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Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
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Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


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Word Problems Integrating Multiple Concepts Numerical Problems from Applied Contexts (e.g., Alligation and Mixture) Solving Miscellaneous Quantitative Problems


General Quantitative Applications



Word Problems Integrating Multiple Concepts

Word problems are a fundamental component of quantitative aptitude sections in competitive examinations. Integrated word problems are designed to assess your ability to identify and apply multiple mathematical concepts within a single scenario. These problems often weave together ideas from different topics such as percentages, profit and loss, ratio and proportion, time and work, speed-distance-time, simple and compound interest, averages, etc.


Nature of Integrated Word Problems:

Solving integrated word problems requires a holistic understanding and the ability to connect seemingly disparate mathematical concepts. Key aspects include:


Strategy for Solving Integrated Word Problems:

A systematic strategy is crucial to tackle these multi-concept problems effectively:

1. Read and Comprehend Thoroughly:

Read the problem statement carefully, perhaps multiple times. Identify exactly what information is given and precisely what question is being asked. Get a clear picture of the scenario.

2. Identify the Concepts and Quantities:

Analyze the language used in the problem to pinpoint the mathematical concepts involved (e.g., percentages indicate likely use of percentage change, profit/loss, discount; mentions of time and distance suggest speed calculations). List the known quantities and the unknown quantities you need to find.

3. Plan the Solution Steps:

Outline the sequence of calculations needed to move from the given information to the required answer. Determine which intermediate values you need to calculate (e.g., first find Marked Price, then Selling Price, then Profit, then Profit Percentage).

4. Translate to Mathematical Terms:

Formulate the necessary equations or expressions based on the identified concepts and your planned steps. Assign variables to unknown quantities if using an algebraic approach.

5. Execute Calculations Step-by-Step:

Perform the calculations according to your plan. Be careful with arithmetic and formula application at each step.

6. Ensure Unit Consistency:

Throughout the calculations, maintain consistency in units. Convert units where necessary (e.g., time in minutes to hours, speed in km/h to m/s) before applying formulas.

7. Review and Verify the Answer:

Once you arrive at a final answer, check if it is in the required format and units. Mentally (or quickly on paper) verify if the answer is reasonable in the context of the problem and if it satisfies all the initial conditions.


Example 1. A shopkeeper bought a mobile phone for $\textsf{₹ } 12000$. He marked up the price by 25% and then sold it after giving a discount of 10%. What is the shopkeeper's profit percentage?

Answer:

This problem integrates concepts of Cost Price (CP), Markup Percentage, Marked Price (MP), Discount Percentage, Selling Price (SP), Profit, and Profit Percentage.

Given:

Cost Price (CP) $= \textsf{₹ } 12000$.

Markup Percentage $= 25\%$ (on CP).

Discount Percentage $= 10\%$ (on MP).

To Find:

Shopkeeper's Profit Percentage.

Solution Steps:

1. Calculate the Marked Price (MP) from CP and Markup %.

2. Calculate the Selling Price (SP) from MP and Discount %.

3. Calculate the Profit amount from SP and CP.

4. Calculate the Profit Percentage based on CP.

Step 1: Calculate Marked Price (MP)

The shopkeeper marked up the price by 25% of the Cost Price. Using the formula $\text{MP} = \text{CP} \times \left(1 + \frac{\text{Markup } \%}{100}\right)$:

MP $= \textsf{₹ } 12000 \times \left(1 + \frac{25}{100}\right) = \textsf{₹ } 12000 \times \left(1 + \frac{1}{4}\right)$

MP $= \textsf{₹ } 12000 \times \frac{5}{4} = \textsf{₹ } \cancel{12000}^{\normalsize 3000} \times \frac{5}{\cancel{4}^{\normalsize 1}} = 3000 \times 5 = 15000$

MP $= \textsf{₹ } 15000$.

Step 2: Calculate Selling Price (SP)

A discount of 10% is given on the Marked Price. Using the formula $\text{SP} = \text{MP} \times \left(1 - \frac{\text{Discount } \%}{100}\right)$:

SP $= \textsf{₹ } 15000 \times \left(1 - \frac{10}{100}\right) = \textsf{₹ } 15000 \times \left(1 - \frac{1}{10}\right)$

SP $= \textsf{₹ } 15000 \times \frac{9}{10} = \textsf{₹ } \cancel{15000}^{\normalsize 1500} \times \frac{9}{\cancel{10}^{\normalsize 1}} = 1500 \times 9 = 13500$

SP $= \textsf{₹ } 13500$.

Step 3: Calculate Profit Amount

Profit is the difference between Selling Price and Cost Price. CP $= \textsf{₹ } 12000$, SP $= \textsf{₹ } 13500$.

Profit $= \text{SP} - \text{CP} = \textsf{₹ } 13500 - \textsf{₹ } 12000 = \textsf{₹ } 1500$.

Step 4: Calculate Profit Percentage

Profit Percentage is calculated on the Cost Price. Using the formula $\text{% Profit} = \left(\frac{\text{Profit}}{\text{CP}} \times 100\right)\%$:

% Profit $= \left(\frac{\textsf{₹ } 1500}{\textsf{₹ } 12000} \times 100\right)\%$

Simplify the fraction and calculate:

$\frac{1500}{12000} = \frac{15}{120} = \frac{1}{8}$

% Profit $= \left(\frac{1}{8} \times 100\right)\% = 12.5\%$

The shopkeeper's profit percentage is $\boldsymbol{12.5\%}$.


Competitive Exam Notes:

Integrated word problems are designed to test your ability to combine concepts. Practice is essential to develop a systematic approach.

  • Identify Concepts: Break down the problem into parts involving different concepts (e.g., markup is % increase on CP, discount is % decrease on MP, profit is difference between SP and CP).
  • Map the Flow: Understand the sequence: CP $\to$ Markup $\to$ MP $\to$ Discount $\to$ SP. Then SP and CP determine Profit/Loss.
  • Step-by-step Calculation: Solve one concept at a time. The result of one step (e.g., MP) becomes the starting point for the next (e.g., calculating discount on MP).
  • Use Formulas: Utilize the formulas for percentage change, profit/loss, and discount to efficiently calculate values at each step.
  • Alternative Approaches: Sometimes, a direct formula (like $\frac{\text{MP}}{\text{CP}} = \frac{100 + \%P}{100 - \%D}$) can be used as a shortcut in integrated problems, but understanding the step-by-step process is fundamental.
  • Verification: Quickly check if the answer makes sense in the context of the problem. A profit percentage of 12.5% seems reasonable given a 25% markup and 10% discount.


Numerical Problems from Applied Contexts (e.g., Alligation and Mixture)

Quantitative aptitude problems are often presented in real-world scenarios to test your ability to translate practical situations into mathematical problems. These problems may involve applying concepts like percentages, ratios, averages, or profit/loss within a specific context. Alligation and Mixture is a classic example of such an applied context. It deals with combining two or more substances with different properties (such as cost, concentration, density, average marks, speed, etc.) and determining the property of the resulting mixture or the ratio in which the substances were combined.


Alligation and Mixture Concepts:

Mixture: A mixture is formed by combining two or more different ingredients or substances.

Alligation: Alligation is a rule or method that helps solve problems related to mixtures. It is particularly useful for:

1. Finding the ratio in which two ingredients with different unit values (cost, percentage, etc.) should be mixed to obtain a mixture with a desired mean unit value.

2. Finding the mean unit value of a mixture when two ingredients are mixed in a given ratio.

The rule of alligation works best when combining two ingredients (though it can be extended). Let's consider mixing two ingredients, Ingredient 1 and Ingredient 2. Assume:

We typically assume that $C_M$ lies between $C_1$ and $C_2$ (i.e., $C_1 < C_M < C_2$ or $C_2 < C_M < C_1$).

The Alligation Rule:

The rule states that the ratio of the quantity of Ingredient 1 to the quantity of Ingredient 2 is inversely proportional to the difference between the mean unit value of the mixture and the unit values of the individual ingredients. Specifically, the ratio is given by the differences subtracted diagonally.

Quantity of Ingredient 1 / Quantity of Ingredient 2 $= (|C_M - C_2|) / (|C_M - C_1|)$

Assuming $C_1 < C_M < C_2$, this becomes:

$\frac{\text{Quantity of Ingredient 1}}{\text{Quantity of Ingredient 2}} = \frac{C_2 - C_M}{C_M - C_1}$

... (i)

Derivation of the Alligation Rule:

Let $Q_1$ be the quantity of Ingredient 1 and $Q_2$ be the quantity of Ingredient 2.

Total quantity of the mixture is $Q_M = Q_1 + Q_2$.

The total value contributed by Ingredient 1 is $Q_1 \times C_1$.

The total value contributed by Ingredient 2 is $Q_2 \times C_2$.

The total value of the mixture is $Q_M \times C_M = (Q_1 + Q_2) \times C_M$.

Based on the principle of conservation (e.g., total cost of ingredients equals total cost of mixture, total amount of pure substance equals total amount in mixture), the sum of the values of the ingredients must equal the value of the mixture:

$\boldsymbol{(Q_1 \times C_1) + (Q_2 \times C_2) = (Q_1 + Q_2) \times C_M}$

Expand the right side:

$(Q_1 \times C_1) + (Q_2 \times C_2) = (Q_1 \times C_M) + (Q_2 \times C_M)$

Rearrange the terms to group $Q_1$ terms on one side and $Q_2$ terms on the other:

$(Q_1 \times C_1) - (Q_1 \times C_M) = (Q_2 \times C_M) - (Q_2 \times C_2)$

Factor out $Q_1$ and $Q_2$:

$\boldsymbol{Q_1 (C_1 - C_M) = Q_2 (C_M - C_2)}$

Rearrange to find the ratio $Q_1 / Q_2$:

$\frac{Q_1}{Q_2} = \frac{C_M - C_2}{C_1 - C_M}$

When using the differences in the ratio, we take the absolute values to represent quantities (which are non-negative). The standard representation in alligation swaps the positions in the numerator and denominator to give positive differences:

$\frac{Q_1}{Q_2} = \frac{|C_M - C_2|}{|C_M - C_1|}$

This leads to the diagrammatic method.

Diagrammatic Representation (Alligation Cross Method):

This is a visual shortcut for applying the alligation rule for two ingredients:

Draw a cross. Place the unit value of the first ingredient ($C_1$) on the top-left and the unit value of the second ingredient ($C_2$) on the top-right. Place the mean unit value of the mixture ($C_M$) in the centre. Subtract diagonally (smaller value from larger) and write the absolute difference at the bottom corners. The values at the bottom corners represent the ratio of the quantities of the ingredients.

Ingredient 1 Value ($C_1$)

Ingredient 2 Value ($C_2$)

Mean Value ($C_M$)

Difference $(|C_M - C_2|)$

Difference $(|C_M - C_1|)$

($\propto$ Quantity of Ingredient 1)

($\propto$ Quantity of Ingredient 2)

The ratio of the quantity of Ingredient 1 to the quantity of Ingredient 2 is $(|C_M - C_2|) : (|C_M - C_1|)$.

$\frac{\text{Quantity of Ingredient 1}}{\text{Quantity of Ingredient 2}} = \frac{|C_M - C_2|}{|C_M - C_1|}$

Note: All values ($C_1, C_2, C_M$) must be in the same units. This method works for any property that is proportional to quantity (like cost, percentage concentration, average, speed, etc.).


Example 1. In what ratio must tea at $\textsf{₹ } 120$ per kg be mixed with tea at $\textsf{₹ } 160$ per kg so that the mixture costs $\textsf{₹ } 135$ per kg?

Answer:

Given: Unit cost of 1st type of tea ($C_1$) $= \textsf{₹ } 120$/kg.

Unit cost of 2nd type of tea ($C_2$) $= \textsf{₹ } 160$/kg.

Desired mean unit cost of the mixture ($C_M$) $= \textsf{₹ } 135$/kg.

We need to find the ratio of the quantity of the first type of tea to the quantity of the second type of tea in the mixture.

Using the Alligation Cross Method:

Cost of 1st Tea ($\textsf{₹}/kg$)

Cost of 2nd Tea ($\textsf{₹}/kg$)

120

160

Mean Cost ($\textsf{₹}/kg$)

135

Difference $(|135 - 160| = 25)$

Difference $(|135 - 120| = 15)$

($\propto$ Qty of 1st Tea)

($\propto$ Qty of 2nd Tea)

The ratio of the quantity of the first type of tea to the quantity of the second type of tea is the ratio of the values at the bottom corners:

Ratio of Quantity 1 : Ratio of Quantity 2 $= 25 : 15$.

Simplify the ratio by dividing both terms by their GCD (5):

Ratio $= \frac{25}{5} : \frac{15}{5} = 5 : 3$.

The two types of tea must be mixed in the ratio $\boldsymbol{5:3}$.


Example 2. A vessel contains 50 litres of milk. 10 litres of water is added to it. What is the percentage of milk in the resulting mixture?

Answer:

Given: Initial quantity of milk $= 50$ litres.

Quantity of water added $= 10$ litres.

We need to find the percentage of milk in the final mixture.

The final mixture is formed by combining the initial milk and the added water.

Total quantity of the mixture $= \text{Quantity of Milk} + \text{Quantity of Water}$

Total Mixture $= 50 \text{ litres} + 10 \text{ litres} = 60 \text{ litres}$.

The quantity of milk in the mixture is the initial quantity of milk, as no milk was removed and water contains no milk.

Quantity of milk in mixture $= 50$ litres.

The percentage of milk in the mixture is the quantity of milk divided by the total quantity of the mixture, multiplied by 100%.

$\text{Percentage of Milk} = \frac{\text{Quantity of Milk in Mixture}}{\text{Total Quantity of Mixture}} \times 100\%$

$\text{Percentage of Milk} = \frac{50 \text{ litres}}{60 \text{ litres}} \times 100\%$

Simplify the fraction and calculate:

$\frac{50}{60} = \frac{5}{6}$

Percentage of Milk $= \frac{5}{6} \times 100\% = \frac{500}{6}\% = \frac{250}{3}\%$

$\frac{250}{3} = 83\frac{1}{3}$

Percentage of Milk $= 83\frac{1}{3}\%$.

The percentage of milk in the resulting mixture is $\boldsymbol{83\frac{1}{3}\%}$.

Alternative Method (Using Alligation):

We are mixing two substances based on their milk percentage:

  • Ingredient 1: Milk (100% milk content). Quantity $= 50$ litres.
  • Ingredient 2: Water (0% milk content). Quantity $= 10$ litres.

Let the percentage of milk in the mixture be $x\%$.

Using the Alligation Cross Method with percentage of milk as the unit value:

Milk (100%)

Water (0%)

100

0

Mean Percentage ($x$)

$x$

The differences are $|x - 0| = x$ and $|x - 100| = 100 - x$ (since $x$ will be between 0 and 100).

Difference $(x - 0) = x$

Difference $(100 - x)$

($\propto$ Qty of 100% Milk)

($\propto$ Qty of 0% Milk)

The ratio of Quantity of 100% Milk : Quantity of 0% Milk (Water) is $x : (100 - x)$.

We are given this ratio is 50 litres : 10 litres $= 5 : 1$.

$\frac{x}{100 - x} = \frac{5}{1}$

Solve for $x$ using cross-multiplication:

$\boldsymbol{x \times 1 = 5 \times (100 - x)}$

$\boldsymbol{x = 500 - 5x}$

Gather $x$ terms on one side:

$\boldsymbol{x + 5x = 500}$

$\boldsymbol{6x = 500}$

$\boldsymbol{x = \frac{500}{6} = \frac{250}{3}}$

So the percentage of milk in the mixture is $x\% = \frac{250}{3}\% = 83\frac{1}{3}\%$. Both methods provide the same result.


Competitive Exam Notes:

Alligation and Mixture problems are common applications of ratio and weighted average concepts. The Alligation Cross Method is a powerful tool for solving these problems quickly.

  • Identify the Unit Value: Determine what property (cost, percentage, etc.) is being mixed. Ensure units are consistent.
  • Alligation Rule: It applies when mixing two quantities with different unit values to get a mixture with a mean unit value. The ratio of quantities is the ratio of the *differences* from the mean value (diagonal differences).
  • Alligation Cross Method: Use the diagram to visually set up and calculate the ratio of quantities: $\frac{\text{Qty}_1}{\text{Qty}_2} = \frac{|C_M - C_2|}{|C_M - C_1|}$.
  • Mean Value Calculation: If ratios and individual values are given, the mean value is the weighted average: $C_M = \frac{(Q_1 C_1) + (Q_2 C_2)}{Q_1 + Q_2}$. This is equivalent to the definition of weighted average.
  • Percentage/Concentration Problems: These are common applications. Treat percentage of a substance as the unit value (e.g., % milk, % acid). The 'other' substance might be 0%.
  • Repeated Dilution/Removal and Replacement: Problems where a part of the mixture is removed and replaced with another substance require step-by-step calculation for each removal/replacement cycle.


Solving Miscellaneous Quantitative Problems

Miscellaneous quantitative problems encompass a wide array of question types that do not always fit into the clearly defined categories like percentages, profit & loss, time & work, etc. These problems draw upon fundamental mathematical skills, logical thinking, and the ability to apply concepts flexibly to solve novel scenarios. They might involve elements of number theory, basic algebra or geometry presented as word problems, pattern recognition in sequences, or puzzles requiring numerical logic.


Characteristics of Miscellaneous Problems:


Approach to Solving Miscellaneous Problems:

A flexible and analytical approach is best suited for these problems:

1. Understand the Problem Completely:

Read the problem statement multiple times to ensure you understand the scenario, all given conditions, and exactly what is being asked. Identify key terms and quantities.

2. Break Down the Problem:

If the problem feels complex, divide it into smaller, more manageable parts. Address each part separately if possible.

3. Identify Underlying Mathematical Principles:

Think about which core mathematical ideas (e.g., ratios, differences, multiples, linear relationships) are relevant to the problem's components.

4. Look for Relationships and Patterns:

Examine the given numbers and conditions for any inherent relationships or patterns that can guide the solution.

5. Plan Your Solution:

Based on your understanding and identified concepts, formulate a plan or a sequence of steps to solve the problem. Decide whether setting up equations, using proportions, or applying a specific theorem is the best approach.

6. Execute the Plan:

Carry out your planned steps, performing calculations carefully. If one approach doesn't seem to be working, be prepared to rethink and try another method.

7. Check and Verify:

Once you arrive at an answer, check if it satisfies all the conditions given in the problem statement and if it makes logical sense in the problem context.


Example 1. The sum of two numbers is 75 and their difference is 25. Find the two numbers.

Answer:

This is a classic word problem that can be solved using simple algebra (system of linear equations).

Given:

Sum of two numbers $= 75$.

Difference of the same two numbers $= 25$.

To Find:

The values of the two numbers.

Solution:

Let the two numbers be $x$ and $y$. Let's assume $x$ is the larger number, so $x > y$.

We can translate the given information into two linear equations:

Equation 1 (Sum): $\boldsymbol{x + y = 75}$

... (i)

Equation 2 (Difference): $\boldsymbol{x - y = 25}$

... (ii)

We can solve this system of equations using the elimination method. Adding equation (i) and equation (ii) will eliminate $y$:

$(x + y) + (x - y) = 75 + 25$

$x + y + x - y = 100$

$\boldsymbol{2x = 100}$

Solve for $x$:

$\boldsymbol{x = \frac{100}{2} = 50}$

Now substitute the value of $x$ (50) into either equation (i) or (ii) to find $y$. Let's use equation (i):

$50 + y = 75$

Solve for $y$:

$\boldsymbol{y = 75 - 50 = 25}$

The two numbers are $\boldsymbol{50}$ and $\boldsymbol{25}$.

Verification:

Sum of the numbers: $50 + 25 = 75$. (Correct)

Difference of the numbers: $50 - 25 = 25$. (Correct)


Example 2. A sum of money triples itself at simple interest in 15 years. In how many years will it become 7 times itself at the same rate of simple interest?

Answer:

This problem involves the concept of Simple Interest (SI) and how principal grows over time.

Given:

A sum of money (Principal, P) triples itself in 15 years at Simple Interest.

Amount ($A_1$) after $T_1 = 15$ years $= 3P$.

To Find:

The time ($T_2$) when the sum becomes 7 times itself at the same simple interest rate.

Amount ($A_2$) $= 7P$.

Solution:

The formula for Simple Interest (SI) is: $\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$, where R is the annual rate.

The Amount (A) is given by $A = P + \text{SI}$.

In the first case, the amount becomes 3 times the principal ($A_1 = 3P$) in $T_1 = 15$ years.

The Simple Interest earned in the first case is $SI_1 = A_1 - P = 3P - P = 2P$.

Substitute $SI_1 = 2P$ and $T_1 = 15$ into the SI formula to find the rate R:

$\boldsymbol{2P = \frac{P \times R \times 15}{100}}$

... (iii)

Assuming $P \neq 0$, we can cancel P from both sides of equation (iii):

$\boldsymbol{2 = \frac{R \times 15}{100}}$

Solve for R:

$\text{R} = \frac{2 \times 100}{15} = \frac{200}{15} = \frac{40}{3}$

$\boldsymbol{R = \frac{40}{3}\% \text{ per annum}}$

Now, in the second case, the amount becomes 7 times the principal ($A_2 = 7P$) at the same rate $R = \frac{40}{3}\%$. We need to find the time $T_2$.

The Simple Interest earned in the second case is $SI_2 = A_2 - P = 7P - P = 6P$.

Substitute $SI_2 = 6P$ and $R = \frac{40}{3}$ into the SI formula with time $T_2$:

$\boldsymbol{6P = \frac{P \times \frac{40}{3} \times T_2}{100}}$

... (iv)

Assuming $P \neq 0$, cancel P from both sides:

$\boldsymbol{6 = \frac{\frac{40}{3} \times T_2}{100}}$

$\boldsymbol{6 = \frac{40 \times T_2}{3 \times 100} = \frac{40 \times T_2}{300} = \frac{4 \times T_2}{30} = \frac{2 \times T_2}{15}}$

Solve for $T_2$:

$\boldsymbol{6 \times 15 = 2 \times T_2}$

$\boldsymbol{90 = 2 \times T_2}$

$\boldsymbol{T_2 = \frac{90}{2} = 45}$ years.

It will take $\boldsymbol{45}$ years for the sum to become 7 times itself at the same rate of simple interest.

Alternative Method (Using Relationship for SI Growth):

In Simple Interest, the interest earned is proportional to the time ($SI \propto T$) for a fixed principal and rate. Also, the total increase in the amount ($A-P$, which is SI) is proportional to the time.

Initial increase: Amount becomes 3P, so SI = 2P. This takes 15 years.

Required increase: Amount becomes 7P, so SI = 6P.

The interest required is 3 times the initial interest ($6P = 3 \times 2P$). Since SI is proportional to time, the time taken will also be 3 times the initial time.

Time taken ($T_2$) $= (\text{Ratio of SIs}) \times T_1 = \left(\frac{6P}{2P}\right) \times 15 \text{ years}$

$\boldsymbol{T_2 = 3 \times 15 = 45}$ years.

This shortcut works because Simple Interest grows linearly. The amount of interest for 7 times the principal (6P) is 3 times the amount of interest for 3 times the principal (2P). So, it will take 3 times the time.


Competitive Exam Notes:

Miscellaneous problems test your core mathematical understanding and problem-solving skills across various topics. Don't be intimidated if a problem doesn't fit a standard template; break it down and use fundamentals.

  • Understand the Question: Identify the core task, given information, and constraints.
  • Break Down Complex Problems: Simplify into smaller, solvable parts.
  • Connect Concepts: Recognize how different mathematical ideas (e.g., percentages and equations, or simple interest properties) come together in the problem.
  • Formulate Equations: Translate verbal descriptions into mathematical relationships.
  • Use Fundamentals: Often, solving involves applying basic arithmetic, algebra (like solving linear equations), or core formulas from standard topics.
  • Look for Patterns/Relationships: In problems involving series or growth, identify the underlying pattern (arithmetic, geometric, linear, exponential).
  • Alternative Methods: For problems spanning concepts (like SI growth), look for shortcuts or proportional relationships that might simplify calculations.