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First-degree equations

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by pullagurla » Thu Sep 02, 2010 10:36 pm
In an increasing sequence of 10 consecutive integers,
the sum of the � rst 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

explain by picking number
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by Rahul@gurome » Thu Sep 02, 2010 11:17 pm
Solution:
Let the integers be x, x+1, x+2, x+3……..x+9.

So x + (x+1) + (x+2) + (x+3) + (x+4) = 560.
Or 5x + (1+2+3+4) = 560.
Or 5x = 550.
Sum of next 5 is (x+5) + (x+6) + (x+7) + (x+8) + (x+9) = 5x + 5+6+7+8+9 = 550 + 35 = 585.

The correct answer is (A).
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by singalong » Tue Dec 13, 2011 4:26 pm
would I be wrong if I take the consecutive integers as (n-2),(n-1),n,(n+1) and (n+2)?
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by neelgandham » Tue Dec 13, 2011 5:14 pm
singalong wrote:would I be wrong if I take the consecutive integers as (n-2),(n-1),n,(n+1) and (n+2)?
No you aren't wrong. If you take the consecutive integers as
n-2, n-1, n, n+1, n+2, n+3, n+4, n+5, n+6, n+7

Sum of first five integers = n-2 + n-1 + n + n+1 + n+2 = 5n = 560
Sum of last five integers = n+3 + n+4 + n+5 + n+6 + n+7 = 5n + 25 = 560 + 25 = 585 !

Clear ? If no, please let me know !
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by GMATGuruNY » Tue Dec 13, 2011 6:20 pm
pullagurla wrote:In an increasing sequence of 10 consecutive integers,
the sum of the � rst 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

explain by picking number
Given evenly spaced numbers:
Average = median.

Thus, the median of the first five integers = 560/5 = 112 -- implying that the first 5 integers are 110,111,112,113,114 and that the last 5 integers are 115,116,117,118,119.

Given evenly spaced numbers:
Sum = number * median.

Thus, the sum of the last 5 integers = 5*117 = 585.

The correct answer is A.
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by rijul007 » Tue Dec 13, 2011 9:14 pm
pullagurla wrote:In an increasing sequence of 10 consecutive integers,
the sum of the � rst 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

explain by picking number
x1,x2,x3,...x10 are consecutive integers

x1+x2+x3+x4+x5= 560
x1 = x6-5
x2 = x7-5
x3 = x8-5
x4 = x9-5
x5 = x10-5
Adding all above expressions
560 = x6 + x7 + x8 + x9 + x10 -25
Sum of last 5 digits = x6 + x7 + x8 + x9 + x10 =560 + 25 = 585

Option A
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