Question 1: Use prime factorization method to find the H.C.F. of the following:

i) 204, 1190       ii) 1445, 1785       iii) 1512, 4212

iv) 280, 315, 385       v) 576, 792, 1512       vi) 1197, 5320, 4389

Note: Prime Factorization Method:

Step 1: Express each one of the given numbers as a product of prime factors

Step 2: The product of terms containing least power of common prime factors gives the HCF of the given numbers.

Answer:

i) 204, 1190  

\begin{array}{r | r}   2 & 204 \\ \hline 2 & 102  \\ \hline 3 & 51 \\ \hline 17 & 17 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm}   \begin{array}{r | r}   2 & 1190 \\ \hline 5 & 595  \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline {\ } & 1 \end{array}

204 = 22 \times 3 \times 17

1190 = 2 \times 5 \times 7 \times 17

Therefore HCF = 2 \times 17 = 34

ii) 1445, 1785  

\begin{array}{r | r}   5 & 1445 \\ \hline 17 & 289  \\ \hline 17 & 17 \\ \hline {\ } & 1 \\ \hline {\ } & {\ } \end{array} \hspace{1.5cm}   \begin{array}{r | r}   5 & 1785 \\ \hline 3 & 357  \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline {\ } & 1 \end{array}

1445 = 5 \times 172

1785 = 5 \times 3 \times 7 \times 17

Therefore HCF = 5 \times 17 = 85

iii) 1512, 4212

\begin{array}{r | r}   2 & 1512 \\ \hline 2 & 756  \\ \hline 2 & 378 \\ \hline 3 & 189  \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\  \hline {\ } & 1 \end{array}  \hspace{1.5cm}   \begin{array}{r | r}   2 & 4212 \\ \hline 2 & 2106  \\ \hline 3 & 1053 \\ \hline 3 & 351  \\ \hline 3 & 117 \\ \hline 3 & 39 \\ \hline 13 & 13 \\  \hline {\ } & 1 \end{array}

1512 = 23 \times 33 \times 7

4212 = 22 \times 34 \times 13

Therefore HCF = 22 \times 33 = 108

iv)  iv) 280, 315, 385

\begin{array}{r | r}   2 & 280 \\ \hline 2 & 140  \\ \hline 2 & 70 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm}   \begin{array}{r | r}   5 & 315 \\ \hline 5 & 63  \\ \hline 5 & 21 \\ \hline 7 & 7 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm}   \begin{array}{r | r}   5 & 385 \\ \hline 7 & 77  \\ \hline 11 & 77  \\ \hline {\ } & 1 \end{array}

280 = 2^3 \times 5 \times 7

315 = 5 \times 3^2 \times 7

385 = 5 \times 7 \times 11

Therefore HCF = 5 \times 7 = 35

v) 576, 792, 1512

\begin{array}{r | r}   2 & 576 \\ \hline 2 & 288  \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 3 & 36 \\ \hline 3 & 12 \\ \hline 2 & 4 \\ \hline 2 & 1 \\  \hline {\ } & 1 \end{array} \hspace{1.5cm}   \begin{array}{r | r}   2 & 792 \\ \hline 2 & 396  \\ \hline 2 & 198 \\ \hline 3 & 99 \\ \hline 3 & 33  \\ \hline 11 & 11  \\ \hline {\ } & 1 \end{array}  \hspace{1.5cm}   \begin{array}{r | r}   2 & 1512 \\ \hline 2 & 756  \\ \hline 2 & 378  \\ \hline 3 & 189  \\ \hline 3 & 63  \\ \hline 3 & 21  \\ \hline 7 & 7  \\ \hline {\ } & 1 \end{array}

576 = 2^7 \times 3^2

792 = 2^3 \times 3^2 \times 11

1512 = 2^3 \times 3^3 \times 7

Therefore HCF = 2^3 \times 3^2 = 72

vi) 1197, 5320, 4389

\begin{array}{r | r}   3 & 1197 \\ \hline 3 & 399  \\ \hline 7 & 133 \\ \hline 19 & 19 \\  \hline {\ } & 1 \end{array} \hspace{1.5cm}   \begin{array}{r | r}   2 & 5320 \\ \hline 2 & 2660  \\ \hline 7 & 1330 \\ \hline 2 & 190 \\ \hline 5 & 95  \\ \hline 19 & 19  \\ \hline {\ } & 1 \end{array}  \hspace{1.5cm}   \begin{array}{r | r}   7 & 4389 \\ \hline 3 & 627  \\ \hline 19 & 209  \\ \hline 11 & 11    \\ \hline {\ } & 1 \end{array}

1197 = 3^2 \times 7 \times 19

5320 = 2^3 \times 7 \times 5 \times 19

4389 = 3 \times 7 \times 19 \times 11

Therefore HCF = 7 \times 19 = 133

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Question 2: Find the HCF of the following numbers by using the long division method:

i) 837, 992            ii) 2923, 3239            ii) 2923, 3239  

iv) 204, 1190, 1445            v) 370, 592, 1036          vi) 1701, 2106, 2754

Answer:

i) 837, 992

\begin{array}{r  r  r  r  r  r}   837) & \overline{992} & (1 &  &  &  \\  & 837 &  &  &  &  \\  & \overline{155)} & \overline{837} & (5 & & \\  & & 775 & & & \\   & & \overline{62)} & \overline{155} & (2 & \\  & & & 124 & & \\  & & & \overline{31)} & \overline{62} & (2 \\  & & & & 62 &  \\  & & & & \overline{0} &   \end{array}

Therefore HCF = 31

ii) 2923, 3239

\begin{array}{r  r  r  r  r  }     2923) & \overline{3239} & (1 &  &   \\  & 2923 &  &  &    \\ & \overline{316)} & \overline{2923} & (9 &  \\ & & 2844 & &  \\  & & \overline{79)} & \overline{316} & (4  \\ & & & 316 &  \\ & & & \overline{0} &  \end{array}

Therefore HCF = 79

iii) 5508, 9282

\begin{array}{r  r  r  r  r  r  r r }   5508) & \overline{9282} & (1 &  &  & & &   \\  & 5508 &  &  &  &  & & \\ & \overline{3774)} & \overline{5508} & (1 & & & & \\ & & 3774 & & & & &  \\ & & \overline{1734)} & \overline{3774} & (2 & & & \\  & & & 3468 & & & &  \\  & & & \overline{306)} & \overline{1734} & (5 & &  \\  & & & & 1530 & & &   \\  & & & & \overline{204)} & \overline{306} & (1 &   \\  & & & &  & 204 & &   \\  & & & & & \overline{102)} & \overline{204} & (2  \\  & & & & &  & 204 &    \\  & & & & &  & \overline{0} &  \end{array}

Therefore HCF = 102

iv) 204, 1190, 1445

\begin{array}{r  r  r  r  r  r}    1190) & \overline{1445} & (1 &  &  &  \\ & 1190 &  &  &  &  \\ & \overline{255)} & \overline{1190} & (4 & & \\ & & 1020 & & & \\  & & \overline{170)} & \overline{255} & (1 & \\ & & & 170 & & \\ & & & \overline{85)} & \overline{170} & (2 \\ & & & & 170 &  \\ & & & & \overline{0} & \end{array}

Therefore HCF of 1445 and 1190 = 85

\begin{array}{r  r  r  r  r  }   85) & \overline{204} & (2 &  &   \\  & 170 &  &  &    \\ & \overline{34)} & \overline{85} & (2 &  \\ & & 68 & &  \\  & & \overline{17)} & \overline{34} & (2  \\ & & & 34 &  \\ & & & \overline{0} &  \end{array}

Therefore HCF = 17

v) 370, 592, 1036

\begin{array}{r  r  r  r  r  }   592) & \overline{1036} & (1 &  &   \\  & 592 &  &  &    \\ & \overline{444)} & \overline{592} & (1 &  \\ & & 44 & &  \\ & & \overline{148)} & \overline{444} & (3  \\ & & & 444 &  \\ & & & \overline{0} &  \end{array}

Therefore HCF of 1036 and 592= 148

\begin{array}{r  r  r  r    }   148) & \overline{370} & (1 &     \\  & 296 &  &      \\ & \overline{74)} & \overline{148} & (2   \\ & & 148 &   \\  & & \overline{0} & \end{array}

Therefore HCF = 74

vi) 1701, 2106, 2754

\begin{array}{r  r  r  r  r  }   2106) & \overline{2754} & (1 &  &   \\  & 2106 &  &  &    \\ & \overline{648)} & \overline{2106} & (3 &  \\  & & 1944 & &  \\  & & \overline{162)} & \overline{648} & (4  \\  & & & 648 &  \\  & & & \overline{0} &   \end{array}

HCF of 2754 and 2106  = 162

\begin{array}{r  r  r  r    }   162) & \overline{1701} & (10 &     \\  & 1620 &  &      \\  & \overline{81)} & \overline{162} & (2   \\  & & 162 &   \\ & & \overline{0} & \end{array}

Therefore HCF = 81

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Question 3: Reduce each of the following fractions to its lowest terms:

i) \frac{682}{868}     ii) \frac{777}{1147}     iii) \frac{1095}{1168}

Answer:

i) \frac{682}{868}

\begin{array}{r  r  r  r  r  r}    682) & \overline{868} & (1 &  &  &  \\ & 682 &  &  &  &  \\ & \overline{186)} & \overline{682} & (3 & & \\ & & 558 & & & \\  & & \overline{124)} & \overline{186} & (1 & \\ & & & 124 & & \\ & & & \overline{62)} & \overline{124} & (2 \\ & & & & 124 &  \\ & & & & \overline{0} & \end{array}

Therefore \frac{682}{868} = \frac{682 \div 62}{868 \div 62} = \frac{11}{14}

ii) \frac{777}{1147}

\begin{array}{r  r  r  r  r  }   777) & \overline{1147} & (1 &  &   \\  & 777 &  &  &    \\ & \overline{370)} & \overline{777} & (2 &  \\  & & 740 & &  \\  & & \overline{37)} & \overline{370} & (10  \\  & & & 370 &  \\  & & & \overline{0} &   \end{array}

Therefore \frac{777}{1147} = \frac{777 \div 37}{1147 \div 37} = \frac{21}{31}

iii) \frac{1095}{1168}

\begin{array}{r  r  r  r    }   1095) & \overline{1168} & (1 &     \\  & 1095 &  &      \\  & \overline{73)} & \overline{1095} & (15   \\  & & 1095 &   \\ & & \overline{0} & \end{array}

Therefore \frac{1095}{1168} = \frac{1095 \div 73}{1068 \div 73} = \frac{15}{16}

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Question 4. Which of the following numbers are co-primes?

i) 18, 25 ii) 62, 81 iii) 69, 92

Answer:

Note: The natural numbers are said to be co-primes if the HCF of the numbers is 1 .
You can calculate the HCF by using any of the two methods…prime factorization or long division method.

i) 18, 25

HCF of 18, 25 = 1

Hence, 18 and 25 are co-primes.

ii) 62, 81

HCF of 62, 81 = 1

Hence, 62 and 81 are co-primes.

iii) 69, 92

HCF of 69, 92 = 23

Hence, 69 and 92 are not co-primes.

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Question 5: Find the greatest number that exactly divides 105, 1001 and 2436 .

Answer:

We need to find the HCF of 105, 1001 and 2436 .

\begin{array}{r  r  r  r  r  r  r r }   1001) & \overline{2436} & (2 &  &  & & &   \\  & 2002 &  &  &  &  & & \\ & \overline{434)} & \overline{1001} & (2 & & & & \\ & & 868 & & & & &  \\ & & \overline{133)} & \overline{434} & (3 & & & \\  & & & 399 & & & &  \\  & & & \overline{35)} & \overline{133} & (3 & &  \\  & & & & 105 & & &   \\  & & & & \overline{28)} & \overline{35} & (1 &   \\  & & & &  & 28 & &   \\  & & & & & \overline{7)} & \overline{28} & (4  \\  & & & & &  & 28 &    \\  & & & & &  & \overline{0} &  \end{array}

\begin{array}{r  r  r    }   7) & \overline{105} & (15      \\  & 105 &    \\   & \overline{0} & \end{array}

Hence the HCF of the three numbers is 7 . That means, 7 is the largest number that would divide all the three numbers perfectly.

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Question 6: Find the largest number which can divide 290, 460 and 552 leaving the remainders 4, 5 and 6 respectively.

Answer:

The largest number should divide (290-4), (460-5) and (552-6) or 286, 455 and 546 perfectly. Now find the HCF of 286, 455 and 546 .

\begin{array}{r  r  r  r    }   455) & \overline{546} & (1 &     \\  & 455 &  &      \\ & \overline{91)} & \overline{455} & (5   \\ & & 455 &   \\  & & \overline{0} & \end{array}

\begin{array}{r  r  r  r    }   91) & \overline{286} & (3 &     \\  & 273 &  &      \\ & \overline{13)} & \overline{91} & (7   \\ & & 91 &   \\  & & \overline{0} & \end{array}

Hence 13 is the number that can divide 290, 460 and 552 leaving the remainders 4, 5 and 6 respectively.

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Question 7: Find the largest number which can divide 1354, 1866 and 2762 leaving the same remainder 10 in each case.

Answer:

The largest number should divide (1354 - 10), (1866 - 10) and (2762 - 10) or 1344, 1856 and 2752 perfectly. Now find the HCF of 1344, 1856 and 2752 .

\begin{array}{r  r  r  r  r  }     1856) & \overline{2752} & (1 &  &   \\  & 1856 &  &  &    \\ & \overline{896)} & \overline{1856} & (2 &  \\ & & 1792 & &  \\  & & \overline{64)} & \overline{896} & (14  \\ & & & 896 &  \\ & & & \overline{0} &  \end{array}

\begin{array}{r  r  r     }   64) & \overline{1344} & (21      \\  & 1344 &         \\  & & \overline{0}  \end{array}

HCF is 64 . Hence 64 is the largest number which can divide 1354, 1866 and 2762 leaving the same remainder 10 in each case.

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Question 8: Find the greatest number that will divide 1305, 4665 and 6905 so as to leave the same remainder in each case.

Answer:

Required number

= HCF of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= HCF of 3360, 2240, 5600 .

\begin{array}{r  r  r  r  r  }     3360) & \overline{5600} & (1 &  &   \\  & 3360 &  &  &    \\ & \overline{2240)} & \overline{3360} & (1 &  \\ & & 2240 & &  \\  & & \overline{1120)} & \overline{2240} & (2  \\ & & & 2240 &  \\ & & & \overline{0} &  \end{array}

\begin{array}{r  r  r     }   1120) & \overline{3360} & (3      \\  & 3360 &    \\   & \overline{0} & \end{array}

HCF is 1120 .

The required number is 112 that will divide 1305, 4665 and 6905 so as to leave the same remainder in each case.

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Question 9: Three pieces of timber 13 m 44 cm, 18 m 56 cm and 27 m 52 cm have to be divided into planks of same lengths. What is the greatest possible length of each plank?

Answer:

The lengths are 1344 cm, 1856 cm and 2752 cm. Find HCF.

\begin{array}{r  r  r  r  r  }     1856) & \overline{2752} & (1 &  &   \\  & 1856 &  &  &    \\ & \overline{896)} & \overline{1856} & (2 &  \\ & & 1792 & &  \\  & & \overline{64)} & \overline{896} & (14  \\ & & & 896 &  \\ & & & \overline{0} &  \end{array}

\begin{array}{r  r  r      }   64) & \overline{1344} & (21      \\  & 1344 &    \\   & \overline{0} & \end{array}

HCF is 64 . Hence greatest length is 64 cm.

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Question 10: An NGO wishes to distribute 1651 pencils and 2032 erasers among poor children in such a way that each child gets the same number of pencils and the same number of erasers. Find the maximum number of children among whom these pencils and erasers are distributed.

Answer:

Take HCF of 1651 and 2032 .

\begin{array}{r  r  r  r  r  }     1651) & \overline{2032} & (1 &  &   \\  & 1651 &  &  &    \\ & \overline{381)} & \overline{1651} & (4 &  \\ & & 1524 & &  \\  & & \overline{127)} & \overline{381} & (3  \\ & & & 381 &  \\ & & & \overline{0} &  \end{array}

HCF = 127 .

The maximum number of children among whom these pencils and erasers are distributed = 127

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Question 11: Find the greatest possible length of a rope which can be used to measure exactly the lengths 5 m 13 cm, 7 m 83 cm and 10 m 80 cm.

Answer:

Lengths are 513 cm, 783 cm and 1080 cm.

HCF of 513, 783 and 1080

\begin{array}{r  r  r  r  r  r  r r }   783) & \overline{1083} & (1 &  &  & & &   \\  & 783 &  &  &  &  & & \\ & \overline{297)} & \overline{783} & (2 & & & & \\ & & 594 & & & & &  \\ & & \overline{189)} & \overline{297} & (1 & & & \\  & & & 189 & & & &  \\  & & & \overline{108)} & \overline{189} & (1 & &  \\  & & & & 108 & & &   \\  & & & & \overline{81)} & \overline{108} & (1 &   \\  & & & &  & 81 & &   \\  & & & & & \overline{27)} & \overline{81} & (3  \\  & & & & &  & 81 &    \\  & & & & &  & \overline{0} &  \end{array}

\begin{array}{r  r  r }   27) & \overline{513} & (19      \\  & 513 &    \\   & \overline{0} & \end{array}

Hence HCF = 27 . Therefore the length of the rope is 27 cm.

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Question 12: Three different containers contain different quantity of mixture of milk and water whose measurements are 403 kg, 465 kg, and 527 kg. What biggest measure must be there to measure all different quantities an exact number of times.

Answer:

Required Number is the HCF of 403, 465 and 527 .

\begin{array}{r  r  r  r  r  }     465) & \overline{527} & (1 &  &   \\  & 465 &  &  &    \\ & \overline{62)} & \overline{465} & (7 &  \\ & & 434 & &  \\  & & \overline{31)} & \overline{62} & (2  \\ & & & 62 &  \\ & & & \overline{0} &  \end{array}

\begin{array}{r  r  r      }   31) & \overline{403} & (13      \\  & 403 &    \\   & \overline{0} & \end{array}

HCF = 31 . Hence 31 liter measure would be required to measure the three given quantities.

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Question 13: Find the least number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.

Answer:

The dimensions of the ceiling are 1517 cm and 902 cm.

The required number of times = HCF of 1517 and 902 .

\begin{array}{r  r  r  r  r  r}   902) & \overline{1517} & (1 &  &  &  \\  & 902 &  &  &  &  \\  & \overline{615)} & \overline{902} & (1 & & \\  & & 615 & & & \\   & & \overline{287)} & \overline{615} & (2 & \\  & & & 574 & & \\  & & & \overline{41)} & \overline{287} & (7 \\  & & & & 287 &  \\  & & & & \overline{0} &   \end{array}

HCF = 41

Dimensions of tiles = 41 cm.

Therefore the number of tiles = \frac{area \ of \ ceiling}{area \ of \ tile} = \frac{1517 \times 902}{41 \times 41} = 814

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Question 14: A school admitted 1190 boys and 204 girls in a year. The authorities decided to divide these students into classes in such a way that each class gets the same number of boys and the same number of girls. Find the maximum number of classes that can be formed.

Answer:

Required number is the HCF of 1190 and 204 .

\begin{array}{r  r  r  r  r  }     204) & \overline{1190} & (5 &  &   \\  & 1020 &  &  &    \\ & \overline{170)} & \overline{204} & (1 &  \\ & & 170 & &  \\  & & \overline{34)} & \overline{170} & (5  \\ & & & 170 &  \\ & & & \overline{0} &  \end{array}

HCF = 34 .

Number of classes = 34

Size of the class = 41 students

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Question 15: In a training camp, there were 195 boys and 143 girls. The coordinator instructed the pupils to form separate teams for boys and girls such that each team includes equal number of pupils. Find the maximum number of pupils that each team can have and also the number of teams so formed.

Answer:

Required number is HCF of 195 and 143 .

\begin{array}{r  r  r  r  r  r}   143) & \overline{195} & (1 &  &  &  \\  & 143 &  &  &  &  \\  & \overline{52)} & \overline{143} & (2 & & \\  & & 104 & & & \\   & & \overline{39)} & \overline{52} & (1 & \\  & & & 39 & & \\  & & & \overline{13)} & \overline{39} & (3 \\  & & & & 39 &  \\  & & & & \overline{0} &   \end{array}

HCF = 13 .

Number of teams = 13 .

Number of pupil in each team = \frac{(195 + 143)}{13} = 26 .