Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 33)
33.
How many natural numbers are there between 23 and 100 which are exactly divisible by 6 ?
Answer: Option
Explanation:
Required numbers are 24, 30, 36, 42, ..., 96
This is an A.P. in which a = 24, d = 6 and l = 96
Let the number of terms in it be n.
Then tn = 96 
a + (n - 1)d = 96
24 + (n - 1) x 6 = 96
(n - 1) x 6 = 72
(n - 1) = 12
n = 13
Required number of numbers = 13.
Discussion:
16 comments Page 2 of 2.
Anil sarode said:
1 decade ago
@Shiv here is the method for you.
Here total 78 numbers are here, check by AP formula.
Tn = a+(n-1) d where a = 23, d =1, Tn = 100, so n = 78. Now simply divide 78/6 = 13.
Here total 78 numbers are here, check by AP formula.
Tn = a+(n-1) d where a = 23, d =1, Tn = 100, so n = 78. Now simply divide 78/6 = 13.
(1)
SHIV said:
1 decade ago
Any other method please.
Avinash said:
1 decade ago
24+(n-1)6 = 96.
= n-1 = 12.
= n = 13.
= n-1 = 12.
= n = 13.
Goms said:
1 decade ago
4*6 = 24
5*6 = 30
6*6 = 36
7*6 = 42
8*6 = 48
9*6 = 54
10*6 = 60
11*6 = 66
12*6 = 72
13*6 = 78
14*6 = 84
15*6 = 90
16*6 = 96
TOTAL = 13.
5*6 = 30
6*6 = 36
7*6 = 42
8*6 = 48
9*6 = 54
10*6 = 60
11*6 = 66
12*6 = 72
13*6 = 78
14*6 = 84
15*6 = 90
16*6 = 96
TOTAL = 13.
Sallick said:
1 decade ago
24+(n-1)6=96
=n-1=12
=n=13
=n-1=12
=n=13
Mahesh Patil said:
1 decade ago
Total numbers between 23 to 100 = 100-23+1 = 78
Numbers Divisible by 6 = 78 / 6 =13 (consider Only Quotient ....not Remainder)
Numbers Divisible by 6 = 78 / 6 =13 (consider Only Quotient ....not Remainder)
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