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Manager  P
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a box contains a total of 56 balls of red and blue color in the ratio  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 54% (02:00) correct 46% (02:07) wrong based on 107 sessions

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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

Originally posted by Turkish on 28 Apr 2015, 10:53.
Last edited by Turkish on 28 Apr 2015, 11:17, edited 1 time in total.
Current Student Joined: 17 Feb 2013
Posts: 6
GMAT 1: 690 Q47 V38 Re: a box contains a total of 56 balls of red and blue color in the ratio  [#permalink]

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R:B = 3:5 for 56 balls in the box
=> 5x + 3x =56 (where x is the multiplier)
=> 8x =56 or x =7.So , R = 3x=21 and B =5x =35.

Now for the ratio to become <2:7 , given that no blue balls are removed
=> 7y =35 (where y is the new multiplier)=> y=5.
So the count of red balls in the box should be < 2y or 10, meaning more than 11 (21 -10) red balls need to be removed from the box.
So answer is 12.
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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: a box contains a total of 56 balls of red and blue color in the ratio  [#permalink]

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1
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Turkish wrote:
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

For find number of balls we should find how much balls contatin one part. For this we should divide total of balls $$56$$ on sum of parts $$3+5 = 8$$
So one part equal to $$7$$ and we have $$3*7 = 21$$ red balls and $$5*7=35$$ blue balls.

We need ratio that will be less than $$\frac{2}{7}$$ by subtracting red balls. So we should bring ratio $$\frac{2}{7}$$ to real situation: $$35$$ blue balls divide by $$7 = 5$$.
By mutiplying ratio $$\frac{2}{7}$$ on $$5$$ we receive actual ratio: $$\frac{10}{35}$$. Task asks about ratio that will be less than $$\frac{10}{35}$$, so we need to receive $$\frac{9}{35}$$.
Initially we have $$21$$ red balls and we can have $$9$$ red balls by subtracting $$12$$ red balls.

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Re: a box contains a total of 56 balls of red and blue color in the ratio  [#permalink]

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1
Turkish wrote:
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

In questions like this, it helps to represent the given information visually in a rough diagram. Let's first represent the initial situation visually: Since the Red: Blue balls are in the ratio 3:5, we can write their number as 3x and 5x respectively.

So, total number of balls = 3x + 5x = 56 (given)
=> 8x = 56
=> x = 7

So, initial number of red balls = 3x = 3*7 = 21
And, initial number of blue balls = 5x = 5*7 = 35

Let's now look at the final situation:

Let y balls be removed from the red balls.

So, the final situation looks like this: So, the final ratio of Red: Blue balls =$$\frac{21-y}{35}$$

The question requires that:

$$\frac{21-y}{35}$$ < $$\frac{2}{7}$$

Multiplying both sides of this inequality with 35 (a positive number) will not change the sign of inequality. We'll get:

21-y < 10

Adding y - 10 to both sides of the inequality, we get:

11 < y

That is, y > 11

Since y denotes the number of balls, y must be a positive integer.

Therefore, the smallest possible value of y is 12.

The takeaway that I hope you draw from this alternate solution is:

In questions that involve multiple quantities and changes done to those quantities, representing the given information visually helps in getting a better understanding of how to solve the question.

Hope you found this little tip useful! Best Regards

Japinder
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Re: a box contains a total of 56 balls of red and blue color in the ratio  [#permalink]

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Turkish wrote:
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.

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Re: a box contains a total of 56 balls of red and blue color in the ratio  [#permalink]

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_________________ Re: a box contains a total of 56 balls of red and blue color in the ratio   [#permalink] 12 May 2019, 06:22
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