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There are only blue, red, and white chips in a bag. There are 4 more 04 Mar 2025, 01:30
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59% (02:51) correct 41% (03:11) wrong based on 191 sessions

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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


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D
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There are only blue, red, and white chips in a bag. There are 4 more 04 Mar 2025, 01:42
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RBW
OriginalW+4W
Changes2W
AfterR+2W = 2R,This gives R = 2W. So, total R would be 4W=4W+W+4+W/2 = 3W+2W

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


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There are only blue, red, and white chips in a bag. There are 4 more 04 Mar 2025, 02:51
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ManifestDreamMBA
RBW
OriginalW+4W
Changes2W
AfterR+2W = 2R,This gives R = 2W. So, total R would be 4W=4W+W+4+W/2 = 3W+2W

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


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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"
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There are only blue, red, and white chips in a bag. There are 4 more 04 Mar 2025, 03:03
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B, R, W

B = R+4
2W+R = 2R i.e.
B = (1/2)*(R+W+B+2W)


i.e. R = 2W
and B = (1/2)*(5W+B)
i.e. B = 5W

i.e. B+R+W = (5W) + 2W + W = 8W

R/Total = (2W)/(8W) =0.25 1/4

Answer: Option D

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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


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There are only blue, red, and white chips in a bag. There are 4 more 04 Mar 2025, 03:29
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The "after" is hypothetical situation. The present total is what I have mentioned and question asks for what the current situation is.
Rohit_842
ManifestDreamMBA
RBW
OriginalW+4W
Changes2W
AfterR+2W = 2R,This gives R = 2W. So, total R would be 4W=4W+W+4+W/2 = 3W+2W

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


­

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"
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There are only blue, red, and white chips in a bag. There are 4 more 23 May 2026, 01:39
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