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# Solved Examples(Set 3) - HCF and LCM

 11. What is the greatest possible length which can be used to measure exactly the lengths 8 m, 4 m 20 cm and 12 m 20 cm? A. 30 cm B. 25 cm C. 20 cm D. 10 cm

Explanation:

Required length
= HCF of 800 cm, 420 cm, 1220 cm
= 20 cm

 12. Which of the following fraction is the largest ? A. $\dfrac{11}{12}$ B. $\dfrac{21}{40}$ C. $\dfrac{5}{6}$ D. $\dfrac{41}{50}$

Explanation:

Solution 1

LCM of 6, 40, 50 and 12 = 600

$\dfrac{5}{6} = \dfrac{500}{600}\\~\\\dfrac{21}{40} = \dfrac{315}{600}\\~\\\dfrac{41}{50} = \dfrac{492}{600}\\~\\\dfrac{11}{12} = \dfrac{550}{600}$

Clearly $\dfrac{550}{600}=\dfrac{11}{12}$ is the largest among these fractions.

Solution 2

$\dfrac{5}{6} = .83\\~\\\dfrac{21}{40} = .52\\~\\\dfrac{41}{50} = .82\\~\\\dfrac{11}{12} = .92$

Clearly $.92=\dfrac{11}{12}$ is the largest among these.

 13. The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers ? A. 13, 156 B. 26, 78 C. 36, 68 D. 39, 52

Explanation:

Let the numbers be $13x$ and $13y~~$ (∵ HCF of the numbers = 13)

$13x × 13y = 2028\\\Rightarrow xy = 12$

co-primes with product 12 are (1, 12) and (3, 4) (∵ we need to take only co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13) (Reference: Co-prime Numbers)

Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)

Given that the numbers are 2 digit numbers.
Hence numbers are 39 and 52

 14. The product of two numbers is 2028 and their HCF is 13. What are the number of such pairs? A. 2 B. 3 C. 4 D. 1

Explanation:

Let the numbers be $13x$ and $13y~~$ (∵ HCF of the numbers = 13)

$13x × 13y = 2028 \\\Rightarrow xy = 12$

co-primes with product 12 are (1, 12) and (3, 4) (∵ we need to take only co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13) (Reference: Co-prime Numbers)

Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)

So, there are 2 pairs of numbers with HCF 13 and product 2028

 15. N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N? A. 5 B. 4 C. 6 D. 3

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.

6905 - 1305 = 5600
6905 - 4665 = 2240
4665 - 1305 = 3360

Hence, the greatest number which divides 1305, 4665 and 6905 and gives the same remainder, N
= HCF of 5600, 2240, 3360
= 1120

Sum of digits in N
= Sum of digits in 1120
= 1 + 1 + 2 + 0
= 4