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For all even integers n, h(n) is defined to be the sum of the even

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For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   Updated on: 31 Aug 2016, 02:18
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For all even integers n, h(n) is defined to be the sum of the even integers between 2 and n, inclusive. What is the value of h(18)/h(10) ?

(A) 1.8
(B) 3
(C) 6
(D) 18
(E) 60
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B

Originally posted by mystiquethinker on 31 Aug 2016, 01:56.
Last edited by Bunuel on 31 Aug 2016, 02:18, edited 1 time in total.
RENAMED THE TOPIC.
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   31 Aug 2016, 05:13
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mystiquethinker wrote:
For all even integers n, h(n) is defined to be the sum of the even integers between 2 and n, inclusive. What is the value of h(18)/h(10) ?

(A) 1.8
(B) 3
(C) 6
(D) 18
(E) 60


CONCEPT: When terms are in Arithmetic Progression (A.P.) i.e. terms are equally spaced then

Mean = Median =(First+Last)/2

and Sum = Mean*Number of terms

h(18) = [(2+18)/2]*9 = 90

h(10) = (2+10)/2]*5 = 30

h(18)/h(10) = (90) / (30) = 3
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   31 Aug 2016, 07:48
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mystiquethinker wrote:
For all even integers n, h(n) is defined to be the sum of the even integers between 2 and n, inclusive. What is the value of h(18)/h(10) ?

(A) 1.8
(B) 3
(C) 6
(D) 18
(E) 60


We can also apply a nice rule that says (a + b)/c = a/c + b/c AND use some estimation, since the answer choices are nicely spread apart.

h(18)/h(10) = (2+4+6+8+10+12+14+16+18)/(2+4+6+8+10)
= (2+4+6+8+10)/(2+4+6+8+10) + (12+14+16+18)/(2+4+6+8+10)
= 1 + (12+14+16+18)/(2+4+6+8+10)
= 1 + some value around 2 [yes, I'm aware that the fraction evaluates to be exactly 2, but we could actually be quite aggressive in our estimation and still reach the correct answer]
≈ 3

Answer:
Show Spoiler
B


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Re: What is the value of h(18)/h(10) ? [#permalink] New post   31 Aug 2016, 02:38
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sum of n terms can be represented by the formula = \(\frac{n*(n+1)}{2}\)

h(10) :- 2+4+6+8+10 = 2(1+2+3+4+5) = 2 * \(\frac{5*6}{2}\) = 30

h(18) :- 2+4+6+8+10+12+14+16+18 = 2(1+2+3+4+5+6+7+8+9) = \(\frac{2 * 9 *10}{2}\) = 90

\(\frac{h(18)}{h(10)}\) = \(\frac{90}{30}\) =3

correct answer -\(B\)
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For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   20 Jul 2017, 13:48
Hi,
Could I apply the following?

\(h(18)/h(10)\) simplified is \(h(9)/h(5)\), so the even numbers are \((2+4+6+8)/(2+4)\) = 3

Thanks
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For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   22 Aug 2017, 09:05
Esguitar wrote:
Hi,
Could I apply the following?

\(h(18)/h(10)\) simplified is \(h(9)/h(5)\), so the even numbers are \((2+4+6+8)/(2+4)\) = 3

Thanks


h(10) = 2+4+6+8+10 = 30
h(5) = 2+4 =6
so h(10) can't = 2 x h(5), you can't simplify like that.
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   22 Aug 2017, 10:46
The number are in AP
we can use n/2(2a + (n-1)d)
we will get sum of n even terms ---( a = 2 , d = 2) =n(n+1)
Now 2 to 18 we have 9 terms
2 to 10 we have 5 terms
9*10 / 5*6
we will get 3
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   24 Aug 2017, 16:05
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mystiquethinker wrote:
For all even integers n, h(n) is defined to be the sum of the even integers between 2 and n, inclusive. What is the value of h(18)/h(10) ?

(A) 1.8
(B) 3
(C) 6
(D) 18
(E) 60


First, notice that h(18) will be the sum of the even integers between 2 and 18 inclusive. We can thus calculate h(18) using the formula sum = avg x quantity.

First, find the average of this evenly spaced set of numbers:

avg = (18 + 2)/2 = 10

Now calculate how many numbers are in this set:

quantity = (18 - 2)/2 + 1 = 9

Now we use the formula sum = avg x quantity:

sum = (10)(9)

sum = 90 = h(18)

Next we can determine h(10) using the same procedure we used to calculate h(18):

avg = (10 + 2)/2 = 6

quantity = (10 - 2)/2 + 1 = 5

sum = 30

Thus, h(18)/h(10) = 90/30 = 3

Answer: B
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   22 Mar 2018, 03:31
Hi Everyone,

I used a different way here

A modified version of \(\frac{(n*(n+1))}{2}\) would be \(\frac{(n*(n+2))}{4}\)

\(h(18)=\frac{(18)(20)}{4} = \frac{360}{4}= 90\)

\(h(10)= \frac{(10)(12)}{4} = \frac{120}{4}= 30\)

\(\frac{h(18)}{h(10)}= \frac{90}{30}= 3\) ===> B
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   23 Mar 2020, 12:54
GMATPrepNow wrote:
mystiquethinker wrote:
For all even integers n, h(n) is defined to be the sum of the even integers between 2 and n, inclusive. What is the value of h(18)/h(10) ?

(A) 1.8
(B) 3
(C) 6
(D) 18
(E) 60


We can also apply a nice rule that says (a + b)/c = a/c + b/c AND use some estimation, since the answer choices are nicely spread apart.

h(18)/h(10) = (2+4+6+8+10+12+14+16+18)/(2+4+6+8+10)
= (2+4+6+8+10)/(2+4+6+8+10) + (12+14+16+18)/(2+4+6+8+10)
= 1 + (12+14+16+18)/(2+4+6+8+10)
= 1 + some value around 2 [yes, I'm aware that the fraction evaluates to be exactly 2, but we could actually be quite aggressive in our estimation and still reach the correct answer]
≈ 3
_______________________________________________________________________

Can I use another nice rule ab/cd for this?
The answer is also a close approximation to B. If h(18)/h(10) = h/h * 18/10 = 9/5.
In the numerator, the even terms would be 2+4+6+8=20.
In the denominator, the even terms would be 2+4=6.
Therefore, 20/6=~3.
Please explain why "yes"/"no". Thank you!
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   21 Dec 2022, 01:07
BrentGMATPrepNow wrote:
mystiquethinker wrote:
For all even integers n, h(n) is defined to be the sum of the even integers between 2 and n, inclusive. What is the value of h(18)/h(10) ?

(A) 1.8
(B) 3
(C) 6
(D) 18
(E) 60


We can also apply a nice rule that says (a + b)/c = a/c + b/c AND use some estimation, since the answer choices are nicely spread apart.

h(18)/h(10) = (2+4+6+8+10+12+14+16+18)/(2+4+6+8+10)
= (2+4+6+8+10)/(2+4+6+8+10) + (12+14+16+18)/(2+4+6+8+10)
= 1 + (12+14+16+18)/(2+4+6+8+10)
= 1 + some value around 2 [yes, I'm aware that the fraction evaluates to be exactly 2, but we could actually be quite aggressive in our estimation and still reach the correct answer]
≈ 3

Answer:
Show Spoiler
B


RELATED VIDEO


Hi Brent BrentGMATPrepNow, is this way of calculation correct too?

n∗(n+1)/2 = (18 * 19)/2 = 171
n∗(n+1)/2 = (10 * 11)/2 = 55

h(18)/h(10) = 171/55 = 34/11 > 30/10 approximately 3
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   21 Dec 2022, 15:35
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Kimberly77 wrote:
Hi Brent BrentGMATPrepNow, is this way of calculation correct too?

n∗(n+1)/2 = (18 * 19)/2 = 171
n∗(n+1)/2 = (10 * 11)/2 = 55

h(18)/h(10) = 171/55 = 34/11 > 30/10 approximately 3


Given: h(n) is defined to be the sum of the even integers between 2 and n, inclusive

So, h(18) = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90
And, h(10) = 2 + 4 + 6 + 8 + 10 = 30

The formula that you are using is for the sum of all integers from 1 to n inclusive (not just the even integers).

So, it's merely a coincidence that you got the correct answer.
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink] New post   24 Dec 2022, 12:20
BrentGMATPrepNow wrote:
Kimberly77 wrote:
Hi Brent BrentGMATPrepNow, is this way of calculation correct too?

n∗(n+1)/2 = (18 * 19)/2 = 171
n∗(n+1)/2 = (10 * 11)/2 = 55

h(18)/h(10) = 171/55 = 34/11 > 30/10 approximately 3


Given: h(n) is defined to be the sum of the even integers between 2 and n, inclusive

So, h(18) = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90
And, h(10) = 2 + 4 + 6 + 8 + 10 = 30

The formula that you are using is for the sum of all integers from 1 to n inclusive (not just the even integers).

So, it's merely a coincidence that you got the correct answer.


Thanks BrentGMATPrepNow for clarification. Noted thanks :please: :thumbsup:
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Re: For all even integers n, h(n) is defined to be the sum of the even [#permalink]
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