CBSE Class 10 - Real Numbers Problems on Euclid's Division Algorithm(2016)

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REAL NUMBERS PROBLEMS ON

EUCLID'S DIVISION ALGORITHM(2016)

**Q1(CBSE 2012): Using Euclid’s division algorithm, find the HCF of 240 and 228.**

Answer: By Euclid’s division algorithm,

⇒ 240 = 228 × 1 + 12

⇒ 228 = 12 × 19 + 0

∴ HCF (240, 228) = 12

**Q2(**

**CBSE 2014)**

**: The length, breadth and height of a room are 8m 25 cm, 6m 75 cm and 4 m 50 cm respectively. Find the length of the longest rod that can measure the three dimensions of the room exactly.**

Answer:

∵ 1m = 100 cm

∴ 8 m 25 cm = 825 cm

6 m 75 cm = 675 cm

4 m 50 cm = 450 cm

The length of the longest rod = HCF(825, 675, 450)

⇒ 825 = 675 × 1 + 150

675 = 150 × 4 + 75

150 = 75 × 2 + 0

∴ HCF(825, 675) = 75

450 = 75 × 6 + 0

∴ HCF(450, 75) = 75

∴ HCF (825, 675, 450) = 75×

Thus, the length of the longest rod is 75 cm.

**Q3(NCERT Exemplar): Write whether every positive integer can be of the form 4q + 2, where q is an integer.Justify your answer.**

Answer: No, every positive integer cannot be expressed as only of the form 4q + 2.

**Justification:**

Let a be any positive integer. Then by Euclid’s division lemma, we have

a = bq + r, where 0 ≤ r < b

Putting b = 4, we get

a = 4q + r, where 0 ≤ r < 4

Hence, a positive integer can be of the form, 4q, 4q + 1, 4q + 2 and 4q + 3.

**Q4(NCERT Exemplar): Use Euclid’s division algorithm to find HCF of 441, 567, 693**

Answer: By Euclid’s division algorithm,

693 = 567 × 1 + 126

⇒ 567 = 126 × 4 + 63

⇒ 126 = 63 × 2 + 0

∴ HCF(441, 63) = 63

∴ HCF (693, 567) = 63

⇒ 441 = 63 × 7 + 0

∴ HCF (693, 567, 441) = 63

**Q5(CBSE 2015): Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.**

Answer: Let x be any positive integer and b = 3.

Applying Euclid’s division lemma, x = 3q + r for some integer q ≥ 0 and r = 0,1,2 because 0 ≤ r < 3

∴ x = 3q or 3q + 1 or 3q + 2

When x = 3q,

(x)

^{2}= (3q)

^{2}= 9q

^{2}

= 3(3q

^{2})

= 3m, where m is a integer

When x = 3q + 1,

= (x)

^{2}= (3q + 1)

^{2}= 9q

^{2}+ 6q + 1

= 3(3q

^{2}+ 2q) + 1

= 3m + 1, where m is a integer

When x = 3q + 2,

(x)2 = (3q + 2)

^{2}= 9q

^{2}+ 12q + 4

= 3(3q

^{2}+ 4q + 1) + 1

= 3m + 1, where m is a integer

Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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