## Number Systems

Problems and Solutions

*(CBSE Class 10)*

**Q1: Find HCF and LCM of 126 and 156 using prime factorization method.**

Answer: HCF of 126 is:

126 ╱╲ 2 63 ╱╲ 3 21 ╱╲ 3 7

∴ 126 = 2 × 3 × 3 × 7 = 2 × 3² × 7

HCF of 156 is:

156 ╱╲ 2 78 ╱╲ 2 39 ╱╲ 3 13

∴ 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

∴ HCF (126, 156) = product of common factors with lowest power

= (2ⁱ × 3ⁱ) = 6

LCM (126, 156) = product of prime factors with highest power

= (2² × 3² × 7 × 13)

= 4 × 9 × 7 × 13

= 3276

∴ HCF = 6 and LCM = 3276 (Answer)

**Q2: If the HCF of 657 and 963 is expressible in the form of 657x + 963x - 15 find x.**

Answer: According to Euclid's Division Lemma

a = bq + r, where 0 ≤ r < b

⇒ 963 = 657 × 1 + 306

⇒ 657 = 306 × 2 + 45

⇒ 306 = 45 × 6 + 36

⇒ 45 = 36 × 1 + 9

⇒ 36 = 9 × 4 + 0

∴ HCF (657, 963) = 9

9 = 657x + 963× (-15)

657x = 9+963×15

= 9+14445

657x = 14454

x = 14454/657

x =

**22**

**Q3: The HCF of two numbers is 23 and thier LCM is 1449. If one of the number is 161, find the other.**

Answer: ∵ (a × b) = HCF(a,b) × LCM(a,b)

Given,

HCF = 23

LCM = 1449

a = 161

b = ?

∴ 161 × b = 23 × 1449

b = (23 × 1449)/161 =

**207**

**Q4: Find the largest possible positive integer that will divide 398, 436, and 542 leaving remainder 7, 11, 15 respectively.**

Answer: Find the HCF of 391, 425 and 527 by Euclid’s algorithm

∴ HCF (425, 391) = 17

Now we have to find the HCF of 17 and 527

527 = 17 × 31 +0

∴ HCF (17,527) = 17

⇒ HCF (391, 425 and 527) = 17 (Answer)

**Q5: Roma and Sakshi walk around a circular track. Roma takes 16 minutes to complete one round while Sakshi completes it in 20 minutes. If both start from the same point and go in same direction, after how much time both will meet at the starting point?**

Answer: Factors of 16 = 2 × 2 × 2 × 2 = 2⁴

Prime factors of 20 = 2² × 5

Require number of minutes = LCM (16, 20)

= 2⁴ × 5

= 16 × 5

=

**80 minutes**(Answer)

**Q6: Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.**

Answer: LCM (24, 15, 36) = 3 × 2 × 2 × 2 × 3 × 5 = 360

The greatest six digit number is = 999999

Divide 999999 by 360

⇒ 999999 = 2777 × 360 + 2777

∴ Q = 2777 , R = 279

Thus, the required number = 999999 – 279 =

**999720**(Answer)

## No comments:

## Post a Comment