Question

How many two digit numbers are divisible by 2?
$$\begin{gathered} (a)\text{ 30} \\ (b)\text{ 40} \\ (c)\text{ 45} \\ (d)\text{ 50} \end{gathered}$$

Answer 1

Hint: In this question use the concept that a two digit number which is divisible by 2 is from 10 to 98 and the series is 10, 12, 14, 16…………………….98. This forms an A.P so use the direct formula for the nth term of an A.P.

Complete step-by-step answer:

As we know all even numbers are divisible by 2.
So the even numbers are (2, 4, 6, 8, 10, 12, 14……………………………..)
Now we have to find out how many two digit numbers are divisible by 2.
So the first two digit number is (10), the second two digit number is (12) and the last two digit number is (98).
So the set of two digit numbers which are divisible by 2 are (10, 12, 14, 16, .... ,98)
Now as we see this set follows the trend of Arithmetic progression (A.P) with first term (a₁ = 10) common difference (d) = (12 – 10) = (14 – 12) = 2 and the last term (a_n) = 98.
So according to the formula of last term of an A.P or n^th term of an A.P
$\Rightarrow a_n=a_1+\left(n-1\right)d$ Where symbols have their usual meaning and n is the number of terms, so substitute the values in this equation we have,
$\Rightarrow 98=10+\left(n-1\right)2$
Now simplify the above equation we have,
$\Rightarrow 98-10=\left(n-1\right)2$
$\Rightarrow 2\left(n-1\right)=88$
Now divide by 2 we have,
$\Rightarrow n-1={\displaystyle \frac{88}{2}}=44$
$\Rightarrow n=44+1=45$
So this is the required number of two digit terms which are divisible by 2.
Hence option (C) is correct.

Note: Terms are said to be in A.P if the common difference that is the difference between the two consecutive terms remains constant all throughout the series. Another very frequently used series is that of G.P, in it the common ratio that is the ratio between two consecutive terms remains constant.