Question 1. Find the L.C.M. of the following numbers using common division method: 

i) 24, 36, 40        ii) 15, 24, 30, 40        iii) 9, 12, 15, 18, 24, 56  

iv) 22,54,108,135,198        v)  576, 672, 720        vi)  1620, 1728, 1890

Answer:

i) 24, 36, 40

\begin{array}{r | r | r | r }   2 & 24 & 36 & 40  \\ \hline 2 & 12 & 18 & 20   \\ \hline 3 & 6 & 9 & 10  \\ \hline 2 & 2 & 3 & 10   \\ \hline 3 & 1 & 3 & 5   \\ \hline 5 & 1 & 1 & 5  \\ \hline {\ } & 1 & 1 & 1  \end{array} 

LCM = 2 \times 2 \times 3 \times 2 \times 3 \times 5 = 360

ii) 15, 24, 30, 40

\begin{array}{r | r | r | r | r }   3 & 15 & 24 & 30 & 40  \\ \hline 5 & 5 & 8 & 10 & 40   \\ \hline 2 & 1 & 8 & 2 & 8 \\ \hline 2 & 1 & 4 & 1 & 4    \\ \hline 2 & 1 & 2 & 1 & 2 \\ \hline {\ } & 1 & 1 & 1 & 1 \end{array} 

LCM = 3 \times 5 \times 2 \times 2 \times 2 = 120

iii) 9, 12, 15, 18, 24, 56

\begin{array}{r | r | r | r | r | r | r  }   3 & 9 & 12 & 15 & 18 & 24  & 56  \\ \hline 3 & 3 & 4 & 5 & 6 & 8  & 56   \\ \hline 2 & 1 & 4 & 5 & 2 & 8  & 56 \\ \hline 2 & 1 & 2 & 5 & 1 & 4  & 28    \\ \hline 2 & 1 & 1 & 5 & 1 & 2  & 14    \\ \hline 7 & 1 & 1 & 5 & 1 & 1  & 7     \\ \hline 5 & 1 & 1 & 5 & 1 & 1  & 1   \\ \hline {\ } & 1 & 1 & 1 & 1  &  1 & 1 \end{array} 

LMC = 3 \times 3 \times 2 \times 2 \times 7 \times 5 = 2520

iv) 22,54,108,135,198

\begin{array}{ r | r | r | r | r | r   }   2 & 22 & 54 & 108 & 135 & 198    \\ \hline 3 & 11 & 27 & 54 & 135 & 99     \\ \hline 3 & 11 & 9 & 18 & 45 & 33   \\ \hline 11 & 11 & 3 & 6 & 15 & 11      \\ \hline 3 & 1 & 3 & 6 & 15 & 1     \\ \hline 2 & 1 & 1 & 2 & 5 & 1    \\ \hline 5 & 1 & 1 & 1 & 5 & 1  \\ \hline {\ } & 1 & 1 & 1 & 1  &  1  \end{array} 

LCM = 2 \times 3 \times 3 \times 11 \times 3 \times 2 \times 5 = 5940

v)  576, 672, 720

\begin{array}{r | r | r | r }   2 & 576 & 672 & 570  \\ \hline 2 & 288 & 336 & 360   \\ \hline 2 & 144 & 168 & 180  \\ \hline 2 & 72 & 84 & 90   \\ \hline 3 & 36 & 42 & 45   \\ \hline 3 & 12 & 14 & 15  \\ \hline 2 & 4 & 14 & 5  \\ \hline 2 & 2 & 7 & 5  \\ \hline 5 & 1 & 7 & 5  \\ \hline 7 & 1 & 7 & 1  \\ \hline {\ } & 1 & 1 & 1  \end{array} 

LCM = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 2 \times 5 \times 7 = 20160

vi)  1620, 1728, 1890

\begin{array}{r | r | r | r }   2 & 1620 & 1728 & 1890 \\   \hline 5 & 810 & 864 & 945 \\  \hline 2 & 162 & 864 & 189 \\  \hline 3 & 81 & 432 & 189 \\  \hline 3 & 27 & 144 & 63 \\  \hline 3 & 9 & 48 & 21 \\  \hline 3 & 3 & 16 & 7 \\  \hline 7 & 1 & 16 & 7 \\  \hline 2 & 1 & 16 & 1 \\  \hline 2 & 1 & 8 & 1 \\  \hline 2 & 1 & 4 & 1 \\  \hline 2 & 1 & 2 & 1 \\  \hline {\ } & 1 & 1 & 1  \end{array} 

LCM = 2 \times 5 \times 2 \times 3 \times 3 \times 3 \times 3 \times 7 \times 2 \times 2 \times 2 \times 2 = 181440

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Question 2: Use the prime factorization method to find the L.C.M. of the following numbers:

i) 72, 192, 240      ii) 56, 72, 84      iii) 48, 72, 108, 144

iv) 168, 180, 330      v) 180, 384, 432     vi) 288, 432, 486

Answer:

i) 72, 192, 240

\begin{array}{r | r  }   2 & 72 \\  \hline 2 & 36 \\  \hline 2 & 18 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1  \end{array} \hspace{1.5cm}   \begin{array}{r | r  }   2 & 192 \\  \hline 2 & 96 \\  \hline 2 & 48 \\  \hline 2 & 24 \\  \hline 2 & 12 \\  \hline 2 & 6 \\  \hline 3 & 3 \\  \hline & 1    \end{array} \hspace{1.5cm}   \begin{array}{r | r  }   2 & 240 \\ \hline 2 & 120 \\  \hline2 & 60 \\  \hline 2 & 30 \\  \hline 3 & 15 \\  \hline 5 & 5 \\  \hline & 1    \end{array} 

72 = 2^3 \times 3^2

192 = 2^6 \times 3

240 = 2^4 \times 3 \times 5

Hence LCM = 2^6 \times 3^2 \times 5 = 2880

ii) 56, 72, 84

\begin{array}{r | r  }   2 & 56 \\  \hline 2 & 28 \\  \hline 2 & 14 \\  \hline 7 & 7 \\  \hline  & 1  \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 72 \\  \hline 2 & 36 \\  \hline 2 & 18 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1  \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 84 \\  \hline 2 & 42 \\  \hline 3 & 21 \\  \hline 7 & 7 \\  \hline & 1  \end{array} 

56 = 2^3 \times 7

72 = 2^3 \times 3^2

84 = 2^2 \times 3 \times 7

Hence LCM = 2^3 \times 3^2 \times 7 = 504

iii) 48, 72, 108, 144

\begin{array}{r | r  }   2 & 48 \\  \hline 2 & 24 \\  \hline 2 & 12 \\  \hline 2 & 6 \\  \hline 3 & 3 \\  \hline & 1    \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 72 \\  \hline 2 & 36 \\  \hline 2 & 18 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1  \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 108 \\  \hline 2 & 54 \\  \hline 3 & 27 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1  \end{array} \hspace{1.5cm} \begin{array}{r | r  }  2 & 144 \\  \hline 2 & 72 \\  \hline 2 & 36 \\  \hline 2 & 18 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1     \end{array} 

48 = 2^4 \times 3

72 = 2^3 \times 3^2

108 = 2^2 \times 3^3

144 = 2^4 \times 3^2

Hence LCM = 2^4 \times 3^3 = 432

iv) 168, 180, 330

\begin{array}{r | r  }   2 & 168 \\  \hline 2 & 84 \\  \hline 2 & 42 \\  \hline 3 & 21 \\  \hline 7 & 7 \\  \hline & 1    \end{array} \hspace{1.5cm} \begin{array}{r | r  }  2 & 180 \\  \hline 2 & 90 \\  \hline 5 & 45 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1      \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 330 \\  \hline 5 & 165 \\  \hline 3 & 33 \\  \hline 11 & 11 \\  \hline & 1     \end{array} 

168 = 2^3 \times 3 \times 7

180 = 2^2 \times 5 \times 3^2

330 = 2 \times 3 \times 5 \times 11

Hence LCM = 2^3 \times 3^2 \times 5 \times 7 \times 11 = 27720

v) 180, 384, 432

\begin{array}{r | r  }   2 & 180 \\  \hline 2 & 90 \\  \hline 5 & 45 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1    \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 384 \\  \hline 2 & 192 \\  \hline 2 & 96 \\  \hline 2 & 48 \\  \hline 2 & 24 \\  \hline 2 & 12 \\  \hline 2 & 6 \\  \hline 3 & 3 \\  \hline & 1      \end{array} \hspace{1.5cm} \begin{array}{r | r  }  2 & 432 \\  \hline 2 & 216 \\  \hline 2 & 108 \\  \hline 2 & 54 \\  \hline 3 & 27 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1     \end{array} 

180 = 2^2 \times 5 \times 3^2

384 = 2^7 \times 3

432 = 2^4 \times 3^3

Hence LCM = 2^7 \times 3^3 \times 5 = 17280

vi) 288, 432, 486

\begin{array}{r | r  }   2 & 180 \\  \hline 2 & 90 \\  \hline 5 & 45 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1    \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 384 \\  \hline 2 & 192 \\  \hline 2 & 96 \\  \hline 2 & 48 \\  \hline 2 & 24 \\  \hline 2 & 12 \\  \hline 2 & 6 \\  \hline 3 & 3 \\  \hline & 1      \end{array} \hspace{1.5cm} \begin{array}{r | r  }   2 & 432 \\  \hline 2 & 216 \\  \hline 2 & 108 \\  \hline 2 & 54 \\  \hline 3 & 27 \\  \hline 3 & 9 \\  \hline 3 & 3 \\  \hline & 1     \end{array} 

180 = 2^2 \times 5 \times 3^2

384 = 2^7 \times 3

432 = 2^4 \times 3^3

Hence LCM = 2^7 \times 5 \times 3^3 =17280

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Question 3. Find the H.C.F and L.C.M of the following:

i) 23 \times 32 \times 5, 22 \times 33 \times 5   and   24 \times 3 \times 53 \times 7  

ii) 22 \times 3 \times 52, 23 \times 32 \times 7     and   2 \times 33 \times 5 \times 7  

iii) 52 \times 7 \times 2, 5 \times 7 \times 32    and   22 \times 3 \times 11  

iv) 22 \times 33 \times 54, 23 \times 32 \times 5     and   2 \times 3 \times 7 \times 52  

Answer:

i) 23 \times 32 \times 5, 22 \times 33 \times 5   and   24 \times 3 \times 53 \times 7  

HCF = 2^2 \times 3 \time 5 = 60

LCM = 2^4 \times 3^2 \times 5^3 \times 7 = 126000

ii) 22 \times 3 \times 52, 23 \times 32 \times 7     and   2 \times 33 \times 5 \times 7

HCF = 2 \times 3 = 6

LCM =  2^3 \times 3^3 \times 5^2 \times 7 =  37800

iii) 52 \times 7 \times 2, 5 \times 7 \times 32    and   22 \times 3 \times 11

HCF = 1

LCM = 2^2 \times 3^2 \times 5^2 \times 7 \times 11 = 69300

iv) 22 \times 33 \times 54, 23 \times 32 \times 5     and   2 \times 3 \times 7 \times 52

HCF = 2 \times  3 \times  5 = 30

LCM = 2^2 \times  3^3 \times  5^4 \times  7 = 945000

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Question 4. Find the H.C.F. and L.C.M of the following fractions:

i) \frac{5}{6} , \frac{5}{9} , \frac{10}{27}            ii) \frac{2}{3} , \frac{8}{15} , \frac{20}{27} , \frac{40}{81}            iii) \frac{3}{7} , \frac{9}{14} , \frac{18}{35} , \frac{15}{28}  

Note:  HCF of a given fraction = \frac{HCF of Numerators}{LCM of Denominators}

  LCM of a given fraction = \frac{LCM of Numerators}{HCF of Denominators}

Answer:

i) \frac{5}{6} , \frac{5}{9} , \frac{10}{27}

HCF of numerators = HCF of 5, 5, 10 = 5

HCF of denominators = HCF of 6, 9, 27 = 3

LCM of numerators = LCM of  5, 5, 10 = 10

LCM of denominators = LCM of  6, 9, 27 = 54

HCF of a given fraction = \frac{5}{54}

LCM of a given fraction = \frac{10}{3}

 ii) \frac{2}{3} , \frac{8}{15} , \frac{20}{27} , \frac{40}{81}

HCF of numerators = HCF of 2, 8, 20, 40 = 10

HCF of denominators = HCF of 3, 15, 27, 81 = 3

LCM of numerators = LCM of  2, 8, 20, 40 = 40

LCM of denominators = LCM of  3, 15, 27, 81 = 405

HCF of a given fraction = \frac{2}{405}

LCM of a given fraction = \frac{40}{3}

iii) \frac{3}{7} , \frac{9}{14} , \frac{18}{35} , \frac{15}{28}

HCF of numerators = HCF of 3, 9, 18, 15 = 3

HCF of denominators = HCF of 7, 14, 35, 28 = 7

LCM of numerators = LCM of  3, 9, 18, 15 = 90

LCM of denominators = LCM of  7, 14, 35, 28 = 140

HCF of a given fraction = \frac{3}{140}

LCM of a given fraction = \frac{90}{7}

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Question 5. Find the smallest number exactly divisible by 28, 36, 42 and 54

Answer:

The required number is the LCM of 28, 36, 42 and 54

\begin{array}{r | r | r | r | r }   2 & 28 & 36 & 42 & 54  \\ \hline 2 & 14 & 18 & 21 & 27   \\ \hline 7 & 7 & 9 & 21 & 27 \\ \hline 3 & 1 & 9 & 3 & 27    \\ \hline 3 & 1 & 3 & 1 & 9 \\ \hline 3 & 1 & 1 & 1 & 3 \\  \hline {\ } & 1 & 1 & 1 & 1 \end{array} 

LCM = 2 \times 2 \times 7 \times 3 \times 3 \times 3 = 756

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Question 6. Find the least number which when divided by 12, 16, 18, 21 and 28 leaves the same remainder 7 in each case.

Answer:

Required number is the LCM of 12, 16, 18, 21  and 28    plus 7  added to it.

\begin{array}{r | r | r | r | r }   2 & 12 & 16 & 18 & 21  \\ \hline 3 & 6 & 8 & 9 & 21   \\ \hline 2 & 2 & 8 & 3 & 7 \\ \hline 2 & 1 & 4 & 3 & 7    \\ \hline 2 & 1 & 2 & 3 & 7 \\ \hline 3 & 1 & 1 & 1 & 7 \\  \hline {7} & 1 & 1 & 1 & 1 \end{array} 

LCM = 2 \times 3 \times 2 \times 2 \times 2 \times 3 \times 7 = 1008 

Hence the number is 1008 + 7 = 1015 

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Question 7. Find the least number which when diminished by 8 , is divisible by each of the numbers 12, 15, 20 and 54 .

Answer:

Required number is the LCM of 12, 15, 20 and 54    plus 8  added to it.

\begin{array}{r | r | r | r | r }   3 & 12 & 15 & 20 & 54  \\ \hline 5 & 4 & 5 & 20 & 18   \\ \hline 2 & 4 & 1 & 4 & 18 \\ \hline 2 & 2 & 1 & 2 & 9    \\ \hline 3 & 1 & 1 & 1 & 9 \\ \hline 3 & 1 & 1 & 1 & 3 \\  \hline {\ } & 1 & 1 & 1 & 1 \end{array} 

LCM = 3 \times 5 \times 2 \times 2 \times 2 \times 3  = 540 

Hence the number is 540 + 8 = 548 

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Question 8. Find the smallest number which when increased by 5 is divisible by each of the numbers 48, 60, 72 and 108 .

Answer:

Required number is the LCM of 48, 60, 72 and 108    minus 5  added to it.

\begin{array}{r | r | r | r | r }   3 & 48 & 60 & 72 & 108  \\ \hline 2 & 16 & 20 & 24 & 36   \\ \hline 2 & 8 & 10 & 12 & 18 \\ \hline 3 & 4 & 5 & 6 & 9    \\ \hline 2 & 4 & 5 & 2 & 3 \\ \hline 5 & 1 & 5 & 1 & 3  \\ \hline 3 & 1 & 1 & 1 & 3  \\  \hline {\ } & 1 & 1 & 1 & 1 \end{array} 

LCM = 3 \times 2 \times 2 \times 3 \times 2 \times 2 \times 5 \times 3   = 2160 

Hence the number is 2160 - 5 = 2155 

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Question 9. Find the greatest number of four digits which is exactly divisible by each one of the numbers 12, 18, 21 and 28 .

Answer:

The required number must be divisible by LCM  of 12, 18, 21 and 28

The LCM = 252

Now the largest four digit number is 9999

On dividing the 9999 by 252 , we get 9999 \div 252 = 39 and remainder of 71

Therefore the number  is ( 9999 - 71)  = 9828

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Question 10. Five bells begin to toll together and toll respectively at intervals of 6,7,8,9 and 12 seconds. After how much time would they toll together again?

Answer:

Required number is the LCM of 6,7,8,9 and 12 

\begin{array}{r | r | r | r | r | r}  2 & 6 & 7 & 8 & 9  & 12  \\ \hline 3 & 3 & 7 & 4 & 9 & 6  \\ \hline 2 & 1 & 7 & 4 & 3  & 2    \\ \hline 2 & 1 & 7 & 2 & 3  & 1 \\ \hline 3 & 1 & 7 & 1 & 3  & 1 \\ \hline 7 & 1 & 7 & 1 & 1  & 1  \\  \hline {\ } & 1 & 1 & 1 & 1 & 1 \end{array} 

LCM = 2 \times 3 \times 2 \times 2 \times 3 \times 7  = 504 

Hence all the bells will toll together again in 504  seconds  or 8  minutes and 24  seconds.

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Question 11. Find the least perfect square number which is divisible by 3, 4, 5, 6 and 8 .

Answer:

Required number is the square of LCM of 3,4,5,6 and 8 

\begin{array}{r | r | r | r | r | r}  3 & 3 & 4 & 5 & 6  & 8  \\ \hline 2 & 1 & 4 & 5 & 2 & 8  \\ \hline 2 & 1 & 2 & 5 & 1  & 4    \\ \hline 2 & 1 & 1 & 5 & 1  & 2 \\ \hline 5 & 1 & 1 & 5 & 1  & 1 \\  \hline {\ } & 1 & 1 & 1 & 1 & 1 \end{array} 

LCM = 3 \times 2 \times 2 \times 2 \times 5   = 120 

The required perfect square is 120 \times 3 \times 2 \times 5 = 3600

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Question 12. Find the least number that should be added to 2500 so that the sum is divisible by each one of the numbers 3, 4, 5 and 6 .

Answer:

LCM of 3, 4, 5 , and 6 = 60

2500 \div 60 = 41 and remainder of 40 .

Required number = 60-40 = 20   so that the sum is divisible by each of the number 3, 4, 5 , and 6 .

The number is 2500 + 20 = 2520

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Question 13. Find the least number of five digits which is exactly divisible by each one of the numbers 12, 15 and 20 .

Answer:

LCM of 12, 15 , and 20 = 60

The smallest 5 digit number is 10000

10000 \div 60 = 166 and remainder of 40 .

Required number = 10000 + ( 60-40) = 10020 so that the sum is divisible by each of the number 3, 4, 5 , and 6 .

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Question 14. Three boys start cycling around a circular park from the same point at the same time and in the same direction. If these boys, each cycling at a constant speed, complete a round in 24 min, 36 min and 54 min respectively, then after what time would they meet again.

Answer:

The boys complete the circle in 24, 36 and 54 minutes respectively.

LCM of 24, 36 and 54 = 216 minutes or 3 hours 36 minutes.

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Question 15. The product of two numbers is 16184 and their L.C.M. is 952 . Find their H.C.F.

Answer:

Note: The product of two given numbers  = Product of HCF and LCM

Therefore 16184 = HCF \times 952 \Rightarrow HCF = 17

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Question 16. The product of two numbers is 591223 and their H.C.F. is 952 . Find their L.C.M.

Answer: 

Note: The product of two given numbers  = Product of HCF and LCM

Therefore 591223 = 952 \times LCM \Rightarrow LCM = 20387

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Question 17. The H.C.F of two numbers is 14 and their L.C.M is 11592 . If one number is 504 , find the other.

Answer:

Note: The product of two given numbers  = Product of HCF and LCM

Therefore Number \times 504 = 14 \times 11592 \Rightarrow Number = 322

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Question 18. The sum of the H.C.F and L.C.M of two numbers is 1150 . If the L.C.M is 45 times the H.C.F. and one of the numbers is 125 , then find the other number.

Answer:

Note: The product of two given numbers  = Product of HCF and LCM

HCF + LCM = 1150

LCM = 45 \times HCF

\therefore 46 HCF = 1150

\Rightarrow HCF = 25   and LCM = 45 \times 125  = 1125

Now Number \times 125 = 25 \times 1125 \Rightarrow Number = 225

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Question 19. The difference between the L.C.M. and H.C.F of two numbers is 294 . If the L.C.M is 15 times the H.C.F and one of the numbers is 105 , find the other.

Answer:

Note: The product of two given numbers  = Product of HCF and LCM

LCM - HCF   = 294

LCM = 15 \times HCF

\therefore 14 HCF = 294

\Rightarrow HCF = 21   and LCM = 15 \times 21  = 315

Now Number \times 105 = 21 \times 315 \Rightarrow Number = 63

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Question 20. The product of two co-prime numbers is 609 . What would be the L.C.M of these numbers?

Answer:

Co-prime are those numbers where HCF is 1 . So if the product of two co-primes is 609 then the numbers are 1 and 609 . Therefore LCM is 609 .