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03 Apr 2025, 00:30
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D
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Difficulty:
55%
(hard)
Question Stats:
65% (01:47) correct
35%
(02:12)
wrong
based on 133
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
A. 9
B. 10
C. 11
D. 12
E. 13
03 Apr 2025, 00:42
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Ratio of red to blue given: 3:5
Let Red balls: 3x
Blue balls: 5x
Total = 8x = 56 (we're given total 56 balls) => x = 7
So there are 21 red balls and 35 blue balls
Let a be the minimum number of red balls to be removed
Red balls after a red balls are removed: 21-a
Blue balls (they remain unchanged) : 35
New Ratio: 21-a/35
We're asked "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"
So 21-a/35 < 2/7 => a > 11
Hence the minimum value of a will be 12
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 is 12
Answer: (D)
Let Red balls: 3x
Blue balls: 5x
Total = 8x = 56 (we're given total 56 balls) => x = 7
So there are 21 red balls and 35 blue balls
Let a be the minimum number of red balls to be removed
Red balls after a red balls are removed: 21-a
Blue balls (they remain unchanged) : 35
New Ratio: 21-a/35
We're asked "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"
So 21-a/35 < 2/7 => a > 11
Hence the minimum value of a will be 12
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 is 12
Answer: (D)
03 Apr 2025, 04:15
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Bunuel
R:B = 3:5
After removing some red balls the ration of R:B = 2:9
i.e. after removal of some red balls the total balls should be multiple of 9 while before it the total balls should be multiple of 8 (3+5) hence it should be greater than 9 and multiple of 8
Also, the blue balls remain same after removal of red ball so red blue should be multiple of 5 as well as multiple of 7
i.e. minimum value of b = 35
If, Total Balls = 56, R= 21, B = 35, Red balls to be removed = x
\(\frac{(21-x)}{35} = \frac{2}{7}\)
i.e. x = 11
But since the ratio should be less than 2/7 hence red ball to be removed should be 1 greater
hence x+1 = 12
Answer: Option D
