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Question

Solve the following problems.

(1) 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?

(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?

(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?

(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?

(5) Given the fractions $\frac{13}{45}$ and $\frac{22}{75}$. write their equivalent fractions with same denominators and add the fractions.

(1) 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?

(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?

(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?

(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?

(5) Given the fractions $\frac{13}{45}$ and $\frac{22}{75}$. write their equivalent fractions with same denominators and add the fractions.

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Solution

(1)

If the children are to stand in rows such that all rows are complete and no children is left out, then the number of children in the school must be in multiples of both 20 and 25. So, the lowest possible number of children in the school is the LCM of 20 and 25.

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

Multiples of 25: 25, 50, 75, 100, 125, 150, 175, 200,...

Common multiples of 20 and 25: 100, 200,...

∴ Lowest of possible number of children in that school = LCM of 20 and 25 = 100

Thus, the lowest of possible number of children in that school are 100.

(2)

Veena wants to make necklaces with an equal number of beads in each such there is no bead left over, so the number of beads with her must be in mutltiples of 16, 24 and 40. Therefore, the least number of beads with her is the LCM of 16, 24 and 40.

Multiples of 16: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240, 256,...

Multiples of 24: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264,...

Multiples of 40: 40, 80, 120, 160, 200, 240, 280,...

Common multiples of 16, 24 and 40: 240, 480,...

∴ Least number of beads with Veena = LCM of 16, 24 and 40 = 240

Thus, the least number of beads with Veena are 240.

(3)

Equal number of laddoos are distributed among the children equally with not a single laddoo was left over, so number of laddoos in the three boxes must be in multiples of 20, 24 and 12. Therefore, the minimum number of laddoos in the three boxes altogether is the LCM of 20, 12 and 24.

Multiples of 20: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300, 320, 340, 360, 380, 400,...

Multiples of 24: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360, 384,...

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, 192, 204, 216, 228, 240, 252, 264, 276, 288, 300, 312, 324, 336, 348, 360, 372, 384,...

Common multiples of 20, 24 and 12: 360, 720,...

∴ Minimum number of laddoos in the three boxes altogether = LCM of 20, 24 and 12 = 360

Thus, the minimum number of laddoos in the three boxes altogether are 360.

(4)

All the three signals would turn green simultaneously again at a time which is a multiple of the three intervals of 60 seconds, 120 seconds and 24 seconds. So, all three signals turn green simultaneously again after a time interval which is the LCM of 60 seconds, 120 seconds and 24 seconds.

Multiples of 60: 60, 120, 180, 240,...

Multiples of 120: 120, 240,...

Multiples of 24: 24, 48, 72, 96, 120, 144, 168,...

Common multiples of 60, 120 and 24: 120, 240,...

∴ All the three signals turn green simultaneously again = LCM of the time intervals 60 seconds, 120 seconds and 24 seconds = 120 seconds

Thus, all the three signals turn green simultaneously again after 120 seconds.

(5)

The given fractions are $\frac{13}{45}$ and $\frac{22}{75}$.

Multiples of 45: 45, 90, 135, 180, 225, 270,...

Multiples of 75: 75, 150, 225,...

Common multiples of 45 and 75: 225, 450,...

∴ LCM of 45 and 75 = 225

Equivalent fraction of $\frac{13}{45}=\frac{13\times 5}{45\times 5}=\frac{65}{225}$ (225 ÷ 45 = 5)

Equivalent fraction of $\frac{22}{75}=\frac{22\times 3}{75\times 3}=\frac{66}{225}$ (225 ÷ 75 = 3)

Also,

$\frac{13}{45}+\frac{22}{75}\phantom{\rule{0ex}{0ex}}=\frac{65}{225}+\frac{66}{225}\phantom{\rule{0ex}{0ex}}=\frac{65+66}{225}\phantom{\rule{0ex}{0ex}}=\frac{131}{225}$

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