GMAT Test 2_PS Number Prop #3

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GMAT Test 2_PS Number Prop #3

by kwah » Tue Apr 03, 2012 7:07 pm
Attached is a question from GMAT Prep Test 2.

Please advise how to achieve the result.

Answer: D

Thanks,
K
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by [email protected] » Tue Apr 03, 2012 7:40 pm
kwah wrote:Attached is a question from GMAT Prep Test 2.

Please advise how to achieve the result.

Answer: D

Thanks,
K
No. of even integers from 40 to 60, inclusive is y = [(60 - 40)/2] + 1 = 11
Sum of even integers from 40 to 60, inclusive = [(40 + 60)/2] * 11 = 50 * 11 = 550

So, x + y = 550 + 11 = 561

The correct answer is D.
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by gmatmath » Fri Apr 06, 2012 1:56 am
to find x, the sum upto n terms is Sn, we need to make use of AP.
the first term 'a' = 40, last term Tn = 60, common difference'd' = 2
hence, number of terms 'n' = ?
Pluging in the formula Tn = a + (n - 1)d, we get,
60 = 40 + 2(n - 1)
==> 20 = 2(n - 1)
==> n - 1 = 10
==> n = 11
there are 11 terms between 40 and 60.
Now, finding the sum of these 11 terms,
Sn = (n/2)[2a + (n - 1)d]
Sn = (11/2)[2*40 +(11-1)*2]
Sn = (11/2)[80 + 20]
Sn = 50 * 11 = 550
So, x = 550


y is the number of even numbers between 40 and 60, which was found to be 'n' = 11.
Hence, x + y = 550 + 11 = 561, which is option D

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by Birottam Dutta » Fri Apr 06, 2012 8:16 am
Number of even terms between 40 and 60, both inclusive:

Nth term = first term + (N-1)* common difference, where N is number of terms.

Here First term is 40 and common difference is 2 and Nth term is 60.

Therefore, 60 = 40 + (N-1)*2, => N=11.

Sum of even terms between 40 and 60, both inclusive:

Sum= N/2 {2* First term + (N-1)* common difference}

N=11 (previously worked out), First term =40, common difference = 2.

Then, Sum= 11/2 {2*40 + (11-1)*2} => Sum= 550

Therefore, Required answer = 550+11 = 561

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by ronnie1985 » Sat Apr 07, 2012 9:38 am
Sum of an AP is No of terms* Avg of the AP
40 to 60 no of terms = (60-40)/2+1 = 11
Sum of AP = 11/2* (40+60) = 550
Sum = 550+11 = 561
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