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Numbers. GMAT paper test

by rb90 » Wed Oct 20, 2010 11:22 pm
How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339

The OA is E
Please can someone explain how to go about this sum?
Thanks in advance :)
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by shovan85 » Thu Oct 21, 2010 12:27 am
rb90 wrote:How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339

The OA is E
Please can someone explain how to go about this sum?
Thanks in advance :)
So all the integers between 324,700 and 458,600 will have the format XXXX13 (tens digit 1 and units digit 3 )

The very first integer satisfying this condition is 324713
The very last integer satisfying this condition is 458513

Find the difference = 458513 - 324713 = 133800

We know very first integer is 324713 ten what is next one?
It will be 324813 then 324913 and so on ... up to 458513. Each difference 100.

Thus total number of integers between 324,700 and 458,600 satisfying the conditions are 133800/100 = 1338

But this excludes the very first integer 324713.

Hence answer is 1338 + 1 = 1339

IMO E
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by frank1 » Thu Oct 21, 2010 12:51 am
rb90 wrote:How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339

The OA is E
Please can someone explain how to go about this sum?
Thanks in advance :)
Ok it is not pure mathematical way of doing it
but i think as i can be done with in 2 minutes not a bad method either
if really stumped in gmat then i would suggest you to do

Know the question stem:what it is asking
it is asking number for having 13 as last digit between x and y and to itimidate you it has given large numbers
now
The math:in every hundered there is 1 such number 113,213,313
and in every thousand there are 10 such numbers
so now we have numbers
324700 ------458600
3+(75+58)x10+6=1339
how ,to make things easier i have divided numbers in 3 parts
3=324713,....813,913
75+58=133 ....number of thousand between them each thousand has 10 such numbers
6=...458013,458113.....458513

It seems long but solution is just this
3+ (75+58)x10 +6=1339

Thanks
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by Brian@VeritasPrep » Thu Oct 21, 2010 9:04 am
Interesting question - and a great example of how the GMAT can make a fairly simple concept look unnecessarily complex!

This one really breaks down to the notion that there is exactly one number that ends in 13 per set of 100 numbers.

From there, you just need to find out how many sets of 100 numbers you have, which means you can forget about the tens and units digits. You have 4586 - 3247 sets of 100, and even that math isn't truly necessary as only one answer choice (E) ends in a 9, so the answer must be 1339.

Now...one other principle that's interesting here - is this set inclusive or exclusive? This is where I've always said that I'd much rather you think it through than try to apply a rule. We're looking for "how many sets of 100 are in this range", and you can prove that with small numbers. How many sets of 100 are between 100 and 200? 2-1 = 1, and you know that there's only one set of 100 in that range, so the add/subtract one ideology doesn't apply.

Why not? There's nothing to include or exclude - neither 324,700 nor 458,600 ends in 13, so you can't exclude or include either end point as part of the set. But, again, you can just use small numbers to prove to yourself what you need to do in this case.
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by shovan85 » Thu Oct 21, 2010 9:53 am
Brian@VeritasPrep wrote: Why not? There's nothing to include or exclude - neither 324,700 nor 458,600 ends in 13, so you can't exclude or include either end point as part of the set. But, again, you can just use small numbers to prove to yourself what you need to do in this case.
Great explanation :)
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by kapur.arnav » Thu Oct 21, 2010 10:29 am
rb90 wrote:How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339

The OA is E
Please can someone explain how to go about this sum?
Thanks in advance :)
Another method:

First term (a) = 324713
Final Term (nth term) = 458513
Common Difference (d) = 100

Apply Formula: nth Term = a + (n-1)d
Ans: 1339
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