# NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.1

Pair of Linear Equations in Two Variables

## Chapter 3 Ex.3.1 Question 1

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

**Solution**

**Video Solution**

**What is Known?**

\(7\) years ago, Aftab was \(7\) times as old as his daughter then and \(3\) years from now, Aftab shall be \(3\) times as old as his daughter will be.

**What is Unknown?**

Represent the situation algebraically and graphically.

**Reasoning: **

Assume the present age of Aftab be \(x\) years and his daughter be \(y\) years then represent their ages \(7\) years later and \(3\) years ago in term of *\(x\)* and \(y.\) Two linear equations can be formed to represent the above situation algebraically.

Using algebraic equation and truth table they can be graphically represented.

**Steps:**

(i) Present age of Aftab \( = x\) years and his daughter \( = y\) years

Therefore, \(7\) years ago, age of Aftab \( = \left( {x - 7} \right)\) years and his daughter \( = \left( {y - 7} \right)\)years

Using this information and applying the known condition that \(7\) years ago, Aftab was \(7\) times as old as his daughter then:

\[\begin{align}x - 7 &= 7(y - 7)\\x - 7 &= 7y - 49\\x - 7y - 7 + 49 &= 0\\x - 7y + 42 &= 0\end{align}\]

After \(3\) years from now, age of Aftab \( = \left( {x + 3} \right)\) years and his daughter \( = \left( {y + 3} \right)\) years and also Aftab will be \(3\) times as old as his daughter. Then mathematically,

\[\begin{align}x + 3 &= 3\left( {y + 3} \right)\\x + 3 &= 3y + 9\\x - 3y + 3 - 9 &= 0\\x - 3y - 6 &= 0\end{align}\]

Algebraic representations, where *\(x\)* and *\(y\)* are present ages of Aftab and his daughter respectively:

\[\begin{align}x - 7y + 42 &= 0\qquad(1)\\x - 3y-6 &= 0\qquad(2)\end{align}\]

Therefore, the algebraic representation is for equation \(1\) is:

\[\begin{align}x - 7y + 42 &= 0\\- 7y &= - x - 42\\7y &= x + 42\\y &= \frac{{x + 42}}{7}\end{align}\]

And, algebraic representation for equation \((2)\) is:

\[\begin{align}x - 3y - 6 &= 0\\- 3y &= - x + 6\\3y &= x - 6\\y &= \frac{{x - 6}}{3}\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

For equation \((1)\)

\(x\) |
\(21\) |
\(28\) |

\(y = \frac{{x + 42}}{7}\) |
\(9\) |
\(10\) |

For equation (2)

\(x\) | \(30\) | \(21\) |

\(y = \frac{{x - 6}}{3}\) | \(8\) | \(5\) |

The graphical representation is as follows.

Unit: \(1 \,\rm{cm}=5\) years.

The answer is \((42, \;12)\)

## Chapter 3 Ex.3.1 Question 2

The coach of a cricket team buys \(3\) bats and 6 balls for \(₹\,3900.\) Later, she buys another bat and \(3\) more balls of the same kind for ₹ \(1300.\) Represent this situation algebraically and geometrically.

**Solution**

**Video Solution**

**What is Known?**

(i) Three bats and six balls for ₹ \(3900\)

(ii) One bat and three balls for ₹ \(1300\)

**What is Unknown?**

Represent the situation geometrically and algebraically

**Reasoning: **

Assuming the cost of one bat as ₹ \(x\) and the cost of one ball as ₹ \(y,\) two linear equations can be formed for the above situation.

**Steps:**

The cost of \(3\) bats and 6 balls is ₹ \(3900.\)

Mathematically:

\[\begin{align}3x + 6y&= 3900\\3(x + 2y)& = 3900\\x + 2y& = 1300\end{align}\]

Also, the cost of \(1\) bat and \(3\) balls is ₹ \(1300.\)

Mathematically:

\[x + 3y = 1300\]

Algebraic representation where *\(x\)* and *\(y\)* are cost of bat and ball respectively.

\[\begin{align}x + 2y &= 1300 \qquad(1)\\x + 3y &= 1300\qquad(2)\end{align}\]

Therefore, the algebraic representation for equation \(1\) is:

\[\begin{align}x + 2y &= 1300\\2y &= 1300-x\\y &= \frac{{1300 - x}}{2}\end{align}\]

And, the algebraic representation for equation \(2\) is:

\[\begin{align}x + 3y &= 1300\\3y &= 1300-x\\y &= \frac{{1300 - x}}{3}\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

\(x\) | \(700\) | \(500\) |

\(y = \frac{{1300 - x}}{2}\) | \(300\) | \(400\) |

\(x\) | \(400\) | \(700\) |

\(y = \frac{{1300 - x}}{3}\) | \(300\) | \(200\) |

The graphical representation is as follows.

Unit: \(1\,\rm{cm} =\) ₹ \(100.\)

The answer is \((1300, \,0)\)

## Chapter 3 Ex.3.1 Question 3

The cost of \(2 \,\rm{kg}\) of apples and \(1 \,\rm{kg}\) of grapes on a day was found to be ₹ \(160\) After a month, the cost of \( 4\,\rm{kg}\) of apples and \( 2\,\rm{kg}\) of grapes is ₹ \(300\) Represent the situation algebraically and geometrically.

**Solution**

**Video Solution**

**What is Known?**

(i) Cost of \(2 \,\rm{kg} \)of apples and \(1 \,\rm{kg} \) of grapes is ₹ \(160\)

(ii) Cost of \(4 \,\rm{kg} \) of apples and \(2 \,\rm{kg} \) of grapes is ₹ \(300.\)

**What is Unknown?**

Represent the situation geometrically and algebraically.

**Reasoning: **

Assuming the cost of \(1 \,\rm{kg} \) apples as ₹ *\(x\)* and the cost of \(1 \,\rm{kg} \) grapes as ₹ \(y,\) two linear equations can be formed for the above situation.

**Steps:**

Let the cost of \(1 \,\rm{kg} \) of apples be *\(x\)* and cost of \(1 \,\rm{kg} \) of grapes be *\(y\)*

Cost kg \(2 \,\rm{kg} \) apples and \(1 \,\rm{kg} \) of grapes is ₹ \(160.\)

Mathematically,

\[2x + y = 160\]

Also, cost kg \(4 \,\rm{kg} \) apples and \(2 \,\rm{kg} \) of grapes is ₹ \(300.\)

Mathematically,

\[\begin{align}4x + 2y &= 300\\2(2x + y) &= 300\\2x\, + \,y\,& = \,150\end{align}\]

Algebraic representation where *\(x\)* and *\(y\)* are the cost of \(1 \,\rm{kg} \) apple and \(1 \,\rm{kg} \) grapes respectively.

\[\begin{align}2x + y &= 160 \,\dots(1)\\2x + y& = 150\,\dots(2)\end{align}\]

Therefore, the algebraic representation is for equation \(1\) is:

\[\begin{align}2x + y &= 160\\y &= 160-2x\end{align}\]

And, the algebraic representation is for equation \(2\) is:

\[\begin{align}2x + y& = 150\\y &= 150-2x\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

\(x\) | \(50\) | \(30\) |

\(y = 160-2x\) | \(60\) | \(100\) |

\(x\) | \(30\) | \(40\) |

\(y = 150-2x\) | \(90\) | \(70\) |

The graphical representation is as follows.

Unit \(= 1\, \rm{cm} = \)₹ \(10\)

Since the lines are parallel hence no Solution