A box contains red and blue balls in the ratio 5 : 3. If 16 random balls are replaced by 16 balls of blue colour, then the ratio of the number of balls of red and blue colour will be 3 : 5. How many balls of blue colour were there in the box initially?
A box contains red and blue balls in the ratio 5 : 3. If 16 random balls are replaced by 16 balls of blue colour, then the ratio of the number of balls of red and blue colour will be 3 : 5. How many balls of blue colour were there in the box initially?
1). 12
2). 16
3). 15
4). 18
1 answers
Let the number of red and blue balls in the box be 5x and 3x respectively.
Given that 16 balls are removed from the box and replaced with 16 other blue balls
⇒ Number of red balls in the box left after this process = 5x – (5/8 × 16)
⇒ 5x – 10
⇒ Number of blue balls in the box left after this process = 3x – (3/8 × 16)
⇒ 3x – 6
Since 16 additional balls were added to the box,
⇒ Total number of blue balls in the box will be (3x – 6 + 16)
⇒ 3x + 10
Given that the ratio of number of red and blue balls after this process = 3/5
⇒ (5x – 10)/(3x + 10) = 3/5
⇒ (5x – 10) × 5 = (3x + 10) × 3
⇒ 25x – 50 = 9x + 30
⇒ 16x = 80
⇒ x = 5
∴ Initial number of blue balls present in the box initially = 3 × 5 = 15