Latest TCS Aptitude Question SOLUTION: 20. How many numbers are there from 100 to 1200 which are not divisible either by 2, 3, or 5?

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20. How many numbers are there from 100 to 1200 which are not divisible either by 2, 3, or 5?

Asked In TCS Bishal Pandey (11 years ago)

Unsolved Read Solution (11)

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Sumit Bhattacharyya  (11 years ago)

tn=a+(n-1)d
For 2=>n=551
For 3=>n=367
For 5=>n=221
For (2*3)=6=>n=184
For (3*5)=15=>n=74
For (5*2)=10=>n=111
For LCM (2,3,5)=30=>n=37
So NUMBERS are=(551+367+221)-(184+74+111)+37
=1139-369+37
=807

Therefore no of numbers not divisible=1200-99-807=294

Is this solution Helpfull? Yes (54) | No (4)  

pratik jitendra malviya  (11 years ago)

Sn=a+(n-1)d
For 2=>n=550
For 3=>n=267
For 5=>n=220
But in 5 Half of the terms will end with 0 hence they are also divisible by 2
So 1098-(550+367+110)=1098-1027=71

Is this solution Helpfull? Yes (1) | No (8)  

Debu Singh  (11 years ago)

tn=a+(n-1)d
For 2=>n=551
For 3=>n=367
For 5=>n=221
For (2*3)=6=>n=184
For (3*5)=15=>n=74
For (5*2)=10=>n=111
For LCM (2,3,5)=30=>n=37
So NUMBERS are=(551+367+221)-(184+74+111)+37
=1139-369+37
=807

Is this solution Helpfull? Yes (1) | No (4)  

Pradeep   (11 years ago)

1/2 of the no.s divisible by 2 so 550 no.s divisible and for5 as already cancelled the
10 20 30 40........by 2 so in every 10no.s one will be divisible for 55 tens. 55 can divisible
So 550-55=495
Then after removing 2 and 5 multiples in every 30 no.s 4 will divisible
For 495 no.s 16*4+2=66
So ans is 495-66=429

Is this solution Helpfull? Yes (1) | No   

Pradeep   (11 years ago)

294 answer

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Priyanka Shekhaw3at  (11 years ago)

what is 1098 here

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Vivek Banga  (11 years ago)

Priyanka..1098 is the total no. of terms from a=100 and tn=1200..

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mritunjay kumar  (11 years ago)

ans--1139...

Is this solution Helpfull? Yes | No (2)  

anand singh  (11 years ago)

n(AUBUC) = n(A) + n(B) + n(C) - n(A intersectn B) - n(B intersectn C) - n(A intersectn C) + n(A intersectn B intersectn C)

n(2U3U5) = n(2) + n(3) + n(5) - n(6) - n(15) - n(10) + n(30)

--> n(2U3U5) = 600 + 400 + 240 - 200 - 80 - 120 + 40 = 880

Hence number of numbers not divisible by any of the numbers 2, 3, and 5

= 1100 - 880 = 220.

Is this solution Helpfull? Yes | No (4)  

Mudassir Fazal  (11 years ago)

total number of terms b/w 100 and 1200 are 1001
out of it 501 are even
remaining terms=500
numbers divisible by 3
3*34=102
3*400=1200
so numbers divisible by 3 are 400-34+1=367
remaining terms=500-367=133
similarly number divisible by 5 221
remainig number 133-221= -88
now adding multiples of 2*3(6),2*5(10),3*5(15) as they are reapetely subtracted
ie
we get 281 as total number not divisible by 2, 3,or 5

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Somobrata Ghosh  (8 years ago)

why - 99 from 1200??

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