## Chapter 9 – HCF – LCM

## Practice set 23

**Write all the factors of the given numbers and list their common factors.**

**(1) 12, 16 **

**Solution:**

Factors of 12 = 1, 2, 3, 4, 6, 12

Factors of 16 = 1, 2, 4, 8, 16

∴ Common factors of 12 and 16 = 1, 2, 4

**(2) 21, 24 **

**Solution:**

Factors of 21 = 1, 3, 7, 21

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

∴ Common factors of 21 and 24 = 1, 3

**(3) 25, 30 **

**Solution:**

Factors of 25 = 1, 5, 25

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

∴ Common factors of 25 and 30 = 1, 5

**(4) 24, 25 **

**Solution:**

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Factors of 25 = 1, 5, 25

∴ Common factor of 24 and 25 = 1

**(5) 56, 72**

**Solution:**

Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

∴ Common factors of 56 and 72 = 1, 2, 4, 8

## Practice set 24

**1. Find the HCF of the following numbers.**

**(1) 45, 30 **

**Solution:**

Factors of 45 = 1, 3, 5, 9,15, 45

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

∴ HCF of 45 and 30 = 15

**(2) 16, 48 **

**Solution:**

Factors of 16 = 1, 2, 4, 8, 16

Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

∴ HCF of 16 and 48 = 16

**(3) 39, 25 **

**Solution:**

Factors of 39 = 1, 3, 13, 39

Factors of 25 = 1, 5, 25

∴ HCF of 39 and 25 = 1

**(4) 49, 56 **

**Solution:**

Factors of 49 = 1, 7, 49

Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

∴ HCF of 49 and 56 = 7

**(5) 120, 144**

**Solution:**

Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Factors of 144 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144

∴ HCF of 120 and 144 = 24

**(6) 81, 99 **

**Solution:**

Factors of 81 = 1, 3, 9, 27, 81

Factors of 99 = 1, 3, 9, 11, 33, 99

∴ HCF of 81 and 99 = 9

**(7) 24, 36 **

**Solution:**

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

∴ HCF of 24 and 36 = 12

**(8) 25, 75 **

**Solution:**

Factors of 25 = 1, 5, 25

Factors of 75 = 1, 3, 5, 15, 25, 75

∴ HCF of 25 and 75 = 25

**(9) 48, 54 **

**Solution:**

Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54

∴ HCF of 48 and 54 = 6

**(10) 150, 225**

**Solution:**

Factors of 150 = 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

Factors of 225 = 1, 3, 5, 9, 15, 25, 45, 75, 225

∴ HCF of 150 and 225 = 75

**2. If large square beds of equal size are to be made for planting vegetables on a plot of land 18 metres long and 15 metres wide, what is the maximum possible length of each bed ?**

**Solution:**

Length of the land = 18 m

Width of the land = 15 m

The maximum length of each bed will be the greatest common factor of 18 and 15.

Factors of 18 = 1, 2, 3, 6, 9, 18

Factors of 15 = 1, 3, 5, 15

∴ HCF of 18 and 15 = 3

∴ The maximum possible length of each bed is 3 metres.

**3. Two ropes, one 8 metres long and the other 12 metres long are to be cut into pieces of the same length. What will the maximum possible length of each piece be ?**

**Solution:**

Length of first rope = 8 m

Length of second rope = 12 m

The maximum length of each piece will be the greatest common factor of 8 and 12.

Factors of 8 = 1, 2, 4, 8

Factors of 12 = 1, 2, 3, 4, 6, 12

∴ HCF of 8 and 12 = 4

∴ The maximum possible length of each piece is 4 metres.

**4. The number of students of Std 6th and Std 7th who went to visit the Tadoba Tiger Project at Chandrapur was 140 and 196 respectively. The students of each class are to be divided into groups of the same number of students. Each group can have a paid guide. What is the maximum number of students there can be in each group? Why do you think each group should have the maximum possible number of students? **

**Solution:**

Number of students of Std 6th = 140

Number of students of Std 7th = 196

The maximum number of students in each group will be the greatest common factor of 140 and 196.

Factors of 140 = 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140

Factors of 196 = 1, 2, 4, 7, 14, 28, 49, 98, 196

∴ HCF of 140 and 196 = 28

∴ Maximum students in each group are 28. Each group should have a maximum number of students so that there will be a minimum number of groups and hence minimum number of paid guides.

**5. At the Rice Research Centre at Tumsar, there are 2610 kg of seeds of the basmati variety and 1980 kg of the Indrayani variety. If the maximum possible weight of seeds has to be filled to make bags of equal weight. What should be the weight of each bag ? How many bags of each variety will there be ?**

**Solution:**

Weight of basmati rice = 2610 kg

Weight of indrayani rice = 1980 kg

The weight of each bag will be the greatest common factor of 2610 and 1980.

Factors of 2610 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 29, 30, 45, 58, 87, 90, 145, 174, 261, 290, 435, 522, 870, 1305, 2610

Factors of 1980 = 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20, 22, 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990, 1980

∴ HCF of 2610 and 1980 = 90

Maximum weight of each bag = 90 kg

Number of bags of basmati rice

= 2610 ÷ 90

= 29

Number of bags of indrayani rice

= 1980 ÷ 90

= 22

∴ Maximum weight of each bag is 90 kg. The number of bags of basmati rice is 29, and the number of bags of indrayani rice is 22.

## Practice set 25

**1. Find out the LCM of the following numbers.**

**(1) 9, 15 **

**Solution:**

Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90

Multiples of 15 = 15, 30, 45

∴ LCM of 9 and 15 = 45

**(2) 2, 3, 5 **

**Solution:**

Multiples of 2 = 2, 4,6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Multiples of 5 = 5, 10, 15, 20, 25, 30

∴ LCM of 2,3 and 5 = 30

**(3) 12, 28**

**Solution:**

Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

Multiples of 28 = 28, 56, 84

∴ LCM of 12 and 28 = 84

**(4) 15, 20 **

**Solution:**

Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120

Multiples of 20 = 20, 40, 60

∴ LCM of 15 and 20 = 60

**(5) 8, 11**

**Solution:**

Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

Multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88

∴ LCM of 8 and 11 = 88

**2. Solve the following problems.**

**(1) On the playground, if the children are made to stand to drill either 20 to a row or 25 to a row, all rows are complete and no child is left out. What is the lowest possible number of children in that school?**

**Solution:**

The lowest possible number of children is equal to the lowest common multiple of 20 and 25.

Multiples of 20 = 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

Multiples of 25 = 25, 50, 75, 100

∴ LCM of 20 and 25 = 100

∴ The least number of students in the school is 100.

**(2) Veena has some beads. She wants to make necklaces with an equal number of beads in each. If she makes necklaces of 16 or 24 or 40 beads, there is no bead left over. What is the least number of beads with her? **

**Solution:**

The least number of beads with Veena is equal to the lowest common multiple of 16, 24 and 40.

Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240, 256, 272, 288

Multiples of 24 = 24, 48, 72, 96, 120, 144, 168, 192, 216, 240

Multiples of 40 = 40, 80, 120, 160, 200, 240

∴ LCM of 16, 24 and 40 = 240

∴ The least number of beads with Veena are 240.

**(3) An equal number of laddoos have been placed in 3 different boxes. The laddoos in the first box were distributed among 20 children equally, the laddoos in the second box among 24 children and those in the third box among 12 children. Not a single laddoo was left over. Then, what was the minimum number of laddoos in the three boxes altogether?**

**Solution:**

The lowest common multiple of 20, 24 and 12 gives the minimum number of laddoos in one box.

Multiples of 20 = 20, 40, 60, 80, 100,120, 140, 160, 180, 200

Multiples of 24 = 24, 48, 72, 96, 120

Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

∴ LCM of 20, 24 and 12 = 120

∴ Minimum number of ladoos in 1 boxes = 120

∴ Minimum number of ladoos in 3 boxes

= 3 × 120

= 360

∴ The minimum number of ladoos in 3 boxes are 360.

**(4) We observed the traffic lights at three different squares on the same big road. They turn green every 60 seconds, 120 seconds and 24 seconds. When the signals were switched on at 8 o’clock in the morning all the lights were green. How long after that will all three signals turn green simultaneously again?**

**Solution:**

All three signals will turn green for lowest common multiple of 60 seconds, 120 seconds and 24 seconds.

Multiples of 60 = 60, 120, 180, 240, 300, 360, 420, 480

Multiples of 120 = 120, 240, 360

Multiples of 24 = 24, 48, 72, 96, 120

∴ LCM of 60, 120 and 24 = 120

Since, 60 seconds = 1 minute

∴ 120 seconds = 2 minutes

∴ The signals will turn green simultaneously again after 120 seconds i.e. 2 minutes.

**(5) Given the fractions \(\large \frac {13}{45}\) and \(\large \frac {22}{76}\) write their equivalent fractions with the same denominators and add the fractions.**

**Solution:**

The lowest common multiple of 45 and 75 gives the same denominator.

Multiples of 45 = 45, 90, 135, 180, 225, 270, 315, 360, 405, 450

Multiples of 75 = 75, 150, 336

∴ LCM of 45 and 75 = 225

\(\large \frac {13}{45}\) = \(\large \frac {13\,×\,5}{45\,×\,5}\) = \(\large \frac {65}{225}\)

\(\large \frac {22}{75}\) = \(\large \frac {22\,×\,3}{75\,×\,3}\) = \(\large \frac {66}{225}\)

Now,

\(\large \frac {13}{45}\) + \(\large \frac {22}{75}\)

= \(\large \frac {65}{225} + \(\large \frac {66}{225}\)

= \(\large \frac {65\,×\,66}{225}\)

= \(\large \frac {131}{225}\)

Equivalent fractions are \(\large \frac {66}{225} and \(\large \frac {65}{225}\) and their sum is \(\large \frac {131}{225}\).