700 level question

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saurabhkamal1981: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Sun Nov 08, 2009 5:03 am

by saurabhkamal1981 » Tue Mar 01, 2011 10:38 am

Source "BTG practice question" - 700 level question

Hi All,

Please help. I did a question from "MGMAT- Number properties" - What is the sum of all the integers from 20 to 100, Inclusive ?

(1) Found the average - 100 + 20 /2 = 120/2 = 60
(2) Count the number of terms - 100 - 20 + 1 = 81
(3) Multiply 60 by 80, 60 * 81 = 4860

Ans. 4860

Came across a question from "BTG practice questions". Applied the same method. But my answer differs from the correct answer.

The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8 + 1+9 + 2+0 + 2+1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

(A) 450
(B) 810
(C) 900
(D) 1000
(E) 1100

Correct Answer:- C

My way:-

(1) 99/2 = 49.5
(2) 99 - 0 + 1 = 100
(3) 49.5 * 100 = 4950 (But this is wrong).

Thanks
Saurabh

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Source: Beat The GMAT — Problem Solving | Tue Mar 01, 2011 10:38 am

arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Tue Mar 01, 2011 10:47 am

Saurabh the difference is between the "sum of the numbers" and "sum of the digits"

for sum of all digits from 0 to 99 we shall have

sum of all units digits + sum of all tens digits:

(10 ones{in one,eleven,twentyone and so on...} + 10 twos + .... + 10 nines) +(10 ones{ten,eleven,twelve and so on...} + 10 twos +....)

(10 +20+30+40+50+60+70+80+90) + (10 +20+30+40+50+60+70+80+90)

=2*10(1+2+3+....9)
=2*10*((9*10)/2)
=2*10*45
=900

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Tue Mar 01, 2011 10:50 am

hi saurabhkamal1981, your approach was mechanical --> you found the sum of all the integers, and not the sum of all the digits of integers from 0 to 99

here your take on the given data could be find 1) the sum of all the units and tens in the integers from 0 to 99; 2) sum up the totals from the first step 1)),

1) we have got 10 units and (0,1,2,3,4 ...9) their sums are [(0+9)/2]*10=45
2) we have got 9 tens (1,2,3,...9) their sums are [(1+9)/2]*9=45

45*10 (tens' totals from 0 to 99) and 45*10 (units' totals) = 900

saurabhkamal1981 wrote:Source "BTG practice question" - 700 level question

Hi All,

Please help. I did a question from "MGMAT- Number properties" - What is the sum of all the integers from 20 to 100, Inclusive ?

(1) Found the average - 100 + 20 /2 = 120/2 = 60
(2) Count the number of terms - 100 - 20 + 1 = 81
(3) Multiply 60 by 80, 60 * 81 = 4860

Ans. 4860

Came across a question from "BTG practice questions". Applied the same method. But my answer differs from the correct answer.

The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8 + 1+9 + 2+0 + 2+1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

(A) 450
(B) 810
(C) 900
(D) 1000
(E) 1100

Correct Answer:- C

My way:-

(1) 99/2 = 49.5
(2) 99 - 0 + 1 = 100
(3) 49.5 * 100 = 4950 (But this is wrong).

Thanks
Saurabh

My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com

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saurabhkamal1981: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Sun Nov 08, 2009 5:03 am

by saurabhkamal1981 » Wed Mar 02, 2011 10:00 am

Thank you very much arora007 and Night reader. Really appreciate for your reply.

Regards
Saurabh

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