number properties

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sud21: Master | Next Rank: 500 Posts; Posts: 344; Joined: Sat Nov 12, 2011 3:21 amFollowed by: 2 members; Thanked: 1 times

by sud21 » Tue Jan 31, 2012 2:57 am

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Source: Beat The GMAT — Problem Solving | Tue Jan 31, 2012 2:57 am

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Tue Jan 31, 2012 2:56 pm

for non-native speakers, note rounded to the nearest tens' not tenths

In set P, 5 integers are EVEN and 15 are ODD. The integers can be odd or even (excluding 0 as non-positive). In set R 5 are rounded DOWN and 15 rounded UP.

question- which could be Sum(numbers in R) - Sum(numbers in P)

So the fixed terms is rounding and assigned around the even/the odd. If the numbers are rounded to the nearest tens' their digits will be 0. We may have different scenarios how the numbers are rounded to the tens, but for all the numbers in set R the digits will be ZERO and their numbers will be added/summed with the digit ZERO too.. Thus, we have some value for the Sum(numbers in P) minus some value with the digit 0, and there are total 20 integers in set P.

Case I) -25 is possible if set P contains 17 even and 3 odd numbers, 30-Sum(set P)=-25, Sum(set P)=55 -> 17*2 + 3*7 (set P contains seventeen 2s, and three 7s, in total 20 numbers in P, three odd numbers) could be

Case II) 75 is not quite possible as set P contains 10 even and 10 odd numbers, 100-Sum(set P)=75, Sum(set P)=35 -> 10*2 + 7*1 + ...

Case III) 145 is possible if set P contains 3 even and 17 odd numbers, 170-Sum(set P)=145, Sum(set P)=25 -> 3*2 + 16*1 + 3 (set P contains three 2s, sixteen 1s, and 3, in total 20 numbers in P, sevetnee odd numbers) could be

Cases possible: I and III only

correct answer e

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MakeUrTimeCount: Master | Next Rank: 500 Posts; Posts: 104; Joined: Sun Dec 11, 2011 1:25 pmFollowed by: 3 members; Thanked: 11 times

by MakeUrTimeCount » Wed Feb 01, 2012 9:39 am

Sorry Pemdas,
I think otherwise.

There is no need to consider 17 odd or 3 even or vice -versa. The question itself says 1/4 of the numbers are even i.e 5 Even and 15 odd.

Odd numbers are rounded up and even are rounded down. So for odd number valus increased and for even it decreases.

Minimum unit digit for odd = 1, Maximum unit digit = 9
Minimum unit digit for Even = 0, Maximum unit digit = 8

Let's consider extreme cases:
For R-P to be minimum: Odd has to be minimum and even has to be maximum.
So on rounding 5 even digits sum decreased by 8*5 = 40
So on rounding 15 odd digits sum decreased by 1*15 = 15
Overall change = 15 - 40 = -25

For R-P to be maximum: Odd has to be maximum and even has to be minimum.
So on rounding 5 even digits sum decreased by 0*5 = 0
So on rounding 15 odd digits sum decreased by 9*15 = 135
Overall change = 135 - 0 = 135

So -25 <= R -P <= 135

Ans D.

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