Minimum number of balls: Problem 2 (harder)

Question

In a bin, if the ratio of blue balls to green balls is 6:8 and the ratio of green balls to red balls is 7:3, what is the minimum number of balls that are not green?

Answer 1

The first ratio, reduced to it lowest terms, is 3:4 (or 4 green balls for every 3 blue balls). The second ratio is in its lowest terms already (3 red for every 7 green balls). How to combine them?

Since green is the common color, we multiply each ratio by a (different) number so the number of green balls is equal in both ratios.

To find the multipliers, we first note that the LCM (leaast common multiple) of 4 and 7 is 28. Since 4 goes into 28 7 times, we rewrite 3:4 as 3x7:4x7 = 21:28.

Similarly, we calculate the factor to multiply the second ratio is 4 (by dividing 28 by 7),and scale the second ratio (7:3) to 28:12.

Since the number of green balls match, we can combine the two ratios as Blue:Green:Red :: 21:28:12. So the fewest number of blue + red balls is 21+12 = 33.