Question 515 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:
4x - 3y + 4 = 0, 4x + 3y - 20 = 0
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 4x - 3y + 4 = 0, 4x + 3y - 20 = 0 Graph of 4x - 3y + 4 = 0: 4x - 3y + 4 = 0 $\Rightarrow\text{y}=\frac{4\text{x}+4}{3}\ \dots(1)$ Thus, we have the following table for equation (1)
On the graph paper plot the points A(-1, 0), B(2, 4) and C(5, 8). Join AB and BC to get the graph line AC. Thus, the line AC is the graph of the equation of 4x - 3y + 4 = 0. Graph of 4x + 3y - 20 = 0: For graph of 4x + 3y - 20 = 0 $\Rightarrow\text{y}=\frac{-\text{4x}+20}{3}\ \dots(2)$ Thus, we have the following table for equation (2)
Now, on the same graph paper plot the points P(-1, 8) and Q(5, 0). The third point B(2, 4) has already been plotted. Join PB and QB to get the line PQ. Thus, line PQ is the graph of the equation 4x + 3y - 20 = 0.
The two graph lines intersect at B(2, 4). $\therefore$ x = 2, y = 4 is the solution of the given system of equations. Clearly, the vertices of $\triangle\text{ABQ}$ formed by these lines and the x-axis are A(-1, 0), B(2, 4) and Q(5, 0) Consider the triangle $\triangle\text{ABQ}:$ Height of the triangle = 4 units and base (AQ) = 6 units Area of triangle Area $=\Big(\frac{1}{2}\times\text{Base}\times\text{height}\Big)\text{sq. units}$ $=\Big(\frac{1}{2}\times4\times6\Big)\text{sq. units}$ Area of $\triangle\text{ABQ}=12\text{sq. }\text{units}$ View full question & answer→Question 525 Marks
Solve for x and y:
$\frac{1}{2(\text{x}+\text{2y)}}+\frac{5}{3(3\text{x}-\text{2y)}}=\frac{-3}{2},$
$\frac{5}{4(\text{x}+\text{2y)}}-\frac{3}{5(3\text{x}-\text{2y)}}=\frac{61}{60}$
Answer$\frac{1}{2(\text{x}+\text{2y)}}+\frac{5}{3(3\text{x}-\text{2y)}}=\frac{-3}{2},$ and $\frac{5}{4(\text{x}+\text{2y)}}-\frac{3}{5(3\text{x}-\text{2y)}}=\frac{61}{60}$ Putting $\frac{1}{\text{x}+\text{2y}}=\text{u},\ \frac{1}{\text{3x}-\text{2y}}=\text{v,}$ we get $\frac{1}{2}\text{u}+\frac{5}{3}\text{v}=-\frac{3}{2}\ \dots(1)$ $\frac{5}{4}\text{u}-\frac{3}{5}\text{v}=\frac{61}{60}\ \dots(2)$ Multiplying (1) by 6 and (2) by 20, we get $\text{3u}+\text{10v}=-9\ \dots(3)$ $\text{25u}-12\text{v}=\frac{61}{3}\ \dots(4)$ Multiplying (3) by 6 and (4) by 5, we get $\text{18u}+\text{60v}=-54\ \dots(5)$ $\text{125u}-60\text{v}=\frac{305}{3}\ \dots(6)$ Adding (5) and (6), we get$\text{143u}=\frac{305}{3}-54=\frac{305-162}{2}=\frac{143}{3}$
$\therefore\text{u}=\frac{1}{3}=\frac{1}{\text{x}+\text{2y}}$
$\therefore\text{x}+\text{2y}=3\ \dots(7)$
Putting value of u in (3), we get 1 + 10v = -9 10v = -10 or v = -1 $\Rightarrow-1=\frac{1}{\text{3x}-\text{2y}}$ 3x - 2y = -1 ...(8) Adding (7) and (8), we get $\text{4x}=2$ $\therefore\text{x}=\frac{1}{2}$ Putting value of x in (7) $\frac{1}2{}+\text{2y}=3$ or $\text{2y}=3-\frac{1}{2}=\frac{5}2{}$ $\therefore\ \text{y}=\frac{5}{4}$ Thus required solution is $\text{x}=\frac{1}{2},\ \text{y}=\frac{5}{4}$
View full question & answer→Question 535 Marks
Solve for x and y:
$\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=0,$
$\text{ax}+\text{by}=(\text{a}^2+\text{b}^2)$
Answer$\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=0$
$\Rightarrow\text{x}=\frac{\text{ay}}{\text{b}}\ \dots(\text{i})$
$\text{ax}+\text{by}=(\text{a}^2+\text{b}^2)\ \dots(\text{ii})$
Substituting (i) in (ii), we get
$\text{a}\Big(\frac{\text{ay}}{\text{b}}\Big)+\text{by}=(\text{a}^2+\text{b}^2)$
$\Rightarrow\Big(\frac{\text{a}^2\text{y}}{\text{b}}\Big)+\text{by}=(\text{a}^2+\text{b}^2)$
$\Rightarrow\text{a}^2\text{y}+\text{b}^2\text{y}=(\text{a}^2\text{b}+\text{b}^3)$
$\Rightarrow\text{y}(\text{a}^2+\text{b}^2)=\text{b}(\text{a}^2+\text{b}^2)$
$\Rightarrow\text{y}=\text{b}$
Substituting in (i), we get x = a
So, x = a and y = b.
View full question & answer→Question 545 Marks
Places A and B are 160km apart on a highway. One car starts from A and another car from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.
AnswerLet X and Y be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.
Then, AB = 160km.
Case 1: When the two cars move in the same direction In this case, let the two cars meet at a point M.
Distance covered by X in 8 hours = 8x km
Distance covered by Y in 8 hours = 8y km
$\therefore$ AM = (8x)km and BM = (8y)km
⇒ AM - BM = AB
⇒ (8x - 8y) = 160
⇒ x - y = 20 ...(i)
Case 2: When the two cars move in the opposite direction.
In this case, let the two cars meet at a point P.
Distance covered by X in 2 hours = 2x km
Distance covered by Y in 2 hours - 2y km
AP = (2x)km and BP = (2y)km
⇒ AP + BP = AB
⇒ (2x + 2y) = 160
⇒ x + y = 80 .....(ii)
Adding (i) and (ii), we get
2x - 100
⇒ x = 50
Substituting x = 50 in (ii), we get
⇒ y = 30.
Hence, the speed of the car starting from A is 50km/h
and that of the one starting from B is 30km/h. View full question & answer→Question 555 Marks
Solve for x and y:
$\frac{3}{\text{x}}-\frac{1}{\text{y}}+\text{9}=0,$
$\frac{2}{\text{x}}+\frac{3}{\text{y}}=\text{5}$ $(\text{x}\neq0,\ \text{y}\neq0).$
AnswerPutting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ the given equations become 3u - v = 9 ...(1) 2u + 3v = 5 ...(2)Multiplying (1) by 3 and (2) by 1, we get
9u - 3v = -27 ...(3)
2u + 3v = 5 ...(4)
Adding (3) and (4), we get
11u = -22
$\Rightarrow\text{u}=\frac{-22}{11}=-2$Putting u = -2 in (1), we get
3 × (-2) - v = -9
⇒ -6 - v = -9
⇒ -v = -9 + 6
⇒ -v = -3
⇒ 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 565 Marks
Solve the following system of equations graphically:
2x + 3y - 4 = 0,
3x - y + 5 = 0
Answer$\text{2x}+\text{3y}-4=0$ $\Rightarrow\text{y}=\frac{4-\text{2x}}{3}$
$\text{3x}-\text{y}+5=0$ $\Rightarrow\text{y}=\text{3x}+5$
Since the two graph intersect at (-1, 2), x = -1 and y = 2 View full question & answer→Question 575 Marks
The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m. If the length is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.
AnswerLet the length of the rectangle be x and the breadth be y.
So, the area of the rectangle = xy
According to the first condition,
(x + 3)(y - 4) = xy - 67
⇒ xy - 4x + 3y - 12 - xy - 67
⇒ 4x + 3y = -55 ...(i)
According to the second condition,
(x - 1)(y + 4) = xy + 89
⇒ xy + 4x - y - 4 = xy + 89
⇒ 4x - y = 93 ...(ii)
Adding (i) and (ii), we get
2y = 38
⇒ y = 19
Substituting y = 19 in (ii), we get x = 28.
Hence, the dimensions of the rectangle are 28m and 19m.
View full question & answer→Question 585 Marks
Draw the graphs of the following equations on the same graph paper:
2x + y = 2, 2x + y = 6
Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.
Hint: The line 2x + y = 2 cuts the x-axis at A(1, 0) and the y-axis at B(0, 2).
The line 2x + y = 6 cuts the x-axis at C(3, 0) and the y-axis at D(0, 6).
Area of trap. ABCD $=\text{ar}(\triangle\text{OCD})-\text{ar}(\triangle\text{OAB})$
$=\Big(\frac{1}{2}\times3\times6\Big)-\Big(\frac{1}{2}\times1\times2\Big)$
$=8\ \text{sq. units}$
Answer$2\text{x}+\text{y}=2$ $\Rightarrow\text{y}=2-\text{2x}$
$\text{2x}+\text{y}=6$ $\Rightarrow\text{y}=6-\text{2x}$
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 595 Marks
Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80km, he pays ₹ 1,330, and travelling 90km, he pays ₹ 1,490. Find the fixed charges and rate per km.
AnswerLet the fixed charges be ₹ x and other charges be ₹ y per km.
According to the given condition,
x + 80y - 1330 ...(i)
x + 90y = 1490 ...(ii)
Subtracting (i) from (ii), we get
10y = 160
⇒ y = 16
Substituting y - 16 in (i), we get
x = 50.
Hence, the fixed charges is ₹ 50 and the other charges is ₹ 16 per km.
View full question & answer→Question 605 Marks
A sailor goes 8km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.
AnswerLet the speed of the sailor in still water be x km/h and the speed of the current be y km/hr.
The upstream speed = (x + y) km/hr
The upstream speed = (x - y) km/hr
We know that, distance = speed × time
When the sail or goes downstream,
$(\text{x}+\text{y})\times\frac{40}{60}=8$
⇒ x + y = 12 ....(i)
When the sailor goes upstream,
(x - y) × 1 = 8
⇒ x - y = 8 ...(ii)
Adding (i) and (ii), we get
2x = 20
⇒ x = 10
Substitutiting in (i), we get
⇒ y = 2.
Hence, the speed of the sailor in still water is 10km/hr
and the speed of the current is 2km/hr.
View full question & answer→Question 615 Marks
The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves ₹ 9000 per month, find the monthly income of each.
AnswerLet the monthly incomes of A and B be ₹ 5x and 4x respectively and let their expenditures be ₹ 7y and 5y respectively.
We know that, savings = income - expenditure
Then, A's monthly savings = ₹ (5x - 7y)
and B's monthly savings = ₹ (4x - 5y)
But, the monthly saving of each is ₹ 9000.
$\therefore$ 5x - 7y = 9000 ...(i)
and 4x - 5y = 9000 ...(ii)
Multiply (i) by 5 and (ii) by 7,
25x - 35y - 45000 ...(iii) and
28x - 35y - 63000 ...(iv)
Subtracting (iii) from (iv), we get
3x = 18000
⇒ x = 6000
Substituting x = 6000 in (i), we get
⇒ y = 3000.
So, A's income = 5x = ₹ 30000
and B's income = 4x = ₹ 24000
Hence, the monthly income of A is ₹ 30000
and that of B is ₹ 24000.
View full question & answer→Question 625 Marks
Solve for x and y:
$px + qy = p - q,$
$qx - py = p + q$
Answer$px + qy = p - q ...(i)$
$qx - py = p + q ...(ii)$
Multiplying (i) by p and (ii) by q and adding, we get
$p^2x + q^2x = p^2- pq + pq + q^2$
$\Rightarrow\text{x}=\frac{\text{p}^2+\text{q}^2}{\text{p}^2+\text{q}^2}$
$\Rightarrow x = 1$
Substituting $x = 1$ in (i), we get
$p + qy = p - q$
$y = -1$
So, $x = 1$ and $y = -1$
View full question & answer→Question 635 Marks
5 chairs and 4 tables together cost ₹ 5,600, while 4 chairs and 3 tables together cost ₹ 4,340. Find the cost of a chair and that of a table.
AnswerLet each chair cost ₹ x and each table cost ₹ y
According to the first condition,
5x + 4y = 5600 ...(i)
According to the second condition,
4x + 3y = 4340 ...(ii)
Multiplying (i) by 3 and (ii) by 4,
We get:
15x + 12y = 16800 and 16x + 12y = 17360
Subracting the above equations, we get
x = 560
Substituting x = 560 in (iii), we get
y = 700
Hence, the cost of each chair is ₹ 560 and the cost of each table is ₹ 700.
View full question & answer→Question 645 Marks
The difference between two numbers is $14$ and the difference between their squares is $448.$ Find the numbers.
AnswerLet the numbers be x and y respectively.
According to the question:
$x - y = 14 ...(1)$
$x^2 - y^2 = 448 ...(2)$
From (1), we get:
$x = 14 + y ...(3)$
Putting $x = 14 + y$ in $(2),$ we get
$(14 + y)^2 - y^2 = 448$
$196 + y^2 + 28y - y^2 = 448$
$196 + 28y = 448$
$28y = 448 - 196$
$\text{y}=\frac{252}{28}$
$y = 9$
Putting $y = 9$ in (1), we get
$x - 9 = 14$
$\Rightarrow x = 14 + 9$
$\Rightarrow x = 23$
Hence, the required numbers are $23$ and $9.$
View full question & answer→Question 655 Marks
Solve the following system of equations graphically:
2x - 5y + 4 = 0,
2x + y - 8 = 0
AnswerOn a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-axis and y-axis, respectively. Given equations are 2x - 5y + 4 = 0 and 2x + y - 8 = 0 Graph of 2x - 5y + 4 = 0: 2x - 5y + 4 = 0 $\Rightarrow\text{y}=\frac{\text{2x}+4}{5}\ \dots(1)$ Thus, we have the following table for 2x - 5y + 4 = 0
On the graph paper plot the points A(-2, 0), B(3, 2) and C(8, 4). Join AB and BC to get the graph line AC. Thus, the line AC is the graph of the equation of 2x - 5y + 4 = 0. Graph of 2x + y - 8 = 0: For graph of 2x + y - 8 = 0 ⇒ y = -2x + 8 ...(2) Thus, we have the following table for 2x + y - 8 = 0
Now, on the same graph paper plot the points P(1, 6) and Q(2, 4). The third point B(3, 2) has already been plotted. Join PQ and QB to get the line PB. Thus, line PB is the graph of the equation 2x + y - 8 = 0.
The two graph lines intersect at B(3, 2). $\therefore$ x = 3, y = 2 is the solution of the given system of equations. View full question & answer→Question 665 Marks
A boat goes 12km upstream and 40km downstream in 8 hours. It can go 16km upstream in 8 hours. It can go 16km upstream and 32km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
AnswerLet the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.
Then,
Speed upstream = (x - y)km/hr
Speed downstream = (x + y) km/hr
Time taken to cover 12km upstream $=\frac{12}{\text{x}-\text{y}}\text{hrs.}$
Time taken to cover 40km downstream $=\frac{40}{\text{x}+\text{y}}\text{hrs.}$
Total time taken = 8hrs.
$\therefore\frac{12}{\text{x}-\text{y}}+\frac{40}{\text{x}+\text{y}}=8$
Again, time taken to cover 16km upstream $=\frac{16}{(\text{x}-\text{y})}$
Time taken to taken to cover 32km downstream $=\frac{32}{(\text{x}+\text{y})}$
Total time taken = 8hrs.
$\therefore\frac{16}{(\text{x}-\text{y})}+\frac{32}{(\text{x}+\text{y})}=8$
Putting $\frac{1}{(\text{x}-\text{y})}=\text{u}$ and $\frac{1}{(\text{x}+\text{y})}=\text{v},$ we get
12u + 40v = 8
3u + 10v = 2 ...(1)
and
16u + 32v = 8
2u + 4y = 1 ...(2)
Multiplying (1) by 4 and (2) by 10, we get
12u + 40v = 8 ...(3)
200 + 40v = 10 ...(4)
Subtracting (3) from (4), we get
$\text{8u}=2$
$\Rightarrow\text{u}=\frac{1}{4}$
Putting $\text{u}=\frac{1}{4}$ in (3), we get
$3\times\frac{1}{4}+\text{10v}=2$
$\Rightarrow\text{10v}=\frac{5}{4}$
$\Rightarrow\text{v}=\frac{1}{8}$
$\text{u}=\frac{1}{4}$
$\Rightarrow\frac{1}{\text{x}-\text{y}}=\frac{1}{4}$
$\Rightarrow\text{x}-\text{y}=4\ \dots(5)$
$\text{v}=\frac{1}{8}$
$\Rightarrow\frac{1}{\text{x}+\text{y}}=\frac{1}{8}$
$\Rightarrow\text{x}+\text{y}=8\ \dots(6)$
On adding (5) and (6), we get
2x = 12
x = 6
Putting x = 6 in (6), we get
6 + y = 8
⇒ y = 8 - 6 = 2
$\therefore$ x = 6, y = 2
Hence, the speed of the boat in still water = 6km/hr and speedot the stream = 2km/hr
View full question & answer→Question 675 Marks
Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solutions:
x - 2y = 6, 3x - 6y = 0
Answer$\text{x}-\text{2y}=6$ $\Rightarrow\text{y}=\frac{\text{x}-\text{6}}{2}$
$\text{3x}-\text{6y}=0$ $\Rightarrow\text{y}=\frac{1}{2}\text{x}$
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 685 Marks
Points A and B are 70km 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
⇒ 7x - 7y = 70
⇒ 7(x - y) = 70
⇒ 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
⇒ x= 40
Putting x = 40 in (1), we get
40 - y = 10
⇒ y = (40 - 10) = 30
$\therefore$ x = 40, y = 30
Hence, the speeds of these cars are 40km/hr and 30km/hr respectively. View full question & answer→Question 695 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}$
$2\text{x} + \text{y} - 6 = 0$ $\Rightarrow\text{y}=6-\text{2x}$
Since the two graph intersect at (2, 2), x = 2 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 705 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 × 4 - 4v = 8
⇒ 36 - 4v = 8
⇒ -4v = 8 - 36
⇒ -4v = -28
⇒ 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 715 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 725 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,
$\frac{90}{100}\text{x}+\frac{97}{100}\text{y}=\frac{95}{100}(21)$
⇒ 90x + 97y = 95(21)
⇒ 90x + 97y = 1995 ...(i)
Since the amount of each solutions adds to 21 liters,
⇒ x + y = 21 ...(ii)
Multiplying (ii) by 90, we get
⇒ 90x + 90y = 1890 ...(iii)
Subtract (iii) from (i).
⇒ 7y = 105
⇒ x = 80
Substituting y = 15 in (ii), we get
⇒ x = 6.
Hence, the amount of each type is 6 liters and 15 liters.
View full question & answer→Question 735 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
$au - bv + 0 = 0 ...(i)$
$ab^2u + a^2bv - (a^2 + b^2) = 0 ...(ii)$
Here, $ a_1 = a, b_1 = -b, c_1 = 0, a_2 = ab^2, b_2 = a^2b$ and $c_2 = -(a^2 + b^2).$
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 745 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 755 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°
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
⇒ (y + 3)° + (4x - 5)° = 180°
⇒ 4x + y = 182 ...(i)
Similarly, $\angle\text{A}+\angle\text{C}=180^\circ$
⇒ (2x + 4)° + (2y + 10)° = 180°
⇒ 2x + 2y = 166
⇒ x + y = 83 ...(i)
Subtracting (ii) from (i), we get
⇒ 3x = 99
⇒ x = 33
Substituting x = 33 in (ii), we get
⇒ 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 765 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)
On the graph paper plot the points A(1, -2), B(0, -5) and C(2, 1). Join AB and AC to get the graph line BC. Thus, the line BC is the graph of the equation of 3x - y = 5. Graph of 6x - 2y = 10: For graph of 6x - 2y = 10 $\Rightarrow\text{y}=\frac{\text{6x}-10}{2}\ \dots(2)$ Thus, we have the following table for equation (2)
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 775 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,
$\frac{18}{24}\text{x}+\frac{12}{24}\text{y}=\frac{16}{24}(120)$
⇒ 18x + 12y = 16(120)
⇒ 3x + 2y = 320 ...(i)
Since the amount of each add up to 120g,
⇒ x + y = 120 ...(ii)
Multiplying (ii) by 2, we get
⇒ 2x + 2y = 240 ...(iii)
Subtract (iii) from (i).
⇒ x = 80
Substituting x = 80 in (ii), we get
⇒ 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 785 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
⇒ 86(x - y) = 86
⇒ x - y = 1 ...(4)
Adding (3) and (4), we get
2x = 6
⇒ x = 3
Putting x = 3 in (3), we get
3 + y = 5
⇒ y = 5 - 3 = 2
$\therefore$ The solution is x = 3, y = 2
View full question & answer→Question 795 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 805 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'OX and a vertical line YOY' representing the x-axis and y-axis, respectively.
The given system equations is 2x - 5y + 4 = 0, 2x + y - 8 = 0
Graph of 2x - 5y + 4 = 0:
2x - 5y + 4 = 0
$\Rightarrow\text{y}=\frac{\text{2x}-4}{5}\ \dots(1)$
Thus, we have the following table for equation (1)
On the graph paper plot the points A(3, 2), B(-2, 0) and C(8, 4).
Join AB and AC to get the graph line BC.
Thus, the line BC is the graph of the equation of 2x - 5y + 4 = 0.
Graph of 2x + y - 8 = 0:
For graph of 2x + y - 8 = 0
⇒ y = -2x + 8 ...(2)
Thus, we have the following table for equation (2)
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 2x + 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 815 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$2\text{x}+\text{y}=2$ $\Rightarrow\text{y}=2-\text{2x}$
$\text{2x}+\text{y}=6$ $\Rightarrow\text{y}=6-\text{2x}$
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 825 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 × 2) - 3v = 1
⇒ 10 - 3v = 1
⇒ -3v = 1 - 10
⇒ -3v = -9
⇒ 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 835 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'OX and a vertical line YOY' representing the x-axis and y-axis, respectively. Given equations are x + 2y + 2 = 0 and 3x + 2y - 2 = 0 Graph of x + 2y + 2 = 0: x + 2y + 2 = 0 $\Rightarrow\text{y}=\frac{-\text{x}-2}{2}\ \dots(1)$ Thus, we have the following table for x + 2y + 2 = 0
On the graph paper plot the points A(-2, 0), B(0, -1) and C(2, -2). Join AB and BC to get the graph line AC. Thus, the line AC is the graph of the equation of x + 2y + 2 = 0. Graph of 3x + 2y - 2 = 0: For graph of 3x + 2y - 2 = 0 $\Rightarrow\text{y}=\frac{-\text{3x}+2}{2}\ \dots(2)$ Thus, we have the following table for 3x + 2y - 2 = 0
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 3x + 2y - 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 845 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)
⇒ x + 5 = 3y + 15
⇒ x - 3y = 10 ...(i)
According to the second condition,
x - 5 = 7(y - 5)
⇒ x - 5 = 7y - 35
⇒ x - 7y = - 30 ...(iii)
Subtracting (ii) from (i), we get
4y = 40
⇒ y - 10
Substituting y = 10 in (i), we get
⇒ 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 855 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 × Quotient + Remainder
3x = y × 4 + 8
3x - 4y = 8 ...(1)
And
5y = x × 3 + 5
-3x + 5y = 5 ...(2)
Adding (1) and (2), we get
y = 13
Putting y = 13 in (1)
3x - 4 × 13 = 8
⇒ 3x = 8 + 52
⇒ 3x = 60
$\Rightarrow\text{x}=\frac{60}{3}$
⇒ x = 20
Hence, the larger and smaller numbers are 20 and 13 respectively.
View full question & answer→Question 865 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 875 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}$
$\text{3x}-\text{6y}=15$ $\Rightarrow\text{y}=\frac{\text{2x}-15}{6}$
Since the two graph of the system of equations is coincident lines, the system has infinitely many solutions. View full question & answer→Question 885 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$
$3\text{x} + 5\text{y} -15 = 0$ $\Rightarrow\text{y}=\frac{15-\text{3x}}{5}$
Since the two graph intersect at (5, 0), x = 5 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 units Base = 8 units Area of triangle $=\frac{1}{2}\times$ base × height $=\frac{1}{2}\times8\times5$ $=20\ \text{sq. units}$ View full question & answer→Question 895 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}$
$\text{5x}-\text{2y}=4$ $\Rightarrow\text{y}=\frac{\text{5x}-4}{2}$
Since the two graph intersect at (2, 3), x = 2 and y = 3 View full question & answer→Question 905 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$
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y = 180 - 20 = 160
⇒ x + y = 40 ....(i)
Also, $\angle\text{B}+\angle\text{D}=(3\text{x}-5)^\circ+(7\text{y}+5)^\circ=180^\circ$
⇒ 3x + 7y = 180 ...(ii)
On multiplying (i) by 3, we get:
⇒ 3x + 3y = 120 ...(iii)
On subtracting (iii) from (ii), we get:
4y = 60
⇒ y = 15
On subtracting y = 15 in (1), we get:
x + 15 = 40
⇒ 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 915 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 $a^2$
$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 925 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 935 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}$
$2\text{x} + 3\text{y} -18 = 0$ $\Rightarrow\text{y}=\frac{18-\text{2x}}{3}$
Since the two graph intersect at (3, 4), x = 3 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 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 945 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 955 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$
⇒ 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
⇒ x + y = 14750 ...(iii)
Subtracting (ii) from (i), we get
x - y = 2250 ...(iv)
Adding (iii) and (iv), we get
2x = 17000
⇒ 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 965 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)
⇒ x° + (3x - 2)° + y° = 180°
⇒ 4x + y = 182 ...(i)
Also, given that
$\angle\text{C}-\angle\text{B}=9^\circ$
⇒ y° - (3x - 2)° = 9°
⇒ y - 3x + 2 = 9
⇒ 3x - y = -7 ...(ii)
Adding (i) and (ii), we get
7x = 175
⇒ x = 25
Substituting x = 25 in (i), we get
⇒ 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 975 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 × 2) - 3v = 1
⇒ 10 - 3v = 1
⇒ -3v = 1 - 10
⇒ -3v = -9
⇒ 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 985 Marks
Solve for x and y:
x + y = 5xy,
3x + 2y = 13xy $(\text{x}\neq0,\ \text{y}\neq0).$
Answerx + 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).
⇒ 3u + 3v = 15 and 2u + 3v = 13
⇒ 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 995 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
Answerx - y + 3 = 0 ⇒ y = x + 3
2x + 3y - 4 = 0 $\Rightarrow\text{y}=\frac{4-\text{2x}}{3}$
Since the two graph intersect at (-1, 2), x = -1 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 1005 Marks
Show graphically that each of the following given systems of equations has infinitely many solutions:
2x + 3y = 6, 4x + 6y = 12
Answer$\text{2x}+\text{3y}=6$ $\Rightarrow\text{y}=\frac{\text{6}-\text{2x}}{3}$
$\text{4x}+\text{6y}=12$ $\Rightarrow\text{y}=\frac{12-\text{4x}}{6}$
Since the two graph of the system of equations is coincident lines, the system has infinitely many solutions. View full question & answer→