How many number of numbers are possible which are greater than10 and less than 1000 in which digit is greater than its left digit? Eg 12,23....but not 11.
1)100 2)120 3)220 4)240

• The answer would be 120
  the numbers b/w 10 to 100 satisfying above condition are:36 i.e 8+7+6+5+4+3+2+1 (Ex: 12,13...19, 23...29,34....39,45...49,56...59,67..69,78,79,89,
  similarly 101 to 200 are= 36-8=28, 201 to 300 are=28-7=21.......700 to 800 is 789 only(1)
  So the total is (8+7+6+5+4+3+2+1)+(7+6+5+4+3+2+1)+(6+5+4+3+2+1)+(5+4+3+2+1)+(4+3+2+1)
  +(3+2+1)+(2+1)+1 =36+28+21+15+10+6+3+1
  =120
  
  
  
• 9 years agoHelpfull: Yes(42) No(1)
• Let us first check for the nos in 10 -100
  10 - 20 : 12, 13, ....19: 8 nos
  20 - 30 : 23, 24.......39 : 7 nos
  30 - 40 : 34,35.......49 : 6 nos
  .
  .
  .
  80-90: 89 : 1 no
  90- 100 : 0 nos
  
  Adding all of them 8+7+6+5+4+3+2+1 = 36
  
  Now from 100 - 200
  We get the first digit as 123. ( all the nos below that do not satisfy the condition)
  
  110 - 120 : 0 nos
  120 - 123 : 7 nos
  .
  .
  .
  180 - 190 : 189 : 1 no only
  
  Adding all we get 7+6+5+4+3+2+1 = 28
  
  We observe a pattern in the series being made. The no decreased from 36 to 28 with a difference of 8 in the next group of ( 200- 300) we get the 1st no as 134. So the series would be decreased by 7. So we would get
  200-300 : 6+5+4+3+2+1 = 21
  
  Summing up : 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120
• 9 years agoHelpfull: Yes(8) No(1)
• Option2nd is correct answer
  Explanation:the number in right should be greater then the number in its left so for number less than 100 there are 36 such numbers and so on on adding all the numbers we will get
  36+28+21+15+10+6+3+1=120
• 9 years agoHelpfull: Yes(7) No(1)
• 12 to 89 = 36
  123 to 189=28
  234 to 289=21
  345 to 389=15
  456 to 489=11
  567 to 589=6
  678 to 689=2
  789 to 789=1
  total=120
• 6 years agoHelpfull: Yes(1) No(0)