Problems on H.C.F and L.C.M - 05

Explanation:-

Let the required numbers be 33a and 33b.
Then 33a + 33b = 528
⇒ a + b = 16
Now co - primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9)
∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9)
The number of such pairs is 4.

Explanation:-

$$\eqalign{ & {\text{Let number are }}a,{\text{ }}b,{\text{ }}c \cr & {\text{ = }}a,{\text{ }}b,{\text{ }}c{\text{ are co - prime numbers}} \cr & {\text{HCF (}}a,b,c{\text{) = 1}} \cr & \Rightarrow \frac{{a \times b}}{{b \times c}} = \frac{{1073}}{{551}} = \frac{{37 \times 29}}{{17 \times 29}} \cr & \Rightarrow \frac{a}{c} = \frac{{37}}{{19}} \cr & \Rightarrow {\text{Common }}b{\text{ factor is cancel out}}{\text{.}} \cr & \therefore a = 37,\,b = 29,\,c = 19 \cr & \therefore {\text{Sum of numbers}} \cr & {\text{ = }}a + b + c \cr & = 37 + 29 + 19 \cr & = 85 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Distance = 5 km}} \cr & {\text{Speed of A = 2}}\frac{1}{2}{\text{ km/hr}} \cr & {\text{Time taken by A}} \cr & {\text{ = }}\frac{5}{2} \times 2 = 5{\text{hours}} \cr & {\text{Speed of B = 3 km/hr}} \cr & {\text{Time taken by B}} \cr & {\text{ = }}\frac{5}{3}{\text{hours}} \cr & {\text{Speed of C = 2 km/hr}} \cr & {\text{Time taken by C}} \cr & {\text{ = }}\frac{5}{2}{\text{hours}} \cr & \frac{{{\text{LCM of numerator}}}}{{{\text{HCF of denominator}}}} \cr & = 2,\,\frac{5}{3},\,\frac{5}{2} \cr & {\text{LCM = }}\frac{{10}}{1} = 10 \cr}$$ They will meet again after 10 hours

Explanation:-

3240 = 2³ × 3⁴ × 5
3600 = 2⁴ × 3² × 5²
HCF = 36 = 2² × 3²
Since H.C.F. is the product of lowest powers of common factors, so the third number must have (2² × 3²) as its factors.
Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 3⁵ and 7² as its factors.
∴ Third number = 2² × 3⁵ × 7²

Explanation:-

LCM (10, 16, 24)
= 5 × 2 × 8 × 3 = 240
⇒ for square number split the LCM into its factors
= 5 × 2 × 2 × 2 × 2 × 3
= 5 × 5 × 2 × 2 × 2 × 2 × 3 × 3
= 3600

Explanation:-

LCM (4, 6, 10, 15)
LCM = 2 × 2 × 3 × 5 = 60
⇒ Least number of six digit = 100000
⇒ Divide 100000 by 60 we get remainder 40
⇒ Least six digit number which is divisible by (4, 6, 10, 15)given number is
[100000 + (60 - 40)] = 100020
∴ N ⇒ 100020 + 2 = 100022
∴ Sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5

Explanation:-

$$\eqalign{ & {\text{Required LCM}} \cr & = \frac{{{\text{LCM of 2,3,4,9}}}}{{{\text{HCF of 3,5,7,13}}}} \cr & = \frac{{36}}{1} \cr & = 36 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Required HCF }} \cr & {\text{ = }}\frac{{{\text{HCF of 9,12,18,21}}}}{{{\text{LCM of 10,25,35,40}}}} \cr & = \frac{3}{{1400}} \cr}$$

Explanation:-

L.C.M. of (21, 36, 66)
= 21 × 12 × 11
= 7 × 3 × 4 × 3 × 11
= 7 × 3 × 2 × 2 × 3 × 11
For perfect square multiply by 7 × 11
So that pairs of number from perfect square
∴ 7 × 7 × 3 × 3 × 2 × 2 × 11 × 11
Required result is
= 213444 (which is a perfect square)

Explanation:-

Given,
H.C.F. = 27
L.C.M. = 2079
one number = 189
Let another number be y
⇒ Product of numbers = L.C.M. × H.C.F.
∴ 189xy = 27 × 2079
y = 297