Once Xavier decides that he would like to tackle his own house construction, he begins to get his plans in order. The lot Xavier purchased many years ago is just perfect for the new, prefabricated house he wants to build for his family. To accomplish this large task, Xavier is going to need to consult with people from all three career clusters that we have considered in this unit—and you’re going to help! This activity will be broken up into two parts, each of which have their own assignment: Part 1 will ask you to consider aspects of Xavier’s building project and complete math problems that help him make decisions about what to build. Part 2 will have you take the answers to the math problems you completed in Part 1 and build a Job Board to keep Xavier organized during his construction project. Let’s walk through this process with Xavier. Part 1: The Math Behind Construction For the design stage, Xavier hires a surveyor, Bradley, to look at his lot and decide what size house he would be able to have on the property. Bradley sees that Xavier’s lot is the shape of a trapezoid, with the boundary line (or back edge of the lot) extending 100 feet from the front property line: image On your math sheet, complete the following problems to help Xavier find out how big a house he can build. Math Problem 1: What is the total area of Xavier’s lot? Use the formula for calculating the area of a trapezoid to help answer your question: A=a+b2c or front of lot+back of lot2length of lot Once Bradley has the total area of the lot, he explains to Xavier that the county only allows for 40 percent of any lot to be covered with buildings. This is in order to preserve the natural surroundings and allow water runoff. Bradley says that if a lot of land has too much house on it, there is little land for rain to drain, and that can cause flooding. Using the total lot area, calculate the available lot coverage that Xavier can use. This will give him an idea of how big of a house to look for. Math Problem 2: What is Xavier’s availabl
Answers
Answer:
1400
Explanation:
The sum is Rs.1400
Step-by-step explanation:
Let the principal be x
Case 1:
Time = 5 years
Rate of interest = 10%
Formula :SI = \frac{P \times T \times R}{100}SI=
100
P×T×R
SI=\frac{x \times 5 \times 10}{100}SI=
100
x×5×10
Case 2 :
Time = 4 years
Rate of interest = 16%
SI = \frac{x \times 4 \times 16}{100}SI=
100
x×4×16
We are given that The simple interest on a certain sum of money for 5 years at 10% per annum is Rs 196 less than the simple interest on the same sum for 4 years at 16% per annum.
\begin{gathered}\frac{x \times 4 \times 16}{100}-\frac{x \times 5 \times 10}{100}=196\\\frac{x}{100} (4 \times 16 - 5 \times 10)=196\\x=1400\end{gathered}
100
x×4×16
−
100
x×5×10
=196
100
x
(4×16−5×10)=196
x=1400
Hence the sum is Rs.1400
#Learn more:
The simple interest on a certain sum of money for 5 years at 10% per annum is Rs 196 less than the simple interest on the same sum for 4 years at 16% per annum. Find the sum.
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