Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I was thrown off by "between 18 and 70". To me, that didn't include 18 or 70, so my initial answer was 10. Does "between" generally include the first and last numbers in a series?

How many integers between 18 and 70 inclusive are divisible by either 7 or 9 but not both?

(A) 11 (B) 12 (C) 13 (D) 14 (E) 15

Ans is B

This is not hard counting. Here is the formula. {( last divisible number - first divisible number)/divisor}+1

Here, between 18 and 70, For 7, last divisible number is 70 first divisible number is 21 so {(70-21)/7}+1 =9

similarly we can find out for 9, which is 5 So 9+5=14

But we have to exclude the number which are divisible by both. Here in this case that number is 63(here also you can use above formula by taking LCM of 7 and 9 as divisor). So remove this common number from both list.

I was thrown off by "between 18 and 70". To me, that didn't include 18 or 70, so my initial answer was 10. Does "between" generally include the first and last numbers in a series?

The question mentions the term "inclusive" which means first and last terms included. _________________

Since 18 and 70 are included, the range of numbers allowed is (70-18) + 1. This becomes 53. Next, divide 53/7, we get 7*7=49 as the closest. Hence only 7 numbers are present.

Similarly for 9, 53/9, we get 9*5=45 as the closest. Hence only 5 numbers are present. 63 is the first number divisible by both 9&7, but its outside the range. So total numbers = 7+5=12 (B). This method is similar to iPinnacle's method but I think its a bit faster.

My Approach, considering multiple of 7 and 9 as two arithmetic progressions, general formula an(last term of the series)= a1(first term of the series) + (n-1) d (distance btw terms) applied:

1) (multiple of 7) 70=21+(n-1)7 (diving by 7) 10=3+n-1 n= 8

2) (multiple of 9) 63=18+(n-1)9 (diving by 9) 7=n+1 n=6

Note: 63 is counted twice and we need to exclude it since the question stem is telling us # of integer divisible by either 7 or 9 (BUT NO BY BOTH)

N=14-2=12 ANSWER IS B

P.S. this approach might not be as useful as other approaches (posted here) on this question considering the small range of values. However, it might be good to keep in mind when you need to consider a larger series of integers btw i.e 322 to 1345 .... my 2 cents, TM

Basic Mathematical Tables & Counting: i) For 7, the numbers lying between 18 and 70 (inclusive) are from: 7*3=21 to 7*10=70 Therefore 10-3 = 7

ii) For 9, the numbers lying between 18 and 70 (inclusive) are from: 9*2=18 to 9*7=63 Therefore 7-2 = 5

Now we simply add them up;

7+5 = 12

@intelindahouse, Not sure if i quite understand your process, when you say 10-3= 7 you are excluding one integer on the table count ... we have 21 28 35 42 49 56 63 70 .... it will be good to say that you are already not considering 63 which is divisible by both (and we need to exclude as the question stem is telling us) same process for the series of 9,

when counting range of integers (inclusive of both start and end integer) always remember to add 1, my two cents, if I misunderstood you I am sorry,

to find total nos between 18 and 70 that which are divisible by 7 :

first count the total nos between the 21 and 70 (including both). It comes out to be 50. so, total nos divisible by 7 = 50/7 + 1= 8 ( adding 1 because 70 gets left out while dividing the range by 7). but we cannot count 63 ( because it is divisible by both 7 and 9) hence total nos divisible by 7 = 7.

similarly for the total nos divisible by 9 :

we select the range 18 to 63. But since 63 will not be counted anyways, so we can just count the nos between 18 and 54.

so the total nos divisible by 9 comes out as (54-18)/9 + 1 = 5

the answer is 7+5 = 12.

it looks lengthy but once you get the idea it can be done really fast!

# of multiples of 7 in the given range is (last-first)/multiple+1=(70-21)/7+1=8; # of multiples of 9 in the given range is (last-first)/multiple+1=(63-18)/9+1=6; # of multiples of both 7 and 9 is 1: 7*9=63. Notice that 63 is counted both in 8 and 6;

So, # of multiples of either 7 or 9 but not both in the given range is (8-1)+(6-1)=12.