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01 Oct 2018, 06:38
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D
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Difficulty:
95%
(hard)
Question Stats:
34% (02:16) correct
66%
(02:09)
wrong
based on 146
sessions
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Two consecutive positive integers are divided by 25. What is the sum of the remainders obtained?
A. The sum of the digits at the last two places of the consecutive number is 49
B. The sum of the remainders obtained is even
Source: Experts Global
A. The sum of the digits at the last two places of the consecutive number is 49
B. The sum of the remainders obtained is even
Source: Experts Global
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
01 Oct 2018, 08:14
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pushpitkc
Let the numbers be x and x+1...
1) The sum of the digits at the last two places of the consecutive number is 49
Any number divisible by 25 will have 00,25,50 or 75 as last two digits..
If the last two numbers are combined equal to 49, the remaining der is 49-25=24
Otherwise both consecutive means x+x+1=49 means last two digits are xy24 and xy25
Remainder is 24
Sufficient
2). The sum of the remainders obtained is even
Consecutive integers means one even and one odd and therefore remainder too will be one odd and one even so combined remainder is odd+even=ODD..
Only case when it is even is when it is even and 0 so 24 and 0
Thus remainder is 24
Sufficient
D
17 Jan 2019, 09:16
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pushpitkc
integer would be divisible by 25 , when last two digits are 00,50,75
#1
last two digits sum is 49
we can say
x+x+1 = 49
x = 24
so
ab24 and ab25 could be two no ; remainder would be 24 ; sufficient
#2:
sum of remainders is even
for a no to be divisible by 25 the last digits have to be 00,24,75 so in the case to get even remainder the last two digits have to be both even 00, 24
sufficient
IMO D
