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(#M40030610) ELITMUS question qant Keep an EYE

how many numbers are there in between 100 to 400 ,in which adding any of two digit then the sum is equal to third digit.

Asked In Elitmus vaibhav kumar (11 years ago)

Unsolved Read Solution (33)

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Read Solution (33) :

Ankit Gupta  (11 years ago)

51 is the right answer 100% suree guys..
**for last digit 1 there is 1 possibility 101 and rearrange it 110 so total-[2] **same for 2 ( 112,121,211,202,220 )---[5]
**same for 3 ( 123,132,213,231,312,321,303,330) --- [8]
**same for 4 (134,143,314,341,224,242)--- [6]
** same for 5 ,6,7,8 and 9 all have [6] possibilities ----- finally (2+5+8+6+6+6+6+6+6 = 51 ) Answer

Is this solution Helpfull? Yes (21) | No (2)  

Md Asif Ashraf  (11 years ago)

let take an arrangement three nos like
|a|b|c|
in this condition is number should be greater than 100 so we can fix a=1 in first condition
(0-9)
|
|1|b|c|->(0-9) we have equn 1+b=c; or 1+c=b or c+b=1; like this arrangement we can get 9*2 no i.e 18
similarly if we fix a=2;
2+b=c; or 2+c=b; or b+c=2; like this we have 8*2+1=17 nos.
similarly fix a=3;
we have eqn 3+b=c;or 3+c=b; or b+c=3;
like this we have 7*2+2=16 nos
so finally we total ans=18+17+16=51; 100% sure

Is this solution Helpfull? Yes (17) | No   

Gaurav Janoti  (11 years ago)

51

100% sure its correct

Is this solution Helpfull? Yes (10) | No (1)  

abc  (11 years ago)

51 is the right answr

Is this solution Helpfull? Yes (5) | No   

Shilpa Jha  (11 years ago)

I think the answer is 26...

Is this solution Helpfull? Yes (3) | No (12)  

vaibhav kumar   (11 years ago)

no reshav ...mainly i have put the question here and there is no any condition that only last digit is sum of other two and the option is
a)27
b)30
c)41
d)51

and if u count then the answer is 51 but if any one know the short trick then please share because if u go for counting then it takes lot of time...

and thanks to all for ur valuable comments

Is this solution Helpfull? Yes (3) | No   

vijay chaudhary  (11 years ago)

51 is correct

Is this solution Helpfull? Yes (2) | No   

Priyanka Sonal  (11 years ago)

The Answer is 51....

Is this solution Helpfull? Yes (2) | No   

vishal rathore  (11 years ago)

101 ,112 , .... 189 ( 1 at 1st place ) total are 9 numbers
202 , 213 , ....279 ( 2 at 1st position ) total are 8 numbers
303 , 314 , ....369 ( 3 at 1st postion ) total are 7 numbers
unit ,tens digit are interchanged then number doubles hence
total = ( 9 + 8 + 7 )*2 = 48

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

Piyush Chandra  (11 years ago)

Can You explain

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seema  (11 years ago)

ans: 33
100-200: 11 nos
201-300: 11 nos
301-400: 11 nos

Is this solution Helpfull? Yes (1) | No (6)  

RAKESH  (11 years ago)

i think there is no no. betn 100 & 400
adding any of two digit gives sum equal to third digit.
ans zero

Is this solution Helpfull? Yes (1) | No (30)  

Mahesh  (11 years ago)

Sorry for the confusion, the following is the solution for numbers 100 to 400

Last digit 1 : First two digits (1,0) 1 possibility
Last digit 2 : First two digits (2,0) (1,1) 2 possibilities
Last digit 3 : First two digits (3,0) (2,1) (1,2) 3 possibilities
Last digit 4 : First two digits (3,1) (2,2) (1,3)
Last digit 5 : First two digits (3,2) (2,3) (1,4)

As the first digit cannot be more than 3, for each of last digits 6,7,8 and 9 also we get three numbers each

So, total possibilities = 27

Answer is 27

Is this solution Helpfull? Yes (1) | No (6)  

Tushant  (11 years ago)

answer is 46

101
110
112
121
123
132
134
143
145
154
156
165
167
176
178
189
198
202
213
220
224
235
242
246
253
257
264
268
275
279
286
297
303
314
325
330
336
341
347
352
358
363
369
374
385
396

Is this solution Helpfull? Yes (1) | No (5)  

Md Inamul Haque  (11 years ago)

Question is not cleared, it must be precised,...."in which adding any of two digit then the sum is equal to third digit".....it tells that we can take any two digits out of three then sumation must be equal to the rest one, which is not possible or it cause problem to understand.

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

ravi gangwar  (11 years ago)

any of two digit......not frst two is condition here.......n sum sld be equal to the thrd......nt neccsry eql to the digit at unit place.......i thnk no such no exist.......

Is this solution Helpfull? Yes (1) | No   

Mahesh  (11 years ago)

The last digit can be any number from 1 to 9.

If the last digit is 1, the first two digits should be (1,0), one possibility
If the last digit is 2, the first two digits should be (2,0),(1,1) two possibilities
If the last digit is 3, the first two digits should be (3,0),(2,1) and (1,2) three possibilities
If the last digit is 4, the first two digits should be (4,0),(3,1), (2,2) and (1,3) two possibilities

Similarly, if the last digit is 9, there 9 possibilities, (9,0) (8,1) (7,2) (6,3)(5,4)(4,5)(3,6)(2,7)(1,8)

So, total number of possibilities = 1+2+3+4+5+6+7+8+9 = 45 numbers

Is this solution Helpfull? Yes | No (4)  

Arun  (11 years ago)

If we go counting wise starting from 100, such nos. are
112
123
134
145
156
167
178
.
..
.
.
235
246
257
268
274
.
.
.
347
358
369

The answr is 22

Is this solution Helpfull? Yes | No (4)  

rishabh kapoor  (11 years ago)

101
112
123
134
145
156
167
178
189
202
213
224
235
246
257
268
279
303
314
325
336
347
358
369

http://www.referralduty.com/index.php?invite=31681

Is this solution Helpfull? Yes | No (1)  

RAKESH  (11 years ago)

101,112,123,134,145,156,167,178,189
202,213,224,235,246,257,268,279
303,314,325,336,347,358,369
total=26

Is this solution Helpfull? Yes | No (7)  

rishabh kapoor  (11 years ago)

rakesh etna e likh deta same as above lolz
copy cat

Is this solution Helpfull? Yes | No (3)  

RAKESH  (11 years ago)

101,112,123,134,145,156,167,178,189
202,213,224,235,246,257,268,279
303,314,325,336,347,358,369
total=9+8+7=24

Is this solution Helpfull? Yes | No (5)  

rishabh kapoor  (11 years ago)

sushant read the condition last digit must be sum of two

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Mahesh  (11 years ago)

Answer is 24

Is this solution Helpfull? Yes | No (1)  

abc  (11 years ago)

@vaibhav
in which elitmus test it was asked?

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vaibhav kumar   (11 years ago)

@just 27 nov

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Naman Arusia  (11 years ago)

Answer is 16+17+16=49

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Aditya Mishra  (11 years ago)

Bhai log, series ban jaega count karna padega, aur koi option nai hae

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sudheer  (11 years ago)

may be 0. there is no number between 100 & 400

Is this solution Helpfull? Yes | No (1)  

vivek tiwari  (11 years ago)

awesome answer ashraf bhai

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praveen  (11 years ago)

visit this for solution http://www.onlineadhyayana.com/ elitmus section...:)

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Ajay Sharma  (11 years ago)

Can anyone plz explain me the solution of Md Asif Ashraf

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Umesh Yadav  (11 years ago)

I think if 100th place is fixed for 1,2 and 3
then for 1 "101" can be arranged in 2 ways.
.
. . 101,112,123,134,145,156,167,178,189=2*9=18
similarly 202,213,224,235,246,257,268,279=2*8=16
303,312,314,325,336,347,358,369=2*8=16
therefore ans=18+16+16=50

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