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02 Aug 2026, 19:00
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D
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A local group working to get signatures for a petition to pass a new ordinance lists all signers in their database as either “residents” or “non-residents”. If the ratio of residents to nonresidents in the database is 3 : 5, how many signers are currently listed as non-residents in the database?
(1) If the next six signers are all residents, the ratio of residents to nonresidents in the database will be 1 : 1
(2) If three signers had been incorrectly assigned and were nonresidents rather than residents, the ratio of residents to nonresidents in the database would be 1: 3 once this was corrected.
(1) If the next six signers are all residents, the ratio of residents to nonresidents in the database will be 1 : 1
(2) If three signers had been incorrectly assigned and were nonresidents rather than residents, the ratio of residents to nonresidents in the database would be 1: 3 once this was corrected.
02 Aug 2026, 21:52
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Nice one, this is a set-up-two-equations-and-let-algebra-do-the-work question.
1. Let residents = 3k and non-residents = 5k right now, since the ratio is 3:5.
2. Statement 1: add six residents, and the ratio flips to 1:1. So 3k + 6 = 5k, which gives 6 = 2k, k = 3. Non-residents = 5k = 15. That's a full number, so sufficient.
3. Statement 2: three people who were logged as residents actually should have been non-residents. So residents become 3k - 3, non-residents become 5k + 3, and that new ratio is 1:3. Set it up: 3(3k - 3) = 5k + 3. That's 9k - 9 = 5k + 3, so 4k = 12, k = 3. Same answer, non-residents = 15. Also sufficient on its own.
Answer: D, each statement alone gets you there.
What I like about this one is that both statements are really just "correct the count and set the new ratio equal," and once you're comfortable translating word problems into a single variable k for the ratio unit, both statements resolve fast. Early on I used to rewrite R and N as two separate variables and solve a messier system. One variable for the ratio unit does the whole job here, and it's worth practicing that shortcut deliberately instead of hoping it shows up on its own under time pressure.
1. Let residents = 3k and non-residents = 5k right now, since the ratio is 3:5.
2. Statement 1: add six residents, and the ratio flips to 1:1. So 3k + 6 = 5k, which gives 6 = 2k, k = 3. Non-residents = 5k = 15. That's a full number, so sufficient.
3. Statement 2: three people who were logged as residents actually should have been non-residents. So residents become 3k - 3, non-residents become 5k + 3, and that new ratio is 1:3. Set it up: 3(3k - 3) = 5k + 3. That's 9k - 9 = 5k + 3, so 4k = 12, k = 3. Same answer, non-residents = 15. Also sufficient on its own.
Answer: D, each statement alone gets you there.
What I like about this one is that both statements are really just "correct the count and set the new ratio equal," and once you're comfortable translating word problems into a single variable k for the ratio unit, both statements resolve fast. Early on I used to rewrite R and N as two separate variables and solve a messier system. One variable for the ratio unit does the whole job here, and it's worth practicing that shortcut deliberately instead of hoping it shows up on its own under time pressure.
02 Aug 2026, 23:01
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ratio of resident and non resident are given i.e 3:5 , R:NR =3:5
from St1 if R has another 6 then ratio becomes 1:1, 3k+6 : 5k = 1:1 so we can calculate each value easily it is sufficient
from St2 if 3h has assigned from R to NR it becomes 3k-3 :5k+3 = 1:3 from here we can also calculate each value so each statement alone is sufficient answer is D
from St1 if R has another 6 then ratio becomes 1:1, 3k+6 : 5k = 1:1 so we can calculate each value easily it is sufficient
from St2 if 3h has assigned from R to NR it becomes 3k-3 :5k+3 = 1:3 from here we can also calculate each value so each statement alone is sufficient answer is D
Bunuel
