Lamp posts are positioned at intervals of 120 m along a road and pots of bougainvillea plants placed at intervals of 96 m. The first pot of bougainvillea plant is placed at the foot of the first lamp post.
a) If the last pot of bougainvillea plant is placed at the foot of a lamp post, find the distance between the first and the last pot of plant.
b) Find the total number of lamp posts and pots of bougainvillea plants within this road segment.
** Need help with b). Thank you in advance.
a)
120=5x3x22
96=3×25
LCM=5x3x25
=480
b)
No. of lamp posts in this segment= 480÷120+1
= 5
No. of pots of bougainvillea plants= 480÷96+1
=6
Total number of lamp posts and pots of bougainvillea plants= 5+6
=11
a) To find lowest common multiple of 120 and 96
Taking multiples of 120 – 120, 240, 360, 480 and so forth, try dividing by 96.
480 is the lowest multiple of 120 that is divisible by 96. Thus the answer is 480m.
b) Total number of lampost = (480/120)+1 = 5
n/b: add one for the first lampost.
LP1 —-120m—-LP2 —-120m—-LP3 —-120m—-LP4 —-120m—-LP5
Total number of bougainvillea plant pot = (480/96)+1 = 6
n/b: add one for the first pot.
Pot1 –96m–Pot2 –96m–Pot3 –96m–Pot4 –96m–Pot5 –96m–Pot6
Total number of lamp posts and pots of bougainvillea plant = 5+6 = 11
a) Find the lowest common multiple between 96 and 120. Answer will be 480.
b) Take 480 divide by 96 = 5.
take 480 divide by 120 = 4.
5+4+2= 11
arosisi
120 240 360 480 –> 5 lamp posts
96 192 288 384 480 –> 6 pots of plant
5 + 6 = 11 (Ans)
