Maths 5 Marks Questions

$\Rightarrow(\text{x}_1-15)+(\text{x}_2-15)+\dots+(\text{x}_\text{n}-15)=-90$

$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})\\-(15+15+15+\dots+15)=-90$

$\Rightarrow\sum\text{x}-\text{15n}=-90\dots(1)$

And $\sum\limits^{\text{i}-1}_{\text{n}}(\text{x}_\text{i}+3)=54$

$\Rightarrow(\text{x}_1+3)+(\text{x}_2+3)+\dots+(\text{x}_\text{n}+3)=54$

$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})-(3+3+3+\dots+3)=54$

$\Rightarrow\sum\text{x}+\text{3n}=54\dots(2)$

By subtracting equation (1) from equation (2), we get

$\sum\text{x}+\text{3n}-\sum\text{x}+\text{15n}=54+90$

$\Rightarrow\text{18n}=144$

$\Rightarrow\text{n}=\frac{144}{18}=8$

Put value of n in equation (1)

$\sum\text{x}-15\times8=-90$

$\Rightarrow\sum\text{x}-120=-90$

$\Rightarrow\sum\text{x}=120-90=30$

$\therefore\text{x}=\frac{\sum\text{x}}{\text{n}}=\frac{30}{8}=3.75$

$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$

$=\frac{ 3+ 4+5}{3}$

$=4$

1. Adding constant term k = 2 in each term.

New numbers are = 5, 6, 7

$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$

$=\frac{ 5+ 6+7}{3}$

$\therefore$ New mean will be 2 more than the original mean.

2. Subtracting constant term k = 2 in each term.

New numbers are = 1, 2, 3

$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$

$=\frac{ 1+ 2+3}{3}$

$\therefore$ New mean will be 2 less than the original mean.

3. Multiplying by constant term k = 2 in each term.

New numbers are = 6, 8, 10

$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$

$=\frac{ 6+ 8+10}{3}$

$=8=4\times2$

$\therefore$ New mean will be 2 times of the original mean.

4. Divide the constant term k = 2 in each term.

New numbers are = 1.5, 2, 2.5.

$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$

$=\frac{ 1.5+ 2+2.5}{3}$

$=2=\frac{4}{2}$

$\therefore$ New mean will be half of the original mean.