A tea packet in a shape of a cuboid were L =8cm, W=5cm,D=12.5cm?
-
(a) calculate the volume of the packet? There are 125 grams of tea in a full packet, Jason has to design a new packet that will contain 100g of tea when it is full (b1)work out the volume of the new packet? (b2) express the weight of the tea in the new packet as a % of the weight of the tea in the packet shown.
-
Answer:
volume = l*w*d = 8*5*12.5 = 500 cm^3 500 cm^2 contains 125 gm tea. so, 1 gm can be contained in = 500/125 so, 100 g needs = 500/125*1`00 = 400 for 100 gms, 400 cm^3 is required. ================================ weight of tea in new pack = 100 g 100 g as % of weight of original pack = 100 / 125 * 100 = 80 %
tycoon at Yahoo! Answers Visit the source
Other answers
(a) volume of the cuboid L X W X D 8 X 5 X 12.5 = 500cm^3 500/125 = 4 4 X 100 400cm^3
Light
Related Q & A:
- How to find out whether a shape has touched another shape?Best solution by Stack Overflow
- Whats a word that starts with "w" that describes a house?Best solution by Yahoo! Answers
- What's a good title for a To Kill a Mockingbird essay?Best solution by wiki.answers.com
- Is it illegal to have a car seat in a car without a base?Best solution by ChaCha
- What is a good ballad from a musical for a tenor?Best solution by Yahoo! Answers
Just Added Q & A:
- How many active mobile subscribers are there in China?Best solution by Quora
- How to find the right vacation?Best solution by bookit.com
- How To Make Your Own Primer?Best solution by thekrazycouponlady.com
- How do you get the domain & range?Best solution by ChaCha
- How do you open pop up blockers?Best solution by Yahoo! Answers
For every problem there is a solution! Proved by Solucija.
-
Got an issue and looking for advice?
-
Ask Solucija to search every corner of the Web for help.
-
Get workable solutions and helpful tips in a moment.
Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.