suresh

• May 30th, 2005

 

55

Mubeen

• Jul 2nd, 2005

 

Can u please work this out as i am thinking the answer is 65. 
Since 1st digit = 0-5 
2nd digit = 0-5 
 

Mubeen

• Jul 2nd, 2005

 

Can u please work this one out as i am thinking the answer is 65. Since 
1st digit = 0-5 
2nd digit = 0-5 
3rd digit = 9 always 
So considering all the combinations u still have two 5's that can combine with every other number except each other so u have missed those 10 combinations. 

kavita kanyal

• Jul 7th, 2005

 

Ans : 55 

Soumya

• Jul 25th, 2005

 

Since the last digit is 9... the sum of the first two digits shud not exceed 9...  
so the possible combinations are only:  
0-9, 
1-8, 
2-7, 
3-6, 
4-5, 
5-4, 
6-3, 
7-2, 
8-1, 
9-0. 
So, totally 10... 
Answer: 10

oktay

• Aug 23rd, 2005

 

our number is xy9 
now if x is 0 then y has 10 options 
now if x is 1 then y has 9 options (except 9) 
now if x is 2 then y has 8 options (except 9 and 8) 
. 
. 
. 
now if x is 9 then y has 1 option (that is 0) 
 
so the answer is 1 + 2 + ... + 10 = 55

vikram_g32

 Profile Answers by vikram_g32 

• Dec 21st, 2005

 

if x bcmes 0 then the number will be 2-digit and not 3-digit(in the question it is mentioned that the lock is a combinayion of three digit number)

gaurav

• Dec 30th, 2005

 

ans will be 55 as the combination with 5 & 5 of first & second digit will lead to sum up of 10 which is greater than 9

pavitra

• Jan 8th, 2006

 

it will not go upto 101+2+3+...........................+9 = 45 (ANSWER)

Vivek S

• Jan 12th, 2006

 

55 is the correct answer

dhanasree

• Feb 8th, 2006

 

from the data --9

the two numbers are choosen such that their sum should not exceed 9

te posibilities are

if one number is 0 there are 9 ways of filling other number such that sum does not exceed 9

thyey are 0,1,2,3,4,5,6,7,8,9

similarly if one number is 1 then 9 chances and so on

so total number of ways=10+9+8+...+1=55

Vinay Karalkar

• Feb 23rd, 2006

 

ANS= 55 ways

Bcoz

0 can combine with 0 to 9 digits to satisfy both conditions in 10 ways

1 can combine with 0 to 8 digits to satisfy both conditions in 9 ways

if try for 2 to 9 same thing then

No. Of ways =10+9+8+....+1=(10*(10+1))/2=55

Think on it

jitusonawane

 Profile Answers by jitusonawane Questions by jitusonawane

• Mar 27th, 2006

 

55 ways

009,019,029,039,049,059,069,079,089,099,109,209,309,409,509,609,709,

809,909,119,129,139,149,159,169,179,189,219,229,239,249,259,269,279,

319,329,339,349,359,369,419,429,439,449,459,519,529,539,549,619,629,639,

719,729,819,

vidhya

• Apr 8th, 2006

 

1. 100First Digit can be 0-9 --> 10 waysSecond digit can be 0-9 --> 10waysLast digit 9 --> One wayIt is -> 10*10*1 = 100 ways

Riya

• Sep 13th, 2006

 

81

vamc

• Oct 22nd, 2006

 

ans is 55

balaganesh k

• Nov 10th, 2006

 

0+0 1+0   2+0     3+0     4+0      5+0    6+0     7+0     8+0   9+0         
  1      1    1 1 1 1 1 1 1 
  2   2 2 2 2 2 2 2
  3   3 3 3 3 3 3
  4   4 4 4 4 4
  5   5 5 5 5
  6   6 6 6
  7   7 7 
  8   8
  9

This is variou combination, for the first two key add together 55

bnreddy

• Jan 30th, 2007

 

the answer is 45.because in the number,first digit fill starts with 1and second digit in between 0-8 only=9 first digit fills start with 2 and second digit in between 0-7 only=8 as it is . .......................3.....................................................0-6..........=7.....................................................................................4....................................................0-5...........=6 5...................................................0-4.............=5 6 0-3.............=4 7 ...........................................0-2............=3 8...................................................0-1.............=2 9...................................................0...............=1total=9+8+7+6+5+4+3+2+1=45 ANSWER.HERE NEVER FORGET THE RULES THAT ARE THREE DIGIT NUMBER SO IT NEVER STARTS WITH ZERO. AND THE NUMBER IS LESS THAN OR EQUAL TO 9.

Ashish taunk

• Apr 1st, 2007

 

ans: 55
see we have the sum of 1st two digits is not more than 9;
so for 1st digit is 0,there are 10 combinations of two digit no(i.e. 00,01,02,03,....,09);
     for 1st digit is 1,there are 9 combinations of two digit no(i.e. 10,11,12,13,....,18);
     for 1st digit is 2,there are 8 combinations of two digit no(i.e. 20,21,22,,....,27);
     for 1st digit is 3,there are 7 combinations of two digit no(i.e. 30,31,32,33,....,36);
     .
     .
     .
      .
     for 1st digit is 7,there are 3 combinations of two digit no(i.e. 70,71,72);
     for 1st digit is 8,there are 2 combinations of two digit no(i.e. 80,81);
     for 1st digit is 9,there is only 1 combination of two digit no(i.e. 90);


so totoal no of combinations we have = 10+9+8+.....+3+2+1 = 10 (10 + 1) / 2 = 55

gurjot

• Apr 4th, 2007

 

look ppl the question says that the sum of first two digits should not exceed 9..
therefore if the first digit is:
0 the possible combinations will be 0,1,2,3,4,5,6,7,8,9....therefore the number of combinations is 10
1 the possible combinations are 0,1,2,3,4,5,6,7,8.. therefore  9
2................8
3................7
4................6
5................5
6.................4
7.................3
8.................2
9..................1

that makes a total of 55 combinations.. so there are 55 possible combinations..

Harish A

• Apr 18th, 2007

 

Answer is 55

kishore vemuri

• Jun 26th, 2007

 

see there
if 1'st is 0 - 10 chances for the second
             1 -  9
             2 -  8
             3 -  7
              ................
      we will have (10+9+8+7+6+5+4+3+2+1)=55

amish021

 Profile Answers by amish021 

• Jul 10th, 2007

 

The lock can be opened in only 1 way, depending upon the code that was used while locking it

pavan kumar karasala

• Jul 24th, 2007

 

55 is the correct answer..............
first two digits sum must be less than or equal.........
0-9
1-8
2-7
.
.
.
.
9-0 these are the digits that are ar equal to 9 =10 digits
 and what abt which are leass than 9 

similarly their total must be equal to 8,7,6,5,4,3,2,1,0.........
so for total of 9 we have 10 digits.......
for total of 8 we have 9 digits..
8,7,6,5,4,3,2,1...........so the total is 55.

karan

• Jul 24th, 2007

 

I feel the answer should be 1 .

as every lock can be opend by only one combination that sets it free!

This is my tought please reply if you agree ! i feel the other stats are simply to misguide.

bonthala narsimha reddy

• Jul 28th, 2007

 

Answer is 45.
because
number is 3 digit number.
so in first digit can fill up with only 1 to 9.
due to second rule , 2nd  digit fill up with only 0 to 8. because 2nd one is less than or equal to 9 only.
so like
series is
109
119
.
.
.
.
189
.
209
.
289
.
909.

ramprasad

• Aug 10th, 2007

 

55 is the correct answer
Since the last digit is 9 there are only 55 ways in which the first two digits can be placed.

uma

• Aug 15th, 2007

 

Let the No. be xy9.
x cannot be 0, because the no will become 2 digit no. x can be from 1-9.
y can be anything between 0-9.
and x+y=9.
so the possibilities are

189
279
369
459
549
639
729
819
909

so the total no of ways are 9.

Preethi

• Aug 17th, 2007

 

Ans is 45
9+8+........0

shrisha89

 Profile Answers by shrisha89 

• Feb 10th, 2010

 

If both combinations are combined answer is 45.

pravesh28

 Profile Answers by pravesh28 Questions by pravesh28

• Mar 27th, 2010

 

55

ans 10*9/2

eqal55

G.Yamini Devi

 Profile Answers by G.Yamini Devi 

• Jul 1st, 2010

 

There are 8 ways for the first digit and 9 ways for the second digit.

therfore 8*9=72.

raulharish

 Profile Answers by raulharish 

• Feb 24th, 2011

 

009 109 209 309 409 509 609 709 809 909   |
019 119 219 319 419 519 619 719 819          |
029 129 229 329 429 529 629 729                 |
039 139 239 339 439 539 639                        |
049 149 249 349 449 549                               |
059 159 259 359 459                                      |
069 169 269 369                                             |
079 179 279                                                    |
089 189                                                           |
099                                                                  |                    total= 55 ways                     

vinoth kumar

• Dec 14th, 2011

 

Answer is ii) and sum of the first two digits is less than or equal to the last digit.

numbers are from 0-9

sakshi

• Sep 9th, 2017

 

The digits might repeat.

Azharuddin Khan

• Oct 29th, 2017

 

Answer will be 55.
As last digit is fixed as 9. so there are only 2 digits which we can set according to first possibility.
Now according to second possibility sum of both remaining digits should not be greater than 9, so
If we put 0 at first place then we can place 10 digits from 0 to 9 at second position, so 10 possibilities.
If we put 1 at first place then we can put 9 digits 0-8 at second place, so 9 possibilities at second position.and so on.
so possible ways are 10+9+8+7+6+5+4+3+2+1=55.