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Re: The sum of four consecutive odd numbers is equal to the sum
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29 Sep 2016, 00:54
VeritasPrepKarishma pls verify this method, you may be able to put it more clearly I guess here's my attempt: SUM = average * number of elements sum depends on average , lets find the average average of 4 consecutive odd integers must be even for eg: 75,77,79,81 here the Avg is the middle no. 78 {Basically, 4 consecutive positive odd integers can be formed if we have an even no.>4 as the average.} With this understanding , We have middle no. of 3 consecutive even no.s(i.e. average) ranging from 102 to 198 (inc.) How many even no.s can be formed using above no.s , lets see 102/4  not integer 104/4 = 26 ( even) 106/4  not integer 108/4  27 ( divisible, but not even!) 110/4  not integer 112/4  28 (even).......... If you see the pattern here we get the number we want at the gap of 8 starting from 104 so 104,112,120,.......192 {take for eg: With 120 we can form 4 consecutive odd integers by using 120 as the average  117,119,121,123} count the no. of middle elements = 192104/8 +1 = 12



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Re: The sum of four consecutive odd numbers is equal to the sum
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29 Sep 2016, 21:56
deepak268 wrote: VeritasPrepKarishma pls verify this method, you may be able to put it more clearly I guess here's my attempt: SUM = average * number of elements sum depends on average , lets find the average average of 4 consecutive odd integers must be even for eg: 75,77,79,81 here the Avg is the middle no. 78 {Basically, 4 consecutive positive odd integers can be formed if we have an even no.>4 as the average.} With this understanding , We have middle no. of 3 consecutive even no.s(i.e. average) ranging from 102 to 198 (inc.) How many even no.s can be formed using above no.s , lets see 102/4  not integer 104/4 = 26 ( even) 106/4  not integer 108/4  27 ( divisible, but not even!) 110/4  not integer 112/4  28 (even).......... If you see the pattern here we get the number we want at the gap of 8 starting from 104 so 104,112,120,.......192 {take for eg: With 120 we can form 4 consecutive odd integers by using 120 as the average  117,119,121,123} count the no. of middle elements = 192104/8 +1 = 12 Yes Deepak, it's perfectly fine. The way I see it, you are saying that the mean of 4 odd integers will be even. This mean will be equal to 3*Mean of even integers/4. Since 3 is not divisible by 4, Mean of even integers should be divisible by 4. So for every such value of mean, you will have a corresponding sequence of 4 odd integers.
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Re: The sum of four consecutive odd numbers is equal to the sum
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27 Feb 2017, 03:08
Let four odd numbers be a3, a1 a+1, a+3 and even numbers be b2, b, b+2 where a and b are even numbers. Therefore 4a =3b. Also it known that 101<b<200. a = 3b/4. Since a is even, b should be the multiple of 8. First number is 104 and last is 192. 192 =104 +(n1)8. n =12. Option A
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Re: The sum of four consecutive odd numbers is equal to the sum
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11 Apr 2017, 08:42
This is a great example of a problem that is most easily solved via pattern recognition. If there are three consecutive even numbers, then the mean of that set equals the median, which is the middle number. The sum of any set is the mean times the number of elements, so in this case, the middle number multiplied by 3. You should be able to notice that as we progress through possible "middle terms" as defined in the problem, the sum of the set increases by 6. Quickly list some of these.
100 102 => 306 104 => 312 106 => 318 108 => 324 110 => 330 112 => 336 114 => 342 116 => 348 118 => 354 120 => 360
Now concerning the second constraint. The sum of the set of odd integers will again be the mean multiplied by the number of terms. The mean of any set of consecutive odd integers will always be an integer. Thus, a number from our list that meets the constraint must be divisible by 4.
Further, the mean of any even set of odd consecutive integers will be the integer exactly between the two middle elements.
1,3,5,7 => 4 17,19 => 18 199,201,203,205 => 202
This number will ALWAYS be even when the number of elements in the set of consecutive odd integers is even. It's a given that there are four elements in the set, so after division by 4, the number must be divisible by 2. The sum of the three consecutive even integers must be divisible by 8 in order for there to be a set of 4 consecutive odd integers that sum to the same amount.
Test the list:
104 => 312 112 => 336 120 => 360
So the "middle term" that satisfies the problem starts at 104 and increments by 8 to a number less that 200.
\(\frac{(192104)}{8} + 1 = 12\)
Answer A



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Re: The sum of four consecutive odd numbers is equal to the sum
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12 Apr 2017, 01:28
Let the sequence of odd numbers ={ 2a3, 2a1, 2a+1, 2a+3 } > sum = 8a Let the sequence of even numbers ={ 2b2, 2b, 2a+2 } > sum = 6b 8a = 6b > b is divisible by 4 <> b = 4x 101 < the middle term of even numbers <200 or 101 < 2b < 200 or 101 < 8x < 200 or 12 < x < 25  there are 12 values of x > A
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Re: The sum of four consecutive odd numbers is equal to the sum
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18 Apr 2017, 10:48
thanks karishma for a highly lucid and easy to understand approach of yours. personally appreciate all your posts on the forums and thank you for your generous contribution to the forum... thanks again



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Re: The sum of four consecutive odd numbers is equal to the sum
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18 Apr 2017, 22:05
vishwash wrote: thanks karishma for a highly lucid and easy to understand approach of yours. personally appreciate all your posts on the forums and thank you for your generous contribution to the forum... thanks again Thanks Vishwash. Good to know that you find the posts useful.
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Re: The sum of four consecutive odd numbers is equal to the sum
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19 Apr 2017, 03:23
Lucky2783 wrote: krishnasty wrote: Hi, pls help me proceed, my style...or tell me wats wrong in my approach:
odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6 hence, 8a+16=6b+6 a= (3b5)/4
..how to proceed from here? odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 = \(8K_1\) even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6= 6(b+1) we have to find number which is both multiple of 6 and 8. lets looks at the minimum sum 102+104+106 = 312 = 75+77+79+81lets looks at maximum sum = 198+200 +202 = 600=197+ 199 + 201 + 203 so 312<=6(b+1) <= 600 and 6(b+1) should be divisible by 6 and 8 . 51<=b<=99, off these values whenever b+1 is multiple of 4 , 6(b+1) will be divisible by 8 and 6 too . 52 is first term and 96 is last , N= 12 . hope it helps . VeritasPrepKarishmaThanks Karishma. Your posts always help and I very well understood the solution after reading your post. As I started solving the question in the above way(that means, I took even and odd numbers as taken by Lucky), my mind keeps on thinking how can this be solved if we take these number. I tried to understand his solution. Please help why did he take the smallest middle number 104(as highlighted in red), the smallest middle number could be 102. Please correct me if I am wrong. I could not think of converting the sum of odd numbers to \(8K_1\). Please tell how can this be solved when the below even and odd numbers are considered?odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) even numbers:(2b)+(2b+2)+(2b+4)



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Re: The sum of four consecutive odd numbers is equal to the sum
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19 Apr 2017, 22:53
AR15J wrote: Lucky2783 wrote: krishnasty wrote: Hi, pls help me proceed, my style...or tell me wats wrong in my approach:
odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6 hence, 8a+16=6b+6 a= (3b5)/4
..how to proceed from here? odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 = \(8K_1\) even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6= 6(b+1) we have to find number which is both multiple of 6 and 8. lets looks at the minimum sum 102+104+106 = 312 = 75+77+79+81lets looks at maximum sum = 198+200 +202 = 600=197+ 199 + 201 + 203 so 312<=6(b+1) <= 600 and 6(b+1) should be divisible by 6 and 8 . 51<=b<=99, off these values whenever b+1 is multiple of 4 , 6(b+1) will be divisible by 8 and 6 too . 52 is first term and 96 is last , N= 12 . hope it helps . VeritasPrepKarishmaThanks Karishma. Your posts always help and I very well understood the solution after reading your post. As I started solving the question in the above way(that means, I took even and odd numbers as taken by Lucky), my mind keeps on thinking how can this be solved if we take these number. I tried to understand his solution. Please help why did he take the smallest middle number 104(as highlighted in red), the smallest middle number could be 102. Please correct me if I am wrong. I could not think of converting the sum of odd numbers to \(8K_1\). Please tell how can this be solved when the below even and odd numbers are considered?odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) even numbers:(2b)+(2b+2)+(2b+4) odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6= 6(b+1) \(8a+16=6b+6\) \(b + 1 = \frac{4*(a+2)}{3}\) So (b + 1) is a multiple of 4. "Given that the middle term of the even numbers is greater than 101 and lesser than 200"  Middle term of even numbers = 2b + 2 The minimum value it can take is 102 so minimum value of b + 1 is 51. But, b+1 needs to be a multiple of 4. Hence, the minimum value of b+1 would be 52 and hence the minimum value of the minimum term would be 104. Hope this clarifies.
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Re: The sum of four consecutive odd numbers is equal to the sum
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20 Apr 2017, 01:28
Option A4 Odd numbers = O, O+2, O+4 & O+6 3 Even numbers = E, E+2 & E+4 Given: 101 < E+2 < 200. \(O + O+2 + O+4 + O+6 = E + E+2 + E+4\) \(4O + 12 = 3E + 6\) \(\frac{4}{3}O + 4 = E + 2\) \(:101 < \frac{4}{3}O + 4 < 200\) \(:97 < \frac{4}{3}O < 196\) \(:\frac{291}{4} < O < \frac{588}{4}\) \(:72.75 < O < 147\) \(:73 <= O <= 145\) [PS  O is an Odd Integer] Total number of O's satisfying above condition =\((145  73)/2 + 1\). But, we require O to be a multiple of 3 as well. As per  \(\frac{4}{3}O + 4 = E + 2\): O also needs to a multiple of 3. So \(O = 3k\) [PS  k is an arbitrary Positive Odd integer]. Now, \(:73 <= O <= 145\) can be written as \(:73 <= 3k <= 145\) \(:\frac{73}{3} <= k <= \frac{145}{3}\) \(: 24.33 <= k <= 48.33\) \(: 25 <= k <= 48\) Total number of k's satisfying above condition \(= (48  25)/2 + 1 = 11.5 + 1 = 12.5\) ~ \(12\)
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Re: The sum of four consecutive odd numbers is equal to the sum
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20 Apr 2017, 05:32
VeritasPrepKarishma wrote: odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 = \(8K_1\) even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6= 6(b+1) VeritasPrepKarishmaThanks Karishma. Your posts always help and I very well understood the solution after reading your post. As I started solving the question in the above way(that means, I took even and odd numbers as taken by Lucky), my mind keeps on thinking how can this be solved if we take these number. I tried to understand his solution. Please help why did he take the smallest middle number 104(as highlighted in red), the smallest middle number could be 102. Please correct me if I am wrong. I could not think of converting the sum of odd numbers to \(8K_1\). Please tell how can this be solved when the below even and odd numbers are considered?odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) even numbers:(2b)+(2b+2)+(2b+4) odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6= 6(b+1) \(8a+16=6b+6\) \(b + 1 = \frac{4*(a+2)}{3}\) So (b + 1) is a multiple of 4. "Given that the middle term of the even numbers is greater than 101 and lesser than 200"  Middle term of even numbers = 2b + 2 The minimum value it can take is 102 so minimum value of b + 1 is 51. But, b+1 needs to be a multiple of 4. Hence, the minimum value of b+1 would be 52 and hence the minimum value of the minimum term would be 104. Hope this clarifies.[/quote] Thanks. You are just amazing.



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Re: The sum of four consecutive odd numbers is equal to the sum
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13 Jun 2017, 08:11
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
check this out, m = odd number n = even number
n is 101<n<200, therefore 3 consecutive even numbers will be n2, n, n+2 and for m it will be m2, m, m+2, m+4
m2+m+m+2+m+4=n2+n+n+2 4m+4=3n m = 3n/4  1
now n is such, 101<n, also, n divisible by 4, and 3n/4 is even so subtracting 1 later will give odd number which would be m (102)*3/4 is not divisible by 4 = exclude (104)*3/4 is both divisible by 4 and even = include (106)*3/4 = exclude (108)*3/4 = exclude (divisible by 4 but odd) (110)*3/4 = exclude (112)*3/4 = include (200)*3/4 = include here, we observe jump of 8 from 104 to 112 therefore, 200104/8 = 98/8 = 12
Ans : A



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The sum of four consecutive odd numbers is equal to the sum
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24 Jul 2017, 07:26
bindiy to be clear the phrase should read "less than 200 . . . "
"the middle term of the even numbers is greater than 101 and lesser than 200 . . ."



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Re: The sum of four consecutive odd numbers is equal to the sum
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24 Jul 2017, 07:51
bindiyajoisher wrote: The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12 (B) 17 (C) 25 (D) 33 (E) 50 Four consecutive odd numbers = \((2x3), (2x1), (2x+1), (2x+3)\). Three consecutive even numbers = \((2y2), 2y, (2y+2)\). \((2x3) + (2x1) + (2x+1) + (2x+3) = (2y2) + (2y) + (2y+2)\) \(8x = 6y\) \(4x = 3y\), so, \(\frac{x}{y} = \frac{3}{4}\). Also, given: \(101 < 2y < 200\). So, \(2y\) must be divisible of \(8\) between the numbers \(101\) and \(200\), possible values are \(104, 112, 120, ... 192\): \(\frac{192104}{8} + 1 = 12\). Ans  A.
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Re: The sum of four consecutive odd numbers is equal to the sum
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07 Aug 2018, 08:25
bubna3number wrote: Is the answer A? Yes!! The correct answer is A.
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Re: The sum of four consecutive odd numbers is equal to the sum &nbs
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