Problem Solving

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

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

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

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

Regards
Saurabh