A relatively popular concept on the GRE math is series – when you have a list of numbers that follow a certain pattern(e.g. consecutive numbers). Usually, the question will ask you to add up all the numbers that are part of the series.
Today, we are not going to go quite so far. Instead, we are going to focus on a step that is far more basic, yet one that often eludes students. Try the following problem:
How many integers are there in Set A, if Set A includes all the numbers 1 – 10?
Really Easy, Right?
The answer seems pretty straightforward – 10. That is, there are ten possible integers: 1, 2, 3… 9, 10.
Now, you are probably thinking that this question is really obvious. Of course, there are ten numbers, if you are counting the numbers 1 – 10. I agree, but try this almost identical question:
How many integers are there in Set A, if Set A includes all the numbers 5 – 15?
The answer is ten, right? Actually, the answer is eleven. And that’s what makes this question so tricky. In answering ten, you probably went something like this – 15 – 5 = 10; therefore, there are ten numbers.
The Twist
Whenever you are counting the total numbers in a list by subtracting, you always have to count an extra number. For instance, with our original question, if you subtract 10 – 1 = 9, you get one fewer page. The reason we always add an extra one is because we want to include the number we are counting. For instance, if I have read pages 1 – 3 in a book, I would want to make sure I count page 1 as one of the pages I read.
With this in mind, now try the following problem:
Set A consists of the digits 50 – 100. How many digits are there in set A?
(A) 49
(B) 50
(C) 51
(D) 99
(E) 100
Solution:
The simple formula for counting the number of elements in a consecutive sequence, like the one above, is L – F + 1. 100 – 50 + 1 = 51.
Practice Time
The next three questions pertain to a 210-page novel:
How many pages are there between 45 and 111, not including either of those pages?
(A) 45
(B) 65
(C) 66
(D) 67
(E) 76
If I am at the top of page 125 of a book, then how many pages will I have to read if I want to read to the end of page 152?
(A) 25
(B) 26
(C) 27
(D) 28
(E) 31
I am printing a book, and, for each digit I print on the page, I have to pay 5 cents (pg. 25 = 10 cents, pg. 134 = 15 cents). How much will it cost to print the page numbers, starting on page 9 and continuing for another 100 pages?
(A) 9.05
(B) 10.15
(C) 10.20
(D) 10.55
(E) 10.60
Answers:
1. B
2. D
3. D
Hey Chris, I was hoping you could post more lessons that are specific to counting (all of the evil inclusive and exclusive stuff included)? I’m having trouble finding a tutorial through the GRE math videos included on my account.
James,
I will definitely be posting more videos on counting/sums! Definitely an area which fills many with trepidation.
As for our product, the lesson video is entitled: “Sums of Sequences”, under “Word Problems.”
Hope that helps!
Thanks Chris..
Can u explain the third example….
I am not sure if you wanted me to explain the question (because it may be confusing) but here it goes…
Each digit on the top of the page corresponds to a certain cost.
If there is one digit (pg. 1-9) it will cost 5 cents. If two digits (10-99) 10 cents. Three digits (100-999) 15 cents. Page 9 costs 5 cents. Page 10-99, which is 90 numbers (L – F + 1) = 9 dollars.
100 more pages after page nine takes us to 109, which is a total of 10 pages. 15 cents each = 1.50.
This gives us 10.55 or answer choice (D).