NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7
Pair of Linear Equations in Two Variables
Question 1
The ages of two friends Ani and Biju differ by \(3\) years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by \(30\) years. Find the ages of Ani and Biju.
Solution
Reasoning:
The difference between the ages of Biju and Ani is \(3\) years. Either Biju is \(3\) years older than Ani or Ani is \(3\) years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be \(30\) years more than that of Cathy’s age.
Steps:
Let the age of Ani and Biju be \(x\) and \(y\) years respectively.
Therefore, age of Ani’s father, Dharam be \(2x\) years
And age of Biju’s sister Cathy be \(\frac{y}{2}\) years
Case (I) When Ani is older than Biju
The ages of Ani and Biju differ by \(3\) years,
\[x - y = 3 \qquad \left( 1 \right)\]
The ages of Cathy and Dharam differs by \(30\) years,
\[\begin{align}2x - \frac{y}{2} &= 30\\4x - y &= 60 \qquad \left( 2 \right)\end{align}\]
Subtracting \((1)\) from \((2),\) we obtain
\[\begin{align}3x &= 57\\x &= 19\end{align}\]
Substituting \(x = 19\) in equation \((1),\) we obtain
\[\begin{align}19 - y &= 3\\y& = 16\end{align}\]
Therefore, Ani is \(19\) years old and Biju is \(16\) years old
Case (II) When Biju is older than Ani.
The ages of Ani and Biju differ by \(3\) years,
\[\begin{align}y - x &= 3\\ - x + y &= 3 \qquad \left( 1 \right)\end{align}\]
The ages of Cathy and Dharam differs by \(30\) years,
\[\begin{align}2x - \frac{y}{2} &= 30\\4x - y &= 60 \qquad (2)\end{align}\]
Adding \((1)\) and \((2),\) we obtain
\[\begin{align}3x &= 63\\x &= 21\end{align}\]
Substituting \(x = 21\) in equation \((1),\) we obtain
\[\begin{align} - 21 + y &= 3\\y &= 24\end{align}\]
Therefore, Ani is \(21\) years old and Biju is \(24\) years old.
Hence, Ani is \(19\) years old and Biju is \(16\) years old or Ani is \(21\) years old and Biju is \(24\) years old.
Question 2
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
\( \begin{bmatrix} \textbf{Hint:} \, x + 100 = 2 ({y - 100}),\\ y + 10 = 6 (x - 10) \end{bmatrix} \)
Solution
Reasoning:
Assume the friends have ₹ \(x\) and ₹ \(y\) with them. Then based on given conditions, two linear equations can be formed which can be easily solved.
Steps:
Let the first friend has ₹ \(x\)
And second friend has ₹ \(y\)
Using the information given in the question,
When second friend gives ₹ \(100\) to first friend;
\[\begin{align}x + 100 &= 2\left( {y - 100} \right)\\x + 100 &= 2y - 200\\x - 2y &= - 300 \qquad \qquad \left( 1 \right)\end{align}\]
When first friend gives ₹ \(10\) to second friend;
\[\begin{align}y + 10 &= 6\left( {x - 10} \right)\\y + 10 &= 6x - 60\\6x - y &= 70 \qquad \qquad \left( 2 \right)\end{align}\]
Multiplying equation \((2)\) by \(2,\) we obtain
\[12x - 2y = 140 \qquad {\rm{ }}\left( 3 \right)\]
Subtracting equation \((1)\) from equation \((3),\) we obtain
\[\begin{align}11x &= 440\\x &= \frac{{440}}{{11}}\\x &= 40\end{align}\]
Substituting \(x = 40\) in equation \((1)\), we obtain
\[\begin{align}40 - 2y &= - 300\\2y &= 40 + 300\\y &= \frac{{340}}{2}\\y &= 170\end{align}\]
Therefore, first friend has ₹ \(40\) and second friend has ₹ \(170\) with them.
Question 3
A train covered a certain distance at a uniform speed. If the train would have been \(10\,\rm{ km/h}\) faster, it would have taken \(2\) hours less than the scheduled time. And if the train were slower by \(10\,\rm{ km/h;}\) it would have taken \(3\) hours more than the scheduled time. Find the distance covered by the train.
Solution
What is Known?
Changes in speed of the train as well in the time.
What is Unknown?
Distance covered by the train.
Reasoning:
Assuming uniform speed of the train be \(x\,{\rm{km/h }}\) and time taken to travel a given distance be \(t\) hours. Then distance can be calculated by;
\[{\text{Distance}} = {\rm{ Speed }} \times {\rm{ Time}}\]
Steps:
Let the uniform speed of the train be \(x\,{\rm{ km/h}}\) and the scheduled time to travel the given distance be \(t\) hours
Then the distance be \(xt\,{\rm{ km}}\)
When the train would have been \(10\,{\rm{ km/h}}\) faster, it would have taken \(2\) hours less than the scheduled time;
\[\begin{align}\left( {x + 10} \right)\left( {t - 2} \right)& = xt\\xt - 2x + 10t - 20 &= xt\\ - 2x + 10t &= 20 \qquad \quad \left( 1 \right)\end{align}\]
When the train were slower by \(10\,{\rm{ km/h,}}\) it would have taken \(2\) hours\] more than the scheduled time;
\[\begin{align}\left( {x - 10} \right)\left( {t + 3} \right) &= xt\\xt + 3x - 10t - 30 &= xt\\3x - 10t &= 30 \qquad \quad \left( 2 \right)\end{align}\]
Adding equations \((1)\) and \((2),\) we obtain
\[x = 50\]
Substituting \(x = 50\) in equation \((1),\) we obtain
\[\begin{align} - 2 \times 50 + 10t &= 20\\- 100 + 10t &= 20\\10t &= 120\\t &= \frac{{120}}{{10}}\\t &= 12
\end{align}\]
Therefore, distance,
\(xt = 50 \times 12 = 600\)
Hence, the distance covered by the train is \(600 \,\rm{km.}\)
Question 4
The students of a class are made to stand in rows. If \(3\) students are extra in a row, there would be \(1\) row less. If \(3\) students are less in a row, there would be \(2\) rows more. Find the number of students in the class.
Solution
What is Known?
Changes in number of students in a row and number of rows.
What is Unknown?
Number of students in the class.
Reasoning:
Assume number of rows equal to \(x\) and number of students in each row be \(y.\) Then the total number of students in the class can be calculated by;
Total number of students \( =\) Number of rows \(\times\) Number of students in each row
Steps:
Let the number of rows be \(x\)
And number of students in each row be \(y\)
Then the number of students in the class be \(xy\)
Using the information given in the question,
Condition 1 If \(3\) students are extra in a row, there would be \(1\) row less
\[\begin{align}\left( {x - 1} \right)\left( {y + 3} \right) &= xy\\xy + 3x - y - 3 &= xy\\3x - y &= 3 \qquad \quad \left( 1 \right)\end{align}\]
Condition 2 If \(3\) students are less in a row, there would be \(2\) rows more
\[\begin{align}\left( {x + 2} \right)\left( {y - 3} \right) &= xy\\xy - 3x + 2y - 6 &= xy\\ - 3x + 2y &= 6 \qquad \quad \left( 2 \right)\end{align}\]
Adding equations \((1)\) and \((2),\) we obtain \(y = 9\)
Substituting \(y = 9\) in equation \((1),\) we obtain
\[\begin{align}3x - 9 &= 3\\3x &= 12\\x &= 4\end{align}\]
Hence, number of students in the class, \(xy = 4 \times 9 = 36\)
Question 5
In \(\Delta ABC \) ,
\(\angle {C} = 3\angle B = 2(\angle A + \angle B)\).
Find the three angles.
Solution
What is Known?
Relation between the angles of the triangle.
What is Unknown?
Measurement of each angles of the triangle.
Reasoning:
Sum of the measures of all angles of a triangle is \(180^\circ.\)
Steps:
Let the measurement of \(\angle A = {x^{\rm{\circ}}}\)
And the measurement of \(\angle B = {y^{\rm{\circ}}}\)
Using the information given in the question,
\(\angle C = 3\angle B = 2\left( {\angle A \!+\! \angle B} \right)\)
\[\begin{align} \Rightarrow 3\angle B &= 2\left( {\angle A + \angle B} \right)\\ \Rightarrow 3y &= 2\left( {x + y} \right)\\ \Rightarrow 3y &= 2x + 2y\\ \Rightarrow \;2x - y &= 0 \qquad \qquad \qquad \left( 1 \right)\end{align}\]
We know that the sum of the measures of all angles of a triangle is \(180^\circ.\)
Therefore,
\[\begin{align}\angle A + \angle B + \angle C &= {180^{\circ}}\\\angle A + \angle B + 3\angle B &= {180^{\circ}} \\&\begin{bmatrix} \because \angle C = 3\angle B \end{bmatrix} \\\angle A + 4\angle B &= {180^{\circ}}\\ x + 4y &= 180 \qquad \quad \left( 2 \right)\end{align}\]
Multiplying equation \((1)\) by \(4,\) we obtain
\[8x - 4y = 0 \qquad \left( 3 \right)\]
Adding equations \((2)\) and \((3),\) we obtain
\[\begin{align}9x &= 180\\x &= 20\end{align}\]
Substituting \(x = 20\) in equation \((1)\), we obtain
\[\begin{align}2 \times 20 - y &= 0\\y &= 40\end{align}\]
Therefore,
\[\begin{align}\angle A &= {x^{\circ}} = {20^{\circ}}\\\angle B &= {y^{\circ}} = {40^{\circ}}\\\angle C &= 3\;\angle B = 3 \times {40^{\circ}} = {120^{\circ}}\end{align}\]
Question 6
Draw graphs of the equations \(5x - y = 5\) and \(3x - y = 3\). Determine the coordinates of the vertices of the triangle formed by these lines and the \(y\) axis.
Solution
Steps:
\[\begin{align}5x - y &= 5\\\Rightarrow \quad y &= 5x - 5\end{align}\]
The solution table will be as follows.
\(x\) |
\(0\) |
\(2\) |
\(y\) |
\(-5\) |
\(5\) |
\[\begin{align}3x - y &= 3\\ \Rightarrow \quad y &= 3x - 3\end{align}\]
The solution table will be as follows.
\(x\) |
\(0\) |
\(2\) |
\(y\) |
\(-3\) |
\(3\) |
The graphical representation of these lines will be as follows.
It can be observed that the required triangle is \(ABC\) formed by these lines and \(y-\)axis.
The coordinates of vertices are \(A( 1, 0),\; B (0, - 3),\; C (0, - 5)\).
Question 7
Solve the following pair of linear equations.
(i) \(\begin{align} px + qy &= p - q\\qx - py &= p + q\end{align}\)
(ii) \(\begin{align} ax + by &= c\\bx + ay &= 1 + c\end{align}\)
(iii) \(\begin{align} \frac{x}{a} - \frac{y}{b} &= 0\\ax + by &= {a^2} + {b^2}\end{align}\)
(iv) \(\begin{align} \begin{bmatrix}(a - b)x + \\(a + b)y \end{bmatrix}\!&= \! {a^2} \! -\! 2ab \! -\! {b^2} \\(a + b)(x + y) &= {a^2} + {b^2}\end{align}\)
(v) \(\begin{align} 152x - 378y &= - 74\\- 378x + 152y &= - 604\end{align}\)
Solution
Steps:
(i)
\(\begin{align} px + qy &= p - q \qquad \ldots \left( 1 \right)\\qx - py &= p + q \qquad \ldots \left( 2 \right)\end{align}\)
Multiplying equation \((1)\) by \(p\) and equation \((2)\) by \(q,\) we obtain
\[\begin{align}{p^2}x + pqy &= {p^2} - pq \qquad \ldots \left( 3 \right)\\{q^2}x - pqy &= pq + {q^2} \qquad \ldots \left( 4 \right)\end{align}\]
Adding equations \((3)\) and \((4),\) we obtain
\[\begin{align}{p^2}x + {q^2}x &= {p^2} + {q^2}\\\left( {{p^2} + {q^2}} \right)x &= {p^2} + {q^2}\\
x &= \frac{{{p^2} + {q^2}}}{{{p^2} + {q^2}}}\\x &= 1\end{align}\]
Substituting \(x = 1\) in equation \((1),\) we obtain
\[\begin{align}p \times 1 + qy &= p – q\\qy &= - q\\y &= - 1\end{align}\]
Therefore, \(x = 1\) and \(y = - 1\)
(ii)
\(\begin{align}ax + by &= c \qquad \quad\;\; \ldots \left( 1 \right)\\
bx + ay &= 1 + c \qquad \ldots \left( 2 \right)\end{align}\)
Multiplying equation \((1)\) by \(a\) and equation \((2)\) by \(b\), we obtain
\[\begin{align}{a^2}x + aby &= ac \qquad \quad\;\; \ldots \left( 3 \right)\\{b^2}x + aby &= b + bc \qquad \ldots \left( 4 \right)\end{align}\]
Subtracting equation \((4)\) from equation \((3),\)
\[\begin{align}\left( {{a^2} – {b^2}} \right)x &= ac – bc – b\\x &= \frac{{c(a – b) – b}}{{{a^2} – {b^2}}}{\rm{ }}\end{align}\]
Substituting \(\begin{align}x = \frac{{c(a – b) – b}}{{{a^2} – {b^2}}} \end{align}\) in equation \((1),\) we obtain
\[\begin{align}ax + by &= c \\ a\left[ {\frac{{c(a – b) – b}}{{{a^2} – {b^2}}}} \right] + by &= c\\\frac{{ac(a – b) – ab}}{{{a^2} – {b^2}}} + by &= c\\ \end{align} \]
\[\begin{align} by &= c - \frac{{ac(a – b) – ab}}{{{a^2} – {b^2}}}\\ by &= \frac{{{a^2}c – {b^2}c – {a^2}c + abc + ab}}{{{a^2} – {b^2}}}\\by &= \frac{{abc – {b^2}c + ab}}{{{a^2} – {b^2}}}\\ by &= \frac{{bc(a – b) + ab}}{{{a^2} – {b^2}}}\\by& = \frac{{b\left[ {c(a – b) + a} \right]}}{{{a^2} – {b^2}}}\\ y &= \frac{{c(a – b) + a}}{{{a^2} – {b^2}}}\end{align}\]
Therefore,
\(\begin{align}x = \frac{{c(a – b) – b}}{{{a^2} – {b^2}}}\end{align}\)
and
\(\begin{align}y=\frac{c(a-b)+a}{{{a}^{2}}-{{b}^{2}}}\end{align}\)
(iii)
\(\begin{align} \frac{x}{a} - \frac{y}{b} &= 0 \qquad \qquad\;\; \ldots \left( 1 \right)\\ax + by &= {a^2} + {b^2} \qquad \ldots \left( 2 \right)\end{align}\)
By solving equation \((1),\) we obtain
\[\begin{align}\frac{x}{a} - \frac{y}{b} &= 0\\x &= \frac{{ay}}{b} \qquad \ldots \left( 3 \right)\end{align}\]
Substituting \(\begin{align}x = \frac{{ay}}{b} \end{align}\) in equation \((2),\) we obtain
\[\begin{align}a \times \left( {\frac{{ay}}{b}} \right) + by &= {a^2} + {b^2}\\
\frac{{{a^2}y + {b^2}y}}{b} &= {a^2} + {b^2}\\\left( {{a^2} + {b^2}} \right)y &= b\left( {{a^2} + {b^2}} \right)\\y &= b\end{align}\]
Substituting \(y = b\) in equation \((3),\) we obtain
\[\begin{align}x &= \frac{{a \times b}}{b}\\x &= a\end{align}\]
Therefore, \(x = a\) and \(y = b\)
(iv)
\(\begin{align} &\begin{bmatrix} (a - b)x +\\ (a + b)y \end{bmatrix} \! = \! {a^2} \! - \! 2ab \! - \! {b^2} \ldots \left( 1 \right) \\ \\ & (a + b)(x + y) \! = \! {a^2} \! + \! {b^2}\ldots \left( 2 \right)\end{align}\)
By solving equation \((2),\) we obtain
\[\begin{align}(a + b)(x + y) &= {a^2} + {b^2}\\ \begin{bmatrix}(a + b)x +\\ (a + b)y \end{bmatrix}&= {a^2} + {b^2} \ldots \left( 3 \right)\end{align}\]
Subtracting equation \((3)\) from \((1),\) we obtain
\[\begin{align}\!\begin{bmatrix}(a – b)x –\\ (a + b)x \end{bmatrix}&=\! \begin{bmatrix} \left( {{a^2} – 2ab – {b^2}} \right) \!-\! \\\left( {{a^2} + {b^2}} \right)\end{bmatrix} \\ \begin{bmatrix} {(a – b) – }\\{(a + b)} \end{bmatrix}x \!& =\! \begin{bmatrix}{a^2} – 2ab – {b^2} – \\{a^2} – {b^2}\end{bmatrix} \\\left[ {a – b – a – b} \right]x \!&=\! - 2ab – 2{b^2}\\ - 2bx \!&= \!- 2b\left( {a + b} \right)\\x\!& = \!\left( {a + b} \right)\end{align}\]
Substituting \(x = \left( {a + b} \right)\) in equation (1), we obtain
\[\begin{align} \begin{bmatrix}(a – b)(a + b) + \\ (a + b)y \end{bmatrix}&= {a^2} \! –\! 2ab \! –\! {b^2} \\ \begin{bmatrix}({a^2} – {b^2}) +\\ (a + b)y \end{bmatrix} &= {a^2} \! – \! 2ab \! – \! {b^2} \end{align} \]
\[\begin{align} (a \! + \! b)y & \! = \! {a^2} \! – \! 2ab \! – \! {b^2} \! – \! ({a^2} \! – \! {b^2}) \\(a \! + \! b)y \! &= \! {a^2} \! – \! 2ab \! – \! {b^2} \! – \! {a^2} \! + \! {b^2} \\y \! &= \! \frac{{ - 2ab}}{{(a \! + \! b)}}\end{align}\]
(v)
\(\begin{align}152x - 378y &= - 74 \;\; \ldots \left( 1 \right)\\ - 378x + 152y &= - 604 \;\; \ldots \left( 2 \right)\end{align}\)
Adding equations \((1)\) and \((2),\) we obtain
\[\begin{align} - 226x - 226y &= - 678\\ - 226\left( {x + y} \right) &= - 678\\x + y &= 3 \qquad \ldots \left( 3 \right)\end{align}\]
Subtracting equation \((2)\) from \((1),\) we obtain
\[\begin{align}530x - 530y &= 530\\530\left( {x - y} \right) &= 530\\x - y &= 1 \qquad \ldots \left( 4 \right)\end{align}\]
Adding equations \((3)\) and \((4),\) we obtain
\[\begin{align}2x &= 4\\x& = 2\end{align}\]
Substituting \(x = 2\) in equation \((3),\) we obtain
\[\begin{align}2 + y &= 3\\y &= 1\end{align}\]
Therefore, \(x = 2\) and \(y = 1\)
Question 8
\(ABCD\) is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.
Solution
What is Known?
Measurement of the angles of the cyclic quadrilateral in terms of \(x\) and \(y.\)
What is Unknown?
Measurement of the angles of the cyclic quadrilateral.
Reasoning:
Pairs of opposite angles of a cyclic quadrilateral are supplementary.
Steps:
We know that the sum of the measures of opposite angles in a cyclic quadrilateral is \(180^\circ.\)
Therefore,
\[\begin{align}\angle A + \angle C &= {180^\circ}\\\left( {4y + 20} \right) + \left( { - 4x} \right) &= 180\\
4y + 20 - 4x &= 180\\ - 4\left( {x - y} \right)& = 160\\x - y &= - 40 \qquad \left( 1 \right)\end{align}\]
And
\[\begin{align}\angle B + \angle D &= {180^\circ}\\\left( {3y - 5} \right) + \left( { - 7x + 5} \right) &= 180\\
3y - 5 - 7x + 5 &= 180\\ - 7x + 3y &= 180\\7x - 3y &= - 180 \quad \left( 2 \right)\end{align}\]
Multiplying equation \((1)\) by \(3,\) we obtain
\[3x - 3y = - 120\qquad \left( 3 \right)\]
Subtracting equation \((3)\) from equation \((2),\) we obtain
\[\begin{align}4x &= - 60\\x &= - 15\end{align}\]
Substituting \(x = - 15\) in equation \((1),\) we obtain
\[\begin{align} - 15 - y &= - 40\\y &= 25\end{align}\]
Therefore,
\[\begin{align}\angle A &= 4 \times 25 + 20 = {120^\circ}\\\angle B &= 3 \times 25 - 5 = {70^\circ}\\
\angle C &= - 4 \times \left( { - 15} \right) = {60^\circ}\\\angle D &= - 7 \times \left( { - 15} \right) + 5 = {110^\circ}\end{align}\]
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