Attached is a question from GMAT Prep Test 2.
Please advise how to achieve the result.
Answer: D
Thanks,
K
GMAT Test 2_PS Number Prop #3
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No. of even integers from 40 to 60, inclusive is y = [(60  40)/2] + 1 = 11kwah wrote:Attached is a question from GMAT Prep Test 2.
Please advise how to achieve the result.
Answer: D
Thanks,
K
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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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 +(111)*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
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 +(111)*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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Number of even terms between 40 and 60, both inclusive:
Nth term = first term + (N1)* 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 + (N1)*2, => N=11.
Sum of even terms between 40 and 60, both inclusive:
Sum= N/2 {2* First term + (N1)* common difference}
N=11 (previously worked out), First term =40, common difference = 2.
Then, Sum= 11/2 {2*40 + (111)*2} => Sum= 550
Therefore, Required answer = 550+11 = 561
Nth term = first term + (N1)* 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 + (N1)*2, => N=11.
Sum of even terms between 40 and 60, both inclusive:
Sum= N/2 {2* First term + (N1)* common difference}
N=11 (previously worked out), First term =40, common difference = 2.
Then, Sum= 11/2 {2*40 + (111)*2} => Sum= 550
Therefore, Required answer = 550+11 = 561
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Sum of an AP is No of terms* Avg of the AP
40 to 60 no of terms = (6040)/2+1 = 11
Sum of AP = 11/2* (40+60) = 550
Sum = 550+11 = 561
40 to 60 no of terms = (6040)/2+1 = 11
Sum of AP = 11/2* (40+60) = 550
Sum = 550+11 = 561
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