I think, we've all seen the accidentally doubling a portion of a mixture problem. Is there a way to do them without making up numbers? Making one up: Mixture X has A,B, and C. Recipe calls for 1/5 of A, 2/7 of B, and the remainder C. While mixing the portion of A is accidentally doubled. What percentage of the mixture is C?

A = .20X B = .29X (approx.) C = .51X Total = X Double A you get Total = 1.20X percentage of C = .51X/1.20X approximately 42.5%

in general two scenarios are possible: 1) keep the mixture amount unchanged 2) change the amount of mixture strange, sometimes we eat with knife and fork - does it really matter how we eat and what we eat on GMAT under 1) apply smart number --> 35 A is 7, B is 5*2, C is 35-7-10 A becomes 14, C is 35-14-10 C/(A+B+C)=C/(35) or 11/35 (31%) under 2) apply smart number --> 35 A is 7, B is 5*2, C is 35-7-10 A becomes 14, C is 18 (still) C/(A+A+B+C)=C/(35+7) or 18/52 (35%) yellowho wrote:I think, we've all seen the accidentally doubling a portion of a mixture problem. Is there a way to do them without making up numbers? Making one up: Mixture X has A,B, and C. Recipe calls for 1/5 of A, 2/7 of B, and the remainder C. While mixing the portion of A is accidentally doubled. What percentage of the mixture is C?

just checked the solution is identical Tani Wolff - Kaplan wrote:A = .20X B = .29X (approx.) C = .51X Total = X Double A you get Total = 1.20X percentage of C = .51X/1.20X approximately 42.5%

Night reader wrote:hi Tani, doubling A we can't get 1.20x since x is responsible for B and C too doubling the percentage violates our A-B-C ratio of three numbers in x Tani Wolff - Kaplan wrote:A = .20X B = .29X (approx.) C = .51X Total = X Double A you get Total = 1.20X percentage of C = .51X/1.20X approximately 42.5% above it's already not .51X @Night reader Your second method is equivalent to Tani's method (change the amount of mixture). Tani is right. You made a typo here : C/(A+A+B+C)=C/(35+7) or 18/52 (35%) Otherwise the answer matches as well.

really? ah such nuisance , appologies Tani anshumishra wrote:Night reader wrote:hi Tani, doubling A we can't get 1.20x since x is responsible for B and C too doubling the percentage violates our A-B-C ratio of three numbers in x Tani Wolff - Kaplan wrote:A = .20X B = .29X (approx.) C = .51X Total = X Double A you get Total = 1.20X percentage of C = .51X/1.20X approximately 42.5% above it's already not .51X @Night reader Your second method is equivalent to Tani's method (change the amount of mixture). Tani is right. You made a typo here : C/(A+A+B+C)=C/(35+7) or 18/52 (35%) Otherwise the answer matches as well.

The point of the question is that A has been doubled in violation of the original proportions and you now have to calculate the new proportions. if you doubled everything the proportions wouldn't change.