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# If x is the sum of the even integers from 200 to 600 inclusive, and y

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Math Expert
Joined: 02 Sep 2009
Posts: 62380
If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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23 Aug 2015, 11:48
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68% (01:59) correct 32% (01:49) wrong based on 245 sessions

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If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401

Kudos for a correct solution.

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Re: If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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23 Aug 2015, 21:48
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Bunuel wrote:
If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401

Ans: D

Solution: y = number of even int from 200 to 600 inc.
(600-200)/2 + 1 = 201=y

x= sum of even int from 200 to 600 inc.
(201/2)[200+600]= 201*400 = x

so x+y = 201+ 201*400 = 201*401
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Re: If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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23 Aug 2015, 16:37
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Bunuel wrote:
If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401

Kudos for a correct solution.

y = number of even integers between 200 and 600 (inclusive) = $$\frac{(600-200)}{2} + 1$$ = 201

Sum of all even integers from 200 to 600 , inclusive =$$\frac{y*[2*200+(y-1)*2]}{2}$$= 201*400 =x

Thus, x+y = 201*400+201 = 201(400+1) = 201*401. D is the correct answer.
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Re: If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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24 Aug 2015, 00:20
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Bunuel wrote:
If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401

Kudos for a correct solution.

IMO : D

Sum of AP = n/2 [ 1st term + last term ]
where n = number of terms.
But given in the question stem n = y
and Sum of AP = x

Sum of AP = y/2 [200+600]
x = y*400 --(i)

y = $$\frac{(400-200)}{2}$$ + 1 = 201

Thus x+y = (201*400) + 201
= 201( 400+1)
= 201*401
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Re: If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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24 Aug 2015, 01:55
1
Bunuel wrote:
If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401

Kudos for a correct solution.

number of integers from 200 to 600 is 401 (600-199)
since the first and the last terms are even, the even numbers are 201 and odd numbers are 200
Sn=sum = (201/2)(200+600) = 201*400 (AP formula)
n=201
Sn + n = 201*400 + 201 = 201*401
Math Expert
Joined: 02 Sep 2009
Posts: 62380
Re: If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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30 Aug 2015, 07:52
Bunuel wrote:
If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

There are various ways of getting the answer here. We will use the concepts we learned last week.

The given sequence is 200, 202, 204, … 600

It is an arithmetic progression. What is the total number of terms here?

You can use one of two methods to get the number of terms here:

Method 1: Using Logic

In every 100 consecutive integers, there are 50 odd integers and 50 even integers. So we will get 50 even integers from each of 200 – 299, 300 – 399, 400 – 499 and 500 – 599 i.e. a total of 50*4 = 200 even integers. Also, since the sequence includes 600, number of even integers = 200 + 1 = 201

Method 2:

Recall that in our arithmetic progressions post, we saw that the last term of a sequence which has n terms will be first term + (n – 1)* common difference.

$$600 = 200 + (n – 1)*2$$

$$n = 201$$

Hence $$y = 201$$ (because y is the number of even integers from 200 to 600)

Let’s go on now. What is the average of the sequence? Since it is an arithmetic progression with odd number of integers, the average must be the middle number i.e. 400.

Notice that since this arithmetic progressions looks like this:

(n – m), … (n – 6), (n – 4), ( n – 2), n, (n + 2), (n + 4), (n + 6), … (n + m)

We can find the middle number i.e. the average by just averaging the first and the last terms.

$$\frac{(n – m) + (n + m)}{2} = \frac{2n}{2} = n$$

$$Average = \frac{(200 + 600)}{2} = 400$$

Sum of all terms in the sequence = x = Arithmetic Mean * Number of terms = 400*201

$$x + y = 400*201 + 201 = 401*201$$

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Re: If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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10 Dec 2016, 01:39
1
Great Question.
Here is my solution to this one.

The number of multiples of 2 between 200 and 600 inclusive => $$\frac{600-200}{2} +1 = 201$$
Hence y=201

Now to get x we can use various methods.
Lets use the method pertaining to sum of an Arithmetic Progression.
x= 201/2 [200+600] = 201*400

Hence x+y=201*400 + 201 => 201[400+1] = 201*401

Hence D

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Re: If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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29 Dec 2016, 00:44
Hey, I would be very thankful if someone could use this example to illustrate the following:

(1) what would be the sum of all the odd integers?
(2) what would be the sum of all consecutive integers?
(3) what would be the sum of the reciprocal of all integers?

Thank you very much!!!!!
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If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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29 Dec 2016, 01:01
1
1
JuliaLu wrote:
Hey, I would be very thankful if someone could use this example to illustrate the following:

(1) what would be the sum of all the odd integers?
(2) what would be the sum of all consecutive integers?
(3) what would be the sum of the reciprocal of all integers?

Thank you very much!!!!!

Hi JuliaLu
Case (1) and Case (2)
I am taking both of them together as both are evenly spaced sets.
There are a number of methods to find the sum of evenly spaced set.

Let see each one of them=>
Consider a series 1,3,5,7,9,11
Method-1=>
Mean of any evenly spaced set = Average of the first and the last term.
Hence mean = 1+11/2 = 6
Mean = Sum/#
Hence Sum = Mean *6 = 6*6 = 36
Method 2=>
Using the formula => Sum =$$\frac{n}{2}[2a+(n-1)d]$$
Here n= number of terms in any evenly spaced set.
a is the first term
d is the common difference
In our example => Sum = 6/2[2*1+(6-1)*2] = 36
Method 3=>
Sum = $$\frac{n}{2}$$[First term +last term] =$$\frac{6}{2}[1+11]$$ = 36

As for your case (3)i don't think sum of reciprocals is asked on the GMAT directly.That is to say they wont ask us to find the sum of a reciprocal series,rather they will just ask the boundary condition.

Regards
Stone Cold

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If x is the sum of the even integers from 200 to 600 inclusive, and y  [#permalink]

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23 Aug 2019, 10:41
Hi everyone,

If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401

Y= ((600-200)/2)+1= 201

X=Y*average=201*400

average=(600+200)/2=400

Now: X+Y= 201+201*400=201*(1+400)=201*401
If x is the sum of the even integers from 200 to 600 inclusive, and y   [#permalink] 23 Aug 2019, 10:41
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