Question

A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of 24. If three times the number of red balls exceeds the number of blue balls by 20, find the number of red and blue balls.

Answer 1

We are given that a bag contains some red and blue balls.

10% of red balls + 20% of blue balls = 24

3 times the number of red balls exceeds the number of blue balls by 20

We need to find the number of red and blue balls.

Let:

x = number of red balls

y = number of blue balls

Step 1:

Equation 1: Percentage Condition

10% of red balls = `10/100x = x/10`

20% of blue balls = `20/100y = y/5`

`x/10 + y/5 = 24`

Multiply the whole equation by 10 to eliminate fractions,

x + 2y = 240

Equation 2: Given Condition on Red and Blue Balls

3x = y + 20

3x − y = 20   ..... [Rearrange it]

Step 2: Solve the Equations

We now have the system of equations,

x + 2y = 240

3x − y = 203

Multiply Equation (2) by 2 to align y terms,

6x − 2y = 40

Add both equations:

(x + 2y) + (6x − 2y) = 240 + 40

7x = 280

x = 40

Substituting x = 40 into Equation (2),

3(40) − y = 20

120 − y = 20

y = 100