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# A certain drive-in movie theater has total of 17 rows of parking space

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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Bunuel wrote:
A certain drive-in movie theater has total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row there are 2 more parking spaces than in the previous row. What is the total number of parking spaces in the movie theater?

A) 412
B) 544
C) 596
D) 632
E) 692

We can see that the 3rd row has 21 + 2(1) = 23 parking spaces, the 4th row has 21 + 2(2) = 25 parking spaces and so on. Therefore, the 17th row has 21 + 2(15) = 51 sparking spaces. Therefore, the total parking spaces of row 2 to row 17 is:

21 + 23 + 25 + … + 51

Since the terms of the above sum are evenly spaced, we can use the formula sum = average x quantity. The average is (21 + 51)/2 = 72/2 = 36. The quantity is 17 - 2 + 1 = 16. Therefore, the sum is 36 x 16 = 576. We still have to add the number of parking spaces in the first row, so the total number of parking spaces in all 17 rows is 576 + 20 = 596.

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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Bunuel wrote:
A certain drive-in movie theater has total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row there are 2 more parking spaces than in the previous row. What is the total number of parking spaces in the movie theater?

A) 412
B) 544
C) 596
D) 632
E) 692

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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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596

first row 20 then other 16 rows form an Ap with first term 21 and last 51.Avg is 36
so 36*16+20=596
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Attached is a visual that should help.
Attachments

Screen Shot 2016-08-10 at 7.45.53 PM.png [ 155.8 KiB | Viewed 72539 times ]

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A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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First Row = 20

Second Row =21

Third Row = 21 + 2

Fourth Row = 21 + 4

Third till Last Row are in a sequence with difference of 2 between each terms. Hence, Sum of all terms in this sequence of 15 terms = N/2 * (2* First Term + {n-1} * Difference) = 15/2 * (2*23 + 14*2) = 555

Sum of all terms = 20 + 21 + 555 = 596
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A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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First row will have 20 parking spaces.

Second column onwards is an AP with Starting term(a) = 21, Common difference(d) = 2, and Number of terms(n) = 16

Sum of n terms(Sn) = $$\frac{n}{2}*(2a + (n-1)d)$$

Substituting the values, we get

Sn = $$\frac{16}{2}*(2*21 + (16-1)2) = 8 * (42+30) = 8 * 72 = 576$$

Total number of parking spaces are 20+576 = 596(Option C)
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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MENNAWALID wrote:

Posted from my mobile device

I believe you are searching for a formula.
This one is a simple AP formula.

Sum= n/2{2a+(n-1)d}
here a=21
n=16
d=2

So now we get = 16/2{2*21+(16-1)*2}
=> 16(21+15)
=> 16*36
=> 576
Add the 20 from the first row to get 596 as your total.
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Hi All,

This question can be solved with a bit of Arithmetic and 'bunching.'

We're told that the first two rows of parking spaces contain 20 and 21 spaces respective, then the following 15 rows increase by 2 each (re: 23, 25, 27, etc.). We're asked for the TOTAL number of parking spaces.

The first two rows total 20+21 = 41 spaces. The remaining 15 rows are the odd numbers from 23 through 51, inclusive. By adding the smallest and largest of those numbers, we get:

23 + 51 = 74

By adding the next smallest and next biggest, we get:

25 + 49 = 74

This pattern is consistent, so we'll end up with 7 "pairs" of numbers that add up to 74 and one number "in the middle" that doesn't 'pair up': 37.

Thus, the total number of spaces is... 41 + 7(74) + 37 = 596

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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Just a tip: when you come to final calculations (or in the beginning) glance at answer choices before making arithmetics. In this case I came up with 72*8+20. As you can see, the last digit has to be 6. The only answer that ends with 6 is C. So I skipped the part with calculations and picked option C. This approach helps to save your energy when you are doing tests and saves you from silly mistakes related to arithmetics.
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Easy Trick
17 rows can be written this way

20
21 51 =72
23 49 =72
25 47 =72
27 45 =72
29 43 =72
31 41 =72
33 39 =72
35 37 =72

20+(72X8)=596

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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Bunuel wrote:
A certain drive-in movie theater has total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row there are 2 more parking spaces than in the previous row. What is the total number of parking spaces in the movie theater?

A) 412
B) 544
C) 596
D) 632
E) 692

Try intuition!

Let's add the first two row first.
The sum is 20+21=41

Now, we have 15 other row. All are evenly spaced! Just write these down.

23, 25, 27, 29, 31, 33, 35, 37, ....
Look there are total fifteen terms here. The middle term is 8th term. If we know the middle term, we can find the sum easily. Here, The middle(8th) term is 37.
So the sum is 37× 15 = be smart!
You don't need to multiply here. Just think about unit digit only.
7× 5 = gives unit digit 5.

We already have sum 41 from first two row.
So the unit digit of the total sum is
41 + ...5 = unit digit 6.

Only option C has unit digit 6.

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A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
KarishmaB IanStewart avigutman BrentGMATPrepNow is there any logical way of solving this question? Because everyone has stuck to a formula based approach such as $$Tn = a + (n-1)d$$ AND sum = avg x number of terms
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Hoozan wrote:
is there any logical way of solving this question? Because everyone has stuck to a formula based approach such as $$Tn = a + (n-1)d$$ AND sum = avg x number of terms

Adding equally spaced sets is standard on the GMAT, so there's nothing wrong with just doing that here. I did it differently (since it's a GMAT problem, you can correctly anticipate a shortcut will be available) -- in our sum, we're adding the odd numbers between 20 and 30, between 30 and 40, and between 40 and 50. Each of those three sums will end in 5 (each equals 5 times the median, and the median ends in 5, or you can just add the units digits), so when we add all of those odd numbers from 20 to 50, we get something ending in 5. We have to add two other numbers, 20 and 51, to complete the sum, so the answer will end in 6, and only one choice is possible.
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Bunuel wrote:
A certain drive-in movie theater has total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row there are 2 more parking spaces than in the previous row. What is the total number of parking spaces in the movie theater?

A) 412
B) 544
C) 596
D) 632
E) 692

As Ian said, sum of AP terms is a standard formula and it is good to know it. It will come in handy in GMAT.

This is what I did:
All 17 rows had at least 20 parkings = 17 * 20 = 340
Other 16 had some extra parkings: 1, 3, 5, 7, 9, 11, 13, 15 ... etc.
These are 16 terms so their average will be between 8th and 9th terms i.e. between 15 and 17 so 16.
Then the sum of extra terms is 16 * 16 = 256

Total parkings = 340 + 256 = 596
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
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Hoozan wrote:
KarishmaB IanStewart avigutman BrentGMATPrepNow is there any logical way of solving this question? Because everyone has stuck to a formula based approach such as $$Tn = a + (n-1)d$$ AND sum = avg x number of terms

Hoozan Pretend all 17 rows have 20 parking spaces each, for a total of 340. Now, add the extra parking spaces from the last sixteen rows: 1+3+5+...+27+29+31. Note that, working from the outside toward the middle, you have eight sums of 32 (1+31, 3+29, 5+27, etc.). This is a trick that's always available to us in any sum of an arithmetic sequence. So the answer is 340 + 8*32.
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
warriorguy wrote:
First Row = 20

Second Row =21

Third Row = 21 + 2

Fourth Row = 21 + 4

Third till Last Row are in a sequence with difference of 2 between each terms. Hence, Sum of all terms in this sequence of 15 terms = N/2 * (2* First Term + {n-1} * Difference) = 15/2 * (2*23 + 14*2) = 555

Sum of all terms = 20 + 21 + 555 = 596

Why is N divided by 2? What does 2 represent?
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Re: A certain drive-in movie theater has total of 17 rows of parking space [#permalink]
Bunuel wrote:
A certain drive-in movie theater has total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row there are 2 more parking spaces than in the previous row. What is the total number of parking spaces in the movie theater?

A) 412
B) 544
C) 596
D) 632
E) 692

(Almost) all of the solutions posted solve for the exact value and then find that value among the answer choices. It's minor, but we can skip a little of the arithmetic (which adds unnecessary steps and is a chance to make a silly mistake).

17 rows. Let's just say that the first is 19 and then we add 2 each time. Yeah, we will be off by 1, but so what? The answer choices are spread out enough that it won't matter.
The ninth row will be the median. The ninth row has 35 spaces.
35*17 = 350+210+35 = 595

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