# Ex.3.5 Q4 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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## Question

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student $$A$$ takes food for $$20$$ days, she has to pay ₹ $$1000$$ as hostel charges whereas a student $$B,$$ who takes food for $$1000$$ days, pays ₹ $$1180$$ as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes $$13$$ when $$1$$ is subtracted from the numerator and it becomes $$14$$ when $$8$$ is added to its denominator. Find the fraction.

(iii) Yash scored $$40$$ marks in a test, getting $$3$$ marks for each right answer and losing $$1$$ mark for each wrong answer. Had $$4$$ marks been awarded for each correct answer and $$2$$ marks been deducted for each incorrect answer, then Yash would have scored $$50$$ marks. How many questions were there in the test?

(iv) Places $$A$$ and $$B$$ are $$100\,\rm{ km}$$ apart on a highway. One car starts from $$A$$ and another from $$B$$ at the same time. If the cars travel in the same direction at different speeds, they meet in $$5$$ hours. If they travel towards each other, they meet in $$1$$ hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by $$9$$ square units, if its length is reduced by $$5$$ units and breadth is increased by $$3$$ units. If we increase the length by $$3$$ units and the breadth by $$2$$ units, the area increases by $$67$$ square units. Find the dimensions of the rectangle.

Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.5 | Question 4

## Text Solution

Reasoning:

Assume one variable equal to $$x$$ and another be $$y.$$ Then based on given conditions, two linear equations can be formed which can be easily solved.

Steps:

(i) Let $$x$$ be the fixed charge of the food and $$y$$ be the charge for food per day.

According to the given information,

When a student A, takes food for 20 days, pays ₹ $$1000$$ as hostel charges.

$x+20y=1000\qquad\ldots \left( 1 \right)$

When a student $$B,$$ who takes food for $$26$$ days, pays ₹ $$1180$$ as hostel charges.

$x+26y=1180 \qquad \ldots \left( 2 \right)$

Subtracting equation $$(1)$$ from equation $$(2),$$ we obtain

\begin{align}6y &= 180\\y &= \frac{{180}}{6}\\y &= 30\end{align}

Substituting $$y = 30$$ in equation $$(1),$$ we obtain

\begin{align}x + 20 \times 30 &= 1000\\x &= 1000 - 600\\x &= 400\end{align}

Equations are $$x + 20y = 1000$$ and $$x + 26y = 1180$$ where $$x$$ is the fixed charge of the food and $$y$$ is the charge for food per day

Hence, fixed charge is ₹ $$400$$

And charge per day is ₹ $$30$$

(ii) Let the numerator be $$x$$ and denominator be $$y,$$ thus the fraction be \begin{align} \frac{x}{y} \end{align}

According to the given information,

When $$1$$ is subtracted from the numerator

\begin{align}\frac{{x - 1}}{y}\, &= \frac{1}{3}\\3x - 3 &= y\\3x - y &= 3 \qquad \ldots (1)\end{align}

When $$8$$ is added to the denominator,

\begin{align}\frac{x}{{y + 8}}\, &= \frac{1}{4}\\4x &= y + 8\\4x - y &= 8 \qquad \ldots (2)\end{align}

Subtracting equation $$(1)$$ from equation $$(2),$$ we obtain

$x = 5$

Putting $$x = 5$$ in equation $$(1),$$ we obtain

\begin{align}3 \times 5-y &= 3\\y &= 15 - 3\\y& = 12 \end{align}

Equations are $$3x - y = 3$$ and $$4x - y = 8$$ where the numerator of the fraction is $$x,$$ and denominator is $$y.$$

Hence, the fraction is \begin{align} \frac{5}{{12}} \end{align}

(iii) Let the number of right answers and wrong answers be $$x$$ and $$y$$ respectively.

Therefore, total number of questions be $$\left( {x + y} \right)$$

According to the given information,

\begin{align}3x - y &= 40 \qquad \ldots (1)\\\\4x - 2y &= 50\\2x - y &= 25 \qquad \ldots (2)\end{align}

Subtracting equation $$(2)$$ from equation $$(1),$$ we obtain

$x = 15 \qquad \ldots (3)$

Substituting this in equation $$(2),$$ we obtain

\begin{align}2 \times 15 - y &= 25\\y &= 30 - 25\\y &= 5\end{align}

Equations are $$3x - y = 40$$ and $$2x - y = 25$$ where the number of right and wrong answers are $$x$$ and $$y$$ respectively.

number of right answers $$= 15$$ and number of wrong answers $$= 5$$

Hence, Total number of questions $$= 20$$

(iv) Let the speed of $$1^\rm{st}$$ car and $$2^\rm{nd}$$ car be $$u \text{ km/h}$$ and $$v\text{ km/h}$$ respectively.

According to the given information,

When the cars travel in the same direction at different speeds, they meet in $$5$$ hours.

Therefore, distance travelled by $$1^\rm{st}$$ car $$= 5u\,{\rm{km }}$$

and distance travelled by $$2^\rm{nd}$$ car $$= 5v\,{\rm{ }}km$$

\begin{align}5u - 5v &= 100\\5\left( {u - v} \right) &= 100\\u - v &= 20 \qquad \ldots (1)\end{align}

When the cars travel towards each other at different speeds, they meet in $$1$$ hour

therefore, distance travelled by $$1^\rm{st}$$ car $$= u\,{\rm{ km}}$$

and distance travelled by 2nd car $$= v\,{\rm{ }}km$$

$u + v = 100 \qquad \ldots (2)$

Adding both the equations, we obtain

\begin{align}2u &= 120\\u &= 60\end{align}

Substituting this value in equation (2), we obtain

\begin{align}60 + v &= 100\\v &= 40\end{align}

Equations are $$u - v = 20$$ and $$u+v=100$$ where the speed of $$1^\rm{st}$$ car and $$2^\rm{nd}$$ car be $$u\text{ km/h}$$ and $$v\text{ km/h}$$ respectively.

Hence, speed of the $$1^\rm{st}$$ car $$= 60\,{\rm{ }}km/h$$ and speed of the $$2^\rm{nd}$$ car $$= 40\,{\rm{ }}km/h$$

(v) Let length and breadth of rectangle be $$x$$ unit and $$y$$ unit respectively.

Then the area of the rectangle be $$xy$$ square units.

According to the question,

When length is reduced by $$5$$ units and breadth is increased by $$3$$ units, area of the rectangle gets reduced by $$9$$ square units;

\begin{align}\left( {x - 5} \right)\left( {y + 3} \right) &= xy - 9\\xy + 3x - 5y - 15 &= xy - 9\\3x - 5y - 6 &= 0 \qquad \ldots (1)\end{align}

When we increase the length by $$3$$ units and the breadth by $$2$$ units, the area increases by $$67$$ square units;

\begin{align}\left( {x + 3} \right)\left( {y + 2} \right) &= xy + 67\\xy + 2x + 3y + 6 &= xy + 67\\2x + 3y - 61 &= 0 \qquad \ldots (2)\end{align}

By cross-multiplication method, we obtain

\begin{align} \frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}&=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\\frac{x}{305-(-18)}&=\frac{y}{-12-(-183)}=\frac{1}{9-(-19)} \\ \frac{x}{323}&=\frac{y}{171}=\frac{1}{19} \\\frac{x}{323}&=\frac{1}{19}, \quad \frac{y}{171}=\frac{1}{19} \\ x&=17,\qquad \,y=9\end{align}

Equations are $$3x - 5y - 6 = 0$$ and $$2x + 3y - 61 = 0$$ where length and breadth of the rectangle are $$x$$ and $$y$$ respectively.

Hence, the length and breadth of the rectangle are $$17$$ units and $$9$$ units respectively.

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