Latest Elitmus Aptitude Question SOLUTION: how many no. will be possible between 100 to 500 such that the sum of two digit is equal to the third(example -312, 2+1=3)?? a)12 b)24 c)66 d)72

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(#M40157242) ELITMUS Numerical Ability question Keep an EYE

how many no. will be possible between 100 to 500 such that the sum of two digit is equal to the third(example -312, 2+1=3)??
a)12
b)24
c)66
d)72

Asked In Elitmus umang agarwal (7 years ago)

Unsolved Read Solution (12)

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Abhinaw kumar  (7 years ago)

is quant question is repeat in elitmus..tell me

Is this solution Helpfull? Yes (28) | No (1)  

pankaj  (7 years ago)

ans:- 66
just like sanjib ans
18 no between 100-199
17 no between 200-299 (211)
16 no between 300-399 (312,321)
15 no between 400-499(413,431,422)

sum is 66

Is this solution Helpfull? Yes (19) | No (3)  

Sanjib Badyakar   (7 years ago)

Ans:-72

100-199
add middle digit......so there will be ...110, 121,132,143,154,165.....total 9 no.
add last digit......101,112,123,134....total 9 no.
so total 18 nos.
200-299
calculate same as before....we will get total of 18 nos.

so in between 100-500
we will get 18*4=72

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Subrata Dutta  (7 years ago)

100 - 199 : 18
200 - 299 : 17
300 - 399 : 16
400 - 499 : 15

Total no. 66

Is this solution Helpfull? Yes (10) | No   

shiv  (7 years ago)

for 123 total permutation 123,132,213,231,312,321
123=6
134=6
145=4 because in 6 case two case 541>500 and 514>500
156=2 because in 6 case four case 651>500 and 615>500 and 516 ,561
167=2
178=2
189=2

235=4
246=4
257=2
268=2
279=2

347=4
358=2
369=2
total=46
and
101,110
202,220
303,330
404,440
total=8
121,221,112
242,224,422
336,363
448,484
total=10


total no of case 46+8+10=66

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Vikash kumar  (7 years ago)

correct answer is 66
explanation;
a b c
1 0 1=(a=b+c)
1 1 0
1 1 2
1 2 1
1 2 3
1 3 2
1 3 4
1 4 3
1 4 5
1 5 4
1 5 6
1 6 5
1 6 7
1 7 6
1 7 8
1 8 7
1 8 9
1 9 8
2 0 2
2 1 1
2 1 3
2 2 0
2 2 4
2 3 1
2 3 5
2 4 2
2 5 3
2 5 7
2 6 4
2 6 8
2 7 5
2 7 9
2 8 6
297
303
312
314
325
321
330
336
341
347
352
358
363
374
385
396
404
......

......
total u got 66 no

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ramakant  (7 years ago)

ans...c..
101,112,123,134,145,156,167,178,189---9+(last 2 digit can be interchanged---so more 9 also) so total 18
202,213,224,235,246,257,268,279----------8+(last 2 digit can be interchanged --so 8 more)+211--so total 17
303,314,325,336,347,358,369------------7+(last 2 digit can be interchanged --so 7 more)+312,321--so total 16
404,415,426,437,448,459-------------------------------------------------------------------------------------------------------------15
505,516,527,538,549.............................................................................................................................................................14
------------------------------------------------------------------------------------------------------------------------------------------------------
total 66
505,516,527,538,549--------------------5
----------------------------------------------------33

Is this solution Helpfull? Yes (2) | No   

sandeep  (7 years ago)

Answer is : c) 66
Number between 100 to 500
apply permutation: abo 2 ways - abo, bao ( o= zero)
101, 202, 303, 404 => 4*2 = 8 ways
apply permutation : aab (according to question in given range for a and b means a, b < 5)
aab , aba, baa - 3 ways
112, 224=> 3*2=6 ways
if b >=5 so 336, 448 -- 2*2=4 ways
if abc permut: 6ways
123, 134, = 6*2 =12 ways
if abc permute : 4 ways ( if you see c>=5)
145, , 235, 246, 347=> 4*4= 16 ways
if abc permuate : 2 ways( if you see b and c both >=5)
156, 167, 178, 189, 257, 268, 279, 358, 369, 459= > 10*2= 20 ways
so total possible number => 20+16+12+4+6+8=66 ans
if we clearly understand it take less time with in 2 or 3 mins

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umang agarwal  (7 years ago)

how you have counted nine no.

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Amutharajan  (7 years ago)

15. 6*22=132-7=125-35=90/6=15

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Pushp Raj  (7 years ago)

2 no between 100-199 (101,110)
3 no between 200-299 (202,220,211)
4 no between 300-399 (303,312,321,330)
5 no between 400-499(404,413,422,431,440)
Total no=14
Explanation:-
minimum possibility of getting 1 is 0+1 and 1+0.
minimum possibility of getting 2 is 0+2 ,2+0 and 1+1.
minimum possibility of getting 3 is 0+3, 3+0,1+2 and 2+1.
minimum possibility of getting 4 is 0+4, 0+4,2+2,1+3 and 3+1 .

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ravi  (7 years ago)

go through options one's digit 3+3=6, ten's digit =6 ans.66

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