i. On the playground, if the children are made to stand for 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?
ii. 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?
iii. 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. What was the minimum number of laddoos in the three boxes altogether?
iv. 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 are 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?
v. Given the fractions $\frac{13}{45}$ and $\frac{22}{75}$ Write their equivalent fractions with same denominators and add the fractions.
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.
ii. 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.
iii. 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 x 120 = 360
∴ The minimum number of ladoos in 3 boxes are 360.
iv. 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.
v. 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
$\begin{aligned} & \frac{13}{45}=\frac{13 \times 5}{45 \times 5}=\frac{65}{225} \\ & \frac{22}{75}=\frac{22 \times 3}{75 \times 3}=\frac{66}{225} \\ \therefore \quad & \frac{13}{45}+\frac{22}{75}=\frac{65}{225}+\frac{66}{225}=\frac{65+66}{225}=\frac{131}{225}\end{aligned}$
$\therefore \quad$ Equivalent fractions of $\frac{13}{45}$ and $\frac{22}{75}$ are $\frac{65}{225}$ and $\frac{66}{225}$ respectively, and the addition of $\frac{13}{45}$ and $\frac{22}{75}$ is $\frac{131}{225}$.