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04 Mar 2025, 01:30
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58% (02:54) correct
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There are only blue, red, and white chips in a bag. There are 4 more blue chips than white chips in the bag. If a number of red chips equal to twice the number of white chips were added, the number of red chips would double, and half the total number of chips would be blue chips. What fraction of the chips in the bag are red?
A. 1/8
B. 1/6
C. 1/5
D. 1/4
E. 1/3
A. 1/8
B. 1/6
C. 1/5
D. 1/4
E. 1/3
04 Mar 2025, 01:42
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| R | B | W | |
| Original | W+4 | W | |
| Changes | 2W | ||
| After | R+2W = 2R,This gives R = 2W. So, total R would be 4W | =4W+W+4+W/2 = 3W+2 | W |
Note, there is no change in the number of blue chips, so
W+4 = 3W+2
W = 1
Which gives,
B = W+4 = 5
R = 2W = 2
R/T = 2/(1+5+2) = 1/4
Bunuel
There are only blue, red, and white chips in a bag. There are 4 more blue chips than white chips in the bag. If a number of red chips equal to twice the number of white chips were added, the number of red chips would double, and half the total number of chips would be blue chips. What fraction of the chips in the bag are red?
A. 1/8
B. 1/6
C. 1/5
D. 1/4
E. 1/3
A. 1/8
B. 1/6
C. 1/5
D. 1/4
E. 1/3
04 Mar 2025, 02:51
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ManifestDreamMBA
| R | B | W | |
| Original | W+4 | W | |
| Changes | 2W | ||
| After | R+2W = 2R,This gives R = 2W. So, total R would be 4W | =4W+W+4+W/2 = 3W+2 | W |
Note, there is no change in the number of blue chips, so
W+4 = 3W+2
W = 1
Which gives,
B = W+4 = 5
R = 2W = 2
R/T = 2/(1+5+2) = 1/4
Bunuel
There are only blue, red, and white chips in a bag. There are 4 more blue chips than white chips in the bag. If a number of red chips equal to twice the number of white chips were added, the number of red chips would double, and half the total number of chips would be blue chips. What fraction of the chips in the bag are red?
A. 1/8
B. 1/6
C. 1/5
D. 1/4
E. 1/3
A. 1/8
B. 1/6
C. 1/5
D. 1/4
E. 1/3
If according to you B is 5 then why isn't your T equal to 10, as it is mentioned in the question "half the total number of chips would be blue chips"
