Extra Questions For Class 10 Maths Chapter 1 | Q1

This CBSE class 10 Maths Extra Practice Question is from Real Numbers. This extra question is a word problem that tests your ability to translate information given in words into mathematical concepts and solve the question.

Question 1 : Karan has 180 blue marbles and 150 red marbles. He wants to pack them into packets containing equal number of marbles of the same colour. What is the maximum number of marbles that each packet can hold?

Video Solution

Explanatory Answer

Given: Number of blue marbles = 180
Number of red marbles = 150
Each packet contains the same number of marbles.

Step 1 - Calculate the number of packets and number of marbles in each packet

Let the maximum number of marbles that each packet can hold be ‘x’.

For Blue marbles: Let the 180 blue marbles be divided into 'a' packets containing 'x' marbles each.
So, 180 = ax.
Because number of marbles and number of packets are integers, x and a are factors of 180.

For Red marbles: Let the 150 red marbles be divided into 'b' packets containing 'x' marbles each.
So, 150 = bx.
Because number of marbles and number of packets are integers, x and b are factors of 150.

∴ x is a factor of 180 and 150. i.e., x is a common factor of 180 and 150.
Because ‘x’ is the maximum number of marbles that each packet contains, x is the HCF of 180 and 150.

Step 2 - Apply Euclid’s Lemma to find HCF

Step 1: Apply Euclid’s Lemma on 180 and 150
180 = 150 × 1 + 30
The remainder is not ‘0’

Step 2: Apply Euclid’s Lemma on 150 and 30
Note: If the remainder is not zero in any step, the divisor of that step becomes the new dividend for the next iteration and the remainder of that step becomes the divisor for the next iteration.
150 = 30 × 5 + 0
The remainder is ‘0’.

If the remainder becomes zero in a step, the process is stopped.
The divisor of the step in which the remainder becomes zero is the HCF.
The divisor of step 2 is 30 and it is the HCF.

So, the maximum number of marbles in each packet is 30.