M23-5-MATHX-SP2-ENG-TZ1-XX

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Keycap Digit Two

Paper 2

Question 1 - MaximumMarks - 5

In a diagram depicting three cities, Jaipur, Lucknow, and Delhi, we observe the following: Lucknow is positioned 500 km directly east of Jaipur, and the angle at Jaipur between the lines to Lucknow and to Delhi is 25 degrees. Additionally, the bearing from Jaipur to Delhi is given as 034 degrees. How might we use this information in a geometric analysis?

Question 1 - Image

Question 1 - A

Calculate the value of $( J \hat{D} L )$, where (J), $( \hat{D} )$, and (L) are specific variables related to the given context.

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Question 1 - A - Answer

To find the angle JDL, we need to consider the bearing of D from J, which is 034&deg;. Since Lucknow (L) lies directly east of Jaipur (J), we can establish that JL is a line pointing due east, which by definition is at 90&deg; from north.

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The bearing is measured in a clockwise direction from north, so we need to calculate the internal angle at J. Subtracting the bearing of D from J from the angle due east gives us:

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90&deg; - 34&deg; = 56&deg;

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Therefore, the angle between JL and JD is 56&deg;. The question states that JLD = 25&deg;. Now, to find JDL, we use the fact that the sum of angles in a triangle is 180&deg;. We can write:

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180&deg; - JLD - JLD = JDL

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180&deg; - 25&deg; - 56&deg; = JDL

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180&deg; - 81&deg; = 99&deg;

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Thus,

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&ang;JDL = 99&deg;.

Question 1 - A - Answer - EC

The candidate correctly identifies that the bearing of D from J is 034&deg;, indicating a clear understanding of how bearings are measured. They have also correctly noted that since L is directly east of J, the line JL would have a bearing of 90&deg; from north.

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In calculating the internal angle at J, the candidate has correctly subtracted the bearing from the eastward bearing, finding that the angle JLD is 56&deg;. This demonstrates a proper understanding of the relationship between bearings and internal angles.

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The use of the triangle angle sum to find &ang;JDL is correctly applied, and the arithmetic is accurate, resulting in the correct answer of 99&deg;. The candidate&rsquo;s method is logical, clear, and follows a concise mathematical approach, thereby meeting the criteria for full marks according to the marking scheme.

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No marks were deducted as the candidate provided a full, clear, and correct solution to the problem posed, with all the required steps shown.

Question 1 - B

How far is it between Lucknow and Delhi?

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This question requires calculating the geographic distance between two major cities in India, namely Lucknow and Delhi. The distance referred to here is typically the shortest path over the earth&#39;s surface between these two urban centers.

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Question 1 - B - Answer

To determine the distance between Lucknow and Delhi using the sine rule, we consider a triangle where the line connecting Lucknow and Delhi represents one side of the triangle, and we have an angle and the length of the side opposite this angle. Assuming that the given angle and side length pertain to this scenario, we can apply the sine rule as follows:

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$\frac{DL}{ \sin(56°)} = \frac{500 }{ \sin(99°)}$

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First, we calculate the sine of the angles:

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sin(56&deg;) &asymp; 0.8290

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sin(99&deg;) &asymp; 0.9848

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Now, we apply the sine rule:

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$\frac{DL }{ 0.8290} = \frac{500 }{ 0.9848}$

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Multiplying both sides by 0.8290 to solve for DL:

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$DL = (\frac{500 }{ 0.9848}) × 0.8290 ≈ 420\ km$

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Therefore, the distance between Lucknow and Delhi is approximately 420 kilometers.

Question 1 - B - Answer - EC

The candidate has successfully applied the sine rule to find the distance between Lucknow and Delhi. The answer shows a clear understanding of the sine rule, which is used to find a side in a triangle when two angles and the side opposite one of those angles are known. The use of the sine values for the given angles is accurate, and the arithmetic is carried out correctly, leading to a final answer of 420 km for the distance between Lucknow and Delhi. This is the expected result as per the marking scheme provided.

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It is important to note that the candidate should ensure that the calculator is set to the correct mode (degrees, in this case) when computing trigonometric functions. As indicated in the marking scheme, an answer of 261 km would suggest the use of a calculator set to radians, which is a common mistake. However, in this instance, the candidate has correctly used degrees, resulting in the correct distance calculation.

Question 2 - MaximumMarks - 6

The question asks us to consider a model that describes how the value of a car decreases over time. Specifically, the value (C) of the car in U.S. dollars is given by a formula: $( C = 40,000 \times (0.91)^t )$, where (t) represents the time in years since January 1, 2023. This type of model is often used in finance to depict depreciation, reflecting how items lose value as they age.

Question 2 - A

What this question is asking is to find out how much value a car loses each year from its original purchase price. This is referred to as the &#39;depreciation rate&#39;. It&#39;s important because it tells us how quickly a car&#39;s value goes down over time, which can impact decisions related to buying, selling, or owning a car.

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Question 2 - A - Answer

The annual rate of depreciation of a car can be determined from the exponential decay function representing the car&rsquo;s value over time. The function provided is C = 40,000(0.91)t, where C represents the current value of the car in US dollars, and t is the time in years since 1 January 2023.

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In this function, 0.91 is the base of the exponent, which indicates that the car retains 91% of its value each year. To find the annual rate of depreciation, we subtract this percentage from 100%, as the total value at the beginning of each year is considered to be 100%. Thus, the annual rate of depreciation is 100% - 91% = 9%.

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Therefore, the car depreciates at an annual rate of 9%.

Question 2 - A - Answer - EC

The candidate&rsquo;s answer correctly identifies the exponential decay model C = 40,000(0.91)t for the depreciation of the car&#39;s value. The key step in their response is recognizing that the coefficient 0.91 represents the proportion of the car&rsquo;s value that is retained each year. By correctly subtracting this from 100%, the candidate has effectively determined the annual rate of depreciation, which is 9%.

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The clear presentation of the calculation and the logical explanation of the steps taken to arrive at the answer demonstrate a strong understanding of the concept of exponential decay in the context of depreciation. The response is concise, correctly formatted, and contains no extraneous information, which aligns well with the expectations of an IB Mathematics AA SL examination.

Question 2 - B

This question asks you to calculate the expected value of a car on a specific future date, which is January 1, 2028. Essentially, it involves projecting the car&rsquo;s depreciation or appreciation over time to arrive at its future value.

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Question 2 - B - Answer

The value of the car is determined by the function (C = 40,000(0.91)t), where (t) represents the number of years since 1 January 2023, and (C) is the value of the car in USD. To find the value of the car on 1 January 2028, we first calculate (t), the number of years from 2023 to 2028. This is (t = 2028 - 2023 = 5) years.

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Substituting (t = 5) into the function, we get:

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[C = 40,000(0.91)5]

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Calculating the value:

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[C = 40,000 &times; 0.915]

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[C = 40,000 &times; 0.624032]

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[C &asymp; 24,961.28]

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Therefore, the value of the car on 1 January 2028 is approximately $24,961.28. However, since the answer should be in dollars, it is reasonable to round this to $25,000.

Question 2 - B - Answer - EC

The candidate correctly identified the need to calculate the number of years from 2023 to 2028, correctly finding (t = 5). This is a crucial step and is often where students make their first mistake. The substitution of (t = 5) into the given depreciation function was done accurately.

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The calculation was carried out correctly, with appropriate use of the exponential function. The candidate demonstrated good numerical skills in handling the calculation, arriving at a precise answer of $24,961.28. It is important to note that the candidate recognized the context of the problem, which involves a real-world financial scenario, and thus rounded the answer to a more practical figure, $25,000. This shows an understanding that in financial contexts, exact cents are often not as relevant as the nearest dollar value.

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Overall, the response shows a clear understanding of exponential decay functions in a real-world context, accurate calculation skills, and the ability to communicate mathematical reasoning effectively. The rounding to the nearest dollar is justified and aligns with the expectations of the IB Mathematics AA SL marking scheme.

Question 2 - C

Alvie has initiated an investment plan to purchase a car on January 1, 2028. He began by depositing $15,000 on January 1, 2023, into a savings account that benefits from an annual interest rate of 3%, compounded annually. Over these five years, Alvie plans not to make any additional contributions or withdrawals from this account.

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On the target date, the total amount Alvie has accumulated in the account will still fall short of the car&rsquo;s price. He will need an additional amount, which we will refer to as (M), to complete the purchase. The task is to determine the exact figure of (M).

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Standard Level

Mathematics AA SL

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English Literature HL

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Question 3 - MaximumMarks - 5

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Question 7 - MaximumMarks - 13

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Question 8 - MaximumMarks - 17

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IB MATH AA SL, Paper 2, May, 2023, TZ1, Solved Past Paper