GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Nov 2019, 00:55

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Of the three digits greater than 700, how many have two digi

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
Joined: 23 May 2008
Posts: 305
Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 20 Nov 2009, 11:07
6
54
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

56% (02:36) correct 44% (02:36) wrong based on 834 sessions

HideShow timer Statistics

Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90
B) 82
C) 80
D) 45
E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?

OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/of-the-three- ... 35188.html
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59200
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 20 Nov 2009, 11:53
19
18
General Discussion
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59200
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 20 Nov 2009, 11:28
3
16
gurpreet07 wrote:
Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90
B) 82
C) 80
D) 45
E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?


Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit;
XYX meaning that the first digit is repeated digit;
and
YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

TOTAL=27+27+26=80

Answer: C.

Hope it helps.
_________________
Senior Manager
Senior Manager
avatar
Joined: 23 May 2008
Posts: 305
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 20 Nov 2009, 11:43
1
1
Bunuel wrote:
gurpreet07 wrote:
Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90
B) 82
C) 80
D) 45
E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?


Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit;
XYX meaning that the first digit is repeated digit;
and
YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

TOTAL=27+27+26=80

Answer: C.

Hope it helps.


thnku....very fast and a efficient method.
Senior Manager
Senior Manager
avatar
Joined: 23 May 2008
Posts: 305
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 20 Nov 2009, 12:07
1
Bunuel wrote:
This can be solved in another way and maybe it's faster:

Numbers between 700 and 999 =299;

Numbers with all distinct digits = 3*9*8=216;

Numbers with all the same digits more than 700 =3, (777, 888, 999);

Total: 299-216-3=80.


Hey Bunuel
thanks man...from where do you get such idea's
just awesome
Intern
Intern
avatar
Joined: 17 Dec 2009
Posts: 20
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 27 Mar 2010, 18:29
Bunuel,

I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand.
Manager
Manager
avatar
Joined: 05 Mar 2010
Posts: 147
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 10 Apr 2010, 04:55
adamsmith2010 wrote:
Bunuel,

I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand.



Adam
in case of XXY, first digit is 7,8 or 9 then so 3 ways
2nd digit has be same as 1st so only one way
you can also write it as 3*1*9 = 27

Hope it helps
_________________
Success is my Destiny
Intern
Intern
avatar
Joined: 29 Mar 2010
Posts: 28
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 14 Apr 2010, 03:10
How have you guys become so confident, predicting :

700-999

XXY setup = t occurrences without a single calculation line etc...

Isnt there any mathematical way of solving it.
_________________
-------------------------------------------------------------------------
Ros.
Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people.
-------------------------------------------------------------------------
Manager
Manager
avatar
Joined: 07 Jan 2010
Posts: 139
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 14 Apr 2010, 06:02
wow Bunuel ...
I tried to formulate a pattern and reach the answer. But it seems I missed something ...well ....
:(
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59200
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 14 Apr 2010, 10:24
3
shrivastavarohit wrote:
How have you guys become so confident, predicting :

700-999

XXY setup = t occurrences without a single calculation line etc...

Isnt there any mathematical way of solving it.


This is how I solved it:

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299.
A. all digits are distinct = 3(first digit can have only three values 7,8, or 9)*9*8=216;
C. all three are alike = 3 (777, 888, 999)

So, 299-216-3=80.

Another way to solve this problem is in my first post.

Hope it's clear.
_________________
Intern
Intern
avatar
Joined: 08 Jun 2010
Posts: 9
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 19 Jun 2010, 22:30
I am getting 78 as the answer. So I went with closest answer 80 (C).

This is how I arrived @ 78:

For numbers greater than 700 we have following possibilities:

a=> 7xx .. Here x being any digit 0-9, excluding 7 and 0 {0 is excluded because number should be >700}
b=> 7x7 .. Here x can be any digit from 0-9, excluding 7. {no 3 digits should be same}
c=> 77x .. Here x can be any digit from 0-9, excluding 7 for same reason as above.

a + b + c = 8 + 9 + 9 = 26

Hence same is true with 8xx , 8x8, 88x and 9xx,9x9, 99x ...

So summing up I got 78 ...

Please help me if I am wrong here. I am unable to figure out where did I miss 2 more numbers. Any pointers ?
Intern
Intern
avatar
Joined: 07 Jun 2010
Posts: 46
Location: United States
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 19 Jun 2010, 23:31
for 700 ==> 711 , 722 , ...... , 799 (except 777) = 8
,707, 717 , 727 , .... , 797 = 9
770, 771 , 772 , ........ , 779 = 9
total = 26

similarly for 800 and 900 series you will have 26 each

so 26 * 3 = 78

and 800 and 900 so 78 + 2 = 80
Manager
Manager
avatar
Joined: 21 Feb 2010
Posts: 167
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 20 Jun 2010, 10:45
Bunuel wrote:
gurpreet07 wrote:
Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90
B) 82
C) 80
D) 45
E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?


Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit;
XYX meaning that the first digit is repeated digit;
and
YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

TOTAL=27+27+26=80

Answer: C.

Hope it helps.

thanks! this is an efficient way to solve it. is this method feasible for this kind of questions--fundamental counting?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59200
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 20 Jun 2010, 10:59
1
Bunuel wrote:
Hope it helps.
tt11234 wrote:
thanks! this is an efficient way to solve it. is this method feasible for this kind of questions--fundamental counting?


I don't really understand your question...

Also there is another, easier way to solve this question, shown in my second post:

3-digit numbers more than 700 = 299;

Numbers with all distinct digits = 3*9*8=216 (first digit can take 3 values - 7, 8, 9; second digit can take 9 values and the third digit 8 values);

Numbers with all the same digits more than 700 = 3, (777, 888, 999);

{all} - {all distinct} - {all same} = {two equal} --> 299-216-3=80.

Answer: C.
_________________
Manager
Manager
avatar
Joined: 09 Jun 2010
Posts: 50
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 30 Jun 2010, 17:41
Thanks Bunuel for the effort
but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8?
Pleas explaing that to me
thanks again
Kaplan GMAT Instructor
User avatar
Joined: 21 Jun 2010
Posts: 66
Location: Toronto
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 30 Jun 2010, 19:44
xmagedo wrote:
Thanks Bunuel for the effort
but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8?
Pleas explaing that to me
thanks again


The question asks about three-digit integers greater than 700. So, all the numbers in the range 701 to 999, inclusive. There will be 999 - 701 + 1 = 299 such numbers in this range. (Whenever you want to count the number of objects in a range, subtract the smallest number from the greatest and add 1. For example, the number of integers in the range of 1 to 6 is 6 -1 +1 = 6).

"distinct" just means different. So, for example, in the number "799", the final two digits are NOT distinct. Here: 798, they are.
Intern
Intern
avatar
Joined: 07 Oct 2011
Posts: 25
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 21 Jan 2012, 20:06
This is the answer posted:

Numbers between 700 and 999 =299;

Numbers with all distinct digits = 3*9*8=216;

Numbers with all the same digits more than 700 =3, (777, 888, 999);

Total: 299-216-3=80.


I've a doubt. Numbers with all distinct digits = 3*9*8=216;

I feel 9 above is not right because if we take one of the 3 digits (7,8 or 9), then we shouldn't take repeated values (how is 9 possible).
Intern
Intern
avatar
Joined: 07 Oct 2011
Posts: 25
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 21 Jan 2012, 20:07
Ok, I got it :)

I forgot to take from 0-9 .. 10 digits

10-1=9
Manager
Manager
avatar
Joined: 20 Dec 2013
Posts: 223
Location: India
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 01 Feb 2014, 20:42
We can fix 2 digits in 3 ways ie 7,8,9 and the third digit can be fixed in 9 ways(10 digits-1 repeating digit)
Total=3*9=27
Since this is possible in 3 different ways(first digit diff from other two,second digit diff from other two and third digit diff from other two) we get 3* 27=81 but we subtract 1 for 700
Ans 80

Posted from my mobile device
Intern
Intern
avatar
Joined: 01 Feb 2016
Posts: 8
Re: Of the three digits greater than 700, how many have two digi  [#permalink]

Show Tags

New post 16 Jun 2016, 11:55
Repeating the explanation from above, the sequence of nos. should be XXY, XYX, YXX. As we need 3 digit nos. greater that 700, the 100th place in can be either of 7, 8 or 9 (3 ways to select 100th place digit)

Once 100th place if fixed, say units place is same as hundred place (XYX). this means that the tens place can be picked in 9 ways (all digits from 0-9 except the place in hundred place) and unit place can be picked in 1 way only (same as hundred place). So total ways for XYX is 3*9*1=27

We need to repeat this for XXY format. This would again generate 3*1*9 ways = 27

We need to repeat this now with YXX which will give 3.9.1 ways = 27. But we need to exclude 700 from this as we need nos greater than 700. This this combination will give 27-1=26 ways

Total ways = 27+27+26=80
GMAT Club Bot
Re: Of the three digits greater than 700, how many have two digi   [#permalink] 16 Jun 2016, 11:55

Go to page    1   2    Next  [ 22 posts ] 

Display posts from previous: Sort by

Of the three digits greater than 700, how many have two digi

  post reply Question banks Downloads My Bookmarks Reviews Important topics  





cron

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne