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12 Jun 2018, 05:55
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I used the following method.
The sum of first 50 even integers is nu of terms* av or 50/2+1=26* 25 (is av)
The sum of first 50 odd integers is nu of terms*av 50/2-1=24*25.
So we have 26*25-24*25=25(26-24)=25*2=50
The sum of first 50 even integers is nu of terms* av or 50/2+1=26* 25 (is av)
The sum of first 50 odd integers is nu of terms*av 50/2-1=24*25.
So we have 26*25-24*25=25(26-24)=25*2=50
04 Aug 2018, 04:04
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First all the answer-choices are non-negative, so figured x would must be equal or bigger to y. Helped in my thinking.
Started with taking 2 (first even integer) + 100 (50th even integer)/2, average=51. Then multiple the number by the amount of even integers=50
51*50.
Then first odd and 50th odd--> 1+99/2=50, 50*50(amount of odd integer)
So (51*50)-(50*50), you dont have to do the calculation, you simply see its gonna be 50 left when (50*50) is subtracted.
Hence, we got our answer :D
Thumbs up if you thought it was a good explanation
Started with taking 2 (first even integer) + 100 (50th even integer)/2, average=51. Then multiple the number by the amount of even integers=50
51*50.
Then first odd and 50th odd--> 1+99/2=50, 50*50(amount of odd integer)
So (51*50)-(50*50), you dont have to do the calculation, you simply see its gonna be 50 left when (50*50) is subtracted.
Hence, we got our answer :D
Thumbs up if you thought it was a good explanation
31 Aug 2018, 23:07
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Hi,
well there are several super explanations given by experts which we must internalize to better our qunat score. However, just an alternate approach.
If we look at set of numbers which we get in x { 2,4,6,8,10,12.......100} and similarly set of numbers y {1,3,5,7,9...99)
now x-y would be {2 -1 + 4-3 + 6-5+ 8-7+....... 100-99}
if we just see the set performs for first 5 even odd we can then infer about the reaming set
So in first 5 even -odd it would be { 2 -1 + 4-3 + 6-5+ 8-7 +10-1}= 5
Now since we have for first 5 (even -odd) the result is 5 , so there are 10 such sets , so the answer must be 10 *5= 50
Probus
well there are several super explanations given by experts which we must internalize to better our qunat score. However, just an alternate approach.
If we look at set of numbers which we get in x { 2,4,6,8,10,12.......100} and similarly set of numbers y {1,3,5,7,9...99)
now x-y would be {2 -1 + 4-3 + 6-5+ 8-7+....... 100-99}
if we just see the set performs for first 5 even odd we can then infer about the reaming set
So in first 5 even -odd it would be { 2 -1 + 4-3 + 6-5+ 8-7 +10-1}= 5
Now since we have for first 5 (even -odd) the result is 5 , so there are 10 such sets , so the answer must be 10 *5= 50
Probus
