Answer = 5 (C)

Total chips = 9 + 7 + 5 = 21
So each stack will have 21/3 = 7 chips (stacks containing equal numbers of chips)

Each stack must contain at least 1 chip of each color

If we keep 1 blue chip, 1 green chip and 5 red chips:

1B 1G 5R => total = 7 chips

5 red chips is the maximum number which can be kept in a stack, as it is impossible to insert more than 5 red chips in one stack because the rule will be broken "each stack contains at least 1 chip of each color".

Three stacks containing equal numbers of chips are to be made from 9 red chips, 7 blue chips, and 5 green chips. If all of these chips are used and each stack contains at least 1 chip of each color, what is the maximum number of red chips in any one stack?

We have 3 stacks. as each stack contains at least 1 chip of each color, we should have at least 1 chip of blue and green in the one stack. We have 9 red chips, thus 9-2 = 7 red chips are the maximum number of red chips.

IMO 5

its a question of min max -
Step 1 GIVE 1 OF EACH TO EACH STACK- Fulfill the min requirement.

Step 2 -Now as we know each stack can have only 7 chips that means to max the num of red chips in stack 1 minimize the number of blue and green chips ( i.e put them in other stacks ) S1 will have only 1 blue and 1 green chip but since we have to maintain an equal of 7 that means you have to give 5 more chips to stack 1

Answer is 5,

to make it easier, first, you give each chip to each stack. we know that it is 21 chips in total, 7 chips a stack.

now, you observe, since there is 3 kinds of chips, you give each chip to each stack, it means 3 chips must occupied in each stack, leaving you only 4 maximum from each color (only red 6 left or blue 4 left, not enough on green chips because green chips will have 2 left)

here is another crucial point, we know the maximum is 4 is actually incomplete, yes, the maximum addition chips is 4, you need to add those chips that are already in the stack, each chips from each color, right?

thus, Answer is 5, 4 additional + 1 already added by the question = 5.

My approach is a bit different, 9RC + 7BC + 5GC, total chips can be 315, LCM of all 3 color chips

Now they will be in 3 stacks of equal number of chips, 315/3 = 105 chips/ stack

You can get this from C

9 * 5 + 7 *5 + 5 * 5 = 105
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