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

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

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 = 9q²
= 3(3q²)
= 3m, where m is a integer
When x = 3q + 1,
= (x)² = (3q + 1)² = 9q² + 6q + 1
= 3(3q² + 2q) + 1
= 3m + 1, where m is a integer
When x = 3q + 2,
(x)2 = (3q + 2)² = 9q² + 12q + 4
= 3(3q² + 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.