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The ratio of speeds of \(A\) is to \(B\) is \(2: 3\) and therefore \(A\) takes 20 minutes less time than \(B\) takes. What is the ratio of time taken by \(A\) and \(B ?\) (a) \(2: 3\) (b) \(2: 5\) (c) \(3: 2\) (d) \(3: 5\)

Short Answer

Expert verified
Answer: (c) \(3: 2\).

Step by step solution

01

Set up the formula

The formula for speed, distance, and time is: Speed = Distance / Time. Let the speeds of A and B be \(2x\) and \(3x\), respectively. Let the time taken by A be \(t_A\) and the time taken by B be \(t_B\). The distance (d) covered by both A and B is the same.
02

Write equations for the distances covered by A and B

From the formula Speed = Distance / Time, we get Distance = Speed * Time. Therefore, Distance covered by A (d) = \(2x * t_A\). And, Distance covered by B (d) = \(3x * t_B\).
03

Write the equation for the given condition

We are given that A takes 20 minutes less time than B. Hence, \(t_A = t_B - 20\).
04

Equate the distances and solve for the ratio of time taken by A and B

Since the distances covered by A and B are the same, we can equate the equations from Step 2: \(2x * t_A = 3x * t_B\). Now replace \(t_A\) with \(t_B - 20\) from Step 3: \(2x * (t_B - 20) = 3x * t_B\). Divide by x: \(2 * (t_B - 20) = 3 * t_B\). Expand: \(2t_B - 40 = 3t_B\). Subtract \(2t_B\) from both sides: \(-40 = t_B\). Now, we know \(t_A = t_B - 20\). Substitute the value of \(t_B\): \(t_A = -40 - 20\). \(t_A = -60\). Since \(t_A\) and \(t_B\) cannot be negative, we assume that the minus sign indicates the ratio and not the actual values. Therefore, the ratio of time taken by A and B is: \(3: 2\). The correct answer is (c) \(3: 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quantitative Aptitude
Quantitative aptitude encompasses the ability to reason logically and solve numerical and mathematical problems. This skill is crucial for various competitive exams, entrance tests, and aptitude assessments where speed and accuracy are vital. In the context of a speed-time ratio problem, like the one in the given exercise, quantitative aptitude involves the comprehension of the concepts of speed, distance, and time, as well as the mastery of ratios and their manipulation.

When faced with such problems, the key is to analyze and synthesize the given information effectively. For instance, converting the ratio of speeds into actionable information to determine the ratio of times taken. It's not just about applying formulas; it's about understanding relationships and developing strategies to simplify and solve equations expediently. The importance of this logical and structured thought process is especially highlighted when misinformation - such as a negative time value, occurs, indicating a need to reassess the contextual meaning of the numbers.
Speed Distance Time
The relationship between speed, distance, and time is a fundamental concept in physics, which also surfaces quite often in mathematical problems under 'quantitative aptitude'. As highlighted in the original exercise step by step solution, the formula connects these three variables: Speed = Distance / Time, which can also be rearranged as Distance = Speed * Time.

To tackle problems involving these elements, it's essential to understand that speed is the rate at which an object covers a distance over a given period of time. When two entities (like A and B in the exercise) have different speeds, and consequently different times to cover the same distance, you can discern the proportional relationship between the time taken and their speeds using the formulas mentioned. Clear comprehension and articulation of these concepts help in simplifying the problem-solving process. Problems may often require an additional step of algebra to express one variable in terms of the others before reaching a solution.
Ratio and Proportion
Ratio and proportion are powerful mathematical concepts used to compare quantities and describe their relationship. They are not just figures; they express the essence of the relative magnitude or frequency between comparable entities.

In the given speed-time ratio problem, understanding the ratio of speeds helps us derive the ratio of times, as the relative speeds dictate how long each entity will take to cover the same distance. Mastery of ratios is crucial when trying to solve such problems as it challenges one not only to use the ratios directly given but also to infer other ratios, like the ratio of times in this case, through logical reasoning and algebraic manipulation. It's important to notice that ratios express a relative measure and do not necessarily represent literal values, which is why the negative solution in the exercise is translated into a valid ratio instead.

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Most popular questions from this chapter

T h e ~ r a t i o ~ o f ~ s p e e d s ~ at which Anil and Mukesh walk is 3: 4. Anil takes 30 minutes more than the time taken by Mukesh in reaching the destination. What is the total distance travelled by each of thom, if the average of speeds of Anil and Mukesh is \(28 \mathrm{~km} / \mathrm{h} ?\) (a) 48 (b) 60 (c) \(17 \mathrm{~km}\) (d) 70

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The length of Lucknow mail is \(120 \mathrm{~m}\) and that of Punjab mail is \(80 \mathrm{~m}\). These two trains are running in the same direction with velocities of \(40 \mathrm{~km} / \mathrm{h}\) and \(50 \mathrm{~km} / \mathrm{h}\) respectively. The time taken by them to cross each other is : (a) \(8 \mathrm{~s}\) (b) \(72 \mathrm{~s}\) (c) \(11.5 \mathrm{~s}\) (d) \(12.5 \mathrm{~s}\)

A man can row \(15 \mathrm{~km} / \mathrm{h}\) in still water and he finds that it takes him twice as much time ro row up than as to row dova the same distance in the river. The speed of the current (in \(\mathrm{km} / \mathrm{h}\) ) is : (a) \(6 \mathrm{~km} / \mathrm{h}\) (b) \(6.5 \mathrm{~km} / \mathrm{h}\) (c) \(4.5 \mathrm{~km} / \mathrm{h}\) (d) \(5 \mathrm{~km} / \mathrm{h}\)

Roorkee express normally reaches its destination at \(50 \mathrm{~km} / \mathrm{h}\) in 30hrs .Find the speed at which it travels to reduce the time by 10hrs? a. 38km/hr b. 76km/hr c.75km/hr d. 60km/hr

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