# Extra Questions For Class 10 Math Chapter 1 | Q2

###### Real Numbers | Highest Common Factor of 2 Numbers

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.

## NCERT Solution to Class 10 Maths

#### With Videos

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.

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###### NCERT Solution to Exercise QuestionsCBSE Class 10 Maths

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