Math, asked by anetashrestha8, 4 months ago

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

Answered by Anonymous
8

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.

Answer is long hopefully you won't forget to follo.w :)

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Answered by durgeshshrivastav205
3

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.

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