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Given ways in solution, is not how you should do it.
Question:
In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 560. What is the sum of the last 5 integers in the sequence?
(A) 585
(B) 580
(C) 575
(D) 570
(E) 565
Approach:
if lets say 5 digits are 1, 2, 3, 4, 5 AND there addition is 15.
How we get digits from 15?
15/5 = 3 consecutive digits means two digits left to 3 and two digits right to 3.
1, 2, 3, 4, 5
Solution:
Given addition is 560.
560/5 = 112, which is middle digit.
Consecution 5 digits for which addition is 560 is 110, 111, 112, 113, 114.
Now they want addition of next 5 digits.
which are: 115 + 116+ 117+ 118+ 119.
Now to sum it up quickly, 5, 6, 7, 8 and 9 are unit digits. and if you want to add them take unit digit of middle number (117) which is 7 and multiple by 5 = 35
Therefore unit digit is 5. now 5*1 = 5+3 = 8, and thousand's place is again five 1s which is 5.
Answer is 585 A.
Question:
In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 560. What is the sum of the last 5 integers in the sequence?
(A) 585
(B) 580
(C) 575
(D) 570
(E) 565
Approach:
if lets say 5 digits are 1, 2, 3, 4, 5 AND there addition is 15.
How we get digits from 15?
15/5 = 3 consecutive digits means two digits left to 3 and two digits right to 3.
1, 2, 3, 4, 5
Solution:
Given addition is 560.
560/5 = 112, which is middle digit.
Consecution 5 digits for which addition is 560 is 110, 111, 112, 113, 114.
Now they want addition of next 5 digits.
which are: 115 + 116+ 117+ 118+ 119.
Now to sum it up quickly, 5, 6, 7, 8 and 9 are unit digits. and if you want to add them take unit digit of middle number (117) which is 7 and multiple by 5 = 35
Therefore unit digit is 5. now 5*1 = 5+3 = 8, and thousand's place is again five 1s which is 5.
Answer is 585 A.
14 Sep 2025, 10:51
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