Problems on H.C.F and L.C.M - 04

Explanation:-

Since HCF is always a factor of LCM, we cannot have three numbers with HCF 35 and LCM 120.

Explanation:-

$$\eqalign{ & {\text{Let HCF = }}x \cr & {\text{LCM = 44}}x \cr & {\text{given HCF + LCM}} \cr & = 44x + x = 45x \cr & 45x = {\text{ }}112{\text{5}} \cr & x = \frac{{1125}}{{45}} = 25 \cr & \therefore {\text{HCF}} = {\text{25}} \cr & {\text{LCM = 25}} \times {\text{44}} \cr & {\text{also given that one number}} \cr & {\text{ = 25}} \cr & {\text{Let another number = }}y \cr & \therefore 25y = 25 \times 25 \times 44 \cr & y = \frac{{25 \times 25 \times 44}}{{25}} \cr & y = 1100 \cr}$$

Explanation:-

LCM of (24, 32, 36, 64)
⇒ 8 × 3 × 4 × 3 × 2 = 576
Required number is 576 - 5 = 571

Explanation:-

$$\eqalign{ & {\text{LCM of }}\left( {9,10,15} \right) \cr & = 3 \times 3 \times 10 \cr & = 90 \cr & \frac{{1936}}{{90}},{\text{remainder = 46}} \cr}$$ Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39

Explanation:-

Let the number be a and b $$\eqalign{ & {\text{Then,}} \cr & a + b = 36\& \cr & ab = 3 \times 105 = 315 \cr & \therefore {\text{Required sum}} \cr & {\text{ = }}\frac{1}{a} + \frac{1}{b} \cr & = \frac{{a + b}}{{ab}} \cr & = \frac{{36}}{{315}} \cr & = \frac{4}{{35}} \cr}$$

Explanation:-

LCM of (5, 10, 12, 15) = 5 × 2 × 6 = 60
smallest number divided by (5, 10, 12, 15)
leaves remainder 2 and when divided by 7 leaves no remainder is $$\eqalign{ & {\text{ = }}\frac{{60{\text{K}} + 2}}{7} = \frac{{4{\text{K}} + 2}}{7} \cr & {\text{At K = 3, }}\frac{{4{\text{K}} + 2}}{7} \cr & \Rightarrow {\text{remainder = 0}} \cr & \therefore {\text{number = 60K + 2 }} \cr & {\text{ = 60}} \times {\text{3 + 2 }} \cr & {\text{ = 182}} \cr}$$

Explanation:-

HCF = 17
Let numbers are = 17x, 17y
LCM = 17xy = 714 (given)
xy = 42
Possible pairs are (1, 42), (2, 21), (3, 14), (6, 7)
Possible numbers are (17, 714), (34, 357), (51, 238), (102, 119)
but given that both numbers are of three digits
∴ numbers are = (102, 119)
∴ sum of numbers = 102 + 119 = 221

Explanation:-

$$\eqalign{ & {\text{Let number are}} \cr & {\text{ = }}2x,3x,4x \cr & {\text{given,}} \cr & {\text{LCM of }}\left( {2 \times 3 \times 2} \right)x = 12x \cr & 12x = 240 \cr & x = 20 \cr & \therefore {\text{numbers are 2}} \times {\text{20 = 40}} \cr & {\text{3}} \times 20 = 60 \cr & {\text{4}} \times 20 = 80 \cr & \therefore {\text{Smaller is 40}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Product of numbers}} \cr & {\text{ = HCF}} \times {\text{LCM}} \cr & \Rightarrow pq = x \times {\text{LCM}} \cr & \Rightarrow {\text{LCM}} = \frac{{pq}}{x} \cr & = \frac{{p\left( {xy} \right)}}{x} = py \cr}$$

Explanation:-

LCM of (12, 15, 18, 27)
⇒ 4 × 3 × 5 × 3 × 3 = 540
largest 4 - digits number = 9999 on dividing by 540 to number = ⁹⁹⁹⁹/₅₄₀
remainder = 279
Required number = 9999 - 279 = 9720