Question 12 Marks
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid $₹ 27$ for a book kept for seven days, while Susy paid $₹ 21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day. Solve the pair of the linear equation obtained by the elimination method.
AnswerSuppose the fixed charge be Rs. x and the extra charge per day be Rs y.
According to the question, Mona paid Rs 27 for a book kept for 7 days,
$\Rightarrow$ $x + 4y = 27$$........(i)$
Tanvy paid $Rs.21$ for a book kept for 5 days,
$\Rightarrow$ $x+ 2y = 21$$.........(ii)$
Subtracting $(ii)$ from $(i)$,
$\Rightarrow 2y = 6$
$\Rightarrow$ $y = 3$
Substituting $y = 3$ in $(ii),$ we get $x = 15$
The fixed charge is $Rs.\ 15$ and the charge per day is $Rs\ 3$.
View full question & answer→Question 22 Marks
Five year hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son.
What are their present ages? Solve by substitution method.
AnswerLet $x$ (in years) be the present age of Jacob's son and $y$ (in years) be the present age of Jacob.
$5$ years hence, it has relation: $(y + 5) = 3(x + 5)$
or, $y + 5 = 3x + 15 3x + 15 - y - 5 = 0$ or, $3 x - y + 10 = 0 .......(i)
5$ years ago, it has relation $(y - 5) = 7(x - 5) y - 5 = 7x - 35$ or,$7x - 35 - y + 5 = 0$ or, $7 x - y - 3 0 = 0 ....(ii)$
From eqn. $(i), y = 3x + 10 ....(iii)$ On substituting the value of y in eqn. $(ii),$
we get $7x-(3x + 10) - 30 = 0 7x - 3x - 10 - 30 = 0$ or, $4x - 40 = 0$ or, $4x = 40 x=10$
On substituting $x = 10$ in eqn. (iii), $y = 3 \times 10 + 10$ y = 30 + 10 $\therefore $ $y = 40$
Hence, the present age of Jacob $= 40$ years and son's age = 10 years
View full question & answer→Question 32 Marks
Solve the pair of linear equations by substitution method:
$\sqrt { 2 } x - \sqrt { 3 } y = 0$
$\sqrt { 3 } x - \sqrt { 8 } y = 0$
AnswerThe given equations are
$\sqrt 2 x - \sqrt 3 y = 0$ $............(i)$
$\sqrt 3 x - \sqrt 8 y = 0$ $.............(ii)$
From equation (i), we obtain:
$x = \frac{{ \sqrt 3 y}}{{\sqrt 2 }}$$ ...(iii)$
Substituting this value in equation (ii), we obtain:
$\sqrt 3 \left( { \frac{{\sqrt 3 y}}{{\sqrt 2 }}} \right) - \sqrt 8 y = 0$
$ \frac{{3y}}{{\sqrt 2 }} - 2\sqrt 2 y = 0$
$y\left( { \frac{3}{{\sqrt 2 }} - 2\sqrt 2 } \right) = 0$
$y = 0$
Substituting the value of y in equation (iii), we obtain:
$x = 0$
$\therefore $ $x = 0, y = 0$
Hence the solution of given equation is $(0,0)$.
View full question & answer→Question 42 Marks
On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } $ and $\frac {c_1}{c_2}$, find out whether the pair of linear equations are consistent, or inconsistent: $ \frac { 3 } { 2 } x + \frac { 5 } { 3 } y = 7,$ $9x − 10y = 14$
AnswerGiven equations are: $ \frac { 3 } { 2 } x + \frac { 5 } { 3 } y = 7$ & $9x - 10y = 14$
Comparing equation $ \frac { 3 } { 2 } x + \frac { 5 } { 3 } y = 7$ with $ a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0$ and 9x − 10y = 14 with $ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $ a _ { 1 } = \frac { 3 } { 2 }$, $ b _ { 1 } = \frac { 5 } { 3 }$, $ c _ { 1 } = - 7 , a _ { 2 } = 9 , b _ { 2 } = - 10 , c _ { 2 } = - 14$
$ \frac { a _ { 1 } } { a _ { 2 } } = \frac { \frac { 3 } { 2 } } { 9 } = \frac { 1 } { 6 }$ and $ \frac { b _ { 1 } } { b _ { 2 } } = \frac { \frac { 5 } { 3 } } { - 10 } = \frac { - 1 } { 6 }$ Here $ \frac { a _ { 1 } } { a _ { 2 } } \neq \frac { b _ { 1 } } { b _ { 2 } }$
Therefore, equations have unique solution. Hence, they are consistent.
View full question & answer→Question 52 Marks
On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } $ and $\frac { c _ { 1 } } { c _ { 2 } }$, find out whether the pair of linear equations are consistent, or inconsistent: $2x − 3y = 8, 4x − 6y = 9.$
AnswerGiven equations are as: $2x − 3y = 8 4x − 6y = 9$
Comparing equation $2x − 3y = 8$
with $ a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0$
and $4x − 6y = 9$ with $ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $a_1=2, b_1=-3, c_1=-8, a_2=4, b_2=-6, c_2=-9$
Here $ \frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }$
$because $ $ \frac { 2 } { 4 } = \frac { - 3 } { - 6 } \neq \frac { - 8 } { - 9 } $
$\Rightarrow \frac { 1 } { 2 } = \frac { 1 } { 2 } \neq \frac { - 8 } { - 9 }$
Therefore, equations have no solution because they are parallel.
Hence, they are inconsistent.
View full question & answer→Question 62 Marks
On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } $ and $\frac { c _ { 1 } } { c _ { 2 } }$, find out whether the pair of linear equations are consistent, or inconsistent: $3x + 2y = 5, 2x − 3y = 7.$
AnswerGiven equations are $3x + 2y = 5 2x− 3y = 7$
Comparing equation $3x + 2y = 5$ with $ a _ { 1 } x + b _ { 2 } y + c _ { 1 } = 0$
and 2x − 3y = 7 with $ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $a_1=3, b_1=2, c_1=-5, a_2=2, b_2=-3, c_2=-7$
$ \frac { a _ { 1 } } { a _ { 2 } } = \frac { 3 } { 2 }$ and $ \frac { b _ { 1 } } { b _ { 2 } } = \frac { 2 } { - 3 }$
Here $ \frac { a _ { 1 } } { a _ { 2 } } \neq \frac { b _ { 1 } } { b _ { 2 } }$
which means equations have unique solution.
Hence they are consistent.
View full question & answer→Question 72 Marks
On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } \text { and } \frac { c _ { 1 } } { c _ { 2 } }$, find out whether the lines representing the pair of linear equations intersect at a point, are parallel or coincide: $6x − 3y + 10 = 0; 2x – y + 9 = 0.$
AnswerGiven equations are $6x − 3y + 10 = 0 2x – y + 9 = 0$
Comparing equation $6x − 3y + 10 = 0$ with $ a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0$
and 2x – y + 9 = 0 with $ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $a_1=6, b_1=-3, c_1=10, a_2=2, b_2=-1, c_2=9$
We have $ \frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }$
because $ \frac { 6 } { 2 } = \frac { - 3 } { - 1 } \neq \frac { 10 } { 9 } \Rightarrow \frac { 3 } { 1 } = \frac { 3 } { 1 } \neq \frac { 10 } { 9 }$
Hence, lines are parallel to each other.
View full question & answer→Question 82 Marks
On comparing the ratios $\frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } }$ and $\frac {c_1}{c_2}$, find out whether the lines representing the pair of linear equations intersect at a point, are parallel or coincident: $9x + 3y + 12 = 0; 18x + 6y + 24 = 0$
AnswerGiven equations are $9x + 3y + 12 = 0 18x + 6y + 24 = 0$
Comparing equation $9x + 3y + 12 = 0$
with $ a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0$
and $18x + 6y + 24 = 0$ with $ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $a_1=9, b_1=3, c_1=12, a_2=18, b_2=6, c_2=24$
We have $ \frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }$
because$ \frac { 9 } { 18 } = \frac { 3 } { 6 } = \frac { 12 } { 24 }$
$ \Rightarrow \frac { 1 } { 2 } = \frac { 1 } { 2 } = \frac { 1 } { 2 }$
Hence, lines are coincident.
View full question & answer→Question 92 Marks
On comparing the ratios $\frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } }$ and $\frac {c_1}{c_2}$, find out whether the lines representing the pair of linear equations intersect at a point, are parallel or coincident: $5x – 4y – 8 = 0; 7x + 6y – 9 = 0.$
Answer$$Given equations are:
$\frac { 4 } { 3 } x + 2y = 8; 2x + 3y = 12$
Compare equation $\frac { 4 } { 3 } x + 2 y = 8$ with $a_1x + b_1y + c_1= 0$ and $2x + 3y = 12$
with $a_2x + b_2y + c_2= 0$, We get, $a _ { 1 } = \frac { 4 } { 3 } ,$ $a_1= \frac { 4 } { 3 }$, $b_1=2, c_1=-8, a_2=2, b_2=3, c_2=-12$
$\frac { a _ { 1 } } { a _ { 2 } } = \frac { \frac { 4 } { 3 } } { 2 } = \frac { 2 } { 3 } , \frac { b _ { 1 } } { b _ { 2 } } = \frac { 2 } { 3 }$ and $\frac { c _ { 1 } } { c _ { 2 } } = \frac { 8 } { 12 } = \frac { 2 } { 3 }$
Here $\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }$
Therefore, the lines have infinitely many solutions.
Hence, they are consistent.
View full question & answer→Question 102 Marks
The ratio of incomes of two persons is $9 : 7$ and the ratio of their expenditures is $4 : 3.$ If each of them manages to save $₹ 2000$ per month, then find their monthly incomes.
AnswerLet us denote the incomes of the two-person by $₹ 9x$ and $₹ 7x$ and their expenditures by $₹ 4y$ and $₹ 3y$ respectively.
Then the equations formed in the situation is given by :
$9x – 4y = 2000 ...(i)$
and $7x – 3y = 2000 ...(2)$
Step $1$: Multiply Equation $(1)$ by $3$ and Equation $(2)$ by $4$ to make the coefficients of $y$ equal. Then, we get the equations:
$27x – 12y = 6000 ...(3)$
$28x – 12y = 8000 ...(4)$
Step $2$: Subtract Equation (3) from Equation $(4)$ to eliminate $y$, because the coefficients of $y$ are the same. So, we get
$(28x – 27x) – (12y – 12y) = 8000 – 6000$
$i.e., x = 2000$
Step $3$: Substituting this value of $x$ in $(1)$, we get
$9(2000) – 4y = 2000$
$i.e., y = 4000$
So, the solution of the equations is $x = 2000, y = 4000$. Therefore, the monthly incomes of the persons are $₹18,000$ and $₹14,000$ respectively.
View full question & answer→Question 112 Marks
Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
$5x - 8y + 1 = 0; ...(i)$
$3x -$ $\frac{24}{5}$y + $\frac{3}{5}$ $= 0 ...(ii)$
Answer$a_1 = 5,\ b_1 = -8,\ c_1 = 1$ and $a_2 = 3,\ b_2 =$ $\frac{-24}{5}$, $c_2 =$ $\frac{3}{5}$
$\frac { a _ { 1 } } { a _ { 2 } }$ = $\frac{5}{3}$ ...(i)
$\frac { b _ { 1 } } { b _ { 2 } }$ =$\frac { - 8 } { - 24 / 5 }$ = $\frac{5}{3}$ $...(ii)$
and $\frac { c _ { 1 } } { c _ { 2 } }$ = $\frac { 1 } { 3 / 5 }$ = $\frac{5}{3}$ $...(iii)$
Form $(i), (ii)$ and $(iii)$
$\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }$
$\therefore$The pair of equations has infinitely many solutions.
View full question & answer→Question 122 Marks
Solve 2x + 3y = 11 and x dot - 2y = - 12 and hence find the value of 'm' for which y = mx + 3.
Answer$2 x+3 y=11$ ...(1)
$x-2 y=-12$ ...(2)
From equation (2), we get $x=2 y-12$.
Substitute this into equation (1):
$\begin{array}{l}2(2 y-12)+3 y=11 \\ 4 y-24+3 y=11 \\ 7 y=35 \\ y=5\end{array}$
Substitute $y=5$ back into $x=2 y-12:$
$x=2(5)-12=10-12=-2$
So, the solution is $x=-2$ and $y=5$.
Now, substitute these values into $y=m x+3:$
$\begin{array}{l}5=m(-2)+3 \\ 2=-2 m \\ m=-1\end{array}$
View full question & answer→