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In the first week of the Year, Nancy saved \$1.  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 75% (01:35) correct 25% (01:51) wrong based on 2859 sessions

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In the first week of the Year, Nancy saved \$1. In each of the next 51 weeks, she saved \$1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

A. \$1,326
B. \$1,352
C. \$1,378
D. \$2,652
E. \$2,756

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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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40
31
Bigred2008 wrote:
In the first week of the Year, Nancy saved \$1. In each of the next 51 weeks, she saved \$1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

A. \$1,326
B. \$1,352
C. \$1,378
D. \$2,652
E. \$2,756

Does anyone have a better way to do this?

The total amount of money will be 1 + 2 + 3 + 4 + ... + 52 (in the 52nd week, she will save \$52)

Sum of first n consecutive positive integers = n*(n+1)/2
Sum = 52*53/2 = 26*53
The product will end with 8 since 6*3 = 18 so answer must be (C)
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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32
3
momentofzen wrote:
How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

If you don't recall the formula, there is one method:

1 + 2 + 3 + 4 + 5 ........ + 26 + 27 + ............ + 48 + 50 + 51 + 52

Sum of 1st & last digit = 1 + 52 = 53

Sum of 2nd & 2nd last digit = 2 + 51 = 53

Sum of 3rd & 3rd last digit = 3 + 50 = 53
.
.
.
.
This is repeated 26 times

So, 26 * 53 = 1378

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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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5
2
Hi,
This can be solved using the formula:

Sum= n/2( a + nd)

n=52

a= first term=1

d= difference =1

Sum=52/2(1+52)
=1378
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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3
For an even list of consecutive numbers where a = the first number and z = the last number and (in this case) a total of 52 consecutive numbers-- (a+z)(.5*number of consecutive integers)---> (1+52)*26=1378
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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1
Simply using summation of first natural numbers formula = n(n+1)/2

=(52*53)/2 = 1378

Hence OA: C
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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12
3
I used:
Sum = (Average of terms) x (# terms)

Average = (1+52) / 2 = 53/2

# terms= 52-1 + 1 = 52

Sum= 53*52/2 = 53*26 = 1378
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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5
14
momentofzen wrote:
How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

The sum of first n consecutive positive integers is given by the formula n(n+1)/2.
For GMAT, it is a good idea to know this formula. It could simplify many calculations. Test prep companies discuss all such useful formulas in their curriculum. The Official Guides only give practice questions.

You should also know the more generic formula of sum of an Arithmetic Progression. From that you can easily derive this formula.

Sum of an Arithmetic Progression will be n*Average where n is the number of terms in the AP and Average will be the average value of the terms. The average value can be found as (First term + Last term)/2

If first term is a, last term is a + (n-1)*d where d is the common difference.

Average = (a + a + (n-1)*d)/2

Sum = n*(a + a + (n-1)*d)/2 = n/2(2a + (n-1)*d)

In case of consecutive integers starting from 1, a = 1 and d = 1

Sum = n/2(2 + (n-1)) = n(n+1)/2

Here are a couple of posts on APs:

http://www.veritasprep.com/blog/2012/03 ... gressions/
http://www.veritasprep.com/blog/2012/03 ... gressions/
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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PareshGmat wrote:
momentofzen wrote:
How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

If you don't recall the formula, there is one method:

1 + 2 + 3 + 4 + 5 ........ + 26 + 27 + ............ + 48 + 50 + 51 + 52

Sum of 1st & last digit = 1 + 52 = 53

Sum of 2nd & 2nd last digit = 2 + 51 = 53

Sum of 3rd & 3rd last digit = 3 + 50 = 53
.
.
.
.
This is repeated 26 times

So, 26 * 53 = 1378

Yes! Definitely better/easier/faster to apply logic than to memorize another formula. This way, you can use the logic even if you don't start at one. Or if you are counting by 2's or 3's. For example, what if Nancy saved \$51 the first week of the year, and increased her savings by \$1 each week after that.

Start at \$51. Know that there are 52 weeks in a year, so the last week she will save 102. So the sequence will be 51, 52, 53... 100, 101, 102.
See how adding the first and last is the same as adding the second and the second-to-last, and the same as the third and third-to-last? All like pairs are \$153.
If there are 52 items, then there are 26 pairs.
\$153 x 26 = \$3978

Counting by 3's:
First week is \$51. Last week will be (52 - 1) x 3 = 153 more, or 51+153=204. So: 51, 54, 57... 198, 201, 204.
First + last: 51 + 204 = \$255.
Times 26 pairs: \$255 x 26 = \$6630

Counting by 3's, odd number of weeks:
Let's say Nancy only saved for 51 weeks, because she bought holiday presents for everyone in the last week.
First week is \$51. Last week will be (51 - 1) x 3 = 150 more, or 51+150=201. So: 51, 54, 57... 195, 198, 201.
First + last: \$51 + \$201 = \$252.
Now, there are 25 pairs + 1. The one is in the very middle. So \$252 x 25 = \$6300. Still have to add that last number that wasn't part of a pair. What was it?
Nancy saved for 51 weeks, so that's 25 weeks on one side, 25 on the other, the middle is the 26th week.
26th week is (26 - 1) x \$3 = \$75 more than week 1, or 51 + 75 = 126.
\$6300 + \$126 = \$6426
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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Puzzler wrote:

Yes! Definitely better/easier/faster to apply logic than to memorize another formula. This way, you can use the logic even if you don't start at one. Or if you are counting by 2's or 3's. For example, what if Nancy saved \$51 the first week of the year, and increased her savings by \$1 each week after that.

It's certainly better to apply logic than just learn up formulas for specific situation because you may find you have no formula for a given situation in the question. Also, you need to know exactly when the formula is applicable for example here you must know that the formula is applicable for first n positive integers only.

That said, n(n+1)/2 is a very basic and useful formula. You should know it for GMAT, not because you may not be able to do a question without it but because you may spend unnecessary amount of time on a question when other people will be able to run away on it with the formula.
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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This is an evenly spaced set since the difference between terms is always 1.

Calculate the number of terms: 52 (last) - 1 (first) + 1 = 52
Calculate the mean: (52+1) / 2 = 26.5
Multiply: 26.5 * 52 = 1,378
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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Bigred2008 wrote:
In the first week of the Year, Nancy saved \$1. In each of the next 51 weeks, she saved \$1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

A. \$1,326
B. \$1,352
C. \$1,378
D. \$2,652
E. \$2,756

Let's first set up the pattern of Nancy's savings. The first week she saved \$1, the second week she saved \$2, the third week she saved \$3, and so forth. Therefore, the total amount of money she will have saved at the end of 52 weeks will be: \$1 + \$2 + \$3 + \$4 + … + \$52. The pattern is obvious, but the arithmetic looks daunting because we need to add 52 consecutive integers. To shorten this task, we can use the formula: sum = average x quantity.

We know that Nancy saved money over the course of 52 weeks, so our quantity is 52.

To determine the average, we add together the first amount saved and the last amount saved and then divide by 2. Remember, this technique only works when we have an evenly spaced set.

The first quantity is \$1 and the last is \$52. Thus, we know:

average = (1 + 52)/2 = 53/2

Now we can determine the sum.

sum = average x quantity

sum = (53/2) x 52

sum = 53 x 26 = 1,378

Note: If we did not want to actually multiply out 26 x 53, we could have focused on units digits in the answer choices. We know that 26 x 53 will produce a units digit of 8 (because 6 x 3 = 18), and the only answer choice that has a units digit of 8 is answer choice C.
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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Bigred2008 wrote:
In the first week of the Year, Nancy saved \$1. In each of the next 51 weeks, she saved \$1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

A. \$1,326
B. \$1,352
C. \$1,378
D. \$2,652
E. \$2,756

Here's another approach (that doesn't require formulas).

So, let's add them in pairs, starting from the outside and working in.
1+2+3+4+...+51+52 = (1+52) + (2+51) + (3+50) + . . .
= 53 + 53 + 53 + ....

How many 53's are there in our new sum?
Well, there are 52 numbers in the sum 1+2+3+..+52, so there must be 26 pairs, which means there are 26 values in our new sum of 53 + 53 + 53 + ....

So, what is (26)(53)?
Fortunately, if we examine the answer choices, we see that we don't even need to calculate (26)(53)

Why not?
Notice that when we multiply (26)(53), the units digit in the product will be 8 (since 6 times 3 equals 18).

Since only 1 answer choice ends in 8, the correct answer must be

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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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Bunuel

I often get confused between n(n-1)/2 and n(n+1)/2 for consecutive integers.

When do we use n(n-1)/2 could you please explain a little.
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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Prashant10692 wrote:
Bunuel

I often get confused between n(n-1)/2 and n(n+1)/2 for consecutive integers.

When do we use n(n-1)/2 could you please explain a little.

Check this thread: https://gmatclub.com/forum/sum-of-n-pos ... 63301.html

Hope it helps.
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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Hi All,

This question describes a sequence of numbers: 1, 2, 3,…..52 and asks you to add them all up.

Adding the numbers up in order would take way too much time to be practical. There are actually a couple of ways to quickly add up these terms; here's one way called "bunching":

When adding up a group of numbers, the order of the numbers does not matter. I can group the numbers into consistent sub-groups:

1+52 = 53
2+51 = 53
3+50 = 53
4+49 = 53
etc.

So every group of 2 terms sums to 53. There are 52 total terms, so that means that there are 26 sets of 2 terms.

26(53) = 1378

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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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First of all, understand that if she saved \$1 each week and saved +\$1 more each week after. We have a number set that starts at 1 till 52. By asking for the total amount - they are asking from us the sum of this number set. How do we calculate this?
Step 1: Count the amount of integers in the set: Last-First/Increment + 1 = 52. (52 weeks)
Step 2: Find the mean of the number set: First + Last / 2 = 53 / 2 = 26.5
Step 3: Multiply the mean * the amount of integers in the set:
52.0 * 26.5 = 1378.00
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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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First week =1

Second week =1+1=2

Third week=2+1=3

Total week =52

So, n(n+1)/2=52*53/2=1378

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Re: In the first week of the Year, Nancy saved \$1.  [#permalink]

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momentofzen wrote:
How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

Its better to apply logic into this:

Sum of n terms/n = Average

In an arithmetic sequence you can tell that (first term + last term)/2 = Average

So Average 1 = Average 2

Sum of n terms/n = (frst term + last term)/2

Sum of n terms = n*(first term + last term)/2

But the last term = first term * (n-1)d ---> d being the common diference between terms

So

Sum of n terms = n*(2*first term + (n-1)d)/2 Re: In the first week of the Year, Nancy saved \$1.   [#permalink] 12 Jul 2020, 15:04

# In the first week of the Year, Nancy saved \$1.   