Latest TCS Aptitude Question SOLUTION: a person starts writing all 4 digits numbers .how many times had he written the digit 2? a)3700 b)32000 c)3200 d)37000

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a person starts writing all 4 digits numbers .how many times had he written the digit 2?
a)3700
b)32000
c)3200
d)37000

Asked In TCS Arun (10 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

Aditya  (10 years ago)

ans:3700
at 1's place=900 times
at 10's place=900 times
at 100's place=900 times
at 1000's place(from 2000 to 2999)=1000 times
total=1000+900+900+900=3700

Is this solution Helpfull? Yes (35) | No   

koushik  (10 years ago)

ans:a
_ _ _ _
2 (10) (10) (10)
_ _ _ _
(9) 2 (10) (10)
_ _ _ _
(9)(10) 2 (10)
_ _ _ _
(9)(10)(10) 2

10*10*10+3(9*10*10)=3700

Is this solution Helpfull? Yes (29) | No (3)  

sriganesh  (10 years ago)

No of two at unit digit is=900
No of two at ten digit is=900
No of two at hundred digit is=900
No of two at thousand digit is=1000
so 2 is repeated in (900+900+900+1000)=3700 is the ans

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

sami wardhan karn  (10 years ago)

numbers with 2 at unit digit=9*10*10=900,
'' '' 2 at tens place=900,
'' '' 2 at hundred place=900,
'' '' 2 at thousand place=10*10*10=1000.therefore,number of times 2 is written=(900*3)+1000=3700

Is this solution Helpfull? Yes (4) | No   

Amrutha Duggirala  (10 years ago)

The 4-Digit Numbers are from 1000 to 9999.
Digit 2 is written in Units Place 900 times; In Ten's Place 900 times; In Hundred's Place 900 times and in Thousands Place 1000 times. So total=1000+900+900+900=3700

Is this solution Helpfull? Yes (3) | No (2)  

Rajesh  (10 years ago)

Ans is 3700.
from 1000 10 1999 consists of 300 digit 2. likewise 9*300(because 1000 to 9999)=2700
also from 2000 t0 2999 consists extra 1000.
therefore 2700+1000=3700

Is this solution Helpfull? Yes (2) | No   

Rajesh  (10 years ago)

Ans is 3700.
from 1000 10 1999 consists of 300 digit 2. likewise 9*300(because 1000 to 9999)=2700
also from 2000 t0 2999 consists extra 1000.
therefore 2700+1000=3700

Is this solution Helpfull? Yes (1) | No   

NARAYANA SWAMY DAMMALA  (10 years ago)

90*20+1000+900=3700
in evry 100 no's 20 2's,so we hve 90 100's,2000-2999 we hve 1000 2's,1200-1299 2200-2299....9200-9299 we hve 900 2's...

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Guru Pranav  (10 years ago)

total=1000+900+900+900=a)3700

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md equbal  (8 years ago)

2001
2002
2003
.
.
.
2999
hence digit 2 will repeat 1000 times at 4'th(1000's) place. Now
1200
1201
1202
.
.
1299 hence 2 repeat 100 times here.replace 1000's place(1) by 2,3,4,5,6,7,8,9.so digit 2 repeat 100*9=900 times at 100's place
similarly for 10's place
1020
1021.
.
.
1029
.
1120
1129.
.
1220
1229
.
.and so on
so the total no=9*10*10=900
similarly for 1's place
1002
1009
.
.
.
so the final ans=1000+900+900+900=3700

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mangaldeep kiran  (8 years ago)

Total number of four digit numbers =9000 (i.e 1000 to 9999 )
We try to find the number of numbers not having digit 2 in them.
Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here '0' cannot be taken
Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832
Total number of numbers having digit 2 in it = 9000-5832 =3168

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