Question

The Population Of A Village Was 7000.If The Male And Female Increased By 7%And 9% .Then Population Will Become 7522 After 1 Year .Find The Number Of Male At Present

Answer 1

Answer:

Step-by-step explanation:

Let the Number of Males At Present be $x$

So, the Number Of Females At Present = 7000 - $x$

New Male Population = $x$ + 7% $x$

                                    = 107%$x$

                                    = $\frac{107 [tex]x$ }{100}[/tex]

New Female Population = (7000 - $x$ )+ 9%(7000 -  $x$ )

                                        = $\frac{763000- 109x}{100}$

Total Population = Male Population + Female Population

                 7522 =  $\frac{107 [tex]x$ }{100}[/tex] + $\frac{763000- 109x}{100}$

                 7522 = $\frac{76300-2x}{100}$

            752200 = $763000 - 2x$

                     2x = 763000 - 752200

      $2x$ = 10800

       $x$  = 5400

     Number Of Male At Present = $x$  = 5400

Answer 2

The Number Of Males At Present Is 5400.

GIVEN

The population of the village at present = 7000

Increase in male population = 7%

Increase in female population = 9%

New population= 7522

TO FIND

Number of males at present.

SOLUTION

We can simply solve the above problem as follows-

Let the present number of males and females be, x and y

so,

x + y = 7000.................(equation 1 )

After a year, the number of males increases by 7% and the number of females increases by 9%.

So,

Number of males after a year will be -

$x + \frac{7x}{100} = \frac{107x}{100}$

Number of females after a year will be-

$y + \frac{9y}{100} = \frac{109y}{100}$

Also, Given that

$\frac{107x}{100} + \frac{109y}{100} = 7522$

or,

107x + 109y = 752200...............(equation 2)

Equation(1)×109 = 109x + 109y = 763000

Equation(2)× 1 = 107x + 109y = 752200

subtracting Equation(2) from Equation(1)

2x = 10,800

x = 5400

Hence, the population of males at present is 5400.

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