Latest Elitmus Aptitude Question SOLUTION: how many numbers are possible between 100 to 999 where middle digit is smaller than other two. e.g. 102?

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(#M40164962) ELITMUS Numerical Ability question Keep an EYE

how many numbers are possible between 100 to 999 where middle digit is smaller than other two. e.g. 102?

Asked In Elitmus Satyajit Pramanik (6 years ago)

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vijay pandey  (6 years ago)

10__=1*1*9=9
20__=1*1*9(when middle digit is 0)+1*1*8(when middle term is 1)=9+8
30__=1*1*9(when middle digit is 0)+1*1*8(when middle term is 1)+1*1*7(when middle term is 2)=9+8+7
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9__=9+8+7+6+5+4+3+2+1

=>9*9+8*8+7*7+6*6+5*5+4*4+3*3+2*2+1*1=285 Ans

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ehsan  (6 years ago)

in case repetition is of allow
10_ last place can be 1,2,3,4,5,6,7,8,9= total 9
2_ in second place we can place either 0,1 when place 0 there 9 option when place 1 so 2,3,4,5,6,7,8,9 there is 8 option = total 9+7
3 _ second place can be 0,1,2 and similar to above total= 9+8+7
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9_ second place 0,1,2,3,4,5,6,7,8 total=9+8+7+6+5+4+3+2+1
so 9 count 9 time , 8 count 8 time and so on
sum=9*9+8*8+7*7+......1*1
this is in A.p so sum=n(n+1)(2n+1)/6
sum=9(9+1)(2*9+1)/6=285

in case repetition is not allowed
sum=8(8+1)(2*8+1)/6=204

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manoj  (6 years ago)

from 201 to 300 it will be 9 only
then again from 301 to 400 it will be 9 and 8
....
901 to 999 it will be 9 and 8 times 8

after calculating these total will be
369

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mohammad arif siddiqui  (6 years ago)

285
add square from 1 to 9

Is this solution Helpfull? Yes (3) | No (1)  

rahul gupta  (6 years ago)

10_ =9
2__ =9+8
3__=9+8+7
4__=9+8+7+6
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;
;
9__=9+8+7+6+5+4+3+2+1
them sum all these=9+17+24+30+35+39+42+44+45=285
ans=285

Is this solution Helpfull? Yes (2) | No (5)  

Avdhesh Sharma  (5 years ago)

1. If middle place is 0 than 1st and third place can be filled by nos 1-9 => total = 9*9
2. Similarly if middle=1 then 1st and 3rd place can be filled by nos 2-8 => total = 8*8
3. Similarly middle=2, total = 7*7
4. Similarly middle=3, total = 6*6
Similarly middle=4, total = 5*5
Similarly middle=5, total = 4*4
Similarly middle=6, total = 3*3
Similarly middle=7, total = 2*2
Similarly middle=8, total = 1*1
Note: We can not have 9 digit at middle position as per the condition given in question.
So the series would be = 1*1+2*2+3*3+4*4+5*5+6*6+7*7+8*8+9*9
For this series we can apply formula = n(n+1)(2n+1) / 6, therefore after putting the values and solving them answer would be = 285.

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