Free Practice Questions for Average in Maths

Let the average of the remaining $2$ numbers is $x$,
$
\begin{aligned}
&9 \times 51=3 \times 49+4 \times 46+2 \times x \\
&2 x=459-147-184 \\
&2 x=459-331 \\
&2 x=128 \\
&x=64
\end{aligned}
$
Hence, Average of the remaining $2$ numbers is $64$.

Average of present age of 4 members $=19+4=23$ years

Total present age of 4 members $=4 \times 23=92$ years
Average present age of 5 members $=19$ years
Total present age of 5 members $=5 \times 19=95$ years
Present age of baby $=95-92=3$ years

According to question,
$
\begin{aligned}
&\Rightarrow \frac{a+b+c}{3}=63 \\
&\Rightarrow a+b+c=63 \times 3=189 \\
&\Rightarrow b+c=189-\mathrm{a}
\end{aligned}
$
Given, $\mathrm{a}=\frac{3}{4}(b+c)$
$
\begin{aligned}
&\Rightarrow \mathrm{a}=\frac{3}{4}(189-a) \\
&\Rightarrow 4 \mathrm{a}=567-3 a \\
&\Rightarrow \mathrm{a}=\frac{567}{7}=81
\end{aligned}
$

$\frac{8+5+3+k+k+1}{5}=7$
$2 k+17=35$
$2 k=18$
$k=9$

Total price of $10$ books $=10 \times 150=1500$
Total price of $8$ books $=8 \times 135=1080$
So, Total price of the remaining $2$ books $=1500-1080=420$
Let the cost price of the 1st book is $100 x$, then
Cost price of the 2nd book $=\frac{140}{100} \times 100 x=140 x$
According to question
$100 x+140 x=420$
$240 x=420$
$x=\frac{420}{240}=\frac{7}{4}$
Hence, Cost price of the costlier book $=140 \times \frac{7}{4}=35 \times 7=245$

ATQ,
Average of $19$ numbers $=58$
Total sum of all the $19$ numbers $=58 \times 19=1102$
Average of the first $12$ numbers $=63.6$
Total sum of the first $12$ numbers $=63.6 \times 12=763.2$
Average of the last $8$ numbers $=55.1$
Total sum of last $8$ numbers $=55.1 \times 8=440.8$
So, $12$th number $=$ Total sum of the first $12$ numbers $+$ Total sum of last $8$ numbers $-$ Total sum of all the $19$ numbers
$12$th number $=763.2+440.8-1102=102$
Average of all numbers excluding 12th number $=\frac{1102-102}{18}=55 \frac{5}{9}$

Average score of $35$ students in class $=46$
Three students marks read as $=2\times33+30=96$
Actual marks of these three students $=2\times23+36=82$
So, these three students get extra marks $=96-82=14$ marks
Now, average of all the students $=46-\frac{14}{35}=46-0.4=45.6$

There will be 5 sunday in the month starting from sunday and 4 saturday.
As per the question,
Let the average of visitors for month $=475$ visitors
So total increase on sundays $=5 \times(502-475)=135$ visitors
And total decrease on other days $=21 \times(475-340)=135 \times 21=2835$ Visitors
Total increase $/$ decrease $=135-2835=-2700$
Average number of visitors per day in the month $=475-\frac{2700}{30}=475-90=385$

Let the total number of persons is $n$,
After 4 persons leave, the number of remaining persons $=n-4$
Number of remaining persons

 $=\frac{\text { total increase or decrease in weight due to leaving of some persons × number of person who left the group }}{\text { average increase or decrease in the weight of the remaining persons }}$
$n-4=\frac{4 \times(80.75-78)}{78-77.725}=\frac{4 \times 2.75}{0.275}=4 \times 10=40$
Hence, $n=40+4=44$

Average weight of 14 boys in a class $=40.25 \mathrm{~kg}$
Average weight of remaining 10 boys in a class $=35.15 \mathrm{~kg}$
Average weight of all the boys in the class $=\frac{40.25 \times 14+35.15 \times 10}{24}=\frac{915}{24}=38.125 \mathrm{~kg}$

ATQ,
Average run rate for a cricketer in 48 innings $=33$ runs
Average run rate for a cricketer in 46 innings $=30$ runs
So, all 46 innings runs less than the average runs of all innings by $=(33-30) \times 46 = 138$ runs
i.e. 138 runs extra will be added in 47 th and 48 th innings,
Total runs of 47 th and 48 th innings $=33 \times 2+138=204$ runs
If highest score more than 144 runs by lowest then,
Lowest score $=\frac{204-144}{2}=30$ runs and Highest score $=204-30=174$ runs

Given that 6 unit = 7 employees

then 12 unit = 14 employees

So the number of total employees is 14.

Alternative method

Let number of Employees = x 

Average Salary of Employee = 18000

Total salary of all Employees =18000x

Average salary of 7 Employees =24000

Total salary of 7 Employees = 24000 7 = 168000

Total salary of the rest of the employees = 12000(x−7) 

18000x=168000 + 12000(x−7)

18000x $-$12000x= 168000 $-$ 84000

6000x = 84000

x=14 

Sum of all the 51 numbers $=51\times52 = 2652$

Sum of the first 28 numbers $=28\times50.5 = 1414$

Sum of the last 24 numbers $=24\times54.5 = 1308$

As the 28th number is counted in both , So

28th number $=(1414+1308)-2652 = 2722-2652 = 70$

So when 28th number is dropped, it will decrease the sum of 50 numbers by $(70-52 = 18)$

So Average of remaining 50 numbers $ = 52 - \frac{18}{50} = 52-0.36 = 51.64 $ 

Given:

Total score = 85%

The scores were 79, 81, 88 and 94 in 4 subjects

Concept Used:

Percentage is used to solve.

Calculation:

Let the marks of 5th subject be 'x'.

Sum of marks of 4 subject = 79 + 81 + 88 + 94 = 342

Total marks of 5 subject = x + 342 

Full marks of each subject = 100

Full marks of total subject = 500

According to the question,

(x + 342) = 85% of 500

⇒ x + 342 = (85 × 500)/100

⇒ x = 425 – 342

⇒ x = 83

∴ The score in 5th subject is 83.

Let the four consecutive even natural numbers are $(x-2),(x),(x+2)$ and $(x+4)$.
Now, according to question -
$
\Rightarrow(x-2)^{2}+(x)^{2}+(x+2)^{2}+(x+4)^{2}=126 \times 4
$
$
\Rightarrow x^{2}+4-4 x+x^{2}+x^{2}+4+4 x+x^{2}+16+8 x=504
$
$
\Rightarrow 4 x^{2}+8 x+24=504
$
$
\Rightarrow 4 x^{2}+8 x-480=0
$
$
\begin{aligned}
&\Rightarrow x^{2}+2 x-120=0 \\
&\Rightarrow x^{2}+12 x-10 x-120=0 \\
&\Rightarrow x(x+12)-10(x+12)=0 \\
&\Rightarrow(x+12)(x-10)=0 \\
&\Rightarrow x=-12 \text { or } 10
\end{aligned}
$
Therefore, four consecutive even natural numbers are $(10-2),(10),(10+2)$ and $(10+4) = 8, 10, 12, 14$.
Required, value of $\frac{(8 \times 14+5 \times 8)}{2}=\frac{112+40}{2}=\frac{152}{2}=76$

Total expenditure of first three months = Average expenditure of first three months $\times 3$
$
\begin{aligned}
&=2400 \times 3 \\
&=7200
\end{aligned}
$
Total expenditure of next five months = Average expenditure of next five months $\times 5$
$
\begin{aligned}
&=3500 \times 5 \\
&=17500
\end{aligned}
$
Total expenditure of remaining four months = Average expenditure of remaining four months $\times 4$
$
\begin{aligned}
&=4800 \times 4 \\
&=19200
\end{aligned}
$
Total expenditure of whole year $=7200+17500+19200=43900$
Total saving during the entire year $=3500$
So total income of a man for whole year $=$ expenditure $+$ saving
$
\begin{aligned}
&=43900+3500 \\
&=47400
\end{aligned}
$
Therefore, required monthly saving $=\frac{47400}{12}=\mathrm{Rs}$. 3950 

Correct average $=70+\frac{(64-54)+(84-74)}{20}$
$=70+\frac{20}{20}=71$

$\begin{aligned} \text { Correct Average } &=75+\frac{(65-56)+(69-38)}{40} \\ &=75+\frac{9+31}{40} \\ &=75+1=76 \end{aligned}$

Given:

A grocer has a sale of Rs. 2,000, Rs. 2,500, Rs. 3,250 and Rs. 4,250 in 4 months.

An average sale of Rs. 3250 is to be achieved in 5 months.

Concept used:

Average, which is the arithmetic mean, is calculated by adding a group of numbers and then dividing by the count of those numbers.

Calculation:

Let the grocer makes a sale of Rs. X in the 5th month.

According to the question,

(2000 + 2500 + 3250 + 4250 + X) ÷ 5 = 3500

⇒ X = 5500

∴ Rs. 5500 should be his sale in the 5th month to get an overall average of Rs. 3,500.

Given:
$
\frac{A+B+C}{3}=78
$
$\Rightarrow A+B+C=234 \ldots . . \mathrm (1)$
$
\frac{C+D+E}{3}=52
$
$\Rightarrow C+D+E=156 \ldots \ldots$ (2)
$
\frac{E+F}{2}=48
$
$
\Rightarrow E+F=96 \ldots . . (3)
$
$
\begin{aligned}
&\frac{E+C}{2}=60 \\
&\Rightarrow E+C=120 \ldots . .( 4)
\end{aligned}
$
adding on (1),(2),(3) and (4)

Now required, average of $A, B, C, D, E$ and $F =\frac{234+156+96-120}{6}=\frac{366}{6}=61$
Hence, option D is correct.