difficult paper test 37

[payam]

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

[abramson]

is the answer (E)?

[emailpa]

Kindly explain this problem

[Harish Dorai]

I guess the answer is (E) and here is the explanation.

The question asks for number of instances between 324700 and 458600 in which the units digit is 3 and the tens digit is 1.

Example: 325713 is a number between 324700 and 458600 which meets the above criteria.

The first such number will be 324713, and the next number 324813, and so on. So if you build a series out of this, it will look like

324713, 324813, 324913, 325013, .......... and the last term is 458513.

So we need to find the number of terms in the above series. In an Arithmetic series, the Formula to calculate the number of terms is = (Last Term - First Term) / Common Difference and then add a 1 to it.

So here it is (458513 - 324713) / 100 and add 1 to it.

This is 1338 + 1 = 1339.

Hope it helps

[Ladan]

I took a simpler approach and got to answer E as well. Here is what I did, it should be right:

From 1 to 100 there is one value that ends in 13. Now using ratios, how many 100s are in 133,900 (the subtration of 458,600 - 324,700)? Remeber since it says between, it's not inclusive, so I didn't add the 1. Therefore, 1/100 = x/133,900 => 100x = 133,900 => x = 133,900/100 = 1,339. Answer E.

I hope this makes sense.

[RonPurewal]

Here's an approach that's arguably simpler that what's already here.

You're looking for the number of entries in the list 324 713, 324 813, ..., 458 513 (as has already been said).

Since these all end with '13,' we can ignore all the 13's, and just treat the problem as though we were counting the following list:
3247, 3248, ..., 4585

This is a direct application of our formula: 4585 - 3247 + 1 = 1339 (E)