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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, 01:11
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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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Re: If x is the sum of all odd numbers between 1001 and 2000, inclusive an [#permalink]
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27 Apr 2018, 01:30
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 inclusivex = 1000*500 + 1 + 3 + 5 +...+ 999 similarly, y is identical to x except for the fact it has only even numbers between 1001 & 2000 inclusivey = 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 outx  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



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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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27 Apr 2018, 03:24
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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



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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, 09:21
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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 SEQUENCESum = (Average)(# of terms) Average: \(\frac{FirstTerm+LastTerm}{2}\)Number of terms: \(\frac{LastTermFirstTerm}{Increment} + 1\)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: \(\frac{19991001}{2}=\frac{998}{2}=(499 +1)=500\)
\(Sum_{x}=(1,500)(500)\)• \(y\) = Sum of all even numbers between \(1001\) and \(2000\), inclusive Average: \(\frac{1002+2000}{2}=1,501\)# of terms: \(\frac{20001002}{2}=(499+1)=500\)
\(Sum_{y}=(1,501)(500)\)• 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
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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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27 Apr 2018, 09:22
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) +... xy= a + (a+2)  (a+1)  (a+3) ... =2*n times Let's calculate n, n= (20001001+1)/4 = 250, so we have 2*250 times = 500. Answer (B)
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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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27 Apr 2018, 13:22
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 SEQUENCESum = (Average)(# of terms) Average: \(\frac{FirstTerm+LastTerm}{2}\)Number of terms: \(\frac{LastTermFirstTerm}{Increment} + 1\)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: \(\frac{19991001}{2}=\frac{998}{2}=(499 +1)=500\)
\(Sum_{x}=(1,500)(500)\)• \(y\) = Sum of all even numbers between \(1001\) and \(2000\), inclusive Average: \(\frac{1002+2000}{2}=1,501\)# of terms: \(\frac{20001002}{2}=(499+1)=500\)
\(Sum_{y}=(1,501)(500)\)• 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 HEREHey generis great explanation you know i thnk i made a mistake in my post formula #5 here https://gmatclub.com/forum/arithmeticp ... l#p2035478to 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



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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, 17:50
dave13 wrote: Hey generis great explanation you know i thnk i made a mistake in my post formula #5 here https://gmatclub.com/forum/arithmeticp ... l#p2035478to 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.
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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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27 Apr 2018, 19:52
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



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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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28 Apr 2018, 05:48
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 SEQUENCESum = (Average)(# of terms) Average: \(\frac{FirstTerm+LastTerm}{2}\)Number of terms: \(\frac{LastTermFirstTerm}{Increment} + 1\)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: \(\frac{19991001}{2}=\frac{998}{2}=(499 +1)=500\)
\(Sum_{x}=(1,500)(500)\)• \(y\) = Sum of all even numbers between \(1001\) and \(2000\), inclusive Average: \(\frac{1002+2000}{2}=1,501\)# of terms: \(\frac{20001002}{2}=(499+1)=500\)
\(Sum_{y}=(1,501)(500)\)• 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 HEREThanks a lot. It's a helpful post. get the clear concept. u have explained it really well.



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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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29 Apr 2018, 18:34
=> \(x = 1001 + 1003 + 1005 + … + 1999\) \(y = 1002 + 1004 + 1006 + … + 2000\) \(x – y = ( 1001 – 1002 ) + ( 1003 – 1004 ) + … + ( 1999 – 2000 )\) \(= (1) + (1) + … + (1)\) \(= (1) * 500\) \(= 500\) Therefore, the answer is B. Answer: B
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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, 16:42
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 Answer: B
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