A box contains a total of 56 balls of red and blue color

[BTGmoderatorDC]

A box contains a total of 56 balls of red and blue color in the ratio of 3: 5. What is the least number of red balls that should be removed from the box, such that the ratio of red and blue balls is less than 2 to 7?

a.9
b. 10
c.11
d.12
e.13

What is the easiest solution in this problem?

OA D

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Hi lheiannie07,

We're told that a box contains a total of 56 balls of red and blue color in the ratio of 3:5. We're asked for the LEAST number of red balls that should be removed from the box, such that the ratio of red and blue balls is LESS than 2 to 7.

To start, we can determine the number of red and blue balls. Since the ratio is 3 red balls for every 5 blue balls, the total number of balls MUST be a multiple of (3+5) = 8. We're told that the total is 56, so that is seven 'groups' of 8 balls...

7(3) = 21 red balls
7(5) = 35 blue balls

Now we have to remove enough red balls that the new ratio of red balls to blue balls becomes LESS than 2:7. Notice that the current number of blue balls is 35 (which is a multiple of 7), so we can use the same math approach from before - but in reverse - to determine the number of red balls we'll need to keep)...

35 blue balls is five 'groups' of 7 blues balls
Five 'groups' of 2 red balls would be 10 red balls

So if we had 10 red balls and 35 blue balls, then the ratio of red to blue would be EXACTLY 2:7. We need it to be LESS than 2:7 though, so we would have to remove one additional red ball...

21 - 11 = 10 red balls - 1 = 9 red balls

Thus, we would have to remove 12 red balls.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Brent@GMATPrepNow]

A box contains a total of 56 balls of red and blue color in the ratio of 3: 5. What is the least number of red balls that should be removed from the box, such that the ratio of red and blue balls is less than 2 to 7?

a.9
b. 10
c.11
d.12
e.13

As Rich has shown, the box ORIGINALLY has 21 RED balls and 35 BLUE balls.

Let x = number of RED balls to be removed from the box.
Once we remove x RED balls, there are 21-x RED balls and 35 BLUE balls.

We want the ratio of red and blue balls to be less than 2 to 7
So, we want: 21-x /35 < 2/7
Multiply both sides of the inequality by 35 to get: 21 - x < 70/7
Simplify: 21 - x < 10
Add x to both sides: 21 < 10 + x
Subtract 10 from both sides to get: 11 < x

If x must be GREATER THAN 11, then the smallest possible value of x is 12

Answer: D

Cheers,
Brent

[Scott@TargetTestPrep]

A box contains a total of 56 balls of red and blue color in the ratio of 3: 5. What is the least number of red balls that should be removed from the box, such that the ratio of red and blue balls is less than 2 to 7?

a.9
b. 10
c.11
d.12
e.13

We an create the equation:

3x + 5x = 56

8x = 56

x = 7

So there are 21 red balls and 35 blue balls. We can let n = the number of red balls to be removed and create the proportion:

(21 - n)/35 < 2/7

7(21 - n) < 70

147 - 7n < 70

77 < 7n

11 < n

The minimum value is 12.

Answer: D