BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

General number properties

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 37
Joined: Sat Apr 24, 2010 8:27 am

General number properties

by GmatTakerNo.1 » Sun Apr 25, 2010 9:37 am
Hey,
I have here some exercises. Sorry for posting them in one thread, but else I will lose the overview.
Can you guys help me figure these out with´two minute approaches?

1. How many positive three-digit integers have an odd digit in both the tens and units place?
a) 25
b) 225
c) 250
d) 450
e) 500

Answer: B
Here I always got 250 which is C.


2. How many 3-digit integers between 100 and 200 have a digit that is the average (arithmetic mean) of the other 2 digits?
a) 1
b) 7
c) 10
d) 11
e) 19

Answer: D


3. What is the value of 1/2 - 1/3 + 1/4 - 1/5 - 1/6 + 1/7
a) 27/140
b) 37/97
c) 49/101
d) 23/41
e) 41/37

Answer: A

4. Which of the following fractions has a decimal equivalent that is a terminating decimal?
a) 10/189
b) 15/196
c) 16/225
d) 25/144
e) 39/128

Answer E

I really would appreciate if you help me out here.
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 am
Thanked: 137 times
Followed by:5 members

by thephoenix » Sun Apr 25, 2010 9:48 am
#1)
we have 5 odd digiys:-1,3,5,7,9
we can make three digits number with the given condition in 9*5*5 ways=225 numbers [for unit digit 5 ways ; for tens digit 5 ways and 9 ways for 100th place]
Top
User avatar
GMAT Instructor
Posts: 509
Joined: Wed Apr 21, 2010 1:08 pm
Location: Irvine, CA
Thanked: 199 times
Followed by:85 members
GMAT Score:750

by tpr-becky » Sun Apr 25, 2010 11:26 am
The first question is a permutation question

There are 9 digits available for the 100's digit, 5 odd digits available for the 10's and 1's digits so the answer is 9*5*5=225.

The second idea is that two of the numbers have to be even or two of the numbers have to be odd with 2 as the average. if this isn't the case, you won't get an integer for the average. the number also has to start with a 2 because it is between 200 and 300

So

2-0 - the average is 1 - there are 2 versions (210, 201)
2--2 the average is 2 - but only 1 version
2-4 - with average 3 - 2 version
2-6 with average 4 - 2 versions
2-8 with average 5 - 2 versions
2 is the average of 1&3 - 2 versions
for a total of 11
Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 63
Joined: Wed Jul 16, 2008 6:43 am
Thanked: 3 times
GMAT Score:680

by moliver » Sun Apr 25, 2010 12:53 pm
tpr-becky wrote:The first question is a permutation question

There are 9 digits available for the 100's digit, 5 odd digits available for the 10's and 1's digits so the answer is 9*5*5=225.

The second idea is that two of the numbers have to be even or two of the numbers have to be odd with 2 as the average. if this isn't the case, you won't get an integer for the average. the number also has to start with a 2 because it is between 200 and 300

So

2-0 - the average is 1 - there are 2 versions (210, 201)
2--2 the average is 2 - but only 1 version
2-4 - with average 3 - 2 version
2-6 with average 4 - 2 versions
2-8 with average 5 - 2 versions
2 is the average of 1&3 - 2 versions
for a total of 11
I think that tpr-becky wanted to say that the number must be between 100 and 200

For the last question (#4) I think that what you mean by saying terminating decimal is that there are a finit number of decimal in the number, didn't you?
In this case the last option is divided by 128 = 2^6
If the number is divided by 5 or 2 it has a finite numbers of decimals.
the answer c) has not a finite decimal because you have there a 225 = 9*5*5

Question #3 is interesting. Different ideas here. What was your approach?
The hard part do all the maths
Group by the numbers who are positive and who are negative.
or something like this for example
if you do 1/2 - 1/3 = 1/6 so now you can simplify with the -1/6
and now I think that the maths are more straight forward

please let me know if this was helpful
Top
Senior | Next Rank: 100 Posts
Posts: 37
Joined: Sat Apr 24, 2010 8:27 am

by GmatTakerNo.1 » Sun Apr 25, 2010 1:18 pm
Thanks for replying.

For question 4 I have the same approach, moliver, but I thought there is maybe a trick.

For question 3 I meant a finite number of decimals by the term "terminating decimal".

Your answers help me a lot, thanks again ;)
Top
Senior | Next Rank: 100 Posts
Posts: 52
Joined: Sat Jan 23, 2010 9:08 pm
Thanked: 2 times

by Ashish8 » Sun Apr 25, 2010 5:09 pm
So exactly what is the strategy for Q4?

Thanks
Top
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Sat Feb 06, 2010 1:55 pm
Thanked: 5 times

by DeepthiRajan » Sun Apr 25, 2010 11:51 pm
If a fraction has to be a terminating decimal, it should have either powers of 2 or 5 in the denominator (eg 2^3 or 5^2)

in case of E, 128 in the denominator can be expressed as 2^7

HTH
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Tue Jan 12, 2010 4:33 pm

by lcd318 » Mon Apr 26, 2010 8:49 am
For number 3, I ballparked it and got the right answer.

"What is the value of 1/2 - 1/3 + 1/4 - 1/5 - 1/6 + 1/7 ?"

= 0.5 - 0.33 + .25 - .20 - .16 + .14

I got roughly .20, or 1/5. Answer A most closely matches that -- the others aren't even close.
Top
Post new topic Post Reply
• Page 1 of 1