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# If x is the sum of all odd numbers between 1001 and 2000, inclusive an

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Math Revolution GMAT Instructor
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If x is the sum of all odd numbers between 1001 and 2000, inclusive an  [#permalink]

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27 Apr 2018, 00:11
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55% (hard)

Question Stats:

64% (01:49) correct 36% (01:20) wrong based on 89 sessions

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[GMAT math practice question]

If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$?

$$A. -1000$$
$$B. -500$$
$$C. 0$$
$$D. 500$$
$$E. 1000$$

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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior PS Moderator Status: It always seems impossible until it's done. Joined: 16 Sep 2016 Posts: 670 Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 27 Apr 2018, 00:30 MathRevolution wrote: [GMAT math practice question] If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$? $$A. -1000$$ $$B. -500$$ $$C. 0$$ $$D. 500$$ $$E. 1000$$ x = 1001 + 1003 + ... + 1999 x = (1000 + 1000 + 1000 + ...+ 1000) + (1 + 3 + 5 + ... + 999) each of the above brackets have 500 terms.. as there are 500 odd numbers between 1001 & 2000 inclusive x = 1000*500 + 1 + 3 + 5 +...+ 999 similarly, y is identical to x except for the fact it has only even numbers between 1001 & 2000 inclusive y = 1000*500 + 2 + 4 + 6 +...+1000 x - y = (1 + 3 + 5 + ... + 999) - (2 + 4 + 6 + ... + 1000) pairing off integers one by one also notw, the two equal terms of 1000*500 in x & y cancel out x - y = (1 - 2) + ( 3 - 4) + ... + (999 - 1000) ... there are 500 brackets in this expression x - y = (-1) + (-1) + ... + (-1) ... 500 times x - y = -500 Hence Option (B) is our answer. Best, Gladi _________________ Regards, Gladi “Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back) Director Joined: 04 Dec 2015 Posts: 739 Location: India Concentration: Technology, Strategy Schools: ISB '19, IIMA , IIMB, XLRI WE: Information Technology (Consulting) Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 27 Apr 2018, 02:24 1 1 MathRevolution wrote: [GMAT math practice question] If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$? $$A. -1000$$ $$B. -500$$ $$C. 0$$ $$D. 500$$ $$E. 1000$$ The pattern which $$1$$ to $$10$$ follows, same will be followed by $$1001$$ to $$2000$$. Lets try using simple set of numbers from $$1$$ to $$10$$. Let $$x$$ be sum of all odd numbers $$= 1 + 3 +5 + 7 + 9 = 25$$ Let $$y$$ be sum of all even numbers $$= 2 + 4 + 6 + 8 + 10 = 30$$ Therefore; $$x - y = 25 - 30 = -5$$ Hence following the pattern from above we can find out for $$1001$$ to $$2000$$ series : $$x - y = -500$$ Answer B Senior SC Moderator Joined: 22 May 2016 Posts: 2359 If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 27 Apr 2018, 08:21 1 2 MathRevolution wrote: [GMAT math practice question] If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$? $$A. -1000$$ $$B. -500$$ $$C. 0$$ $$D. 500$$ $$E. 1000$$ Another way: Find the sums. No need to multiply fully. The answer should be clear from the setup. SUM OF ARITHMETIC SEQUENCE Sum = (Average)(# of terms) Average: $$\frac{FirstTerm+LastTerm}{2}$$ Number of terms: $$\frac{LastTerm-FirstTerm}{Increment} + 1$$ Increment for evens/odds is 2 x, first term is 1001, last term is 1999 y, first term is 1002, last term is 2000 $$x$$ = Sum of all odd numbers between $$1001$$ and $$2000$$, inclusive Average: $$\frac{1001+1999}{2}=\frac{3000}{2}=1500$$ # of terms: $$\frac{1999-1001}{2}=\frac{998}{2}=(499 +1)=500$$ $$Sum_{x}=(1,500)(500)$$ $$y$$ = Sum of all even numbers between $$1001$$ and $$2000$$, inclusive Average: $$\frac{1002+2000}{2}=1,501$$ # of terms:$$\frac{2000-1002}{2}=(499+1)=500$$ $$Sum_{y}=(1,501)(500)$$ Value of $$x – y$$? $$Sum_{x}=(1,500)(500)$$ $$Sum_{y}=(1,501)(500)$$ There is one extra factor of 500* in $$y$$ $$x < y$$ Value of $$(x – y)= -500$$ Answer B *1,500 factors of 500 vs. 1,501 factors of 500 $$Sum_{x}=(1,500)(500)$$ $$Sum_{y}=(1,501)(500)=(1,500+1)(500)$$ An excellent summary/review of sequences on GMAT club is HERE _________________ Life’s like a movie. Write your own ending. Kermit the Frog Senior Manager Joined: 29 Dec 2017 Posts: 386 Location: United States Concentration: Marketing, Technology GMAT 1: 630 Q44 V33 GMAT 2: 690 Q47 V37 GMAT 3: 710 Q50 V37 GPA: 3.25 WE: Marketing (Telecommunications) Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 27 Apr 2018, 08:22 MathRevolution wrote: [GMAT math practice question] If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$? $$A. -1000$$ $$B. -500$$ $$C. 0$$ $$D. 500$$ $$E. 1000$$ Let 1001 = a, then x = a + (a+2) +... y= (a+1) + (a+3) +... x-y= a + (a+2) - (a+1) - (a+3) ... =-2*n times Let's calculate n, n= (2000-1001+1)/4 = 250, so we have -2*250 times = -500. Answer (B) VP Joined: 09 Mar 2016 Posts: 1287 Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 27 Apr 2018, 12:22 generis wrote: MathRevolution wrote: [GMAT math practice question] If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$? $$A. -1000$$ $$B. -500$$ $$C. 0$$ $$D. 500$$ $$E. 1000$$ Another way: Find the sums. No need to multiply fully. The answer should be clear from the setup. SUM OF ARITHMETIC SEQUENCE Sum = (Average)(# of terms) Average: $$\frac{FirstTerm+LastTerm}{2}$$ Number of terms: $$\frac{LastTerm-FirstTerm}{Increment} + 1$$ Increment for evens/odds is 2 x, first term is 1001, last term is 1999 y, first term is 1002, last term is 2000 $$x$$ = Sum of all odd numbers between $$1001$$ and $$2000$$, inclusive Average: $$\frac{1001+1999}{2}=\frac{3000}{2}=1500$$ # of terms: $$\frac{1999-1001}{2}=\frac{998}{2}=(499 +1)=500$$ $$Sum_{x}=(1,500)(500)$$ $$y$$ = Sum of all even numbers between $$1001$$ and $$2000$$, inclusive Average: $$\frac{1002+2000}{2}=1,501$$ # of terms:$$\frac{2000-1002}{2}=(499+1)=500$$ $$Sum_{y}=(1,501)(500)$$ Value of $$x – y$$? $$Sum_{x}=(1,500)(500)$$ $$Sum_{y}=(1,501)(500)$$ There is one extra factor of 500* in $$y$$ $$x < y$$ Value of $$(x – y)= -500$$ Answer B *1,500 factors of 500 vs. 1,501 factors of 500 $$Sum_{x}=(1,500)(500)$$ $$Sum_{y}=(1,501)(500)=(1,500+1)(500)$$ An excellent summary/review of sequences on GMAT club is HERE Hey generis great explanation you know i thnk i made a mistake in my post -formula #5 here https://gmatclub.com/forum/arithmetic-p ... l#p2035478 to find number of terms it says we need add up first and last terms, whereas in your post we subtract - can you please explain is the formula #5 incorrect in my post or these are diffrent things Enjoy the weekend Senior SC Moderator Joined: 22 May 2016 Posts: 2359 If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 27 Apr 2018, 16:50 dave13 wrote: Hey generis great explanation you know i thnk i made a mistake in my post -formula #5 here https://gmatclub.com/forum/arithmetic-p ... l#p2035478 to find number of terms it says we need add up first and last terms, whereas in your post we subtract - can you please explain is the formula #5 incorrect in my post or these are diffrent things Enjoy the weekend dave13 , I checked, and posted a reply. Number 5's formula in your post was incorrect. In # 7 of that same post, however, the formula is correct. _________________ Life’s like a movie. Write your own ending. Kermit the Frog Manager Joined: 02 Jan 2016 Posts: 124 Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 27 Apr 2018, 18:52 MathRevolution wrote: [GMAT math practice question] If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$? $$A. -1000$$ $$B. -500$$ $$C. 0$$ $$D. 500$$ $$E. 1000$$ So, X = sum of all odd numbers and Y = sum of all even numbers. Also X starts from 1001 and ends at 1999 and Y starts at 1002 and ends at 2000. So the pattern is "1001 (x1) - 2000 (y1)" ... up until "1999 - 2000", notice the difference will always be only -1 Therefore, "Sum of X" - Sum of Y" will be negative. Now we just need to find out No. of Terms; # of terms: 1999−1001/2 = 998/2= (499+1)=500 note no. of terms will be same in case of X as well as Y. Thus 500 * -1 = 500 Director Joined: 31 Oct 2013 Posts: 977 Concentration: Accounting, Finance GPA: 3.68 WE: Analyst (Accounting) Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 28 Apr 2018, 04:48 generis wrote: MathRevolution wrote: [GMAT math practice question] If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$? $$A. -1000$$ $$B. -500$$ $$C. 0$$ $$D. 500$$ $$E. 1000$$ Another way: Find the sums. No need to multiply fully. The answer should be clear from the setup. SUM OF ARITHMETIC SEQUENCE Sum = (Average)(# of terms) Average: $$\frac{FirstTerm+LastTerm}{2}$$ Number of terms: $$\frac{LastTerm-FirstTerm}{Increment} + 1$$ Increment for evens/odds is 2 x, first term is 1001, last term is 1999 y, first term is 1002, last term is 2000 $$x$$ = Sum of all odd numbers between $$1001$$ and $$2000$$, inclusive Average: $$\frac{1001+1999}{2}=\frac{3000}{2}=1500$$ # of terms: $$\frac{1999-1001}{2}=\frac{998}{2}=(499 +1)=500$$ $$Sum_{x}=(1,500)(500)$$ $$y$$ = Sum of all even numbers between $$1001$$ and $$2000$$, inclusive Average: $$\frac{1002+2000}{2}=1,501$$ # of terms:$$\frac{2000-1002}{2}=(499+1)=500$$ $$Sum_{y}=(1,501)(500)$$ Value of $$x – y$$? $$Sum_{x}=(1,500)(500)$$ $$Sum_{y}=(1,501)(500)$$ There is one extra factor of 500* in $$y$$ $$x < y$$ Value of $$(x – y)= -500$$ Answer B *1,500 factors of 500 vs. 1,501 factors of 500 $$Sum_{x}=(1,500)(500)$$ $$Sum_{y}=(1,501)(500)=(1,500+1)(500)$$ An excellent summary/review of sequences on GMAT club is HERE Thanks a lot. It's a helpful post. get the clear concept. u have explained it really well. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6824 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink] ### Show Tags 29 Apr 2018, 17:34 => $$x = 1001 + 1003 + 1005 + … + 1999$$ $$y = 1002 + 1004 + 1006 + … + 2000$$ $$x – y = ( 1001 – 1002 ) + ( 1003 – 1004 ) + … + ( 1999 – 2000 )$$ $$= (-1) + (-1) + … + (-1)$$ $$= (-1) * 500$$ $$= -500$$ Therefore, the answer is B. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an  [#permalink]

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30 Apr 2018, 15:42
MathRevolution wrote:
[GMAT math practice question]

If $$x$$ is the sum of all odd numbers between $$1001$$ and $$2000$$, inclusive and $$y$$ is the sum of all even numbers between $$1001$$ and $$2000$$, inclusive, what is the value of $$x – y$$?

$$A. -1000$$
$$B. -500$$
$$C. 0$$
$$D. 500$$
$$E. 1000$$

We se that x = 1001 + 1003 + 1005 + … + 1999 and y = 1002 + 1004 + 1006 + … + 2000. Thus,

x - y = (1001 + 1003 + 1005 + … + 1999) - (1002 + 1004 + 1006 + … + 2000)

x - y = (1001 - 1002) + (1003 - 1004) + (1005 - 1006) + … + (1999 - 2000)

x - y = (-1) + (-1) + (-1) + … + (-1)

We see that x - y is the sum of some quantity of -1’s. We need to determine the number of -1’s we are adding. Notice that there are 500 odd numbers from 1001 to 1999, inclusive. Likewise, there are 500 even numbers from 1002 to 2000, inclusive. When we pair a number from the ‘x’ sum and a number from the ‘y’ sum, we have 500 pairs. Thus, we are adding 500 -1’s. Therefore,

x - y = 500(-1)

x - y = -500

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Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an &nbs [#permalink] 30 Apr 2018, 15:42
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