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In a certain quiz that consists of 10 questions, each question after
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28 Mar 2016, 09:50
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In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the quiz are worth a total of 360 points, how many points is the third question worth? A. 18 B. 24 C. 26 D. 32 E. 44
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Re: In a certain quiz that consists of 10 questions, each question after
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28 Mar 2016, 10:11
lpetroski wrote: In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the quiz are worth a total of 360 points, how many points is the third question worth? A. 18 B. 24 C. 26 D. 32 E. 44 Hi, If we look at the increase, it tells us that it is a arithmetic expression..let the first number be x.. so \(SUM of 1st 10= x+4+x+x+8+....x+4*9\).. so \(SUM = 10x+4(1+2+..+9)=360\) 10x+180=360.. 10x=180 or x=18.. so 3rd number= 18+4*2=26C
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Re: In a certain quiz that consists of 10 questions, each question after
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28 Mar 2016, 19:17
Equation is like this:
\(10x + ( 4 + 8 + 12 + ... + 32) = 360\)
to find sum of the evenly spaced sequence, use average * number of the members in the sequence
\(10x + (20*9) = 360\)
\(10x + 180 = 360\) \(x = 18\)
So, \(x + 8 = 26\)




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Re: In a certain quiz that consists of 10 questions, each question after
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12 Apr 2016, 04:34
Aves wrote: Equation is like this:
\(10x + ( 4 + 8 + 12 + ... + 32) = 360\)
to find sum of the evenly spaced sequence, use average * number of the members in the sequence
\(10x + (20*9) = 360\)
\(10x + 180 = 360\) \(x = 18\)
So, \(x + 8 = 26\) chetan went the orthodox way. But Aves chose the short cut and ended up creating confusion in middle of the solution. Quote: to find sum of the evenly spaced sequence, use average * number of the members in the sequence i understand that very well. What you did further was like you forgot what you said above.



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Re: In a certain quiz that consists of 10 questions, each question after
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05 May 2016, 09:19
Aves wrote: Equation is like this:
\(10x + ( 4 + 8 + 12 + ... + 32) = 360\)
to find sum of the evenly spaced sequence, use average * number of the members in the sequence
\(10x + (20*9) = 360\)
\(10x + 180 = 360\) \(x = 18\)
So, \(x + 8 = 26\) Thanks for the detailed explanation, helped a lot. But the red 32 has to be a 36, right?



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Re: In a certain quiz that consists of 10 questions, each question after
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22 May 2016, 12:23
Aves wrote: Equation is like this:
\(10x + ( 4 + 8 + 12 + ... + 32) = 360\)
to find sum of the evenly spaced sequence, use average * number of the members in the sequence
\(10x + (20*9) = 360\)
\(10x + 180 = 360\) \(x = 18\)
So, \(x + 8 = 26\) Good shortcut  will use this in the future. I did it the long way but still got the question right in about 90 seconds x x+4 x+8 x+12 x+16 x+20 x+24 x+28 x+32 x+36 10x+180=360 10x=180 x=18 3rd question = x+8 = 18+8 = 26 Answer C



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Re: In a certain quiz that consists of 10 questions, each question after
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01 Aug 2016, 12:18
I found a shortcut for the same. Since we know that there are 10 questions on the quiz and all questions are evenly placed by 4 points, DO NOT choose the points in the first question as x.
INSTEAD, choose it as x16. So, the numbers will be: 1) x16 2) x12 3) x8 4) x4 5) x 6) x+4 7) x+8 8) x+12 9) x+16 10) x+20
Adding all the above 10 will lead to a simple sum of 10x + 20 (16 & 16 will lead to a sum of 0, same for 12& +12 so on & so forth)
SO, 10x + 20 = 360 10x= 340 x= 34 3rd question = x8= 348 = 26.
Hope it helps.



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Re: In a certain quiz that consists of 10 questions, each question after
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09 Sep 2016, 12:28
Dondarrion wrote: Aves wrote: Equation is like this:
\(10x + ( 4 + 8 + 12 + ... + 32) = 360\)
to find sum of the evenly spaced sequence, use average * number of the members in the sequence
\(10x + (20*9) = 360\)
\(10x + 180 = 360\) \(x = 18\)
So, \(x + 8 = 26\) Good shortcut  will use this in the future. I did it the long way but still got the question right in about 90 seconds x x+4 x+8 x+12 x+16 x+20 x+24 x+28 x+32 x+36 10x+180=360 10x=180 x=18 3rd question = x+8 = 18+8 = 26 Answer C I did it this way also. The quick way to do this math is x + 4(sum of 1 through 9) = x + 4(45) = x + 180 (Sum of 1 through 9) is found by: Average of first and last term * number of terms = 1 + 9 = 10/2 = 51 through 9 = 9 terms 9*5 = 45



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Re: In a certain quiz that consists of 10 questions, each question after
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27 Sep 2016, 12:12
lpetroski wrote: In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the quiz are worth a total of 360 points, how many points is the third question worth? A. 18 B. 24 C. 26 D. 32 E. 44 This is an Arithmetic Progression (A.P.) with terms => a, a+4, a+8, ...., a + (n1) *4 a => First Term of Series n => Number of the term l => Last Term So, l = a + (101) * 4 = a + 9*4 = a + 36 Sum of an A.P. Series = \(\frac{n*(a+l)}{2}\) => Standard Formula Therefore, \(360 = \frac{n*(a+l)}{2} => 360 = \frac{10*(a+a+36)}{2}\) => 360 = 5 * (2a + 36) => 72 = 2a + 36 => 2a = 36 => a = 18 Therefore, 3rd Term = a + 8 = 26
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Re: In a certain quiz that consists of 10 questions, each question after
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08 Sep 2017, 20:28
I got this in a similar approach but we can calculate and get answer in 50 Secs. Since the seq is in AP evenly spaced the sum= Avg* N Seq is x,x+4, x+2*4,x+3*4, x+4*4,x+5*4,x+6*4,x+7*4,x+8*4,x+9*4 ( don't need to write the entire seq just get the idea how seq is going to form ) and so on the average of evenly spaced set of 10 numbers is average of 5th and 6th value. Since each term is increment of 4 the mid value between 5 and 6 will be increment of 2 from 5th value so fifth value is x+4*4= x+16 so average is x+18 so (x+18)*10=360 x+18=36 or x=18 and third value is x+2*4=18+8= 26 The trick in above question is identifying the evenly spaced seq, then its average or median value and finding x.
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Re: In a certain quiz that consists of 10 questions, each question after
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14 Feb 2018, 23:32
since we have 10 items on this list, starting with n and ending with n+36, I used empowergmat's bunching method: (note that this is useful for lists that have consecutive forms, in this case with +4 increments) method: Add the first and Last number, then multiply it by half of the number of the items, in this case 5 (given our 10 item list) so, (n + n + 36) * 5 should equal to 360 > 2n + 36 = 72 > 2n = 36 > n = 18 since we are looking for 3rd item, that's n + 8, which is 18 + 8 = 26 EMPOWERgmatRichC please let me know if I used the method correctly... not sure if I had ever used bunching in a sequence question like this, but it def appeared to have worked.



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Re: In a certain quiz that consists of 10 questions, each question after
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15 Feb 2018, 04:36
This is an Arithmetic progression question.
n/2 * [2a + (n  1) *d] = 360 10/2 * [2a + (10  1)* 4)] = 360 a = 18
So, first question is 18 marks.
So, third question = 26



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Re: In a certain quiz that consists of 10 questions, each question after
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19 Feb 2018, 10:43
lpetroski wrote: In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the quiz are worth a total of 360 points, how many points is the third question worth? A. 18 B. 24 C. 26 D. 32 E. 44 The first question is worth x, the 2nd worth x + 4, the 3rd worth x + 2(4) = x + 8, ..., the 10th is worth x + 9(4) = x + 36. Since we have a set of evenlyspaced numbers, the average is (x + x + 36)/2 = (2x + 36)/2 = x + 18. Since sum = average * quantity we have: 360 = (x + 18) * 10 36 = x + 18 18 = x The first question is worth x = 18 points. So the third question is worth x + 8 = 18 + 8 = 26 points. Answer: C
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In a certain quiz that consists of 10 questions, each question after
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18 Mar 2018, 18:33
lpetroski wrote: In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the quiz are worth a total of 360 points, how many points is the third question worth? A. 18 B. 24 C. 26 D. 32 E. 44 10x + 4 (1 + 2 + 3 + ....9) = 360 =) 10x + 4 * [n(n+1)]/2 = 360 =) 10x + 4 * [(9 * 10)] / 2 = 360 =) x = 18 third question = (18 +8) =26 = C the answer thakns



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Re: In a certain quiz that consists of 10 questions, each question after
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26 Jul 2018, 00:51
There is no need for any formulas here. Sum of 10 consecutive integers = 360. Average = 36. Average of a set of consecutive integers = Median of the set. Median is the number between 5 and 6, and its equals 36.
Since the difference between terms is 4, therefore, 5th term is 34 and 6th term is 38. 5th term = 34 4th term = 30 3th term = 26



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Re: In a certain quiz that consists of 10 questions, each question after
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27 Jul 2018, 10:23
lpetroski wrote: In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the quiz are worth a total of 360 points, how many points is the third question worth? A. 18 B. 24 C. 26 D. 32 E. 44 If we assume first question is worth "x" points, the proceeding questions would be worth x+4, x+8, x+12,.... x+36 points. So, the sum of 10 questions would be 10x plus the sum of 4  36 inclusive, with multiples of 4 (4, 8, 12, 16, 20, 24, 28, 32, 36). The sum of 436 can be calculated with (4+36)/4*(16+20)/2 = 180. (The first and last numbers in the list divided by 4, times the average of the numbers (16+20)/2=18) Therefore, the total sum would be 10x + 180 = 360, x= 18. Finally the third question was worth x+8 = 18+6 = 26 points.



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Re: In a certain quiz that consists of 10 questions, each question after
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18 Dec 2018, 22:28
Let the first question is worth ‘a’ points. Then, Second question is worth a + 4 points, third of worth (a + 4 + 4) = a + 8 points. So, the sum of the series becomes a + a + 4 + a + 8 + a + 12 ……… upto 10 terms = 10a + 4 + 8 + 12 + ……. upto 9 terms Sum = 10a + Sum of A.P with first term 4 and common difference 4. Sum = 10a + 180 = 360 a = 18 Third question is worth a + 8 points = 18 + 8 = 26 points.



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Re: In a certain quiz that consists of 10 questions, each question after
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10 Feb 2019, 14:02
Hi, here are my two cents for this question First if we recognize that this is a AP with common difference 4 then we can use the concept of corresponding term Using the concept of corresponding terms , we can say that average f corresponding terms * Number of terms =sum of all terms Here is what i mean So Number of terms is 10 \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\), \(a_6\), \(a_7\), \(a_8\), \(a_9\), \(a_{10}\) Then corresponding terms will be \(a_1\) & \(a_{10}\) , \(a_2\) & \(a_9\) , \(a_3\) & \(a_8\) , \(a_4\) & \(a_7\) , \(a_5\) & \(a_6\) Since we are asked to find the value of \(a_3\) i will take the corresponding pair of \(a_3\) & \(a_8\) Also since its an AP series we have \(a_8\)= \(a_3\)+20 So 360= 10 * \(\frac {a_3 + a_8}{2}\) so substituting relevant information we have 360= 10 * \(\frac {2(a_3) + 20}{2}\) 72= 2\(a_3\) + 20 2\(a_3\) = 52 \(a_3\)= 26 Probus
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