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Given that the sum of the odd integers from 1 to 99 is 2,500 (i.e. 1
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09 Dec 2021, 02:15
09 Dec 2021, 02:15
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72% (00:41) correct
27% (01:23) wrong based on 11 sessions
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Given that the sum of the odd integers from 1 to 99 is 2,500 (i.e. 1 + 3 + 5 +.. + 99 = 2,500), what is the sum of the even integers from 2 to 100?
(A) 2,550
(B) 2,600
(C) 2,750
(D) 3,000
(E) 5,000
(A) 2,550
(B) 2,600
(C) 2,750
(D) 3,000
(E) 5,000
Re: Given that the sum of the odd integers from 1 to 99 is 2,500 (i.e. 1
[#permalink]
09 Dec 2021, 09:34
09 Dec 2021, 09:34
1
2+4+6+8+10+...+100 = x (eqn 1) 50 distinct numbers
1+3+5+7+9+....+99 =2500 (eqn 2) 50 distinct numbers
Eqn 1 - eqn 2
1(50 times) = x - 2500
x = 50 + 2500
= 2550
Ans - A
Posted from my mobile device
1+3+5+7+9+....+99 =2500 (eqn 2) 50 distinct numbers
Eqn 1 - eqn 2
1(50 times) = x - 2500
x = 50 + 2500
= 2550
Ans - A
Posted from my mobile device
gmatclubot
Re: Given that the sum of the odd integers from 1 to 99 is 2,500 (i.e. 1 [#permalink]
09 Dec 2021, 09:34
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