A rectangular cartoon is 80cm×60cm×40cm
i.)how many packets of shops can each of 10cm×5cm×4cm be kept inside the carton
ii.)by how many centimetres should the height of the carton be increased to keep 1200 pkts of shops?
Answers
Answer:
Dimensions of cartoon = 80x60x40
Dimensions of packets = 10x5x4
no.of packets inside cartoon = 80x60x40/(10x5x4)
= 960 packets can be kept.
Height increased to keep 1200 packets is
let height be Z
Now,
Zx80x60/(10x4x5)= 1200
24Z= 1200
Z= 1200/24= 50cm
New height = 50
old height = 40
change = Old - New = 10cm.(-10 actually but that is fully ignored here, as we only want to know change not it's sign)
height increased should be 10cm.
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the increased be height of the carton is 10cm in the keep 1200 pkts of shops.
Step-by-step explanation:
i) volume of the carton= 80cm×60cm×40cm
= 192000cm^3
Again,
the volume of packet shops=10cm×5cm×4cm
= 200cm^3
the no. of volume of packet of shops and packet of carton = volume of packet of shops (v) \volume of packet of shops (v) =19200\200
= 960cm
ii)
Let the increased be height of the carton = x
height of the carton = (40 + x) cm
volume of the carton = 1200 x volume of each shops
80cm×60cm×( 40 + x)cm = 1200 ×200
192,000 + 4800x = 240,000
4800x = 48,000
:' x = 48000\ 4800
x = 10cm
the increased be height of the carton is 10cm in the keep 1200 pkts of shops.