Answer

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Hint: In this question, we first need to know about the basic definitions of algebra. Then calculate the time difference between A and B and then calculate the distance B can travel in that time which gives us the value of start that A can give B.

Complete step-by-step answer:

Let us look at the basic definitions of algebra.

LINEAR EQUATIONS:

Equation: A statement of equality of two algebraic expressions involving two or more unknown variables is called equation.

Linear Equation: An equation involving the variables in maximum of order 1 is called a linear equation. Graph of a linear equation is a straight line.

Linear equation in one variable is of the form \[ax+b=0\].

Linear equation in two variables is of the form \[ax+by+c=0\] .

Solution of an Equation- A particular set of values of the variables, which when substituted for the variables in the equation makes the two sides of the equation equal, is called the solution of the equation.

Simultaneous Linear Equation: A set of linear equations in two variables is said to form a system of simultaneous linear equations, if both equations have the same solution.

Consistency of Simultaneous Linear Equation: If a system of simultaneous linear equations has at least one solution, then the system of linear equations is called consistent.

Inconsistency of Simultaneous Linear Equation: If a system of simultaneous linear equations has at least no solution, then the system of linear equations is called inconsistent.

Now, let us assume that the time difference between A and B as t and the distance B can travel in t as x.

conversion of minutes to seconds:

1 minute = 60 seconds.

\[\begin{align}

& \Rightarrow t=5\times 60-\left( \left( 4\times 60 \right)+54 \right) \\

& \Rightarrow t=300-\left( 240+54 \right) \\

& \Rightarrow t=300-294 \\

& \therefore \text{t}=6s \\

\end{align}\]

conversion of kilometre to metre:

1 kilometre = 1000 metre

Given that B can travel 1000 m in 300 sec.

\[\begin{align}

& \Rightarrow \dfrac{x}{6}=\dfrac{1000}{300} \\

& \Rightarrow x\times 3=10\times 6 \\

& \Rightarrow x=\dfrac{60}{3} \\

& \therefore x=20m \\

\end{align}\]

Hence, the correct option is (c).

Note:

Instead of relating the time and distance using the cross multiplication method we can also say that the velocity of the person remains the same. So, by using the velocity, distance and time relation we can get an equation for velocity in both the cases and equate them which also gives the same result.

Conversion from minutes to seconds and kilometre to metre should be done appropriately because neglecting any of the values or wrong conversion changes the equation so formed. Then when we solve for the unknown value we get different results.

Complete step-by-step answer:

Let us look at the basic definitions of algebra.

LINEAR EQUATIONS:

Equation: A statement of equality of two algebraic expressions involving two or more unknown variables is called equation.

Linear Equation: An equation involving the variables in maximum of order 1 is called a linear equation. Graph of a linear equation is a straight line.

Linear equation in one variable is of the form \[ax+b=0\].

Linear equation in two variables is of the form \[ax+by+c=0\] .

Solution of an Equation- A particular set of values of the variables, which when substituted for the variables in the equation makes the two sides of the equation equal, is called the solution of the equation.

Simultaneous Linear Equation: A set of linear equations in two variables is said to form a system of simultaneous linear equations, if both equations have the same solution.

Consistency of Simultaneous Linear Equation: If a system of simultaneous linear equations has at least one solution, then the system of linear equations is called consistent.

Inconsistency of Simultaneous Linear Equation: If a system of simultaneous linear equations has at least no solution, then the system of linear equations is called inconsistent.

Now, let us assume that the time difference between A and B as t and the distance B can travel in t as x.

conversion of minutes to seconds:

1 minute = 60 seconds.

\[\begin{align}

& \Rightarrow t=5\times 60-\left( \left( 4\times 60 \right)+54 \right) \\

& \Rightarrow t=300-\left( 240+54 \right) \\

& \Rightarrow t=300-294 \\

& \therefore \text{t}=6s \\

\end{align}\]

conversion of kilometre to metre:

1 kilometre = 1000 metre

Given that B can travel 1000 m in 300 sec.

\[\begin{align}

& \Rightarrow \dfrac{x}{6}=\dfrac{1000}{300} \\

& \Rightarrow x\times 3=10\times 6 \\

& \Rightarrow x=\dfrac{60}{3} \\

& \therefore x=20m \\

\end{align}\]

Hence, the correct option is (c).

Note:

Instead of relating the time and distance using the cross multiplication method we can also say that the velocity of the person remains the same. So, by using the velocity, distance and time relation we can get an equation for velocity in both the cases and equate them which also gives the same result.

Conversion from minutes to seconds and kilometre to metre should be done appropriately because neglecting any of the values or wrong conversion changes the equation so formed. Then when we solve for the unknown value we get different results.

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