X bought a new car and went for a trip. There was a slight defect with the odometer (which shows the distance travelled) where the digit 9 was missing. At the end of the trip the odometer read 001245. What is the exact distance traveled by X in the new car? (Assume that the odometer read 000000 when X bought it)
A) 755
B) 931
C) 932
D) 1000
E)1300
Numbers
This topic has expert replies
-
- Legendary Member
- Posts: 966
- Joined: Sat Jan 02, 2010 8:06 am
- Thanked: 230 times
- Followed by:21 members
-
- Senior | Next Rank: 100 Posts
- Posts: 56
- Joined: Thu Jul 07, 2011 2:16 am
- Thanked: 2 times
Imo c.
001245 in base 9: 5+36+162+729=932
001245 in base 9: 5+36+162+729=932
Last edited by aplavakarthik on Fri Sep 16, 2011 11:47 pm, edited 1 time in total.
- cans
- Legendary Member
- Posts: 1309
- Joined: Mon Apr 04, 2011 5:34 am
- Location: India
- Thanked: 310 times
- Followed by:123 members
- GMAT Score:750
There was a slight defect with the odometer (which shows the distance travelled) where the digit 9 was missing.
Didn't understand the question..
Didn't understand the question..
If my post helped you- let me know by pushing the thanks button ![Wink ;)](./images/smilies/wink.png)
Contact me about long distance tutoring!
[email protected]
Cans!!
![Wink ;)](./images/smilies/wink.png)
Contact me about long distance tutoring!
[email protected]
Cans!!
-
- Senior | Next Rank: 100 Posts
- Posts: 56
- Joined: Thu Jul 07, 2011 2:16 am
- Thanked: 2 times
the meter should show the reading 9 after travelling 9km right, but it ll show 10 as the digit 9 is missing. the repetition sequence ll be 1 2 3 4 5 6 7 8 0(9 is missing). did u get it cans?
-
- Legendary Member
- Posts: 966
- Joined: Sat Jan 02, 2010 8:06 am
- Thanked: 230 times
- Followed by:21 members
Its an analog meter which readings from 0-9 ( 9 is missing though)in each of units,ten,hundreds and thousands place.
WHEN the units place starts from 0 and increases to 9, and back to 0 again while the 10's place becomes one and so on.
after 0008, the meter reads 0010 and so on
WHEN the units place starts from 0 and increases to 9, and back to 0 again while the 10's place becomes one and so on.
after 0008, the meter reads 0010 and so on
cans wrote:There was a slight defect with the odometer (which shows the distance travelled) where the digit 9 was missing.
Didn't understand the question..
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
The odometer SKIPS OVER every integer with a digit of 9 but INCLUDES these integers in the total distance.shankar.ashwin wrote:X bought a new car and went for a trip. There was a slight defect with the odometer (which shows the distance travelled) where the digit 9 was missing. At the end of the trip the odometer read 001245. What is the exact distance traveled by X in the new car? (Assume that the odometer read 000000 when X bought it)
A) 755
B) 931
C) 932
D) 1000
E)1300
Thus, only integers WITHOUT a digit of 9 should be included in the total distance.
Thus, to determine the actual distance driven, we need to count the positive integers between 1 and 1245, inclusive, that DO NOT have a digit of 9.
From 001 to 999:
Number of options for the units digit = 9. (Any digit 0-8)
Number of options for the tens digit = 9.
Number of options for the hundreds digit = 9.
To combine these options, we multiply:
9*9*9 = 729.
Since 000 does not represent an actual distance, we lose one option:
729-1 = 728.
From 1000 to 1199:
Number of options for the units digit = 9.
Number of options for the tens digit = 9.
Number of options for the hundreds digit = 2. (Must be 0 or 1).
To combine these options, we multiply:
9*9*2 = 162.
From 1200-1245:
Total number of integers = biggest-smallest + 1 = 1245-1200+1 = 46.
Integers with a digit of 9 are 1209, 1219, 1229, and 1239.
Subtracting these bad integers from the total, we get:
46-4 = 42.
Actual distance driven = 728+162+42 = 932.
The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
- Master | Next Rank: 500 Posts
- Posts: 150
- Joined: Thu May 05, 2011 10:04 am
- Thanked: 5 times
- Followed by:4 members
I tried a reverse approach. Please tell me where I'm missing the count.
From 0 - 999
There are 300 9s
From 1000-1099
20 9s
From 1100-1245
4 9s
Total 300+20+4=324. So we have to reduce 324 extra recordings which means 1245-324=931.
Can you please point where did I go wrong?
From 0 - 999
There are 300 9s
From 1000-1099
20 9s
From 1100-1245
4 9s
Total 300+20+4=324. So we have to reduce 324 extra recordings which means 1245-324=931.
Can you please point where did I go wrong?
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
We need to subtract from the total every integer that includes a digit of 9.msr4mba wrote:I tried a reverse approach. Please tell me where I'm missing the count.
From 0 - 999
There are 300 9s
From 1000-1099
20 9s
From 1100-1245
4 9s
Total 300+20+4=324. So we have to reduce 324 extra recordings which means 1245-324=931.
Can you please point where did I go wrong?
You seem to be counting every APPEARANCE of the digit 9.
This method will lead to overcounting integers that contain more than one digit of 9.
The following is an accurate accounting of every integer between 1 and 1245, inclusive, that includes at least one digit of 9.
000-899:
Within every set of 100 integers, there will be 19 integers with a digit of 9:
9,19,29,39,49,59,69,79,89,90-99.
Thus, from 000 to 899, the total number of integers with a digit of 9 = 9*19 = 171.
900-999:
Number of integers with a digit of 9 = 100.
1000-1199:
Since there are two sets of 100 integers, the total number of integers with a digit of 9 = 2*19 = 38.
1200-1245:
Integers with a digit of 9 = 1209,1219,1229,1239 = 4.
Total number of integers with a digit of 9 = 171+100+38+4 = 313.
Thus, the total number of integers without a digit of 9 = 1245-313 = 932.
Last edited by GMATGuruNY on Sun Sep 18, 2011 3:30 am, edited 1 time in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
- Junior | Next Rank: 30 Posts
- Posts: 21
- Joined: Tue Aug 21, 2007 7:41 pm
- Thanked: 1 times
Regular numbering systems would have a base of 10, which means that we can write
1245= 1 X 10^3 + 2 X 10^2 + 4 X 10^1 + 5 X 10^0 = 1245
Unit place have a power of zero, tenth having one and so on
Since one of the digit is missing, so the numbering systems changes to base of 9
1245= 1 X 9^3 + 2 X 9^2 + 4 X 9^1 + 5 X 9^0 = 932
1245= 1 X 10^3 + 2 X 10^2 + 4 X 10^1 + 5 X 10^0 = 1245
Unit place have a power of zero, tenth having one and so on
Since one of the digit is missing, so the numbering systems changes to base of 9
1245= 1 X 9^3 + 2 X 9^2 + 4 X 9^1 + 5 X 9^0 = 932