NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3

Question 18

Step 1: Understand the condition.

A number leaves remainder 3 when divided by 4 if it looks like:

\(4k + 3\), where \(k\) is an integer (whole number).

Step 2: Find the smallest number greater than 10 of this form.

Check numbers just after 10:

• 11 ÷ 4 = 2 remainder 3
• So, first number = 11.

Step 3: Find the largest number less than 300 of this form.

Check numbers near 300:

• 299 ÷ 4 = 74 remainder 3
• So, last number = 299.

Step 4: Notice the sequence.

The numbers are: 11, 15, 19, 23, …, 299.

This is an arithmetic progression (AP) with:

• First term \(a = 11\)
• Common difference \(d = 4\)
• Last term \(l = 299\)

Step 5: Use the formula to find the number of terms in AP:

\(n = \dfrac{l - a}{d} + 1\)

\(n = \dfrac{299 - 11}{4} + 1\)

\(n = \dfrac{288}{4} + 1 = 72 + 1 = 73\)

Final Answer: There are 73 numbers between 10 and 300 that leave remainder 3 when divided by 4.