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# Solved Examples(Set 6) - HCF and LCM

 26. What is the greatest number which divides 24, 28 and 34 and leaves the same remainder in each case? A. 2 B. 1 C. 3 D. 4

Explanation:

If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number.

34 - 24 = 10
34 - 28 = 6
28 - 24 = 4

Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2

 27. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. How many times will they ring together in 60 minutes ? A. 31 B. 15 C. 30 D. 16

Explanation:

LCM of 4, 8, 10, 12, 15 and 20 = 120

120 seconds = 2 minutes

Hence all the six bells will ring together in every 2 minutes

Hence, number of times they will ring together in 60 minutes $=1+\dfrac{60}{2}=31$

 28. What is the least number which when divided by 8, 12, 15 and 20 leaves in each case a remainder of 5 ? A. 125 B. 112 C. 117 D. 132

Explanation:

LCM of 8, 12, 15 and 20 = 120

Required Number = 120 + 5 = 125

 29. The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. What is the largest number? A. 282 B. 322 C. 312 D. 299

Explanation:

The HCF of a group of numbers will be always a factor of their LCM.

HCF is the product of all common prime factors using the least power of each common prime factor.

LCM is the product of highest powers of all prime factors.

HCF of the two numbers = 23

Since HCF will be always a factor of LCM, 23 is a factor of the LCM.
Given that other two factors in the LCM are 13 and 14.

Hence factors of the LCM are 23, 13, 14

So, numbers can be taken as (23 × 13) and (23 × 14)
= 299 and 322

Hence, largest number = 322

Note

We have seen that the numbers are (23 × 13) and (23 × 14).

If we write the numbers as the product of prime factors,
first number = (23 × 13)
second numbers = (23 × 14) = (23 × 2 × 7)

HCF is the product of all common prime factors using the least power of each common prime factor
= 23

LCM is the product of highest powers of all prime factors
= 23 × 13 × 2 × 7 = 23 × 13 × 14

Clearly we get HCF as 23 and the factors in the LCM as 13, 14 and 23.

Hence every conditions are satisfied.

 30. What is the smallest number which when diminished by 12, is divisible 8, 12, 22 and 24? A. 272 B. 268 C. 276 D. 264

Explanation:

Required Number
= (LCM of 8, 12, 22 and 24) + 12
= 264 + 12 = 276

 31. What is the HCF of $\dfrac{1}{3}$, $\dfrac{2}{3}$ and $\dfrac{1}{4}$ ? A. $\dfrac{1}{4}$ B. $\dfrac{1}{3}$ C. $\dfrac{2}{3}$ D. $\dfrac{1}{12}$

Explanation:

HCF of fractions = $\dfrac{\text{HCF of Numerators}}{\text{LCM of Denominators}}$

HCF of $\dfrac{1}{3}$ , $\dfrac{2}{3}$ and $\dfrac{1}{4}$

$=\dfrac{\text{HCF (1, 2, 1)}}{\text{LCM (3, 3, 4)}}\\~\\=\dfrac{1}{12}$

 32. What is the LCM of $\dfrac{2}{3}$, $\dfrac{5}{6}$ and $\dfrac{4}{9}$ ? A. $\dfrac{3}{10}$ B. $\dfrac{10}{3}$ C. $\dfrac{3}{20}$ D. $\dfrac{20}{3}$

Explanation:

LCM of fractions $=\dfrac{\text{LCM of Numerators}}{\text{HCF of Denominators}}$

LCM of $\dfrac{2}{3}$, $\dfrac{5}{6}$ and $\dfrac{4}{9}$
$=\dfrac{\text{LCM (2, 5, 4)}}{\text{HCF (3, 6, 9)}}\\~\\=\dfrac{20}{3}$