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usually it can be done by just checking the range GMAT does not want us to remember the long formula/method to cslculate std deviation

Take the 10 common and you’ll have a list of consecutive integers 1,2...19. Use n(n+1)/2 10*190 =1900

You can use the formula for Sum of n terms: Sn= n/2 , n=19, a=10, d=10

First to calculate no. of digits that are multilple of any given no. we used a formula Hightest divisible no - lowest divisible no./given no.. This how I got the value 19. As you can see the set of values are in airthmetic sequence so, here we can apply sum of nos. formula used in A.R. so, here we go : sum of the terms = n/2(first digit + last digit) = 19/2(190+10)=1900. Hope it helps!

Check this very hard Algebra/Word question: https://gmatclub.com/forum/a-shoe-store ... 98695.html

Check out the question of the day: what-is-the-least-nine-digit-positive-integer-that-contains-each-of-428065.html consider-the-sum-428067.html

I strongly recommend you to post your CAT related queries to a better sub-forum. Here is the link for the same : cat-india-common-admissions-test-399/

I did give CAT 2023 although without preparation. But still will be able to answer your queries. and as posted by poojaarora1818 , you can make use of the link given.

Hi @kingbucky- I was trying to solve question 1 but I don’t think any of the options (given) are divisible by 99. What am I missing?

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