Problems On H_C_F And L_C_M — Part 2 MCQs with Explanations

4 × 27 × 3125 = 2 2 × 3 3 × 5 5 8 × 9 × 25 × 7 = 2 3 × 3 2 × 5 2 × 7 16 × 81 × 5 × 11 × 49 = 2 4 × 3 4 × 5 × 11 × 7 2 ∴ H.C.F = 2 2 × 3 2 × 5 = 180

Since all the numbers formed are even, 2 is a common factor. Also, H.C.F. of two of the numbers i.e., 48 and 490, is 2. So, the H.C.F. of all the numbers formed is 2.

LCM of (15, 18, 21, 24) ⇒ 5 × 3 × 6 × 7 × 4 = 2520 ⇒ In such type of questions,we take the difference between given number and remainder of that number. Number Remainder ⇒ (15 - 11) = 4 ⇒ (18 - 14) = 4 ⇒ (21 - 17) = 4 ⇒ (24 - 20) = 4 It will be same always Now, Largest 4 digit number is 9999 ⇒ On dividing 9999 by LCM (2520) we get remainder ⇒ 2439 Subtract remainder from 9999 we get largest 4 digit number, which is divisible by given number = 9999 - 2439 = 7560 (9999 but required number gives difference on dividing) So, our required number = 7560 - 4(difference) = 7556

989 - 5 = 984 1327 - 7 = 1320 Subtract the remainder from the number. HCF = (984, 1320) = 24 for greatest number take HCF of the numbers.

75 litres, 45 litres For maximum capacity take HCF (75, 45) = 15

H.C.F. of 18 and 25 is 1 So, they are co-primes

2 22 - 54 - 108 - 135 - 198 3 11 - 27 - 54 - 135 - 99 3 11 - 9 - 18 - 45 - 33 3 11 - 3 - 6 - 15 - 11 11 11 - 1 - 2 - 5 - 11 1 - 1 - 2 - 5 - 1 L.C.M. = 2 × 3 × 3 × 3 × 11 × 2 × 5 = 5940

Let number be = x, y x : y = 3 : 4 (given) H.C.F. = 4 ∴ Number are = x = 4 × 3 = 12 y = 4 × 4 = 16 L.C.M. of (12, 16) = 4 × 3 × 4 = 48

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. ∴ 189 x y = 27 × 2079 y = 297

$$\eqalign{ & {\text{HCF of fractions}} \cr & {\text{ = }}\frac{{{\text{HCF of Numerators}}}}{{{\text{LCM of Denominators}}}} \cr & {\text{HCF of }}\frac{a}{b}{\text{,}}\frac{c}{d}{\text{,}}\frac{e}{f} \cr & = \frac{{{\text{HCF of }}a,c,e}}{{{\text{LCM of }}b,d,f}} \cr} $$

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

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

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

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

Given numbers are 1.08, 0.36 and 0.90 HCF of 108, 36 and 90 is 18 ∴ HCF of given numbers = 0.18

3240 = 2 3 × 3 4 × 5 3600 = 2 4 × 3 2 × 5 2 HCF = 36 = 2 2 × 3 2 Since H.C.F. is the product of lowest powers of common factors, so the third number must have (2 2 × 3 2 ) as its factors. Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 3 5 and 7 2 as its factors. ∴ Third number = 2 2 × 3 5 × 7 2

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

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

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.

Let the numbers be 13a and 13b Then, 13a × 13b = 2028 ab = 12 Now, co - primes with product 12 are (1, 12) and (3, 4) So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4) Clearly, there are 2 such pairs

HCF of numbers = 7 Let the numbers are 7x and 7y LCM = 7xy 7xy = 140 (given) xy = 20 ⇒ Possible co-prime factors of xy = (1, 20), (4, 5) ⇒ Numbers are between 20 and 45 ∴ Required number are = 4 × 7 = 28 and 5 × 7 = 35 ⇒ Sum of numbers are = 28 + 35 = 63

LCM of (4, 6, 8, 9) = 2 × 2 × 3 × 2 × 3 = 72 ∴ Required result should be = 72

Let the numbers be 13x and 15x Then, their LCM = 195x So, 195x = 39780 x = 204 ∴ The numbers are 2652 and 3060

Let the numbers be 2x, 3x, 5x and 7x respectively Then their LCM = (2 × 3 × 5 × 7)x = 210x.[∴ 2, 3, 5, 7 are prime numbers] So, 210x = 630 or x = 3 ∴ The numbers are 6, 9, 15 and 21 Required difference = 21 - 6 = 15

LCM of (3, 4, 5, 6, 7, 8) = 3 × 4 × 5 × 7 × 2 = 840 ⇒ divided 10000 by LCM ⇒ $$\frac{{10000}}{{840}},$$ we get remainder = 760 Now two possibilities are = 10000 - 760 = 9240 or 10000 + (840 - 760) = 10080 So, nearest number is = 10080

HCF = 16 ∴ Let number are 16x and 16y ⇒ 16xy = 480 ⇒ xy = 30 ∴ Possible pairs = (1, 30), (2, 15), (6, 5) Possible numbers are = (16, 480), (32, 240), (80, 96) ∴ (80, 96) is the answer in the given options of 2 digits.

10 - 9 = 1 9 - 8 = 1 8 - 7 = 1 ∴ LCM of (10, 9, 8) = 5 × 2 × 9 × 4 = 360 ∴ Required result = 360 - 1 = 359

$$\eqalign{ & {\text{Other number}} \cr & {\text{= }}\left( {\frac{{12 \times 336}}{{84}}} \right)\cr & = 48 \cr} $$

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

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

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

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

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

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

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

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

$$\eqalign{ & {\text{Let HCF be }}h{\text{ and LCM be }}l \cr & {\text{Then }}l = {\text{ }}12h{\text{ and}} \cr & {\text{ }}l + h{\text{ }} = {\text{ }}403 \cr & \therefore 12h + h = 403{\text{ or }}h = {\text{ }}31 \cr & {\text{So }}l{\text{ = }}\left( {403 - 31} \right) = 372 \cr & {\text{Hence other number}} \cr & {\text{ = }}\left( {\frac{{31 \times 372}}{{93}}} \right) = 124 \cr} $$

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

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