Latest Elitmus Aptitude Question SOLUTION: How many numbers in the set X if it has only odd number between 100 and 300 and divisible by 3........?

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(#M40152376) ELITMUS Numerical Ability question Number Systems Keep an EYE

How many numbers in the set X if it has only odd number between 100 and 300 and divisible by 3........?

Asked In Elitmus sujith (8 years ago)

Unsolved Read Solution (21)

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Read Solution (21) :

Utpal kumar  (8 years ago)

the series is- 105,111,117------------297
here
a=105, d=6,
as Tn=a+(n-1)d
so,
297=105+(n-1)6
=>n-1=32
so, n=33

ANS=33

Is this solution Helpfull? Yes (36) | No (9)  

Neeraj kumar  (8 years ago)

the series is:105,111.........297
a=105,d=6,tn=297
tn=a+(n-1)d
297=105+(n-1)6
n=33.

Is this solution Helpfull? Yes (7) | No (3)  

kundan prasad  (8 years ago)

Ans:33

Is this solution Helpfull? Yes (3) | No   

Pankaj Kumar  (8 years ago)

the series would be 102,105.......297
297=102+(n-1)3 =>
n=66
ANS- 66

Is this solution Helpfull? Yes (2) | No (18)  

Pragati Priya  (8 years ago)

105, 111.......,297
297=105+(n-1)6
n= 33

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

Arpit sah  (8 years ago)

the series would be 102, 105,108 and so on the last being 300 so no of terms wouldbe (300-102/3)+1 (Last term -first term/common difference)+1 and so the answer is 67

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

Mukesh  (8 years ago)

The Series Starts From 102 And Ends At 300.
So, Tn = a + (n-1)*d
Tn=300, a= 102, d= 3
300 = 102 + (n-1)*3
=> (n-1) = (300 - 102)/3 = 66
=> n=66 +1 = 67
Hence, n= 67 ::::::::::::::::: :)

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Mukesh  (8 years ago)

Sorry, My Mistake ---- Answer Is 33 = 67/2 (For Odd And Even)... So Odd = 33 And Even = 34...::::::: :)

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rajveer singh  (8 years ago)

If both 100 and 300 included then answer would be 67 terms....

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

alok  (8 years ago)

33 becoz i am saying

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Mukesh  (8 years ago)

Dude !! Question Is To Find Odd numbers So Ans is [67/2] = 33

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kumar saurav  (8 years ago)

8 numbers will be there
111,117,135,153,159,171.177,195,

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AKASH TOMER  (8 years ago)

first number=105 last=297 diff=6
total no=last-first/diff+1
33

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avinash kumar jha  (8 years ago)

term=105,111,117............,297
no. of term=297-105/6=32+1=33

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hriday  (8 years ago)

no. of odd terms b/w 100 to 300 =100
105,107,.......297;
l=a+(n-1)d; =>n=33

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BHAURAO SAMDARE  (8 years ago)

answer is 34

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subham kumar shaw  (8 years ago)

the answer is 67 since 300 is also included

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kishor kumar sahu  (8 years ago)

The first odd no b/w 100-300 is 105 it divisible by 3. the serise is - 105,111,117.........upto 297.so
first term=105 last term=297 and d=6 (d is the difference between terms of the arithmetic progression)
(last term) - (first term)/difference 1.
now, (297-105)/6+1 =>34+1=35 is ans

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Vikash kumar  (7 years ago)

series like start is 105,111,117,...........297..(1)
not start 102,105 bcz it is also divisible by 6
so now equation 1 behaves like A.P.
we know that
tn=a+(n-1)d
297=105+(n-1)6
n=33 . is the correct answer.

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Vikash kumar  (7 years ago)

correct answer is 33
EXPLANATION:
x={105,111,117,...............297} it all number is divisible by 3
accn to arithmetic progression
a=105 ,d =6, tn =297
we know that
tn = a+(n-1)d
297=105+(n-1)6
192=(n-1)6
32=n-1
n=33

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Digesh Singh Bisht  (1 year ago)

No. divided by 3 between are (102,105,108,111,114,117,........... , 297, 300)
but we need only odd numers.
so,
the series is:105,111.........297
a=105,d=6,last number(l)=297
l=a+(n-1)d
297=105+(n-1)6
n=33 answer

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