Free Practice Questions for Maths Subject, Attempt Now!

Let seven consecutive positive integers be $x, x+1, x+2, x+3, x+4, x+5$, and $x+6$ and next five odd numbers are $x+7, x+9, x+11, x+13$, and $x+15$
Sum of seven consecutive integers = $7 \times$ Average
$
\begin{aligned}
(x+x+1+x+2+x+3+x+4+x+5+x+6) &=259 \times 7 \\
7 x+21 &=259 \times 7 \\
x+3 &=259 \\
x &=256
\end{aligned}
$
Now, sum of all 12 integers $=x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+9+x+11+x+13+x+15$ $=12 x+76$
Required average $=\frac{12 x+76}{12}$
$
\begin{aligned}
&=x+\frac{76}{12} \\
&=256+6.33=262.33
\end{aligned}
$

Total age of 40 students $=12 \times 40=480$ years
Total age of 40 students and their teacher $=13 \times 41=533$ years
So, Age of their teacher $=533-480=53$ years
Hence, The teacher will retire after $(55-53)=2$ years.

If the greatest number is excluded from a few consecutive numbers, the average decreases by $0.5$.
For Example : Let consecutive numbers are 1,2 and 3 , then
Average $=\frac{1+2+3}{3}=\frac{6}{3}=2$
and If we exclude 3 from it then
Average $=\frac{1+2}{2}=\frac{3}{2}=1.5$
The average decreases by $0.5$
Hence, If we exclude two greatest numbers from 11 consecutive numbers, the average will decrease by 1 .

$\begin{aligned} \text { The correct average } &=46+\frac{142-42}{10} \\ &=46+\frac{100}{10} \\ &=46+10=56 \end{aligned}$

We know,
$
\begin{aligned}
&\text { Average }=\frac{\text { Sum of all the observations }}{\text { total observations }} \\
&\begin{array}{l}
12.5=\frac{x+2+x+5+x+8+x+11+x+14+x+17}{6} \\
75=6 x+57 \\
x=3
\end{array}
\end{aligned}
$
Average of first three observations,
Average $=\frac{5+8+11}{3}=8$

Let the fifth number be $x$.
As per the question,

$\text { Average }=\frac{\text { Sum of all the observations }}{\text { total observations }}$ (i)

$\frac{5}{6} x=\frac{\text { Sum of first four numbers }}{4}$ (ii)

put the the value of (ii) in (i) equation.

$26=\frac{x+\frac{20}{6} x}{5}$

$26 \times 5=\frac{26 x}{6} $

$x=30$

Let the total number of children is 3 , then

$66 \frac{2}{3} \%$ of $3=\frac{200}{300} \times 3=2$
Let the average of the remaining children is $x$
According to question,
$
\begin{aligned}
&2 \times 13+x=3 \times 14.5 \\
&x=43.5-26 \\
&x=17.5
\end{aligned}
$
Hence, the average of the remaining children is $17.5$.

Current average $=15+1=16$
Age of teacher $=$ Current average $+$ No. of students $\times$ Increase in average
Age of teacher $=16+26 \times 1=16+26=42$ years

As per the question,
$
\begin{aligned}
&7=\frac{4+6+8+12+x}{5} \\
&35=30+x \\
&x=5
\end{aligned}
$
Now,
$
\begin{aligned}
&9=\frac{5+9+13+15+y}{5} \\
&45=42+y \\
&y=3
\end{aligned}
$
So, $2 x-3 y=10-9=1$

As per the question,
Average weight of the students of section $\mathrm{A}=55 \mathrm{~kg}$
Total sum of weight of the student of section $A=55 \times 38=2090 \mathrm{~kg}$
Average weight of the students of section $B=32 \mathrm{~kg}$
Total sum of weight of the student of section $B=32 \times 42=1344 \mathrm{~kg}$
So, average weight of all the students in the class $=\frac{2090+1344}{80}$
$=42.925 \mathrm{~kg}$

Total present age of family $=10 \times 29=290$
Total age of family when the twins born $=290-5 \times 10=240$
Hence, Average age of family when twins were born $=\frac{240}{8}=30$

Total change in weight $=5(58-56)=10 \mathrm{~kg}$
Let the number of students originally was $\mathrm{n}$,
Number of students after increase $=\frac{\text { Total change }}{\text { Change in average }}$
$
\begin{aligned}
&n+5=\frac{10}{0.5} \\
&n+5=20 \\
&n=15
\end{aligned}
$

As per the question,
Average of 19 numbers $=46$
Sum of all the numbers $=46 \times 19=874$
Average of the first 5 numbers $=49$
Sum of the first 5 numbers $=49 \times 5 = 245$
Average of the next 6 numbers $=54$
Sum of the next 6 numbers $=54 \times 6 = 324$
Sum of the remaining numbers $=874-245-324 =305$
Average of the remaining numbers $=\frac{305}{8}=38.1$

Numbers between 50 and 7 which are divisible by 6 =12,18,24,30,36,42 and 48
Average $=\frac{a+l}{2}=\frac{12+48}{2} \\$
$=\frac{60}{2}=30$
Prime Numbers between 10 and $45=11,13,17,19,23,29,31, 37, 41$ and 43
Average $=\frac{11+13+17+19+23+29+31+37+41+43}{10}$
$=\frac{264}{10}=26.4$
Required difference $=30-26.4=3.6$

The sum of 12 numbers $=38 \times 12=456$
Sum of the first three numbers $=33 \times 3=99$
Sum of the next four numbers $=\frac{71}{2} \times 4=142$
Sum of the next two numbers $=40 \times 2=80$
The total sum of 9 numbers $=99+142+80=321$
Sum of the remaining three numbers $=456-321=135$
Now let the tenth numbers be $x$
So, 11 th number is $=x+4$
12 th number is $=x+5$
According to question
$x+x+4+x+5=135$
$3 x=126$
$x=\frac{126}{3}=42$
12th number $=x+5=42+5=47$

Let $x,(x+2),(x+4),(x+6),(x+8),(x+10),(x+12)$ be the seven odd numbers.
According to question
Average $\quad=\frac{x+(x+2)+(x+4)+(x+6)+(x+8)+(x+10)+(x+12)}{7}$
$
\begin{array}{cc}
79 & =\frac{7 x+42}{7} \\
7 x & =79 \times 7-42 \\
7 x & =553-42 \\
x & =\frac{511}{7}=73
\end{array}
$

Number are : $73,75,77,79,81,83,85$
Average of first five odd number $=\frac{73+75+77+79+81}{5}$
$
=\frac{385}{5}=77
$

Number of Sundays $=4$
Remaining days $=26$
Required Average $=\frac{4 \times 240+26 \times 150}{30}$
$
\begin{aligned}
&=\frac{960+3900}{30} \\
&=32+130=162
\end{aligned}
$

The numbers which are divisible by 3 are $15,18,21,24,27,30,33,36,39,42$
Required average $=\frac{a+l}{2}=\frac{15+42}{2}=28.5$

Let the total number of students in a group be $\mathrm{x}$.
Total score of group $=64 \mathrm{x}$
Total score of $20 \%$ students $=\mathrm{x} \times \frac{20}{100} \times 91=\frac{91 x}{5}$
Total score of $25 \%$ students $=\mathrm{x} \times \frac{25}{100} \times 31=\frac{31 x}{4}$
Total score of $55 \%$ students $=64 \mathrm{x}-\left(\frac{91 x}{5}+\frac{31 x}{4}\right)=38.3 \mathrm{x}$
Average score of the remaining students $=\frac{38.3 x}{\frac{55 x}{100}}=\frac{3830}{55}=69.2$

Required difference = 4:3