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Alia and Huda share some sweets in the ratio 7:3. Alia gives 3 sweets to Huda and
now the ratio is 5:3. How many sweets did each have initially?

Respuesta :

BrainLifting

Answer:

Alia=28

Huda=12

Step-by-step explanation:

Let the constant of proportionality be x

Hence, no. of sweets Alia had initially = 7x

No. of sweets Huda had initially = 3x

No. of sweets Alia has now= 7x-3

No. of sweets Huda has now= 3x+3

We know that, the ratio of the no. of sweets Alia has now to no. of sweets Huda has now = 5:3

Hence,

[tex](7x-3) : (3x+3)=5:3\\By\ equalizing\ the\ ratios,\ we\ get,\\ 3(7x-3)=5(3x+3)\\21x-9=15x+15\\21x-15x=15+9\\6x=24\\x=4[/tex]

As we now got the constant of proportionality to be 4,

No. of sweets Alia had initially = 7*4=28

No. of sweets Huda had initially= 3*4=12

quasarJose

Given parameters:

Original ratio number of sweets of Alia to Huda = 7:3

Final ratio  = 5:3

Alia gives 3 sweets

Huda receives 3 sweets

To solve this problem, we have to derive an algebraic equation.

Let us assume that the total number of sweets = x

  So;

  The ratio originally is;

                               7x : 3x

Alia has 7x sweets

Huda has 3x

Now Alia has (7x -3)sweets after giving out 3

        Huda has (3x + 3)sweets after receiving 3

So;

           7x -3  :  3x + 3  = 5 : 3

                    [tex]\frac{7x - 3}{3x + 3} = \frac{5}{3}[/tex]

       3(7x -3) = 5(3x + 3)

       21x - 9  = 15x + 15

       21x -15x = 15 + 9

          6x  = 24

            x = 4

Initially, Alia has 7x sweets  = 7 x 4  = 28 sweets

             Huda has 3x sweets = 3 x 4  = 12 sweets

Therefore, Alia has 28 sweets and Huda 12 sweets initially.

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