Word Math problems

[silveryhair]

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Can someone please show how to solve the following math questions. Thanks!

A carton which is 1.5 feet wide by 2 feet long by one foot high contains cans which are 6 inches high and 8 inches in diameter. Approxiamately how much wasted space is there after the cartons are put in?

Ans: 1111 cubic inches

A slope is increased so that it is now equivalent to a 45 degree angle. If one degree was 20% of the original slope, what percent of increase has occurred?

Ans: 800%

 

[jameszones]

hmmm im not sure about the first one but for the second one:

if 1 degree is 20% of the original slope, then the original slope was 5 degrees because 20% of 5 is equal to 1. you could also do 1/.20 and get the same answer. since the new slope is 45 degrees, there was a 40 degree increase from the original slope (45 - 5 = 40). (40 degree increase / 5 degrees originally) x 100% = 800% increase.

i hope that made sense....lol

 

[Phish]

I can't answer the first question without knowing how many cans are in the carton.

For the second question:
Start off with the last thing they say in the question. If one degree is 20% of the original angle, then we can find out the value of the original angle simply: 1/x = 20/100 (factor) 1/x = 2/10 (factor) 1/x = 1/5 (cross multiply) x = 5.
Now we know that the original angle is 5 degrees, and they tell us that it increased to 45. 45 - 5 = 40 degree increase. Now we have to find out the percent increase from the old angle: 40/5 = x/100 (cross multiply) 2000 = 5x (divide) x = 800% increase.

 

[mashinka]

For the first one can't you just figure out how many cans fit into into the carton, find their volume and then subtract that from the total volume of the carton. So convert the dimensions of the carton into inches to get 16 x 24 x 12. Since the diameter of the cans is 8, you can have 2 cans along the width of the carton and 3 cans along the length. And two cans stacked on top of each other due to the height, so in total 2 x 3 x 2 = 12 cans can fit in the carton.
Now just find the volume of one can, multiply that by 12 and subtract from volume of the carton. I don't have a calculator on me so I can't do the calculations.
Does that give you the right answer?

 

[silveryhair]

Awesome Thanks guys! I tried plugging in the values mashinka, no luck 🙁

 

[DQLEUCD]

Can someone please show how to solve the following math questions. Thanks!

A carton which is 1.5 feet wide by 2 feet long by one foot high contains cans which are 6 inches high and 8 inches in diameter. Approxiamately how much wasted space is there after the cartons are put in?

Ans: 1111 cubic inches

A slope is increased so that it is now equivalent to a 45 degree angle. If one degree was 20% of the original slope, what percent of increase has occurred?

Ans: 800%

1st one if you draw it out, it'll be easier. So you can stack 3 cans to it's width and 4 cans to it's length. With it's 6 inches high for each can, you can fit 2 rows in the carton. so your looking at 12x2 which is 24 cans. Find the volume of 1 can x 24 and subtract it from the volume of the carton.

Volume for a cylinder = (pi)r^2 x H
Volume of the carton = L x W x H

 

[NewBee12]

for the first problem:
first convert the dimensions of the box to inches and you get: L =24 in W =18in and H = 12in
since each can has a diameter of 8in you can have 3 along the length (3x8=24) and 2 along the width ( 8x2 = 16) that gives you 6 cans. and since the height is 12in you can stack 6 more on top for a total of 12 cans.
Now the volume of each can is : pi(4^2)(6) = 96pi
the volume of 12 cans is 12x96pi = 1152pi in^3
the volume of the cartoon is : 18x24x12 = 5184 in^3
the wasted space is : 5184 - 1152pi = 1566 in^3
I dont see how you can do this without a calculator and i dont see how the answer is 1111 in^3
hope this helps.

 

[Pat Kelly]

where did you get question 1 from? I know how to solve it but the question is really stupid/tricky and probably won't appear on the dat

 

[DQLEUCD]

where did you get question 1 from? I know how to solve it but the question is really stupid/tricky and probably won't appear on the dat

It's just more work then tricky, probably considered a hard problem on the DAT's so you never know.