Extra Questions For Class 10 Math Chapter 1 | Q2

This CBSE class 10 Maths Extra Practice Question is from Real Numbers. This word problem tests your understanding of computing HCF of two numbers that leave the same remainder.

Question 2 : Find the largest number that will divide 382 and 710 and leaves a remainder 13 in each case.

Video Solution

Explanatory Answer

Given: The two numbers are 382 and 710
Let 'x' be the largest number that divides 382 and 710, leaving 13 as remainder in both cases.

Step 1 - Decoding the given information

Remainder of \\frac{\text{382}}{\text{x}}) is 13
∴ 382 – 13 = 369 is divisible by x.
Similarly, remainder of \\frac{\text{710}}{\text{x}}) is 13
∴ 710 – 13 = 697 is divisible by x.

So, x divides both 369 and 697.
Or, x is a factor common of both 369 and 697.
Because x is the largest such number, x is the HCF of 369 and 697.

Step 2 - Use Euclid's Division Lemma to find HCF of 369 and 697

Step 1: Apply Euclid's Lemma on 697 with 369 as divisor
697 = 369 × 1 + 328
The remainder is not zero.
Step 2: Apply Euclid's Division Lemma on 369 with 328 as divisor
369 = 328 × 1 + 41
The remainder is not zero.
Step 3: Apply Euclid's Lemma on 328 with 41 as divisor
328 = 41 × 8 + 0
The remainder is zero.
∴ The divisor of this step, 41 is the HCF of 369 and 697

The largest number that divides 382 and 710 leaving remainder 13 in each case is 41.