4}+S_{4-7})-S_{1-7}\\[6pt]\textbf{Solution}\\S_{1-7}=47\times 7=329\\S_{1-4}=36\times 4=144\\S_{4-7}=55.25\times 4=221\\W_4=(144+221)-329\\W_4=365-329=36\\[6pt]\textbf{Final Answer}\\\boxed{36\text{ kg}}$$​​

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$$\textbf{Given}\\\text{Average of }8\text{ numbers}=39.5\\\text{After adding 2 numbers, total}=10\\\text{Average decreases by }2.5\\\text{Ratio of new numbers}=2:1\\[6pt]\textbf{Concept Used}\\\text{Sum}=\text{Average}\times \text{Number of terms}\\[6pt]\textbf{Formula Used}\\S=n\times \text{Average}\\S_{\text{new}}=S_{10}-S_8\\[6pt]\textbf{Solution}\\S_8=8\times 39.5=316\\\text{New average}=39.5-2.5=37\\S_{10}=10\times 37=370\\S_{\text{new}}=370-316=54\\\text{Let new numbers be }2k\text{ and }k\\2k+k=54\Rightarrow 3k=54\Rightarrow k=18\\\text{Greater number}=2k=2\times 18=36\\[6pt]\textbf{Final Answer}\\\boxed{36}$$

Exam Hall Method: 

​$$\textbf{Given:}\\W_{\text{old}}=800\text{ kg}\\\text{One student of }30\text{ kg is removed}\\\text{One teacher of }70\text{ kg is added}\\\Delta \bar{w}=15\dfrac{5}{8}\%\text{ of initial number of students}\\[6pt]\textbf{Concept Used:}\\\bar{w}=\frac{W}{n}\\[6pt]\textbf{Formula Used:}\\\bar{w}=\frac{W}{n},\qquad \Delta \bar{w}=\bar{w}_{\text{new}}-\bar{w}_{\text{old}}\\15\dfrac{5}{8}\%=\frac{125}{8}\cdot\frac{1}{100}=\frac{125}{800}=\frac{5}{32}\\[6pt]\textbf{Solution:}\\\text{Let initial number of students be }n.\\\bar{w}_{\text{old}}=\frac{800}{n}\\W_{\text{new}}=800-30+70=840\\n_{\text{new}}=n-1+1=n\\\bar{w}_{\text{new}}=\frac{840}{n}\\\Delta \bar{w}=\frac{840}{n}-\frac{800}{n}=\frac{40}{n}\\\frac{40}{n}=\frac{5}{32}n\\40\cdot 32=5n^2\\1280=5n^2\\n^2=256\\n=16\\\bar{w}_{\text{old}}=\frac{800}{16}=50\text{ kg}\\[6pt]\textbf{Final Answer:}\\\boxed{50\text{ kg}}$$​​

Information Given in the Question:
Average weight of A, B, C, D, E = 73 kg
Number of persons = 5
So, total weight of A to E = 73 × 5 = 365 kg
Person F’s weight is 25 kg less than the weight of A
We don’t know the exact weight of A
Asked: New average weight of 6 persons (A to F)
Concept/Formula Used in the Question:
Total Weight = Average × Number of Persons
New Total = Old Total + F’s weight
New Average = New Total / 6
But since F's weight = A - 25, and A’s weight is unknown, we can’t find a numeric value
Detailed Explanation:
Let A’s weight be x kg

Then F’s weight = x - 25 kg

Total of all 6 = 365 (from A to E) + x – 25 = x + 340

New average =$$\frac{x + 340}{6}$$​

Since x is unknown, exact numerical average cannot be calculated

Exam Hall Method:

Formula Used:
Total or Sum = Average × total Numbers
Explanation:
Sum of 5 number = 391.4 × 5 = 1957
Sum of first 3 numbers = 329 × 3 = 987
Sum of last 3 numbers = 395 × 3 = 1185
∴ Third number = (987+ 1185) – 1957 = 215
Sum of first and 3rd number = 196.5 × 2 = 393
First number = 393 – 215 = 178

Exam Hall Approach

Let five consecutive odd number series be x, x + 2, x + 4, x + 6, x + 8
Then, five consecutive even number will be x – 9, x – 7, x – 5, x – 3, x – 1
Average of odd number series = (x + 4)
Avg. of even number series = (x – 5)
Required difference = x + 4 – (x – 5) = 9

Information Given:
Average age of x persons = 60 years
Two persons of 52 and 68 years leave the group
Two new persons of y years and 72 years join the group
Average age increases by 1 year (new average = 61 years)
x is a perfect square
54 < y < 64
Asked: Find the value of y

Formula Used:
Initial total age = 60 × x
New total age = (60 × x) - 52 - 68 + y + 72
New average = (New total age) / x = 61

Explanation: 

Total age of x persons initially = 60 x years.
Total age of x persons finally
= 60x - 52 - 68 + y + 72 = 60x + y - 48
= 60x + y - 48 = 61x
y-48=x
As 54 < y < 64, 6 < x < 16
As x is a perfect square, x = 9 and y = 57