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Real Numbers
(NCERT Exemplar Solutions Ex 1.1)

**If the HCF of 65 and 117 is expressible in the form 65m − 117, then the value of m is**

(A) 4

(B) 2

(C) 1

(D) 3

Answer: (B) 2

Explanation:

Using Euclid’s division algorithm,

b = aq + r, 0 ≤ r < a [∵ dividend = divisor × quotient + remainder]

⇒ 117 = 65 × 1 + 52

⇒ 65 = 52 × 1 + 13

⇒ 52 = 13 × 4 + 0

∴ HCF (65, 117) = 13 .... (i)

Also, given that, HCF (65, 117) = 65m – 117 ... (ii)

From Equations (i) and (ii), we get

65m − 117 = 13

⇒ 65m = 130

⇒ m = 2

Question 5:

**The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is**

(A) 13

(B) 65

(C) 875

(D) 1750

Answer: (A) 13

Explanation:

Since, 5 and 8 are the remainders of 70 and 125, respectively.

Thus, after subtracting these remainders from the numbers, we have the numbers

65 = (70 – 5),

117 = (125 − 8), which is divisible by the required number.

Now, required number = HCF of 65, 117 [Since we need the largest number]

For this, 117 = 65 × 1 + 52 [∵ dividend = divisor × quotient + remainder]

⇒ 65 = 52 × 1 + 13

⇒ 52 = 13 × 4 + 0

∴ HCF = 13

Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.

Question 6:

**If two positive integers a and b are written and a = x³y² and b = xy³, where x, y are the prime numbers, then HCF (a, b) is**

(A) xy

(B) xy²

(C) x³y³

(D) x²y²

Answer: (B) xy²

Explanation:

Given that,

a = x³y² = x × x × x × y × y

b = xy³ = x × y × y × y

∴ HCF of a and b

= HCF (x³y² , xy³)

= x × y × y = xy²

[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]

**☛See also**

CH 1: Real Numbers (MCQ)

CH 1: Real Numbers (Study Points)

CH 1: Real Numbers (NCERT Ex 1.1)

CH 1: Real Numbers (Euclid's Division Lemma - Q & A)

CH 1: Problems on Euclid's Division Algorithm

CH 1: Real Numbers (Problems and Answers)

CH 1: Real Numbers (NCERT Ex 1.2)

CH 1: Real Numbers (Exemplar Ex 1.1 Q1-3)