Problems On H.C.F And L.C.M - Aptitude MCQ Questions and Solutions with Explanations

$${\text{LCM = }}\frac{{324}}{3} = 108$$

$$\eqalign{ & {\text{LCM of }}\left( {3,5,6,8,10,12} \right) \cr & = 3 \times 5 \times 2 \times 4 \cr & = 120 \cr & {\text{Required number is }} \cr & \frac{{120K + 2}}{{22}} = \frac{{10K + 2}}{{22}} \cr & {\text{at }}K{\text{ = 2,}}\frac{{10K + 2}}{{22}} \cr & \Rightarrow {\text{Remainder = 0}} \cr & {\text{The given condition satisfied}} \cr & {\text{ = }}120K + 2 \cr & = 240 + 2 \cr & = 242 \cr} $$

$$\eqalign{ & {\text{Let HCF = }}x \cr & {\text{LCM = 20}}x \cr & {\text{sum of HCF + LCM = 2520}} \cr & \Rightarrow {\text{ }}x{\text{ }} + {\text{ }}20x{\text{ }} = {\text{ }}2520 \cr & \Rightarrow 21x = 2520 \cr & \Rightarrow x = 120 \cr & {\text{HCF = 120}} \cr & {\text{LCM = 120}} \times {\text{20 = 2400}} \cr & {\text{One number = 480}} \cr & {\text{Let another number = }}y \cr & y \times {\text{ }}480 = 120 \times 2400 \cr & y = \frac{{120 \times 2400}}{{480}} = 600 \cr} $$

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

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

$$\eqalign{ & {\text{Let the numbers be }}x{\text{ and 4}}x \cr & {\text{Then ,}}x \times 4x = 84 \times 21 \cr & \Leftrightarrow {x^2}{\text{ = }}\left( {\frac{{84 \times 21}}{4}} \right) \cr & \Leftrightarrow x{\text{ = 21 }} \cr & {\text{Hence larger number}} \cr & {\text{ = 4}}x \cr & {\text{ = 84}} \cr} $$

$$\eqalign{ & {\text{Let number are}} \cr & {\text{ = }}2x,3x,4x \cr & {\text{given,}} \cr & {\text{LCM of }}\left( {2 \times 3 \times 4} \right)x = 24x \cr & 24x = 240 \cr & x = 10 \cr & \therefore {\text{numbers are 2}} \times {\text{10 = 20}} \cr & {\text{3}} \times 10 = 30 \cr & {\text{4}} \times 10 = 40 \cr & \therefore {\text{Smaller is 20}} \cr} $$

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

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
$$\eqalign{ & {\text{ = }}\frac{{60{\text{K}} + 2}}{7} = \frac{{4{\text{K}} + 2}}{7} \cr & {\text{At K = 3, }}\frac{{4{\text{K}} + 2}}{7} \cr & \Rightarrow {\text{remainder = 0}} \cr & \therefore {\text{number = 60K + 2 }} \cr & {\text{ = 60}} \times {\text{3 + 2 }} \cr & {\text{ = 182}} \cr} $$

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