Question

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?

Answer 1

Alia=28

Huda=12

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,

`(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`

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

Answer 2

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

`(7x - 3)/(3x + 3) = (5)/(3)`

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.