Question

the population of a town increased by 6% every year if the present population is 15900, then find its population a year ago ​

Answer 1

Answer:

Population increases =6%

x + 6x/100 = 15900

106x/100 = 15900

x = 15900*100/106

x = 15000

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• The population of a town increased by 6% every year if the present population is 15900, then find its population a year ago

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

• Let the Population of the Town be 100 a year ago, Then,
• Increase in the Population = 6% of 100 ${ \boxed{\: \frac{6}{100} \: \times \: 100 \: }}$ = 6

$\therefore \bf \: Present \: population \: = \: 106$

If the Present Population is 106, Population a year ago = 100

If the Present Population is 1 , Population a year ago = $\frac{100}{106}$

If Present Population is 15900, Population a year ago = ${ \boxed{\: \frac{100}{106} \: \times \:15900 \: }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\frac{100}{106} \: \times \:15900 \: }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\frac{100}{ \cancel \green{106}} \: \times \: \cancel \green{15900 }\: }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{100 \: \times \: 150 }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{Population \: = \: 15000 }}}$

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$\begin{gathered}\begin{gathered} \therefore\: \sf{ The \: population \: of \: the \: town \: a \: year \: ago \: was \: = \bf\underline {\underline { 15000}}}\\\end{gathered}\end{gathered}$

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More For Knowledge:-

$\underbrace\red{\boxed{ \sf \blue{ Important \: Rules \:For \: Conversion \: }}}$

• To convert fraction into a percentage we multiply of fraction by 100.
• To convert a ratio into a percentage we write it as a fraction and multiply it by 100.
• To convert a decimal into a percentage we shift the decimal point two places to the right.
• To convert a percentage into a fraction we drop persent sign ( % ) and divide the remainder by 100.

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