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Question: How many whole numbers are between 437 and 487? \(\left( A \right)\) 50 \(\left( B \right)\) 49 ...

How many whole numbers are between 437 and 487?
(A)\left( A \right) 50
(B)\left( B \right) 49
(C)\left( C \right) 51
(D)\left( D \right) None of these

Explanation

Solution

Hint – In this question first find out the first, second and last whole number between 437 and 487, then check out which series it will makes whether it is A.P, G.P or H.P, then apply the formula of nth term of the series and calculated the number of terms in the series so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Whole number
Whole numbers are those which start from zero (i.e. basic counting numbers) 0, 1, 2, 3,............., 437, 438, ......, 486, 487, 488......... Up to so on, it does not include negative integers and also not include irrational numbers, i.e. it only includes positive integers starting from zero.
So we have to find out the number of whole numbers between 437 and 487
So the first, second and last whole number between 437 and 487 are 438, 439 and 486.
So the series is
438, 439, 440,.................., 486
So as we see this will follow the rule of A.P (arithmetic progression) with first term (a) = 438, common difference (d) = (439 – 438) = (440 – 439) = 1 and last term (an) is = 486.
Let n be the number of terms of this series.
Now as we know the nth term of an A.P is given as
⇒an=a+(n−1)d\Rightarrow {a_n} = a + \left( {n - 1} \right)d, where symbols have their usual meaning.
Now substitute the values in the above equation we have,
⇒an=a+(n−1)d\Rightarrow {a_n} = a + \left( {n - 1} \right)d
⇒486=438+(n−1)1\Rightarrow 486 = 438 + \left( {n - 1} \right)1
Now simplify this we have,
⇒486−438=(n−1)\Rightarrow 486 - 438 = \left( {n - 1} \right)
⇒48=(n−1)\Rightarrow 48 = \left( {n - 1} \right)
⇒n=48+1=49\Rightarrow n = 48 + 1 = 49
So there are 49 whole numbers between 437 and 487.
Hence option (B) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that the formula of the nth terms of an arithmetic progression which is written above then using this formula substitute the values of the first term, common difference and nth term of the series and simplify as above we will simply get the number of terms in the series.