Question

\(A\) and \(B\) can do a piece of work in 12 days, \(B\) and \(C\) in 15 days \(\mathrm{C}\) and \(\mathrm{A}\) in 20 days. In how many days can \(\mathrm{C}\) alone do it? (a) 60 (b) 50 (c) 25 (d) 24

Short answer

Answer: (a) 60 days.

Step-by-step solution

Write given information as equations

According to the problem statement: - A and B can do the work in 12 days, so \(A + B = 1/12\). - B and C can do the work in 15 days, so \(B + C = 1/15\). - C and A can do the work in 20 days, so \(C + A = 1/20\).

Add all three equations together

We'll now add all three equations to find the sum of the work done by A, B, and C: \((A + B) + (B + C) + (C + A) = 1/12 + 1/15 + 1/20\)

Simplify the equation

On simplifying the equation by adding A, B, and C, we get: \(2(A + B + C) = 1/12 + 1/15 + 1/20\) Divide both sides of the equation by 2: \(A + B + C = \frac{1}{2}\left(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\right)\) Now find the lcm of 12, 15, and 20 which is 60. So we can rewrite the equation as: \(A + B + C = \frac{1}{2}\left(\frac{5}{60}+\frac{4}{60}+\frac{3}{60}\right)\) Add the fractions in the parentheses: \(A + B + C = \frac{1}{2}\left(\frac{12}{60}\right)\) Now simplifying the equation: \(A + B + C = \frac{1}{10}\)

Find C's work done per day

We know that \(A + B + C = \frac{1}{10}\) and \(C + A =\frac{1}{20}\). Subtracting \(C + A =\frac{1}{20}\) from \(A + B + C = \frac{1}{10}\) to find B's work done per day: \(\frac{1}{10} -\frac{1}{20}= B\) \(B = \frac{1}{20}\) Substitute the value of B in the equation \(B + C = \frac{1}{15}\): \(\frac{1}{20} + C = \frac{1}{15}\) To solve for C, subtract \(\frac{1}{20}\) from both sides: \(C = \frac{1}{15} - \frac{1}{20}\) Find the lcm of 15 and 20 which is 60: \(C = \frac{4}{60} - \frac{3}{60}\) \(C = \frac{1}{60}\)

Find the days taken by C alone to complete the work

We have found that C's work done per day is \(\frac{1}{60}\). To find the number of days C alone takes to complete the work, we need to find the reciprocal of this fraction: Days taken by C alone = \(\frac{1}{C}\) Days taken by C alone = \(\frac{1}{\frac{1}{60}}\) Days taken by C alone = 60. So, C alone will take 60 days to finish the work. The correct answer is (a) 60.

Key concepts

Time and Work Aptitude

Time and work problems are a staple of quantitative aptitude tests and often feature in competitive exams and hiring assessments. These questions assess an individual’s ability to logically determine how long a task will take, often involving multiple workers or machines with varying rates of efficiency.

In such problems, including the exercise provided, the total work is typically considered as a single unit or job. The efficiency of workers is expressed in terms of the fraction of the work they can complete in a given time frame, commonly a day. Understanding the relationship between time, work, and the combined effort of multiple entities is crucial.

When tackling these problems, it's key to establish the work rate of each contributing factor—individuals or machines—and the combined rates when they work together. This forms the foundation for constructing and solving the necessary equations to find the answer.

Quantitative Aptitude

Quantitative aptitude encompasses a broad range of mathematical skills and problem-solving abilities. It includes arithmetic, algebra, geometry, and data interpretation, all of which form part of most standardized testing mechanisms used in academic and professional settings.

The exercise presented illustrates the practical application of quantitative aptitude in solving real-world problems. Cultivating this skill requires a methodical approach to mathematical concepts, as well as practice in analyzing and interpreting numerical data. A strong quantitative aptitude allows individuals to approach complex scenarios systematically and derive logical solutions efficiently.

Math Problem Solving

Math problem solving is an intellectual exercise that requires a combination of cognitive skills and systematic strategies. Key steps in solving math problems include understanding the problem, devising a plan, carrying out the plan, and finally reviewing the solution.

In the context of the given exercise, the problem-solving process involved creating equations from given information, manipulating these equations to determine individual contributions, and using logical reasoning to reach the solution. Critical to effective math problem solving is the ability to translate word problems into mathematical expressions, and then simplify and solve these expressions systematically.

LCM and Fraction Operations

The Least Common Multiple (LCM) is extensively used in solving time and work problems to handle multiple time frames and work rates. In essence, the LCM provides a common denominator that allows for the easy addition or subtraction of fractions representing different work rates or time periods.

In the exercise in question, LCM facilitated the combination of work rates of A, B, and C to determine their collective work rate. Similarly, the subtraction of fractions was streamlined using LCM to isolate C's work rate, exemplifying LCM's utility in fraction operations. Mastering LCM operations is a valuable skill for any student, reducing complex calculations to simpler arithmetic and ensuring that fractions can be easily compared and combined.