Problems on H.C.F and L.C.M - 03

Explanation:-

For HCF of fractions take HCF of numerators and LCM denominators
HCF of 3, 5, 6 = 1
LCM of 3, 6, 7 = 84
Hence, HCF of fractions
=¹/₈₄

Explanation:-

Number of books in each stack = HCF of 336, 240 and 96 = 48.
Hence, total number of stacks $$\eqalign{ & {\text{ = }}\left( {\frac{{336 + 240 + 96}}{{48}}} \right) \cr & = \frac{{672}}{{48}} \cr & = 14 \cr}$$

Explanation:-

For the least number of tiles, the size of the ltie must be the maximum.
Maximum size of the tile = HCF of 1517 cm and 902 cm = 41 cm.
Hence, required number of tiles $$\eqalign{ & {\text{ = }}\frac{{{\text{Area of celling}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{1517 \times 902}}{{41 \times 41}}} \right) \cr & = 814 \cr}$$

Explanation:-

For the least number of tiles, the size of the ltie must be the maximum.
Maximum size of the tile = HCF of 1517 cm and 902 cm = 41 cm.
Hence, required number of tiles $$\eqalign{ & {\text{ = }}\frac{{{\text{Area of celling}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{1517 \times 902}}{{41 \times 41}}} \right) \cr & = 814 \cr}$$

Explanation:-

For the least number of bottles, the capacity of each bottle must be maximum.
Capacity of each bottle = HCF of 403 litres, 465 litres and 496 litres = 31 litres.
Hence,
required number of bottles $$\eqalign{ & {\text{ = }}\left( {\frac{{403 + 465 + 496}}{{31}}} \right) \cr & = \frac{{1364}}{{31}} \cr & = 44 \cr}$$

Explanation:-

LCM of left(24, 36, 54)
⇒ 12 × 2 × 3 × 3 = 216 sec
⇒ They will change simultaneously after every 216 seconds
⇒ ²¹⁶/₆₀ ⇒ 3³⁶/₆₀ ⇒ 3 minute 36 seconds
⇒ They change 1^st at 10:15:00 AM
So, again they change at = 10:18:36 AM

Explanation:-

LCM = 120 (given)
LCM is the product of one common factor and other different factors of the given numbers.
∴ Factorize the given LCM = 120 $$\frac{{\,\,\,{\text{2}} \times {\text{2}} \times {\text{3}} \times {\text{5}} \times {\text{2}}\,\,\,}}{{4\left( {3 \times 5 \times 2} \right)}}$$ = Here 4 is common factor (common factor is the HCF of the given number
∴ HCF = 4
So, for the given numbers the HCF should be multiple of 4
⇒ Hence go through options which is not a multiple of 4 is 35
Hence answer is 35

Explanation:-

HCF = 13 (given)
Let number are 13x & 13y respectively $$\eqalign{ & {\text{Also given}} \cr & 13x \times 13y = 2028 \cr & 13 \times 13 \times xy = 2028 \cr & xy = \frac{{2028}}{{13 \times 13}} = 12 \cr & \therefore {\text{Possible pairs are}} \cr & {\text{ = }}\left( {1,12} \right)\left( {3,4} \right) \cr}$$ Only two pair are possible

Explanation:-

Required capacity = HCF of 120 litres and 56 litres
= 8 litres = 8000 cc [∴ 1 litres = 1000 cc]

Explanation:-

LCM = 520
HCF = 4
one number = 52
Let other number is y
52y = 4 × 520
⇒ y = 40