A box contains some balls of black, blue and red colours in the ratio ...
Let the number of black, blue and red coloured balls in the box be 3x, 8x and 7x respectively.
According to the question, 7x – 3x = multiple of 12 and 9
The multiple of 12 and 9 must be in the form of 36k.
So, 7x – 3x = 36k Or 4x = 36k Or x = 36k/4 = 9k
For x to be a natural number and have the minimum possible value, k should be equal to 1.
So, x = 9 × 1 = 9
Thus, the total number of balls in the box = 3x + 8x + 7x = 18x = 18 × 9 = 162
Hence, option (c) is correct.
View all questions of this testA box contains some balls of black, blue and red colours in the ratio ...
Understanding the Problem:
The given ratio of black, blue, and red balls is 3 : 8 : 7. We need to find the minimum possible number of total balls in the box. The number of red balls is greater than black balls by a number that is a multiple of both 12 and 9.
Solution:
To find the minimum possible number of total balls, we need to consider the least common multiple (LCM) of 12 and 9, which is 36.
Let's assume the number of black balls is 3x. Then, the number of blue balls is 8x, and the number of red balls is 7x.
Given that the number of red balls is greater than the number of black balls by a number that is a multiple of 36:
7x - 3x = 4x = 36k, where k is an integer.
The minimum value of x that satisfies this condition is 9. Therefore, the minimum possible number of total balls in the box is:
3x + 8x + 7x = 18x = 18 * 9 = 162
Therefore, the minimum possible number of total balls contained in the box is 162, which corresponds to option (c).