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May 25, 2022 12:15 pm
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An entrance test consists of 20 questions. Each question after the first is worth 2 points more than the preceding question. If the total questions are worth a total of 400 points, how many points is the fourth question worth?
(A) 5
(B) 7
(C) 11
(D) 19
(E) 38
(A) 5
(B) 7
(C) 11
(D) 19
(E) 38
May 25, 2022 2:41 pm
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Bunuel
If the answer choices were closer together, maybe I'd prefer the Algebra. But we can get rid of three answer choices by spending maybe 15 seconds using a little logic. It's a bit of a combination of Ballparking and PITA (Plugging In The Answers).
Could it be A? If the fourth question is worth 5 points, the third is worth 3, the second is worth 1, and the first is worth...wait a minute...um, -1? Wrong.
Could it be E? If the fourth question is worth 38 points and the ones after it are worth more, we are surely going to be over a total of 400 within the next maybe nine questions. Wrong.
Could it be D? If the fourth question is worth 19, questions 1, 2, and 3 are worth a little less, questions 5, 6, and 7 are worth a little more, and then questions 8-20 would clearly push us over a total of 400. Wrong.
We only have B and C. Let's just try B.
1. 1
2. 3
3. 5
4. 7
...
20. 39
That's ten pairs of 40, which is a total of 400. Is that what we wanted? Yep.
Answer choice B.
ThatDudeKnowsPITA
ThatDudeKnowsBallparking
Nov 09, 2023 6:52 pm
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Sequence theory tells us that this problem can be described by the first term (T1) plus 2 times the nth term (2n) minus 2, or, T1 + 2n - 2. (i.e., if the sequence was 0,2,4,6,8..., then you could get to term n = 4 using the equation 0 + 2*4-2 = 0 + 8 - 2 = 6 (4th term))
Now that you have the formula for the sequence, you know 400 = SUM of T1 + 2n - 2, where n goes up to 20.
Thus, 400 = (-2)*20 + 20*T1 + SUM of 2n up to 20.
The sum of 2n up to 20 is 2 + 4 + 6 + 8 +...+ 36 + 38 + 40. You might notice that 2 + 40 = 42, 4 + 38 = 42, 6 + 36 = 42, 8 + 34 = 42, and so on.
You can therefore, use the following formula: 42*10 (as the number 42 can be added up to 10 times, or 20/2). Therefore, SUM of 2n = 420
Plugging everything in, we know that 400 = (-2)*20 + 20T1 + 420 --> 400 = -40 + 20T1 + 420 --> 400 + 40 - 420 = 20T1 --> 20 = 20T1 --> T1 = 1
The first term is 1. Second term is, therefore, 3 (1+2). Third term is 5 (3+2) and Fourth term, and answer, is 7 (5+2). Thus, B
Now that you have the formula for the sequence, you know 400 = SUM of T1 + 2n - 2, where n goes up to 20.
Thus, 400 = (-2)*20 + 20*T1 + SUM of 2n up to 20.
The sum of 2n up to 20 is 2 + 4 + 6 + 8 +...+ 36 + 38 + 40. You might notice that 2 + 40 = 42, 4 + 38 = 42, 6 + 36 = 42, 8 + 34 = 42, and so on.
You can therefore, use the following formula: 42*10 (as the number 42 can be added up to 10 times, or 20/2). Therefore, SUM of 2n = 420
Plugging everything in, we know that 400 = (-2)*20 + 20T1 + 420 --> 400 = -40 + 20T1 + 420 --> 400 + 40 - 420 = 20T1 --> 20 = 20T1 --> T1 = 1
The first term is 1. Second term is, therefore, 3 (1+2). Third term is 5 (3+2) and Fourth term, and answer, is 7 (5+2). Thus, B
