Imo c.

001245 in base 9: 5+36+162+729=932

There was a slight defect with the odometer (which shows the distance travelled) where the digit 9 was missing.
Didn't understand the question..

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?

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

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..

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

The odometer SKIPS OVER every integer with a digit of 9 but INCLUDES these integers in the total distance.
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.

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?

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?

We need to subtract from the total every integer that includes a digit of 9.
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.

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