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Updated on: 31 Dec 2023, 15:51
Originally posted by GMATCoachBen on 26 Dec 2023, 20:36.
Last edited by GMATCoachBen on 31 Dec 2023, 15:51, edited 4 times in total.
Last edited by GMATCoachBen on 31 Dec 2023, 15:51, edited 4 times in total.
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Video solution here (3:26):
Key Habit for Speed on Data Sufficiency:
We often don't need to actually solve.
[b]We just need to know that we CAN answer the question.
This can save a lot of time on problems with harder calculations.
With 2 UNIQUE LINEAR equations and 2 unknowns (or 1 equation & 1 unknown, 3 equations and 3 unknowns, etc.), we can solve.
Caveats:
⚠️Watch out when it's asking for a "Combo" of two variables. For example, if the question asks for the ratio "X/Y", we can solve with the one equation 2x = 3y. We get "two birds with one stone."
⚠️The equations must be UNIQUE. A common trap is where one equation is just a multiple of the other. For example, x + 2y = 3 and 3x + 6y = 9 are really just the same equation!
⚠️Quadratics often have 2 different solutions for 1 equation. For example, \(x^2 = 9\) has 2 different solutions: x=3 OR x=-3.
Key Habit for Speed on Data Sufficiency:
We often don't need to actually solve.
[b]We just need to know that we CAN answer the question.
This can save a lot of time on problems with harder calculations.
With 2 UNIQUE LINEAR equations and 2 unknowns (or 1 equation & 1 unknown, 3 equations and 3 unknowns, etc.), we can solve.
Caveats:
⚠️Watch out when it's asking for a "Combo" of two variables. For example, if the question asks for the ratio "X/Y", we can solve with the one equation 2x = 3y. We get "two birds with one stone."
⚠️The equations must be UNIQUE. A common trap is where one equation is just a multiple of the other. For example, x + 2y = 3 and 3x + 6y = 9 are really just the same equation!
⚠️Quadratics often have 2 different solutions for 1 equation. For example, \(x^2 = 9\) has 2 different solutions: x=3 OR x=-3.
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2023-12-26 19_35_23-No name - 26 December 2023 (3).mp4 - VLC media player.png [ 5.06 MiB | Viewed 582 times ]
30 Dec 2023, 01:09
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My two cents, since we know the post discount price for adults and children (4950 & 4320) respecfully.
S1: Tells us that there were 6 adults (29700/4950) = 6 --- children could be >=4 since the question tells us there were at least 10 people - (I)
S2: Tells us that children's total tickets price was 4860 more than adullts - Not enough info (I)
S1+S2 : Take S2's 4860 and add it to S1's 29700, you get 34560 and divide by the children's post discount price (34560/4320) = 8 children
Hope it's the right way to think
S1: Tells us that there were 6 adults (29700/4950) = 6 --- children could be >=4 since the question tells us there were at least 10 people - (I)
S2: Tells us that children's total tickets price was 4860 more than adullts - Not enough info (I)
S1+S2 : Take S2's 4860 and add it to S1's 29700, you get 34560 and divide by the children's post discount price (34560/4320) = 8 children
Hope it's the right way to think
28 Jan 2025, 18:54
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
