A cake is divided into 3 portions one for you, one for your friend and one for your neighbor. If you get twice as much as your friend, and your friend gets twice as much as your neighbor. What proportion of the cake do you get?
Please Explain to me how to solve this problem
thanks
4x/(7x) = 4/7
How did you get these numbers?
neighbour ---- x
your friend --- 2x
you ---------- 4x
so you get 2x/7x = .....
If the neighbor gets x, then the friend gets 2x and you get 4x
Thanks Alot
To solve this problem, we need to establish a relationship between the portions of cake that you, your friend, and your neighbor receive based on the given information.
Let's assume that the portion of cake received by your neighbor is x. According to the problem, your friend gets twice as much as your neighbor, so your friend receives 2x. Similarly, you receive twice as much as your friend, so you get 2 * (2x), which simplifies to 4x.
It's important to note that the total cake is divided into three equal portions, so the sum of the portions received by all three individuals must add up to the entire cake. In other words:
x + 2x + 4x = 1
Adding the portions together, we get 7x = 1. To find the value of x, we can divide both sides of the equation by 7:
x = 1/7
Now that we have the value of x, we can determine the portion of the cake that you receive. You get 4 times the portion of your neighbor, so your portion would be 4 * (1/7), which simplifies to 4/7.
Therefore, you receive 4/7 of the cake as your portion.
In summary, to solve the problem:
1. Assume x as the portion received by your neighbor.
2. Determine the portions received by your friend and yourself by applying the given relationships (2x for your friend and 4x for yourself).
3. Set up an equation by summing up all three portions to equal 1 (since the cake is divided into three equal parts).
4. Solve the equation to find the value of x.
5. Calculate your own portion by multiplying your friend's portion by 4.
6. Simplify the fraction to attain the final proportion of the cake you receive.