Maths Problem Solution

Dr. Small will take 18 years to disappeare completely after losing 2 inches per year. So if it takes Dr. Samll 18 years to dissapere, Ms Tall will grow...
18 x (2/3) = 12 inches
So Ms Tall has grown 12 inches in 18 years.
96+12= 108 inches
Ms Tall will be 108 inches tall by the time Dr Small has disappeared completely.

The Shrinking Doctor

Dr. Small is 36 inches tall and Ms. Tall is 96 inches tall. If Dr. Small shrinks 2 inches per year and Ms. Tall grows 2/3 of an inch per year, how tall will Ms. Tall be when Dr. Small disappears altogether?

On the acorn problem

Johnathan has given the right solution to the acorns problem, so congratulations! you have gathered 2 extra credits for you grade this marking period. However, I'd like to make a remark about a mathematical method to solve the problem because Johnathan solution seems to come up by the guess and try method, which is perfectly valid though, at times, can be a little annoying and innefficient.



This problem can be solved by the following reasoning:



Chip has X acorns and Dale must have X+2 because then, when Dales gives one to Chip both will have the same, this is: X+1, as the first part of the problem states.

then when Dale receives one acorn from Chip he will have double than his friend. can be expressed by the following equation:

2(X-1)=X+2+1

Hence: 2x-2=X+3 and solving for X results: X=5

So Chip has 5 acorns and Dale must have 7.

maths probleme solution

Chip had 5 acorns, and Dale had 7 acorns. If Dale gives Chip 1 acorn, they both have 6 acorns. And if Chip gives Dale 1 acord Chip will now have 4 acorns and Dale will have 8 (ie, double the number of acorns that Chip has) Between them they had 12 acorns.

MATH PROBLEM

Chip said to Dale, "if you give me one acorn, then we will have an equal number of acorns," Dale replied with delight, If you give me one acorn, then I will have double the number you have! What was the total number of acorns they had?

Solution problem number 2

What you have to do to solve this problem is to know the formula of the perimeter of a circle.
This formula is 2r·π. So, 2r=4 π=3,14...
The solution is:
15cm+15cm+4·π= 30cm+4π=42,46cm is longitude of the belt.

MATHS PROBLEM

Jim has a broken fan belt on his car. The belt goes around 2 pulleys, whose centers are 15 cm apart, and each pulley is 4 cm in diameter. How long should the belt be?

Solution to the Fish Tank Problem.

As the 60 cm edge is always horizontal, we can forget about it and reduce the problem to a two-dimensional question. When the tank is tilted the water shape against the front wall forms a triangle. When the tank is at rest upon its base the water shape is a rectangle. To keep constant the water volume the triangle and the rectangle must have the same area, thus:

area rectangle = area triangle

from the problem data:

100·water height = (40·50)/2

Hence:

water height = (40·50)/(2·100) = 10

all units are expressed in cm so the answer will be 10 cm.

Congratulations Jonathan!!!

Solution to the Maths problem

First thing to do, is to half the length of the fish tank so that the water reaches exactly half of the newly formed cube. We then find out the volume of the cube, and we know that we have exactly, half the volume of the cube. To find the volume we multiply all 3 sides in metres (0.4m, 0.5m & 0.6m) which gives us 0,12cubic metres. Half will be the volume we have (ie, 0.06cubic metres).
As we know that 1cubic metre is 1000 litres of liquid, we know that 0,06cubic metres is 60 litres.
Now that we know the amount of water in out cube, we can resolve the new equation with the only unknow details as the height of the water.

MATHS PROBLEM

An orthoedric fish tank measures 100 cm long, 60 cm, 40 cm deep. When tilted to rest on a 60 cm edge, the water level reaches the midpoint of the base as shown in the picture. If we return the tank to the horizontal position, what will be the depth of the water?

Solve this problem and publish the solution. If you are the first to publish a right solution you will get an extra point in your Maths grade.