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23 Jan 2019, 02:16
Kudos
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C
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Difficulty:
15%
(low)
Question Stats:
81% (01:02) correct
19%
(01:07)
wrong
based on 121
sessions
History
Date
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Not Attempted Yet
In a series of six consecutive integers, the sum of the first three integers is 93. What is the sum of the last three integers?
A 96
B 99
C 102
D 105
E 108
A 96
B 99
C 102
D 105
E 108
23 Jan 2019, 06:08
Kudos
Bookmarks
six consecutive integers
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)
the sum of the first three integers is 93
x+(x+1)+(x+2)=33
3x+3=93
x=30
What is the sum of the last three integers?
(x+3)+(x+4)+(x+5)=?
(33)+(34)+(35)=102
C
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)
the sum of the first three integers is 93
x+(x+1)+(x+2)=33
3x+3=93
x=30
What is the sum of the last three integers?
(x+3)+(x+4)+(x+5)=?
(33)+(34)+(35)=102
C
Archit3110
Major Poster
23 Jan 2019, 07:40
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Bookmarks
Bunuel
x x+1 x+2 x+3 x+4 x+5
x+ x+1+x+2 = 93
x= 30
so
3x+12 ; 90+12 ; 102 IMO C
