Question 1014 Marks
Points $A$ and $B$ are $70\ km$ apart on a highway. A car starts from $A$ and another car starts from $B$ simultaneously. If they travel in the same direction, they meet in $7$ hours. But, if they travel towards each other, they meet in $1$ hour. Find the speed of each car.
AnswerLet P and Q be the cars starting from $A$ and $B$ respectively and let their speeds be $x \ km/hr$ and $y \ km/hr$ respectively.
Case: I
when the cars P and Q move in the same direction.
Distance covered by the car $P$ in $7$ hours $= 7x km$
Distance covered by the car $Q$ in $7$ hours $= 7y km$
Let the cars meet at point M.
$\therefore AM = 7x km$ and $BM = 7y km$
$\therefore AM - BM = AB$
$\Rightarrow 7x - 7y = 70$
$\Rightarrow 7(x - y) = 70$
$\Rightarrow x - y = 10 ...(1)$
Case: II
When the cars $P$ and $Q$ move in opposite directions
Distance covered by $P$ in $1$ hour $= x km$
Distance covered by $Q$ in $1$ hour $= y km$
In this case let the cars meet at a point N.
$\therefore AN = x km$ and $BN = y km$
$\therefore AN + BN = AB$
$x + y = 70 ...(2)$
Adding $(1)$ and $(2)$, we get
$2x = 80$
$\Rightarrow x= 40$
Putting $x = 40$ in $(1)$, we get
$40 - y = 10$
$\Rightarrow y = (40 - 10) = 30$
$\therefore x = 40, y = 30$
Hence, the speeds of these cars are $40 km / hr$ and $30 km / hr$ respectively. View full question & answer→Question 1024 Marks
Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:
$x - 2y + 2 = 0, 2x + y - 6 = 0$
Answer$\text{x}-2\text{y}+ 2 = 0$
$\Rightarrow\text{y}=\frac{\text{x}+2}{2}$
| $x:$ |
$-2$ |
$2$ |
| $y:$ |
$0$ |
$2$ |
$2\text{x} + \text{y} - 6 = 0$
$\Rightarrow\text{y}=6-\text{2x}$
| $x:$ |
$3$ |
$1$ |
| $y:$ |
$0$ |
$4$ |
Since the two graph intersect at $(2,2)$,
$x=2 \text { and } y=2$
The vertices of the triangle formed by these lines and the $x$-axis are $(2,2),(3,0)$ and $(-2,0)$. So, height of the triangle $=$ distance from $(2,2)$ to $x$-axis
$= 2$ units
Base $= 5$ units
Area of triangle $=\frac{1}{2}\times$ base × height
$=\frac{1}{2}\times5\times2$
$=5\ \text{sq. units}$ View full question & answer→Question 1034 Marks
Solve for $x$ and $y$
$\frac{9}{\text{x}}-\frac{4}{\text{y}}=8,$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}=\text{101}$ $(\text{x}\neq0,\ \text{y}\neq0).$
AnswerPutting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ the given equations become
$9u - 4v = 8 ...(1)$
$13u + 7v = 101 ...(2)$
Multiplying $(1)$ by $7$ and $(2)$ by $4$, we get
$63u - 28v = 56 ...(3)$
$52u + 28v = 404 ...(4)$
Adding $(3)$ and $(4)$, we get
$115u = 460$
$\Rightarrow\text{u}=\frac{460}{115}=4$
Putting $u = 4$ in $(1),$ we get
$9 \times 4 - 4v = 8$
$\Rightarrow 36 - 4v = 8$
$\Rightarrow -4v = 8 - 36$
$\Rightarrow -4v = -28$
$\Rightarrow v = 7$
Now, $u = 4$
$\Rightarrow\frac{1}{\text{x}}=4$
$\Rightarrow\text{x}=\frac{1}{4}$
and, $v = 7$
$\Rightarrow\frac{1}{\text{y}}=7$
$\Rightarrow\text{y}=\frac{1}{7}$
$\therefore$ The solution is $\text{x}=\frac{1}{4}$ and $\text{y}=\frac{1}{7}$
View full question & answer→Question 1044 Marks
If twice the son's age in years is added to the father's age, the sum is $70$ years. But, if twice the father's age is added to the son's age, the sum is $95$ years. Find the age of the father and son.
AnswerLet the present ages of the mother and her son be $x$ and $y$ respectively.
According to the given question:
$x + 2y = 70 ...(1)$
and
$2x + y = 95 ...(2)$
Multiplying $(1)$ by $1$ and $(2) $ by $2$, we get
$x + 2y = 70 ...(3)$
$4x + 2y = 190 ...(4)$
Subtracting $(3)$ from $(4)$, we get
$3x = 120$
$\Rightarrow\text{x}=\frac{120}{3}=40$
Putting $x = 40$ in $(1),$ we get
$40 + 2y = 70$
$⇒ y = 15$
$\therefore x = 40, y = 15$
Hence, the ages of the mother and the son are $40$ years and $15$ years respectively.
View full question & answer→Question 1054 Marks
$90 \%$ and $97 \%$ pure acid solutions are mixed to obtain $21$ litres of $95 \%$ pure acid solution. Find the quantity of each type of acids to be mixed to form the mixture.
AnswerLet the $x$ litres of $90 \%$ and $y$ litres of $97 \%$ pure acid solutions be mixed.
According to the given condition,
$\frac{90}{100}\text{x}+\frac{97}{100}\text{y}=\frac{95}{100}(21)$
$\Rightarrow 90x + 97y = 95(21)$
$\Rightarrow 90x + 97y = 1995 ...(i)$
Since the amount of each solutions adds to $21$ liters,
$\Rightarrow x + y = 21 ...(ii)$
Multiplying $(ii)$ by $90$, we get
$\Rightarrow 90x + 90y = 1890 ...(iii)$
Subtract $(iii)$ from $(i).$
$\Rightarrow 7y = 105$
$\Rightarrow x = 80$
Substituting $y = 15$ in $(ii),$ we get
$\Rightarrow x = 6.$
Hence, the amount of each type is $6$ liters and $15$ liters.
View full question & answer→Question 1064 Marks
Solve the following systems of equations by using the method of cross multiplication:
$\frac{\text{a}}{\text{x}}-\frac{\text{b}}{\text{y}}=0$
$\frac{\text{ab}^2}{\text{x}}+\frac{\text{a}^2\text{b}}{\text{y}}=\text{a}^2+\text{b}^2,$ where $\text{x}\neq0$ and $\text{y}\neq0$
AnswerSubstituting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ in the given equations, we get
$ a u-b v+0=0 \ldots \text { (i) } $
$ a b^2 u+a^2 b v-\left(a^2+b^2\right)=0 \ldots \text { (ii) }$
$\text { Here, } a_1=a, b_1=-b, c_1=0, a_2=a b^2, b_2=a^2 b \text { and } c_2=-\left(a^2+b^2\right) \text {. }$
By cross multiplication, we have:
$\frac{\text{u}}{\text{b}_1\text{c}_2-\text{b}_2\text{c}_1}=\frac{\text{v}}{\text{c}_1\text{a}_2-\text{c}_2\text{a}_1}=\frac{1}{\text{a}_1\text{b}_2-\text{a}_2\text{b}_1}$
$\Rightarrow\frac{\text{u}}{(-\text{b})[-(\text{a}^2+\text{b}^2)]-(\text{a}^2\text{b})(0)}=\frac{\text{v}}{(0)(\text{a}^2\text{b})-(-\text{a}^2-\text{b}^2)(\text{a})}\\=\frac{1}{(\text{a})(\text{a}^2\text{b})-(\text{ab}^2)(-\text{b})}$
$\Rightarrow\frac{\text{u}}{\text{b}(\text{a}^2-\text{b}^2)}=\frac{\text{v}}{\text{a}(\text{a}^2+\text{b}^2)}=\frac{1}{\text{ab}(\text{a}^2+\text{b}^2)}$
$\Rightarrow\text{u}=\frac{\text{b}(\text{a}^2+\text{b}^2)}{\text{ab}(\text{a}^2+\text{b}^2)},\ \text{v}=\frac{\text{a}(\text{a}^2+\text{b}^2)}{\text{ab}(\text{a}^2+\text{b}^2)}$
$\Rightarrow\text{u}=\frac{1}{\text{a}},\ \text{v}=\frac{1}{\text{b}}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\text{a}},\ \frac{1}{\text{y}}=\frac{1}{\text{b}}$
$\Rightarrow\text{x}=\text{a},\ \text{y}=\text{b}$
Hence, $x = a$ and $y = b$ is the required solution.
View full question & answer→Question 1074 Marks
Solve for $x$ and $y$
$\frac{44}{\text{x}+\text{y}}+\frac{30}{\text{x}-\text{y}}=10,$
$\frac{55}{\text{x}+\text{y}}-\frac{40}{\text{x}-\text{y}}=13$
AnswerPutting $\frac{1}{\text{x}+\text{y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}$ in the equation, we get
$44u + 30v = 10 ...(1)$
$55u + 40v = 13 ...(2)$
Multiply $(1)$ by $4$ and $(2)$ by $3$, we get
$176u + 120v = 40 ...(3)$
$165u - 120v = 39 ...(4)$
Subtracting $(4)$ from $(3)$, we get
$\text{11u}=1$
$\text{u}=\frac{1}{11}$
Putting $\text{u}=\frac{1}{11},$ in $(1)$, we get
$44\times\frac{1}{11}+\text{30v}=10$
$\Rightarrow4+\text{30v}=10$
$\Rightarrow\text{30v}=10-4$
$\Rightarrow\text{30v}=6$
$\Rightarrow\text{v}=\frac{6}{30}=\frac{1}{5}$
Now, $\text{u}=\frac{1}{11}$
$\Rightarrow\frac{1}{\text{x}+\text{y}}=\frac{1}{11}$
$\Rightarrow\text{x}+\text{y}=11\ \dots(5)$
and $\text{v}=\frac{1}{5}$
$\Rightarrow\frac{1}{\text{x}-\text{y}}=\frac{1}{5}$
$\Rightarrow\text{x}-\text{y}=5\ \dots(6)$
Adding $(5)$ and $(6)$, we get
$\text{2x}=16$
$\Rightarrow\text{x}= \frac{16}{2}=8$
Putting $x = 8$ in $(5)$, we get
$8 + y = 11$
$⇒ y = 11 - 8 = 3$
$\therefore$ the solution is $x = 8$, and $y = 3$
View full question & answer→Question 1084 Marks
In a cyclic quadrilateral $ABCD$, it is given that $\angle\text{A}=(\text{2x}+4)^\circ,\ \angle\text{B}=(\text{y}+3)^\circ,$ $\angle\text{C}=(\text{2y}+10)^\circ$ and $\angle\text{D}=(\text{4x}-5)^\circ$ Find the four angles.
AnswerGiven that in a cyclic quadrilateral $ABCD$,
$\angle\text{A}=(\text{2x}+4)^\circ,\ \angle\text{B}=(\text{y}+3)^\circ,$
$\angle\text{C}=(\text{2y}+10)^\circ$ and $\angle\text{D}=(\text{4x}-5)^\circ$
We know that,
Opposite angles of a quadrilateral sum upto $180^\circ $
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow (y + 3)^\circ + (4x - 5)^\circ = 180^\circ $
$\Rightarrow 4x + y = 182 ...(i)$
Similarly, $\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow (2x + 4)^\circ + (2y + 10)^\circ = 180^\circ $
$\Rightarrow 2x + 2y = 166$
$\Rightarrow x + y = 83 ...(i)$
Subtracting $(ii)$ from $(i)$, we get
$\Rightarrow 3x = 99$
$\Rightarrow x = 33$
Substituting $x = 33$ in $(ii)$, we get
$\Rightarrow y = 50$
Hence, the angles of $ABCD$ are
So, $\angle\text{A}=70^\circ,\ \angle\text{B}=53^\circ,$
$\angle\text{C}=110^\circ$ and $\angle\text{D}=127^\circ$
View full question & answer→Question 1094 Marks
Show graphically that each of the following given systems of equations has infinitely many solutions:
$3x - y = 5, 6x - 2y = 10$
AnswerOn a graph paper, draw a horizontal line $X'OX$ and a vertical line $YOY'$ representing the $x-$axis and $y-$axis, respectively.
The given system equations is $3x - y = 5, 6x - 2y = 10$
Graph of $3x - y = 5:$
$3x - y = 5$
$⇒ y = 3x - 5 ...(1)$
Thus, we have the following table for equation $(1)$
|
$x:$
|
$1$ |
$0$
|
$2$
|
|
$y:$
|
$-2$
|
$-5$
|
$1$
|
On the graph paper plot the points $A(1,-2), B(0,-5)$ and $C(2,1)$. Join $A B$ and $A C$ to get the graph line $B C$.
Thus, the line $B C$ is the graph of the equation of $3 x-y=5$.
Graph of $6 x-2 y=10$ :
For graph of $6 x-2 y=10$
$\Rightarrow\text{y}=\frac{\text{6x}-10}{2}\ \dots(2)$
Thus, we have the following table for equation $(2)$
|
$x:$
|
$0$
|
$1$
|
$2$
|
|
$y:$
|
$-5$
|
$-2$
|
$1$
|
These points are the same as obtained above.
From the graph, it is clear that these two lines coincide.
Both equations represents same graph.
Hence, these lines have infinitely many solutions. View full question & answer→Question 1104 Marks
A jeweller has bars of $18-$carat gold and $12-$carat gold. How much of each must be melted together to obtain a bar of $16-$carat gold, weighing $120 g?$ (Given: Pure gold is $24-$carat).
AnswerLet the amount of $18-$carat gold and $12-$car at gold to be melted be $x$ g and $y$ g respectively.
According to the given condition,
$\frac{18}{24}\text{x}+\frac{12}{24}\text{y}=\frac{16}{24}(120)$
$\Rightarrow 18x + 12y = 16(120)$
$\Rightarrow 3x + 2y = 320 ...(i)$
Since the amount of each add up to $120g,$
$\Rightarrow x + y = 120 ...(ii)$
Multiplying $(ii)$ by $2$, we get
$\Rightarrow 2x + 2y = 240 ...(iii)$
Subtract $(iii)$ from $(i)$.
$\Rightarrow x = 80$
Substituting $x = 80$ in $(ii)$, we get
$\Rightarrow y = 40$.
Hence, the amount of 18-carat gold is $80g$ and the amount of $12-$carat gold is $40g.$
View full question & answer→Question 1114 Marks
Solve for x and y:
$217x + 131y = 913,$
$131x + 217y = 827$
AnswerThe given equations are:
$217x + 131y = 913 ...(1)$
$131x + 217y = 827 ...(2)$
Adding $(1)$ and $(2)$, we get
$348x + 348y = 1740$
$348(x + y) = 1740$
$x + y = 5 ...(3)$
Subtracting $(2)$ from $(1)$, we get
$86x - 86y = 86$
$\Rightarrow 86(x - y) = 86$
$\Rightarrow x - y = 1 ...(4)$
Adding $(3)$ and $(4)$, we get
$2x = 6$
$\Rightarrow x = 3$
Putting $x = 3$ in $(3)$, we get
$3 + y = 5$
$\Rightarrow y = 5 - 3 = 2$
$\therefore$ The solution is $x = 3, y = 2$
View full question & answer→Question 1124 Marks
Solve the following systems of equations by using the method of cross multiplication:
$\frac{\text{x}}{6}+\frac{\text{y}}{15}=4,$
$\frac{\text{x}}{3}-\frac{\text{y}}{15}=\frac{\text{19}}{4}$
AnswerThe given equations may be written as:
$\frac{\text{x}}{6}+\frac{\text{y}}{15}-4=0\ \dots(\text{i})$
$\frac{\text{x}}{3}-\frac{\text{y}}{15}-\frac{\text{19}}{4}=0\ \dots(\text{ii})$
Here, $\text{a}_1=\frac{1}{6},\ \text{b}_1=\frac{1}{15},\ \text{c}_1=-4,$ $\text{a}_2=\frac{1}{3},\ \text{b}_2=-\frac{1}{12}$ and $\text{c}_2=-\frac{19}{4}$
By cross multiplication, we have:
$\therefore\frac{\text{x}}{\big[\frac{1}{15}\times\big(-\frac{19}{4}\big)-\big(-\frac{1}{12}\big)\times(-4)\big]}=\frac{{\text{y}}}{\big[(-4)\times\frac{1}{3}-\big(\frac{1}{6}\big)\times\big(-\frac{19}{4}\big)\big]}\\=\frac{1}{\big[\frac{1}{{6}}\times\big(\frac{-1}{12}\big)-\frac{1}{3}\times\frac{1}{15}\big]}$
$\Rightarrow\frac{\text{x}}{\big(-\frac{19}{60}-\frac{1}{3}\big)}=\frac{\text{y}}{\big(-\frac{4}{3}+\frac{19}{24}\big)}=\frac{1}{\big(-\frac{1}{72}-\frac{1}{45}\big)}$
$\Rightarrow\frac{\text{x}}{\big(-\frac{39}{60}\big)}=\frac{\text{y}}{\big(-\frac{13}{24}\big)}=\frac{1}{\big(\frac{13}{360}\big)}$
$\Rightarrow\text{x}=\Big[\Big(-\frac{39}{60}\Big)\times\Big(-\frac{360}{13}\Big)\Big]=18,$ $\text{y}=\Big[\Big(-\frac{13}{24}\Big)\times\Big(-\frac{360}{13}\Big)\Big]=15$
Hence, $x = 18$ and $y = 15$ is the required solution. View full question & answer→Question 1134 Marks
Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:
$2x - 5y + 4 = 0, 2x + y - 8 = 0$
AnswerOn a graph paper, draw a horizontal line $X^{\prime} O X$ and a vertical line $Y O Y^{\prime}$ representing the $x$-axis and $y$-axis, respectively. The given system equations is $2 x-5 y+4=0,2 x+y-8=0$
Graph of $2 x-5 y+4= 0$ :
$2 x-5 y+4=0$
$\Rightarrow y=\frac{2 x-4}{5} \ldots(1)$
Thus, we have the following table for equation $(1)$
|
$x:$
|
$3$ |
$-2$
|
$8$
|
|
$y:$
|
$2$
|
$0$
|
$4$
|
On the graph paper plot the points $A (3,2), B (-2,0)$ and $C (8,4)$. Join $A B$ and $A C$ to get the graph line $B C$.
Thus, the line $B C$ is the graph of the equation of $2 x-5 y+4=0$.
Graph of $2 x + y - 8 = 0$ :
For graph of $2 x+y-8=0$
$\Rightarrow y=-2 x+8 \ldots(2)$
Thus, we have the following table for equation $(2)$
|
$x:$
|
$3$
|
$1$
|
$2$
|
|
$y:$
|
$2$
|
$6$
|
$4$
|
Now, on the same graph paper plot the points $P(1,6)$ and $Q(2,4)$.
The third point $A(3,2)$ has already been plotted.
Join $PA.$
Thus, line $PA$ is the graph of the equation $2 x+y-8=0$.
On extending the graph lines on both sides, we find that these graph lines intersect the $y$-axis at the point $R(0,8)$ and $S(0,0.8)$ View full question & answer→Question 1144 Marks
Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solutions:
$2x + y = 2, 2x + y = 6$
Answer$2\text{x}+\text{y}=2$
$\Rightarrow\text{y}=2-\text{2x}$
| $x:$ |
$0$ |
$1$ |
| $y:$ |
$2$ |
$0$ |
$\text{2x}+\text{y}=6$
$\Rightarrow\text{y}=6-\text{2x}$
| $x:$ |
$0$ |
$3$ |
| $y:$ |
$6$ |
$0$ |
Since the two graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent. View full question & answer→Question 1154 Marks
Solve for $x$ and $y:$
$\frac{5}{\text{x}}-\frac{3}{\text{y}}=1,$
$\frac{3}{\text{2x}}+\frac{2}{\text{3y}}=\text{5}$ $(\text{x}\neq0,\ \text{y}\neq0).$
AnswerPutting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ the given equations become
$5u - 3v = 1 ...(1)$
$\frac{3\text{u}}{2}+\frac{\text{2v}}{3}=5$
$\frac{\text{9v}+4\text{v}}{6}=5$
$9u + 4v = 30 ...(2)$
Multiplying $(1)$ by $4$ and $(2)$ by $3$, we get
$20u - 12v = 4 ...(3)$
$27u + 12v = 90 ...(4)$
Adding $(3)$ and $(4),$ we get
$47u = 94$
$\Rightarrow\text{u}=\frac{94}{47}=2$
Putting $u = 2$ in $(1)$, we get
$(5 \times 2) - 3v = 1$
$\Rightarrow 10 - 3v = 1$
$\Rightarrow -3v = 1 - 10$
$\Rightarrow -3v = -9$
$\Rightarrow v = 3$
Now, $u = 2$
$\Rightarrow\frac{1}{\text{x}}=2$
$\Rightarrow\text{x}=\frac{1}{2}$
and, $v = 3$
$\Rightarrow\frac{1}{\text{y}}=3$
$\Rightarrow\text{y}=\frac{1}{3}$
$\therefore$ The solution is $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}{3}$
View full question & answer→Question 1164 Marks
Solve the following system of equations graphically:
$x + 2y + 2 = 0,$
$3x + 2y - 2 = 0$
AnswerOn a graph paper, draw a horizontal line $X^{\prime} O X$ and a vertical line $YOY'$ representing the $x$-axis and $y$-axis, respectively.
Given equations are $x+2 y+2=0$
and $3 x+2 y-2=0$
Graph of $x+2 y+2=0$ :
$x+2 y+2=0$
$\Rightarrow\text{y}=\frac{-\text{x}-2}{2}\ \dots(1)$
Thus, we have the following table for $x + 2y + 2 = 0$
|
$x:$
|
$-2$ |
$0$
|
$2$
|
|
$y:$
|
$0$
|
$-1$
|
$-2$
|
On the graph paper plot the points $A (-2,0), B (0,-1)$ and $C (2,-2)$. Join $A B$ and $B C$ to get the graph line $A C$.
Thus, the line $A C$ is the graph of the equation of $x+2 y+2=0$.
Graph of $3 x+2 y-2=0$ :
For graph of $3 x+2 y-2=0$
$\Rightarrow y=\frac{-3 x+2}{2} \ldots(2)$
Thus, we have the following table for $3 x+2 y-2=0$
|
$x:$
|
$0$
|
$2$
|
$4$
|
|
$y:$
|
$1$
|
$-2$
|
$-5$
|
Now, on the same graph paper plot the points $P (0,1)$ and $Q (4,-5)$. The third point $C(2,-2)$ has already been plotted.
Join $PC$ and $QC$ to get the line $PQ.$
Thus, line $PQ$ is the graph of the equation $3 x+2 y-2=0$.
The two graph lines intersect at $C(2, -2).$
$\therefore$ $x = 2, y = -2$ is the solution of the given system of equations. View full question & answer→Question 1174 Marks
Five years hence, a man's age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.
AnswerLet the present age of the man be $x$ years, and his son's age be $y$ years.
Accroding to the first condition,
$x + 5 = 3(y + 5)$
$\Rightarrow x + 5 = 3y + 15$
$\Rightarrow x - 3y = 10 ...(i)$
According to the second condition,
$x - 5 = 7(y - 5)$
$\Rightarrow x - 5 = 7y - 35$
$\Rightarrow x - 7y = - 30 ...(iii)$
Subtracting $(ii)$ from $(i)$, we get
$4y = 40$
$\Rightarrow y - 10$
Substituting $y = 10$ in $(i)$, we get
$\Rightarrow x = 40$
So, the present age of the man is $40$ years, and that of his son is $10$ years
View full question & answer→Question 1184 Marks
If three times the larger of two numbers is divided by the smaller, we get $4$ as the quotient and $8$ as the remainder. If five times the smaller is divided by the larger, we get $3$ as the quotient and $5$ as the remainder. Find the numbers.
AnswerLet the larger number be $x$ and smaller be $y$ repectively.
We know,
Dividend = Divisor $\times$ Quotient $+$ Remainder
$3x = y \times 4 + 8$
$3x - 4y = 8 ...(1)$
And
$5y = x \times 3 + 5$
$-3x + 5y = 5 ...(2)$
Adding $(1)$ and $(2)$, we get
$y = 13$
Putting $y = 13$ in $(1)$
$3x - 4 \times 13 = 8$
$\Rightarrow 3x = 8 + 52$
$\Rightarrow 3x = 60$
$\Rightarrow\text{x}=\frac{60}{3}$
$\Rightarrow x = 20$
Hence, the larger and smaller numbers are $20$ and $13$ respectively.
View full question & answer→Question 1194 Marks
Solve the following systems of equations by using the method of cross multiplication:
$\text{7x}-\text{2y}=3,$
$\text{11x}-\frac{3}{2}\text{y}=8$
AnswerThe given equations may be written as:
$\text{7x}-\text{2y}-3=0\ \dots(\text{i})$
$\text{11x}-\frac{3}{2}\text{y}-8=0\ \dots(\text{ii})$
Here, $a_1=7, b_1=-2, c_1=-3, a_2=11, b_2=-\frac{3}{2}$ and $c_2=-8$
By cross multiplication, we have:
$\therefore\frac{\text{x}}{\big[(-2)\times(-8)-\big(\frac{3}{2}\big)\times(-3)\big]}=\frac{{\text{y}}}{[(-3)\times11-(-8)\times7]}=\frac{1}{\big[7\times\big(\frac{-3}{2}\big)-11\times(-2)\big]}$
$\Rightarrow\frac{\text{x}}{\big(16-\frac{9}{2}\big)}=\frac{\text{y}}{(-33+56)}=\frac{1}{\big(-\frac{21}{2}+22\big)}$
$\Rightarrow\frac{\text{x}}{\big(\frac{23}2{}\big)}=\frac{\text{y}}{23}=\frac{1}{\big(\frac{23}{2}\big)}$
$\Rightarrow\text{x}=\frac{\frac{23}{2}}{\frac{23}{2}}=1,\ \text{y}=\frac{23}{\frac{23}{2}}=2$
Hence, $x = 1$ and $y = 2$ is the required solution. View full question & answer→Question 1204 Marks
Show graphically that each of the following given systems of equations has infinitely many solutions:
$x - 2y = 5, 3x - 6y = 15$
Answer$\text{x}-\text{2y}=5$
$\Rightarrow\text{y}=\frac{\text{x}-\text{5}}{2}$
| $x:$ |
$1$ |
$3$ |
| $y:$ |
$-2$ |
$-1$ |
$\text{3x}-\text{6y}=15$
$\Rightarrow\text{y}=\frac{\text{2x}-15}{6}$
| $x:$ |
$5$ |
$-1$ |
| $y:$ |
$0$ |
$-3$ |
Since the two graph of the system of equations is coincident lines, the system has infinitely many solutions. View full question & answer→Question 1214 Marks
Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:
$x - y - 5 = 0, 3x + 5y - 15 = 0$
Answer$\text{x}-\text{y}-5 = 0$
$\Rightarrow\text{y}=\text{x}-5$
| $x:$ |
$2$ |
$5$ |
| $y:$ |
$-3$ |
$0$ |
$3\text{x} + 5\text{y} -15 = 0$
$\Rightarrow\text{y}=\frac{15-\text{3x}}{5}$
| $x:$ |
$0$ |
$5$ |
| $y:$ |
$3$ |
$0$ |
Since the two graph intersect at $(5,0)$,
$x=5 \text { and } y=0$
The vertices of the triangle formed by these lines and the $y$-axis are $(5,0),(0,3)$ and $(0,-5)$.
So, height of the triangle $=$ distance from $(5,0)$ to $y$-axis
$=5 \text { units }$
$\text { Base }=8 \text { units }$
Area of triangle $=\frac{1}{2}\times$ base × height
$=\frac{1}{2}\times8\times5$
$=20\ \text{sq. units}$ View full question & answer→Question 1224 Marks
Solve the following system of equations graphically:
$3x + 2y = 12,$
$5x - 2y = 4$
Answer$\text{3x}+\text{2y}=12$
$\Rightarrow\text{y}=\frac{12-\text{3x}}{2}$
|
$x:$
|
$0$
|
$2$
|
|
$y:$
|
$6$
|
$3$
|
$\text{5x}-\text{2y}=4$
$\Rightarrow\text{y}=\frac{\text{5x}-4}{2}$
|
$x:$
|
$0$
|
$2$
|
|
$y:$
|
$-2$
|
$3$
|
Since the two graph intersect at $(2, 3),$
$x = 2$ and $y = 3$ View full question & answer→Question 1234 Marks
Find the angles of a cyclic quadrilateral $ABCD$ in which:
$\angle\text{A}=(\text{4x}+20)^\circ,$ $\angle\text{B}=(\text{3x}-5)^\circ,$ $\angle\text{C}=(\text{4y})^\circ$ and $\angle\text{D}=(\text{7y}+5)^\circ.$
AnswerGiven:
In a cyclic quadrilateral $ABCD$, we have:
$\angle\text{A}=(4\text{x}+20)^\circ$
$\angle\text{B}=(3\text{x}-5)^\circ$
$\angle\text{C}=(4\text{y})^\circ$
$\angle\text{D}=(7\text{y}+5)^\circ$
$\angle\text{A}+\angle\text{C}=180^\circ$ and $\angle\text{B}+\angle\text{D}=180^\circ$ [Since $ABCD$ is a cyclic quadrilateral]
Now, $\angle\text{A}+\angle\text{C}=(4\text{x}+20)^\circ+(4\text{y})^\circ=180^\circ$
$\Rightarrow 4x + 4y + 20 = 180$
$\Rightarrow 4x + 4y = 180 - 20 = 160$
$\Rightarrow x + y = 40 ....(i)$
Also, $\angle\text{B}+\angle\text{D}=(3\text{x}-5)^\circ+(7\text{y}+5)^\circ=180^\circ$
$\Rightarrow 3x + 7y = 180 ...(ii)$
On multiplying $(i)$ by $3$, we get:
$\Rightarrow 3x + 3y = 120 ...(iii)$
On subtracting $(iii)$ from $(ii)$, we get:
$4y = 60$
$\Rightarrow y = 15$
On subtracting $y = 15$ in $(1),$ we get:
$x + 15 = 40$
$\Rightarrow x = (40 - 15) = 25$
Therefore, we have:
$\angle\text{A}=(4\text{x}+20)^\circ=(4\times25+20)^\circ=120^\circ$
$\angle\text{B}=(3\text{x}-5)^\circ=(3\times25-5)^\circ=70^\circ$
$\angle\text{C}=(4\text{y})^\circ=(4\times15)^\circ=60^\circ$
$\angle\text{D}=(7\text{y}+5)^\circ=(7\times15+5)^\circ=(105+5)^\circ=110^\circ$
View full question & answer→Question 1244 Marks
Solve for $x$ and $y:$
$\frac{\text{bx}}{\text{a}}+\frac{\text{ay}}{\text{b}}=\text{a}^2+\text{b}^2,$
$\text{x}+\text{y}=\text{2ab}$
Answer$\frac{\text{bx}}{\text{a}}-\frac{\text{ax}}{\text{b}}+\text{a}^2+\text{b}^2$
By taking $L.C.M.$, we get
$\frac{\text{b}^2\text{x}+\text{a}^2\text{y}}{\text{ab}}=\text{a}^2+\text{b}^2$
$b^2x + a^2y = ab(a^2 + b^2) ...(1)$
$x + y = 2ab ...(2)$
Multiplying $(1)$ by $1$ and $(2)$ by a2
$b^2x + a^2y = a^3b + ab^3 ...(3)$
$a^2x + a^2y = 2a^3b ...(4)$
Subtracting $(4)$ from $(3)$, we get
$b^2x - a^2x = a^3b + ab^3 - 2a^3b$
$x(b^2 - a^2) = ab^3 - a^3b$
$x(b^2 - a^2) = ab(b^2 - a^2)$
$\therefore\ \text{x}=\frac{\text{ab}(\text{b}^2-\text{a}^2)}{(\text{b}^2-\text{a}^2)}=\text{ab}$
Substituting $x = ab$, in $(3)$, we get
$b^2(ab) + a^2y = a^3b + ab^3$
$b^3a + a^2y = a^3b + ab^3$
$a^2y = a^3b + ab^3 - b^3a$
$a^2y = a^3b$
$\Rightarrow\text{y}=\frac{\text{a}^3\text{b}}{\text{a}^3}=\text{ab}$
$\therefore$ solution is $x = ab, y = ab$
View full question & answer→Question 1254 Marks
Solve the following systems of equations by using the method of cross multiplication:
$3x + 2y + 25 = 0,$
$2x + y + 10 = 0$
AnswerThe given equations are:
$3x + 2y + 25 = 0 ...(i)$
$2x + y + 10 = 0 ...(ii)$
Here, $a_1=3, b_1=2, c_1=25, a_2=2, b_2=1$ and $c_2=10$
By cross multiplication, we have:
$\therefore\frac{\text{x}}{[2\times10-25\times1]}=\frac{{\text{y}}}{[25\times2-10\times3]}=\frac{1}{[3\times1-2\times2]}$
$\Rightarrow\frac{\text{x}}{(20-25)}=\frac{\text{y}}{(50-30)}=\frac{1}{(3-4)}$
$\Rightarrow\frac{\text{x}}{(-5)}=\frac{\text{y}}{(20)}=\frac{1}{(-1)}$
$\Rightarrow\text{x}=\frac{-5}{-1}=5,\ \text{y}=\frac{20}{-1}=-20$
Hence, $x = 5$ and $y = -20$ is the required solution. View full question & answer→Question 1264 Marks
Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:
$2x - 3y + 6 = 0, 2x + 3y - 18 = 0$
Answer$2\text{x}-3\text{y}+6 = 0$
$\Rightarrow\text{y}=\frac{2\text{x}+6}{3}$
| $x:$ |
$-3$ |
$0$ |
| $y:$ |
$0$ |
$2$ |
$2\text{x} + 3\text{y} -18 = 0$
$\Rightarrow\text{y}=\frac{18-\text{2x}}{3}$
| $x:$ |
$0$ |
$3$ |
| $y:$ |
$6$ |
$4$ |
Since the two graph intersect at $(3,4)$,
$x=3 \text { and } y=4$
The vertices of the triangle formed by these lines and the $y$-axis are $(3,4),(0,6)$ and $(0,2)$.
So, height of the triangle $=$ distance from $(3,4)$ to $y$-axis
$=3 \text { units }$
Base $=4$ units
Area of triangle $=\frac{1}{2}\times$ base × height
$=\frac{1}{2}\times4\times3$
$=6\ \text{sq. units}$ View full question & answer→Question 1274 Marks
Solve for $x$ and $y:$
$\frac{35}{\text{x}+\text{y}}+\frac{14}{\text{x}-\text{y}}=19,$ $\frac{14}{\text{x}+\text{y}}+\frac{35}{\text{x}-\text{y}}=37$
AnswerWe have:
$\frac{35}{\text{x+y}}+\frac{14}{\text{x}-\text{y}}=19$ and $\frac{14}{\text{x+y}}+\frac{35}{\text{x}-\text{y}}=37$
Taking $\frac{1}{\text{x+y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}.$
$35u + 14v - 19 = 0 ....(i)$
$14u + 35v - 37 = 0 ...(ii)$
Here, $a_1=35, b_1=14, c_1=-19, a_2=14, b_2=35, c_2=-37$
By cross multiplying, we have:
$\therefore\frac{\text{u}}{[14\times(-37)-35\times(-19)]}=\frac{\text{v}}{[(-19)\times14-(-37)\times(35)]}=\frac{1}{[35\times35-14\times14]}$
$\Rightarrow\frac{\text{u}}{-518+665}=\frac{\text{v}}{-266+1295}=\frac{1}{1225-196}$
$\Rightarrow\frac{\text{u}}{147}=\frac{\text{v}}{1029}=\frac{1}{1029}$
$\Rightarrow\text{u}=\frac{147}{1029}=\frac{1}{7},\ \text{v}=\frac{1029}{1029}=1$
$\Rightarrow\frac{1}{\text{x+y}}=\frac{1}{7},\ \frac{1}{\text{x}-\text{y}}=1$
$\therefore (x + y) = 7 ...(iii)$
And, $(x - y) = 1 ...(iv)$
Again, the equation $(iii)$ and $(iv)$ can be written as follows:
$x + y - 7 = 0 ...(v)$
$x - y - 1 = 0 ...(vi)$
Here $a_1=1, b_1=1, c_1=-7, a_2=1, b_2=-1, c_2=-1$
By multiplication, we have:
$\Rightarrow\frac{\text{x}}{[1\times(-1)-(-1)\times(-7)]}=\frac{\text{y}}{[(-7)\times1-(-1)\times1]}=\frac{1}{[1\times(-1)-1\times1]}$
$\Rightarrow\frac{\text{x}}{-1-7}=\frac{\text{y}}{-7+1}=\frac{1}{-1-1}$
$\Rightarrow\frac{\text{x}}{-8}=\frac{\text{y}}{-6}=\frac{1}{-2}$
$\Rightarrow\text{x}=\frac{-8}{-2}=4,\ \text{y}=\frac{-6}{-2}=3$
Hence, $x = 4$ and $y = 3$ is the required solution. View full question & answer→Question 1284 Marks
A man invested an amount at $10\%$ per annum and another amount at $8\%$ per annum simple interest. Thus, he received $₹ 1350$ as annual interest. Had he interchanged the amounts invested, he would have received $₹ 45$ less as interest. What amounts did he invest at different rates?
AnswerLet the amounts invested at $10 \%$ and $8 \%$ p.a. be Rs. $x$ and Rs. $y$ respectively.
Then, $SI$ on Rs. $x$ at $10 \%$ p.a. for $1$ year $=\text{Rs. }\Big(\frac{\text{x}\times10\times1}{100}\Big)$
$=\text{Rs. }\frac{\text{x}}{10}$
and $SI$ on Rs. $y$ at $8 \%$ p.a. for $1$ year $=\text{Rs. }\Big(\frac{\text{y}\times8\times1}{100}\Big)$
$=\text{Rs. }\frac{\text{2y}}{25}$
Total $SI = Rs. 1350$
$\therefore\frac{\text{x}}{10}+\frac{\text{2y}}{25}=1350$
$\Rightarrow 5x + 4y = 67500 ...(i)$
$SI$ on Rs.$x$ at $8 \%$p.a. for $1$ year $=\text{Rs. }\Big(\frac{\text{x}\times8\times1}{100}\Big)$
$=\text{Rs. }\frac{\text{2x}}{25}$
and $SI$ on Rs. $y$ at $10 \%$ p.a. for $1$ year $=\text{Rs. }\Big(\frac{\text{y}\times10\times1}{100}\Big)$
$=\text{Rs. }\frac{\text{y}}{10}$
Total $SI - Rs. 1350 - Rs 45 = Rs 1305$
$\therefore\frac{2\text{x}}{25}+\frac{\text{y}}{10}=1305$
$4x + 5y = 65250 ...(ii)$
Adding $(i)$ and $(ii)$, we get
$9x + 9 = 132750$
$\Rightarrow x + y = 14750 ...(iii)$
Subtracting $(ii)$ from $(i)$, we get
$x - y = 2250 ...(iv)$
Adding $(iii)$ and $(iv)$, we get
$2x = 17000$
$\Rightarrow x = 8500$
Substituting $x = 8500$ in $(iii)$, we get
$y = 6250$
Hence, amount invested at $10 \%=$ Rs. $8500$
and amount invested at $8 \%=$ Rs. $6250$
View full question & answer→Question 1294 Marks
In a $\triangle\text{ABC},\ \angle\text{A}=\text{x}^\circ,$ $\angle\text{B}=(\text{3x}-2)^\circ,\ \angle\text{C}=\text{y}^\circ$ and $\angle\text{C}-\angle\text{B}=9^\circ$ Find the three angles.
AnswerIn a $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ ...(Angle Sum Property)
$\Rightarrow x^\circ + (3x - 2)^\circ + y^\circ = 180^\circ $
$\Rightarrow 4x + y = 182 ...(i)$
Also, given that
$\angle\text{C}-\angle\text{B}=9^\circ$
$\Rightarrow y^\circ - (3x - 2)^\circ = 9^\circ $
$\Rightarrow y - 3x + 2 = 9$
$\Rightarrow 3x - y = -7 ...(ii)$
Adding $(i)$ and $(ii)$, we get
$7x = 175$
$\Rightarrow x = 25$
Substituting $x = 25$ in $(i)$, we get
$\Rightarrow y = 82$
So, $\angle\text{A}=25^\circ,\ \angle\text{B}=(3\text{x} - 2)^\circ=73^\circ$ and $\angle\text{C}=82^\circ$
View full question & answer→Question 1304 Marks
Solve for $x$ and $y:$
$\frac{5}{\text{x}}-\frac{3}{\text{y}}=1,$
$\frac{3}{\text{2x}}+\frac{2}{\text{3y}}=\text{5}$ $(\text{x}\neq0,\ \text{y}\neq0).$
AnswerPutting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ the given equations become
$5u - 3v = 1 ...(1)$
$\frac{3\text{u}}{2}+\frac{\text{2v}}{3}=5$
$\frac{\text{9v}+4\text{v}}{6}=5$
$9u + 4v = 30 ...(2)$
Multiplying $(1)$ by $4$ and $(2)$ by $3$, we get
$20u - 12v = 4 ...(3)$
$27u + 12v = 90 ...(4)$
Adding $(3)$ and $(4)$, we get
$47u = 94$
$\Rightarrow\text{u}=\frac{94}{47}=2$
Putting $u = 2$ in $(1)$, we get
$(5 \times 2) - 3v = 1$
$\Rightarrow 10 - 3v = 1$
$\Rightarrow -3v = 1 - 10$
$\Rightarrow -3v = -9$
$\Rightarrow v = 3$
Now, $u = 2$
$\Rightarrow\frac{1}{\text{x}}=2$
$\Rightarrow\text{x}=\frac{1}{2}$
and, $v = 3$
$\Rightarrow\frac{1}{\text{y}}=3$
$\Rightarrow\text{y}=\frac{1}{3}$
$\therefore$ The solution is $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}{3}$
View full question & answer→Question 1314 Marks
Solve for $x$ and $y:$
$x + y = 5xy,$
$3x + 2y = 13xy$
$(\text{x}\neq0,\ \text{y}\neq0).$
Answer$x + y = 5xy$ and $3x + 2y = 13xy$
Dividing throughtout by xy, we get
$\frac{1}{\text{y}}+\frac{1}{\text{x}}=5$ and $\frac{3}{\text{y}}+\frac{2}{\text{x}}=13$
Put $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$
So, we get
$u + v = 5 ...(i)$ and $2u + 3v = 13 ...(ii$)
Multiply $(i)$ by $3$ and subtract it from $(ii).$
$\Rightarrow 3u + 3v = 15$ and $2u + 3v = 13$
$\Rightarrow u = 2$
Substituting $u = 2$ in $(i)$, we get $v = 3$
$\Rightarrow\frac{1}{\text{x}}=2$ and $\frac{1}{\text{y}}=3$
$\Rightarrow\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}{3}$
View full question & answer→Question 1324 Marks
Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:
$x - y + 3 = 0, 2x + 3y - 4 = 0$
Answer$x - y + 3 = 0$
$⇒ y = x + 3$
| $x:$ |
$0$ |
$-1$ |
| $y:$ |
$3$ |
$2$ |
$2x + 3y - 4 = 0$
$\Rightarrow\text{y}=\frac{4-\text{2x}}{3}$
| $x:$ |
$-4$ |
$-1$ |
| $y:$ |
$4$ |
$2$ |
Since the two graph intersect at $(-1,2)$,
$x=-1 \text { and } y=2$
The vertices of the triangle formed by these lines and the $x$-axis are $(-3,0),(2,0)$ and $(-1,2)$. So, height of the triangle $=$ distance from $(-1,2)$ to $x$-axis $=2$ units
Base $=5$ units
Area of triangle $=\frac{1}{2}\times$ base × height
$=\frac{1}{2}\times5\times2$
$=5\ \text{sq. units}$ View full question & answer→Question 1334 Marks
Show graphically that each of the following given systems of equations has infinitely many solutions:
$2x + 3y = 6, 4x + 6y = 12$
Answer$\text{2x}+\text{3y}=6$
$\Rightarrow\text{y}=\frac{\text{6}-\text{2x}}{3}$
| $x:$ |
$-3$ |
$3$ |
| $y:$ |
$4$ |
$0$ |
$\text{4x}+\text{6y}=12$
$\Rightarrow\text{y}=\frac{12-\text{4x}}{6}$
| $x:$ |
$-6$ |
$0$ |
| $y:$ |
$6$ |
$2$ |
Since the two graph of the system of equations is coincident lines, the system has infinitely many solutions. View full question & answer→Question 1344 Marks
The sum of two numbers is $16$ and the sum of their reciprocals is $\frac{1}{3}.$ Find the numbers.
AnswerLet the two numbers be $x$ and $y$ respectively.
Accroding to the given question:
$\therefore x + y = 16 ...(1)$
And
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{3}\ \dots(2)$
From $(2),$
$\frac{\text{x}+\text{y}}{\text{xy}} =\frac{1}{3}$ or $\frac{16}{\text{xy}}=\frac{1}{3} [x + y = 16]$
$\therefore xy = 48$
We know,
$ (x-y)^2=(x+y)^2-4 x y$
$ =16^2-4 \times 48=256-192=64$
$\therefore x - y = 8 ...(3)$
Adding $(1)$ and $(3),$ we get
$2x = 24$
$\therefore x = 12$
Putting $x = 12$ in $(1),$
$y = 16 - x$
$= 16 - 12$
$= 4$
$\therefore$ The required numbers are $12$ and $4$
View full question & answer→Question 1354 Marks
Solve the following systems of equations by using the method of cross multiplication:
$6x - 5y - 16 = 0,$
$7x - 13y + 10 = 0$
AnswerThe given equations are:
$6x - 5y - 16 = 0 ...(i)$
$7x - 13y + 10 = 0 ...(ii)$
Here, $a_1=6, b_1=-5, c_1=-16, a_2=7, b_2=-13$ and $c_2=10$
By cross multiplication, we have:
$\therefore\frac{\text{x}}{[(-5)\times10-(-16)\times(-13)]}=\frac{{\text{y}}}{[(-16)\times7-10\times6]}=\frac{1}{[6\times(-13)-(-5)\times7]}$
$\Rightarrow\frac{\text{x}}{(-50-208)}=\frac{\text{y}}{(-112-60)}=\frac{1}{(-78+35)}$
$\Rightarrow\frac{\text{x}}{(-258)}=\frac{\text{y}}{(-172)}=\frac{1}{(-43)}$
$\Rightarrow\text{x}=\frac{-258}{-43}=6,\ \text{y}=\frac{-172}{-43}=4$
Hence, $x = 6$ and $y = 4$ is the required solution. View full question & answer→Question 1364 Marks
Solve for $x$ and $y:$
$2\text{x}-\frac{3}{\text{y}}=\text{9},$
$\text{3x}+\frac{7}{\text{y}}=\text{2}$ $(\text{y}\neq0).$
AnswerPutting $\frac{1}{\text{y}}=\text{v}$ the given equations become
$2x - 3v = 9 ...(1)$
$3x + 7v = 2 ...(2)$
Multiplying $(1)$ by $7$ and $(2)$ by $3$, we get
$14x - 21v = 63 ...(3)$
$9x + 21v = 6 ...(4)$
Adding $(3)$ and $(4)$, we get
$23x = 69$
$\Rightarrow\text{x}=\frac{39}{13}=3$
Putting $x = 3$ in $(1),$ we get
$2 \times 3 - 3v = 9$
$-3v = 9 - 6$
$\Rightarrow -3v = 3$
$\Rightarrow v = -1$
$\Rightarrow\frac{1}{\text{y}}=-1$
$\Rightarrow y = -1$
$\therefore$ The solution is $x = 3$ and $y = 1$
View full question & answer→Question 1374 Marks
If $1$ is added to both of the numerator and denominator of a fraction, it becomes $\frac{4}{5}.$ If however, $5$ is subtracted from both numerator and denominator, the fraction $\frac{1}{2}.$ Find the fraction.
AnswerLet the required fraction be $\frac{\text{x}}{\text{y}}.$
Then, we have:
$\frac{\text{x}+1}{\text{y}+1}=\frac{4}{5}$
$\Rightarrow 5(x + 1) = 4(y + 1)$
$\Rightarrow 5x + 5 = 4y + 4$
$\Rightarrow 5x - 4y = -1 ...(i)$
Again, we have:
$\frac{\text{x}-5}{\text{y}-5}=\frac{1}{2}$
$\Rightarrow 2(x - 5) = 1(y - 5)$
$\Rightarrow 2x - 10 = y - 5$
$\Rightarrow 2x - y = 5 ...(ii)$
On multiplying $(ii)$ by $4$, we get:
$\Rightarrow 8x - 4y = 20 ....(iii)$
On subtracting $(i)$ from $(iii)$, we get:
$3x = (20 - (-1)) = 20 + 1 = 21$
$\Rightarrow 3x = 21$
$\Rightarrow x = 7$
On substituting $x = 7$ in $(i)$, we get:
$5 \times 7 - 4y = -1$
$\Rightarrow 35 - 4y = -1$
$\Rightarrow 4y = 36$
$\Rightarrow y = 9$
$\therefore x = 7$ and $y = 9$
Hence, the required fraction is $\frac{7}{9}.$
View full question & answer→