Question:
The number of multiples of 4 lying between 12 and 250 is :
The number of multiples of 4 lying between 12 and 250 is :
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An alternate shortcut to find the number of multiples of $k$ in an open interval $(A, B)$ is:
Number of multiples $= \lfloor \frac{B - 1}{k} \rfloor - \lfloor \frac{A}{k} \rfloor$
Here, $A = 12$, $B = 250$, and $k = 4$.
Number of multiples $= \lfloor \frac{249}{4} \rfloor - \lfloor \frac{12}{4} \rfloor = 62 - 3 = 59$.
This formula is extremely fast and avoids setting up an A.P. equation entirely!
Number of multiples $= \lfloor \frac{B - 1}{k} \rfloor - \lfloor \frac{A}{k} \rfloor$
Here, $A = 12$, $B = 250$, and $k = 4$.
Number of multiples $= \lfloor \frac{249}{4} \rfloor - \lfloor \frac{12}{4} \rfloor = 62 - 3 = 59$.
This formula is extremely fast and avoids setting up an A.P. equation entirely!
Updated On: Jul 7, 2026
- 59
- 59.5
- 60
- 61
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
This question is based on the topic of "Arithmetic Progressions (A.P.)".
We are asked to find the total number of multiples of 4 that lie strictly *between* 12 and 250.
The word "between" is crucial because it excludes the boundary values 12 and 250 from our count, even if they themselves are multiples of 4.
Step 2: Key Formula or Approach:
The multiples of 4 form an Arithmetic Progression (A.P.) because there is a constant difference of $4$ between consecutive terms.
The $n^{\text{th}}$ term of an A.P. is given by the formula:
\[ a_n = a + (n - 1)d \] where:
• $a$ is the first term of the progression.
• $d$ is the common difference.
• $n$ is the number of terms.
• $a_n$ is the last term of the progression.
By identifying the first and last multiples of 4 within the given open interval $(12, 250)$, we can solve for $n$.
Step 3: Detailed Explanation:
• The interval is strictly between 12 and 250, represented as $12 \lt x \lt 250$.
• The first multiple of 4 strictly greater than 12 is:
\[ a = 16 \]
• To find the last multiple of 4 strictly less than 250, divide 250 by 4:
\[ \frac{250}{4} = 62.5 \] This indicates that $62 \times 4 = 248$ is the largest multiple of 4 below 250.
Thus, our last term is:
\[ a_n = 248 \]
• The common difference for multiples of 4 is:
\[ d = 4 \]
• Substitute these values into the $n^{\text{th}}$ term formula:
\[ 248 = 16 + (n - 1) \cdot 4 \]
• Subtract 16 from both sides to isolate the term containing $n$:
\[ 248 - 16 = (n - 1) \cdot 4 \] \[ 232 = (n - 1) \cdot 4 \]
• Divide both sides by 4:
\[ n - 1 = \frac{232}{4} \] \[ n - 1 = 58 \]
• Add 1 to both sides to find $n$:
\[ n = 58 + 1 = 59 \]
• Therefore, there are exactly 59 multiples of 4 lying between 12 and 250.
Step 4: Final Answer:
The total number of multiples of 4 is 59, which corresponds to Option (A).
This question is based on the topic of "Arithmetic Progressions (A.P.)".
We are asked to find the total number of multiples of 4 that lie strictly *between* 12 and 250.
The word "between" is crucial because it excludes the boundary values 12 and 250 from our count, even if they themselves are multiples of 4.
Step 2: Key Formula or Approach:
The multiples of 4 form an Arithmetic Progression (A.P.) because there is a constant difference of $4$ between consecutive terms.
The $n^{\text{th}}$ term of an A.P. is given by the formula:
\[ a_n = a + (n - 1)d \] where:
• $a$ is the first term of the progression.
• $d$ is the common difference.
• $n$ is the number of terms.
• $a_n$ is the last term of the progression.
By identifying the first and last multiples of 4 within the given open interval $(12, 250)$, we can solve for $n$.
Step 3: Detailed Explanation:
• The interval is strictly between 12 and 250, represented as $12 \lt x \lt 250$.
• The first multiple of 4 strictly greater than 12 is:
\[ a = 16 \]
• To find the last multiple of 4 strictly less than 250, divide 250 by 4:
\[ \frac{250}{4} = 62.5 \] This indicates that $62 \times 4 = 248$ is the largest multiple of 4 below 250.
Thus, our last term is:
\[ a_n = 248 \]
• The common difference for multiples of 4 is:
\[ d = 4 \]
• Substitute these values into the $n^{\text{th}}$ term formula:
\[ 248 = 16 + (n - 1) \cdot 4 \]
• Subtract 16 from both sides to isolate the term containing $n$:
\[ 248 - 16 = (n - 1) \cdot 4 \] \[ 232 = (n - 1) \cdot 4 \]
• Divide both sides by 4:
\[ n - 1 = \frac{232}{4} \] \[ n - 1 = 58 \]
• Add 1 to both sides to find $n$:
\[ n = 58 + 1 = 59 \]
• Therefore, there are exactly 59 multiples of 4 lying between 12 and 250.
Step 4: Final Answer:
The total number of multiples of 4 is 59, which corresponds to Option (A).
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