Page No 3.106:
Answer:1
Let the length and breadth of the rectangle be and units respectively
Then, area of rectangle square units
If length is increased and breadth reduced each by units, then the area is reduced by square units
$\left(x+2\right)\left(y-2\right)=xy-28\phantom{\rule{0ex}{0ex}}\Rightarrow xy-2x+2y-4=xy-28\phantom{\rule{0ex}{0ex}}\Rightarrow -2x+2y-4+28=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2x+2y+24=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-2y-24=0$
Therefore,
Then the length is reduced by unit and breadth is increased by units then the area is increased by square units
$\left(x-1\right)\left(y+2\right)=xy+33\phantom{\rule{0ex}{0ex}}\Rightarrow xy+2x-y-2=xy+33\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-y-2-33=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-y-35=0$
Therefore, $2x-y-35=0.....\left(ii\right)$
Thus we get the following system of linear equation
$2x-2y-24=0\phantom{\rule{0ex}{0ex}}2x-y-35=0$
By using cross multiplication, we have
and
The length of rectangle is units.
The breadth of rectangle is units.
Area of rectangle =lengthbreadth,
square units
Hence, the area of rectangle is square units
Page No 3.106:
Answer:2
Let the length and breadth of the rectangle be and units respectively
Then, area of rectangle =square units
If length is increased by meters and breadth is decreased by meters when the area of a rectangle remains the same
Therefore,
If the length is decreased by meters and breadth is increased by meters when the area remains unaffected, then
Thus we get the following system of linear equation
By using cross-multiplication, we have
and
Hence, the length of rectangle ismeters
The breadth of rectangle is meters
Page No 3.106:
Answer:3
Let the length and breadth of the rectangle be and units respectively
Then, area of rectangle = square units
If the length is increased by meters and breath is reduced each by square meters the area is reduced by square units
Therefore,
Then the length is reduced by meter and breadth is increased by meter then the area is increased by square units
Therefore,
Thus, we get the following system of linear equation
By using cross multiplication we have
and
Hence, the length of rectangle ismeter,
The breath of rectangle is meter.
Page No 3.106:
Answer:4
Let the income of be Rs and the income of be Rs.further let the expenditure of be and the expenditure of be respectively then,
Saving of =
Saving of =
Solving equation and by cross- multiplication, we have
The monthly income of =
The monthly income of
Hence the monthly income of is Rs
The monthly income of is Rs
Page No 3.106:
Answer:5
Let the money with A be Rs x and the money with B be Rs y.
If A gives Rs 30 to B, Then B will have twice the money left with A, According to the condition we have,
If B gives Rs 10 to A, then A will have thrice as much as is left with B,
By multiplying equation with 2 we get,
By subtracting from we get,
By substituting in equation we get
Hence the money with A be and the money with B be
Page No 3.106:
Answer:6
Let us take the A examination room will be x and the B examination room will be y
If 10 candidates are sent from A to B, the number of students in each room is same. According to the above condition equation will be
If 20 candidates are sent from B to A, the number of students in A is double the number of students in B, then equation will be,
By subtracting the equationfromwe get,
Substituting in equation, we get
Hence candidates are in A examination Room,
candidates are in B examination Room.
Page No 3.107:
Answer:7
A man can alone finish the work in days and one boy alone can finish it in days then
One mans one days work =
One boys one days work=
2men one day work=
7boys one day work=
Since 2 men and 7 boys can finish the work in 4 days
Again 4 men and 4 boys can finish the work in 3 days
Putting and in equation and we get
By using cross multiplication we have
Now,
and
Hence, one man alone can finish the work in and one boy alone can finish the work in _{.}
Page No 3.107:
Answer:8
Let , and
$\angle C-\angle B={9}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C={9}^{\xb0}+\angle B\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C={9}^{\xb0}+3{x}^{\xb0}-{2}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C={7}^{\xb0}+3{x}^{\xb0}$
Substitute in above equation we get ,
${y}^{\xb0}={7}^{\xb0}+3{x}^{\xb0}$
$\angle A+\angle B+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{\xb0}+(3{x}^{\xb0}-{2}^{\xb0})+({7}^{\xb0}+3{x}^{\xb0})={180}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow 7{x}^{\xb0}+{5}^{\xb0}={180}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow 7{x}^{\xb0}={180}^{\xb0}-{5}^{\xb0}={175}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{\circ}=\frac{{175}^{\circ}}{{7}^{\circ}}={25}^{\circ}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle A={x}^{\circ}={25}^{\circ}\phantom{\rule{0ex}{0ex}}\angle B=(3x-2{)}^{\circ}=3({25}^{\circ})-{2}^{\circ}={73}^{\circ}\phantom{\rule{0ex}{0ex}}\angle C=({7}^{\xb0}+3{x}^{\xb0})={7}^{\circ}+3(25{)}^{\circ}={82}^{\circ}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle A={25}^{\circ},\angle B={73}^{\circ},\angle C={82}^{\circ}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{answer}.$
Page No 3.107:
Answer:9
We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral, angles and and angles and pairs of opposite angles
Therefore and
Taking
By substituting and we get
Taking
By substituting and we get
By multiplying equation by we get
By subtracting equation from we get
By substituting in equation we get
The angles of a cyclic quadrilateral are
Hence, the angles of cyclic quadrilateral ABCD are .
Page No 3.107:
Answer:10
Let take first class full of fare is Rs and reservation charge is Rs per ticket
Then half of the ticket as on full ticket =
According to the given condition we have
Multiplying equation by 2 we have
Subtracting from we get
Putting in equation we get
Hence, the basic first class full fare is
The reservation charge is .
Page No 3.107:
Answer:11
We have to prove that the triangle is right
Given and
Sum of three angles in triangle are
By solving with we get,
Multiplying equation by 3 we get
Subtracting equation from
Substituting in equation we get
Angles and are
A right angled triangle is a triangle in which one side should has a right angle that is in it.
Hence, The triangle ABC is right angled
Page No 3.107:
Answer:12
Let the fixed charges of car be per km and the running charges be km/hr
According to the given condition we have
Putting in equation we get
Therefore, Total charges for travelling distance of km
= Rs
Hence, A person have to pay for travelling a distance of km.
Page No 3.107:
Answer:13
Let the fixed charges of hostel be and the cost of food charges be per day
According to the given condition we have,
Subtracting equation from equation we get
Putting in equation we get
Hence, the fixed charges of hostel is _{.}
The cost of food per day is _{.}
Page No 3.107:
Answer:14
Let perimeter of rectangular garden will be .if half the perimeter of a garden will be
When the length is four more than its width then
Substituting in equation we get
Putting in equation we get
Hence, the dimensions of rectangular garden are and
Page No 3.107:
Answer:15
We know that the sum of supplementary angles will be.
Let the longer supplementary angles will be.
Then,
If larger of supplementary angles exceeds the smaller by degree, According to the given condition. We have,
Substitute in equation, we get,
Put equation, we get,
Hence, the larger supplementary angle is ,
The smaller supplementary angle is.
Page No 3.107:
Answer:16
1 women alone can finish the work in days and 1 man alone can finish it in days .then
One woman one day work=
One man one days work =
2 women’s one days work=
5 man’s one days work =
Since 2 women and 5 men can finish the work in 4 days
3 women and 6 men can finish the work in 3 days
Putting and in equation and we get
By using cross multiplication we have
Now ,
Hence, the time taken by 1 woman alone to finish the embroidery is,
The time taken by 1 man alone to finish the embroidery is .
Page No 3.107:
Answer:17
Let the strike money of first cock-owner be and of second cock-owner be respectively. Then we have,
For second cock-owner according to given condition we have,
By subtracting from, we have,
Putting in equation we get,
Hence the stake of money first cock-owner is and of second cock-owner is respectively.
Page No 3.107:
Answer:18
Let be the notes of and notes will be
If Meena ask for and notes only, then the equation will be,
Divide both sides by then we get,
If Meena got 25 notes in all then the equation will be,
By subtracting the equation from we get,
Substituting in equation, we get
Therefore and
Hence, Meena has notes of and notes of
Page No 3.107:
Answer:19
Let take right answer will beand wrong answer will be .
Hence total number of questions will be
If yash scored marks in atleast getting marks for each right answer and losing mark for each wrong answer then
If 4 marks awarded for each right answer and 2 marks deduced for each wrong answer the he scored 50 marks
By multiplying equation by 2 we get
By subtracting from we get
Putting in equation we have
Total number question will be
Hence, the total number of question is .
Page No 3.107:
Answer:20
Let the number of students be and the number of row be .then,
Number of students in each row
Where three students is extra in each row, there are one row less that is when each row has students the number of rows is
Total number of students =no. of rowsno. of students in each row
If three students are less in each row then there are rows more that is when each row has
Therefore, total number of students=Number of rowsNumber of students in each row
Putting in and equation we get
Adding and equation we get
Putting in equation we get
Hence, the number of students in the class is.
Page No 3.107:
Answer:21
21. Let the money with first person be and the money with the second person be. Then,
If first person gives to second person then the second person will become twice as rich as first person, According to the given condition, we have,
if second person gives to first person then the first person will becomes six times as rich as second person, According to given condition, we have,
Multiplying equation by we get,
By subtracting from, we get
Putting in equation, we get,
Hence, first person’s capital will be ,
Second person’s capital will be .
Page No 3.108:
Answer:1
The given system of equations is
.
For the equations to have no solutions
By cross multiplication we get,
Hence, the value of k is when system equations has no solution.
Page No 3.108:
Answer:2
The given systems of equations are
For the equations to have infinite number of solutions,
By cross Multiplication we get,
Hence the value of k is when equations has infinitely many solutions.
Page No 3.108:
Answer:22
We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral angles and and angles and pairs of opposite angles
Therefore
and
By substituting and we get
Divide both sides of equation by 4 we get
By substituting and we get
By multiplying equation by 3 we get
By subtracting equation from we get
By substituting in equation we get
The angles of a cyclic quadrilateral are
Hence the angles of quadrilateral are
Page No 3.109:
Answer:3
The given system of equations are
For the equations to have no solutions,
If we take
Therefore the value of k is10.
Hence, correct choice is.
Page No 3.109:
Answer:4
The given system of equations are,
Here,
By cross multiply we get
Therefore the value of k is 6,
Hence, the correct choice is.
Page No 3.109:
Answer:5
The given systems of equations are
For the equations to have infinite number of solutions,
Here ,
Let us take
By cross multiplication we get,
Now take
By cross multiplication we get,
Substitute in the above equation
Substitute in equation we get,
Therefore and.
Hence, the correct choice is.
Page No 3.109:
Answer:6
The given equations are
For the equations to have infinite number of solutions
Let us take
Hence, the value of when the pair of linear equations has infinitely many solutions.
Page No 3.109:
Answer:7
The given equations are
Every solution of the second equation is also a solution of the first equation.
Hence, there are, the system equation is consistent.
Page No 3.109:
Answer:8
The given linear pair of equations are
If then
Hence, the number of solutions of the pair of linear equation is.
Therefore, the equations have no solution.
Page No 3.109:
Answer:1
The given system of equations are
for unique solution
Here
By cross multiply we get
Hence, the correct choice is.
Page No 3.109:
Answer:2
The given system of equations are
For the equations to have infinite number of solutions,
Here,
Therefore
By cross multiplication of we get,
And
Therefore the value of k is 6
Hence, the correct choice is .
Page No 3.110:
Answer:6
The given system of equations is inconsistent,
If the system of equations is in consistent, we have
Therefore, the value of k is2.
Hence, the correct choice is .
Page No 3.110:
Answer:7
Given the system of equations has
We know that intersecting lines have unique solution
Here
Therefore intersecting lines, have unique solution
Hence, the correct choice is
Page No 3.110:
Answer:8
Given the system of equations are
For the equations to have infinite number of solutions,
By cross multiplication we have
Divide both sides by 2. we get
Hence, the correct choice is .
Page No 3.110:
Answer:9
The given system of equation is
If then the equation have no solution.
By cross multiply we get
Hence, the correct choice is.
Page No 3.110:
Answer:10
The given system of equations are
For coincident lines , infinite number of solution
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{b2}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{\left(a+b\right)}=\frac{-3}{-\left(a+b-3\right)}=\frac{7}{4a+b}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(a+b-3\right)=3\left(a+b\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+2b-6=3a+3b\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+2b-3a-3b=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow -a-b=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow a+b=-6---\left(i\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3\left(4a+b\right)=7\left(a+b-3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 12a+3b=7a+7b-21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 5a-4b=-21---\left(ii\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{multiply}\mathrm{equation}\left(i\right)\mathrm{by}5,\mathrm{we}\mathrm{get}5a+5b=-30---\left(iii\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{subtract}\left(ii\right)\mathrm{from}\left(iii\right),\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5a+5b\right)-\left(5a-4b\right)=-30+21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 5a+5b-5a+4b=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 9b=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow b=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{substitute}b=-1\mathrm{in}\mathrm{equation}\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}a+\left(-1\right)=-6\phantom{\rule{0ex}{0ex}}\Rightarrow a=-6+1=-5$
Option A.:
Option B:
$5a+b=0\phantom{\rule{0ex}{0ex}}5\left(-5\right)+\left(-1\right)=-25-1=-26\ne 0$
Option.C:
a - b = 0
-5 - (-1) = -4 $\ne $0
None of the option satisfies the values.
Page No 3.110:
Answer:11
If a pair of linear equations in two variables is consistent, then its solution exists.
∴The lines represented by the equations are either intersecting or coincident.
Hence, correct choice is.
Page No 3.110:
Answer:12
Given the area of the triangle formed by the line
If in the equation either A and B approaches infinity, The line become parallel to either x axis or y axis respectively,
Therefore
Area of triangle
Hence, the correct choice is .
Page No 3.110:
Answer:13
Given and
We have plotting points as when
Therefore, area of
Area of triangle is square units
Hence, the correct choice is _{.}
Page No 3.110:
Answer:14
The given system of equations
For the equations to have infinite number of solutions
If we take
And
Therefore, the value of k is 6.
Hence, the correct choice is .
Page No 3.111:
Answer:15
The given systems of equations are
If
Here
Hence, the correct choice is .
Page No 3.111:
Answer:16
Given and
We have plotting points as when
Therefore, area of
Area of triangle is square units
Hence, the correct choice is
Page No 3.111:
Answer:17
Given and
If We have plotting points as
Therefore, area of
Area of triangle is square units
Hence, the correct choice is
Page No 3.12:
Answer:1
Let no. of ride is and no. of Hoopla is .He paid Rs 20 for ride and for Hoopla.
The cost of ride is Rs and cost of Hoopla is Rs.then
The number of Hoopla is the half number of ride, then
Hence algebraic equations are and
Now, we draw the graph for algebraic equations.
Page No 3.12:
Answer:2
Let age of Aftab is years and age of his daughter is years. Years ago his age wastimes older as her daughter was. Then
Three years from now, he will be three times older as his daughter will be, then
Hence the algebraic representation are and
Page No 3.12:
Answer:3
The given equation are and.
In order to represent the above pair of linear equation graphically, we need
Two points on the line representing each equation. That is, we find two solutions
of each equation as given below:
We have,
Putting we get
Putting we get
Thus, two solution of equation are
We have
Putting we get
Putting we get
Thus, two solution of equation are
8 | ||
6 |
Now we plot the pointand and draw a line passing through
These two points to get the graph o the line represented by equation
We also plot the points and and draw a line passing through
These two points to get the graph O the line represented by equation
We observe that the line parallel and they do not intersect anywhere.
Page No 3.12:
Answer:4
Gloria is walking the path joining and
Suresh is walking the path joining and
0 | 4 | |
5 | 0 |
The graphical representations are
Page No 3.12:
Answer:5
(i) Given equation are: 5x + 4y + 8 = 0
7x + 6y − 9 = 0
We have And
Thus the pair of linear equation is intersecting.
(ii) Given equation are: 9x + 3y + 12 = 0
18x + 6y + 24 = 0
We have
Thus the pair of linear is coincident lines.
(iii) Given equation are:
We have
Thus the pair of line is parallel lines.
Page No 3.12:
Answer:6
(i) Given the linear equation are:
We know that intersecting condition:
Where
Hence the equation of other line is
(ii) We know that parallel line condition is:
Where
Hence the equation is
(iii) We know that coincident line condition is:
Where
Hence the equation is
Page No 3.12:
Answer:7
Let the cost of 1 kg of apples be Rs x.
And, cost of 1 kg of grapes = Rs y
According to the question, the algebraic representation is
For ,
The solution table is
x | 50 | 60 | 70 |
y | 60 | 40 | 20 |
For 4x + 2y = 300,
The solution table is
x | 70 | 80 | 75 |
y | 10 | –10 | 0 |
The graphical representation is as follows.
Page No 3.29:
Answer:1
The given equations are:
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 3 | |
3 | 0 |
Draw the graph by plotting the two points from table.
Graph of the equation:
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 6 | |
12/5 | 0 |
Draw the graph by plotting the two points from the table.
The two lines intersect at point P.
Hence, and is the solution.
Page No 3.29:
Answer:2
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 5 | |
0 |
Draw the graph by plotting the two points from table.
Graph the equation
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 5 | |
10/3 | 0 |
Draw the graph by plotting the two points from table.
The two lines intersects at point B
Hence is the solution
Page No 3.29:
Answer:3
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | –1/3 | |
–1 | 0 |
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we ge
Putting in equationwe get
Use the following table to draw the graph.
0 | –4 | |
8/3 | 0 |
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence is the solution.
Page No 3.29:
Answer:4
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
0 | 7/2 | |
–7/3 | 0 |
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence is the solution.
Page No 3.29:
Answer:5
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence is the solution.
Page No 3.29:
Answer:6
The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Plotting the two points equation (i) can be drawn.
Graph of the equation….
Putting in equation, we get:
Putting x=2 in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
We see that the two lines are parallel, so they won’t intersect
Hence there is no solution
Page No 3.29:
Answer:7
The given equations are
Putting in equation we get:
Putting in equation we get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence is the solution
Page No 3.29:
Answer:8
The given equations are:
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
The two lines intersect at points P.
Hence, and is the solution.
Page No 3.29:
Answer:9
The given equations are:
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equationwe get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence, and is the solution.
Page No 3.29:
Answer:10
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equationwe get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersects at points P
Hence, and is the solution.
Page No 3.29:
Answer:11
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Answer:12
The given equations are
Putting in equation we get:
Putting in equationswe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Graph of the equation….
Putting in equations we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Answer:13
The given equations are
Putting in equation we get:
Putting in equationswe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equations we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Answer:14
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations are coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Answer:15
The given equations are
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here we see that the two lines are parallel
Hence the given system of equations has no solution.
Page No 3.29:
Answer:16
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the two points _{.}
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here the two lines are parallel and so there is no point in common
Hence the given system of equations has no solution.
Page No 3.29:
Answer:17
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here two lines are parallel and so don’t have common points
Hence the given system of equations has no solution.
Page No 3.29:
Answer:18
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the two points_{.}
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here, the two lines are parallel.
Hence the given system of equations is inconsistent.
Page No 3.29:
Answer:19
(i) Draw the 3 lines as given by equations
By taking x=1 = 1 cm on x−axis
And y =1=1cm on y−axis
Clearly from graph points of intersection three lines are
(−4,2) , (1,3), (2,5)
(ii) Draw the 3 lines as given by equations
By taking x=1 = 1 cm on x−axis
And y =1=1cm on y−axis
From graph point of intersection are (0,0) (10/3,0) (5/3,5/3)
Page No 3.29:
Answer:20
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
0 | 5/4 | |
–5/2 | 0 |
Draw the graph by plotting the two points from table.
It has unique solution.
Hence the system of equations is consistent
Page No 3.29:
Answer:21
(i) The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at point .
Hence the equations have unique solution.
(ii) The equations of graphs is
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the two points .
Graph of the equation
Putting in equation we get.
Putting in equation we get.
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines are coincident.
Hence the equations have infinitely much solution.
Hence the system is consistent
Page No 3.29:
Answer:22
(i) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
−2 | 8 | |
0 | _{4} |
_{The two points satisfying (ii) can be listed in a table as,}
4 | 2 | |
0 | 4 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 3, y = 2.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(ii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
4 | 6 | |
0 | –3 |
_{The two points satisfying (ii) can be listed in a table as,}
3 | 2 | |
5.5 | 3 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = 3.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(iii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
3 | 1 | |
5 | 9 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 5 | |
0 | 4 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 4, y = 3.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(iv) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
5 | 7 | |
1 | 0 |
_{The two points satisfying (ii) can be listed in a table as,}
2 | 1 | |
0 | –2 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 3, y = 2.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.(v) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
1 | 3 | |
2 | –4 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 4 | |
–3 | 3 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = −1.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(vi) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
1 | 3 | |
–3 | 1 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 5 | |
–2 | 2 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = −1.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
Page No 3.30:
Answer:23
(i) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | _{1} | |
0 | _{1} |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 3 | |
2 | 6 |
_{The two points satisfying (iii) can be listed in a table as,}
0 | 6 | |
6 | 6 |
Now, graph of equations (i), (ii) and (iii) can be drawn as,
It is seen that the coordinates of the vertices of the obtained triangle are
(ii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | 2 | |
0 | 2 |
_{The two points satisfying (ii) can be listed in a table as,}
3 | –3 | |
1 | –1 |
_{The two points satisfying (iii) can be listed in a table as,}
3 | 5 | |
5 | 3 |
Now, graph of equations (i), (ii) and (iii) can be drawn as,
It is seen that the coordinates of the obtained triangle are
Page No 3.30:
Answer:24
The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at. The region enclosed by the lines represented by the given equations and x−axis are shown in the above figure
Hence, and is the solution.
(ii) The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
0 | ||
2 | 0 |
The graph of (i) can be obtained by plotting the two points.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
− |
Draw the graph by plotting the two points from table.
The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.
Hence, and is the solution.
(iii) The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.
Hence, and is the solution.
Page No 3.30:
Answer:25
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Draw the graph by plotting the two points from table.
The intersection point is P(3, 2)
Three points of the triangle are.
Hence the value of and
Page No 3.30:
Answer:26
The given equations are
Putting in equation we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Now, Required area = Area of shaded region
Required area = Area of PBD
Required area =
Required area =
Required area =
Hence the area =
Page No 3.30:
Answer:27
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the points (0, 4/3), (−1, 0).
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Hence is the solution of the given equations.
Now,
Required area = Area of PBD
Required area =
Required area =
Required area =
Hence, the area =
Page No 3.30:
Answer:28
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Hence is the solution of the given equations
The area enclosed by the lines represented by the given equations and the y−axis is shaded region in the figure
Now, Required area = Area of shaded region
Required area = Area of PAC
Required area =
Required area =
Required area =
Hence the required area is sq. unit
Page No 3.30:
Answer:29
(i)
The given equations are
_{The two points satisfying (i) can be listed in a table as,}
4 | _{0} | |
_{−2} | _{6} |
_{The two points satisfying (ii) can be listed in a table as,}
4 | 6 | |
_{3} | _{4} |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
(ii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | 2 | |
2 | 2 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 3 | |
–4 | 4 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 3, y = 4.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
(iii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | _{−1} | |
_{2.5} | _{3} |
_{The two points satisfying (ii) can be listed in a table as,}
4 | –5 | |
4 | –2 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 1, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
Solution is missing
(iv) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
–2 | 7 | |
_{4} | –2 |
_{The two points satisfying (ii) can be listed in a table as,}
–1 | 3 | |
1 | 3 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 1, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
Page No 3.30:
Answer:30
The given equations are
_{The two points satisfying (i) can be listed in a table as,}
−3 | 6 | |
_{0} | _{6} |
The two points satisfying (ii) can be listed in a table as,
0 | 9 | |
6 | _{0} |
The two points satisfying (iii) can be listed in a table as,
−1 | 8 | |
_{2} | _{2} |
Now, graph of equations (i), (ii) and (iii) can be drawn as,
It is seen that the coordinates of the vertices of the obtained triangle are
_{∴Area of ΔABC = }
Page No 3.30:
Answer:31
The given equations are:
Putting in equation (i) we get:
Putting in equation (i) we get:
Use the following table to draw the graph
Draw the graph by plotting the two points from table.
Putting in equationwe get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Hence is the solution of the given equations.
The area enclosed by the lines represented by the given equations and the y−axis
Now,
Required area = Area of PCA
Required area =
Required area =
Required area =
Hence the required area is
Page No 3.31:
Answer:32
The given equations are:
Putting in equation we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
x |
0 |
5 |
y |
0 |
Draw the graph by plotting the two points from table
Putting in equation (ii) we get:
Putting in equation (ii) we get:
Use the following table to draw the graph.
x |
0 |
5 |
y |
3 |
0 |
Draw the graph by plotting the two points from table.
The three vertices of the triangle are.
Hence the solution of the equation is and
Page No 3.31:
Answer:33
The given equations are:
Putting in equation (i) we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
–3 |
Draw the graph by plotting the two points from table.
Hence the vertices of the required triangle are.
Now,
Required area = Area of PCA
Required area =
Required area =
Hence the required area is
Page No 3.31:
Answer:34
(i) Let the number of girls be x and the number of boys be y.
According to the question, the algebraic representation is
x + y = 10
x − y = 4
For x + y = 10,
x = 10 − y
x | 5 | 4 | 6 |
y | 5 | 6 | 4 |
For x − y = 4,
x = 4 + y
x | 5 | 4 | 3 |
y | 1 | 0 | −1 |
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines intersect each other at point (7, 3).
Therefore, the number of girls and boys in the class are 7 and 3 respectively.
(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.
According to the question, the algebraic representation is
5x + 7y = 50
7x + 5y = 46
For 5x + 7y = 50,
x | 3 | 10 | − 4 |
y | 5 | 0 | 10 |
7x + 5y = 46
x | 8 | 3 | − 2 |
y | − 2 | 5 | 12 |
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines intersect each other at point (3, 5).
Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.
(iii) Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :
y = 2x − 2 … (i)
y = 4x − 4 … (ii)
The graphs of the equations (i) and (ii) can be drawn by finding two solutions for each of the equations. They are given in the following table.
x | 2 | 0 |
y = 2x − 2 | 2 | −2 |
Hence, the graphic representation is as follows.
The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.
Page No 3.31:
Answer:35
The given equations are:
Putting in equation (i) we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation (ii) we get:
Putting in equation (ii) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at points of y−axis.
Hence, and is the Solution.
(ii) The equations are:
Putting in equation (1) we get:
Putting in equation (1) we get:
Use the following table to draw the graph:
Draw the graph by plotting the two points from table.
Putting in equation (2) we get:
Putting in equation (2) we get:
Use the following table to draw the graph.
x | ||
Draw the graph by plotting the two points from table.
Two lines intersect at points of y−axis.
Hence and is the solution.
Page No 3.31:
Answer:36
The given equations are
Putting in equation (i) we get:
Putting in equationwe get:
Use the following table to draw the graph.
x | ||
The graph of (i) can be obtained by plotting the two points.
Putting in equation (ii) we get:
Putting in equation (ii) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of lines represented by the equations meet y−axis at respectively.
Page No 3.31:
Answer:37
(i) For intersecting lines,
Equation of another intersecting line to the given line is−
Since, condition for intersecting lines and unique solution is−
(ii) For parallel lines,
Equation of another parallel line to the given line is−
Since, condition for parallel lines and no solution is−
(iii) For co−incident lines,
Equation of another coincident line to the given line is−
Since, condition for coincident lines and infinite solution is−
Page No 3.43:
Answer:1
The given equations are:
Multiply equation by 2 and equation by 15, and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.43:
Answer:2
The given equations are:
…
…
Multiply equation by 2 and equation by, and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.43:
Answer:3
The given equations are:
…
…
Multiply equation by 2 and equation by, and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.43:
Answer:4
The given equations are:
…
Subtract (ii) from (i) we get
Put the value of in equation we get
Hence the value of and .
Page No 3.43:
Answer:5
The given equations are:
…
…
Multiply equation by and equation by and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.43:
Answer:6
The given equations are:
…
…
Multiply equation by and add both equations we get
Put the value of in equation we get
Hence the value of and .
Page No 3.43:
Answer:7
The given equations are:
…
…
Multiply equation by and equation by and add both equations we get
Put the value of in equation we get
Hence the value of and .
Page No 3.43:
Answer:8
The given equations are:
…
…
Multiply equation by and equation by and add both equations we get
Put the value of in equation we get
Hence the value of and .
Page No 3.44:
Answer:9
The given equations are:
…
…
Multiply equation by and subtract equations, we get
Put the value of in equation, we get
Hence the value of x and y are and
Page No 3.44:
Answer:10
The given equations are:
…
…
Multiply equation by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:11
The given equations are:
…
…
Multiply equation by and equation by and subtract equation (ii) from (i), we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:12
The given equations are:
$3x-\frac{y+7}{11}+2=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 3x-\frac{\left(y+7\right)}{11}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{33x-y-7}{11}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 33x-y-7=88\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 33x-y=95........\left(1\right)$
$2y+\frac{x+11}{7}=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{14y+x+11}{7}=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y+x+11=70\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y+x=59\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow x+14y=59.........\left(2\right)$
Multiply equation (1) by , we get
$462x-14y=1330......\left(3\right)$
adding (2) and (3), we get$\left(x+14y\right)+\left(462x-14y\right)=59+1330\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 463x=1389\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow x=3$
$\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3+14y=59\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y=59-3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y=56\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow y=4\phantom{\rule{0ex}{0ex}}$
Hence the value of x and y are and
Page No 3.44:
Answer:13
The given equations are:
…
…
Multiply equation by and by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:14
The given equations are:
…
…
Multiply equation by and by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:15
The given equations are:
…
…
Multiply equation by and add both equations we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:16
The given equations are:
…
…
Multiply equation by and by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
$\frac{1}{2\times {\displaystyle \frac{1}{2}}}+\frac{1}{3y}=2\phantom{\rule{0ex}{0ex}}\frac{1}{3y}=2-1\phantom{\rule{0ex}{0ex}}\frac{1}{3y}=1\phantom{\rule{0ex}{0ex}}y=\frac{1}{3}\phantom{\rule{0ex}{0ex}}$
Hence the value of and $y=\frac{1}{3}$
Page No 3.44:
Answer:17
The given equations are:
Adding both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:18
The given equations are:
…
…
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.44:
Answer:19
The given equations are:
…
…
Multiply equation by and add both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:20
The given equations are:
Multiply equation by and subtract (ii) from equation (i), we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.44:
Answer:21
The given equations are:
…
…
Multiply equation by and equation by, add both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:22
The given equations are:
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:23
The given equations are:
Let and then equations are
…
…
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in second equation, we get
Hence the value of and.
Page No 3.44:
Answer:24
The given equations are:
Add both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Answer:25
The given equations are:
Let and then equations are
…
…
Multiply equation by and subtracting (ii) from (i), we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in second equation, we get
Hence the value of and .
Page No 3.44:
Answer:26
The given equations are:
Let and then equations are
…
…
Multiply equation by and equation by and add both equations, we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in first equation, we get
Hence the value of and .
Page No 3.45:
Answer:27
The given equations are:
Let and then equations are
…
…
Multiply equation by and add both equations, we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in first equation, we get
Hence the value of and .
Page No 3.45:
Answer:28
The given equations are:
Let and then equations are
…
…
Multiply equation by and equation by add both equations, we get
Put the value of in equation, we get
Then
…
…
Add both equations, we get
Put the value of in equation we get
Hence the value of and .
Page No 3.45:
Answer:29
The given equations are:
Let and then equations are
…
…
Add both equations, we get
Put the value of in equation, we get
Then
Hence the value of and .
Page No 3.45:
Answer:30
The given equations are:
…
…
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Answer:31
The given equations are:
…
…
Add both equations we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Answer:32
The given equations are:
Multiply equation by and add both equations, we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Answer:33
The given equations are:
Let and then equations are
…
…
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Then
…
…
Add both equations, we get
Put the value of in equation we get
Hence the value of and .
Page No 3.45:
Answer:34
The given equations are:
…
…
Multiply equation by and equation by, add both equations, we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Answer:35
The given equations are:
…
…
Multiply equation by and equation by, and subtract (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Answer:36
The given equations are:
…
…
Multiply equation by and equation by and subtract (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Answer:37
The given equations are:
…
First of all we find the value of
Put the value of in equation, we get
Put the value of and in equation in we get
Multiply equation by and add equations and, we get
Put the value of in equation, we get
Put the value of and in equation we get
Hence the value of, and.
Page No 3.45:
Answer:38
The given equations are:
…
First of all we find the value of
Put the value of in equation, we get
Put the value of and in equation in we get
Put the value of and in equation, we get
Hence the value of, and.
Page No 3.45:
Answer:39
The given equations are:
Let and then equations are
…
…
Multiply equation by and equation by add both equations, we get
Put the value of in equation, we get
Then
…
…
Add both equations, we get
Put the value of in equation we get
Hence the value of and
Page No 3.45:
Answer:40
The given equations are:
…
…
Multiply equation by and equation by and subtract (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Answer:41
The given equations are:
…
…
Multiply equation by and equation by 3 and add both equations we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Answer:42
The given equations are:
Let and then equations are
…
…
Multiply equation by and add both equations, we get
Put the value of in equation, we get
Then
Hence the value of and .
Page No 3.45:
Answer:43
The given equations are:
Let and then equations are
…
…
Multiply equation by and equation by and add both equations, we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in equation we get
Hence the value of and .
Page No 3.45:
Answer:44
The given equations are:
Let and then equations are
…
…
Multiply equation by and add both equations, we get
Put the value of in equation, we get
Then
…
…
Add both equations, we get
Put the value of in equation we get
Hence the value of and
Page No 3.45:
Answer:45
The given equations are:
…
…
Multiply equation by and add both equations we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Answer:46
The given equations are:
…
…
Multiply equation by and equation by, add both equations we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Answer:47
The given equations are:
…
…
Multiply equation by and equation by and add both equations we get
Put the value of in equation, we get
Hence the value of and
Page No 3.56:
Answer:1
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
By cross multiplication method we get
and
Hence we get the value of and
Page No 3.56:
Answer:2
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
By cross multiplication method we get
Also
Hence we get the value of and
Page No 3.56:
Answer:3
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Also
Hence we get the value of and
Page No 3.56:
Answer:4
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Hence we get the value of and
Page No 3.56:
Answer:5
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
Let
By cross multiplication method we get
So
We know that
Hence we get the value of and
Page No 3.56:
Answer:6
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
T
Thereforeand
Hence we get the value of and
Page No 3.56:
Answer:7
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
, and
Hence we get the value of and
Page No 3.56:
Answer:8
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
And
Hence we get the value of and
Page No 3.56:
Answer:9
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
So for x we have
And
Hence we get the value of and
Page No 3.56:
Answer:10
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
And
Hence we get the value of and
Page No 3.56:
Answer:11
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
And
Hence we get the value ofand
Page No 3.56:
Answer:12
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
Rewriting the equation again
By cross multiplication method we get
And
We know that
Adding equation (3) and (4)
Substituting value of x in equation (3) we get
Hence we get the value ofand
Page No 3.56:
Answer:13
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
Rewriting the equation again
By cross multiplication method we get from eq. (1) and eq. (2)
And
We know that
Hence we get the value of and
Page No 3.56:
Answer:14
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
And
Hence we get the value of and
Page No 3.56:
Answer:15
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Now consider
And
Hence we get the value of and
Page No 3.56:
Answer:16
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Consider the following to calculate x
And
Hence we get the value of and
Page No 3.56:
Answer:17
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
And
Hence we get the value of and
Page No 3.56:
Answer:18
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Consider the following for x
And
Hence we get the value of and
Page No 3.56:
Answer:19
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Consider the following for x
Now for y
Hence we get the value of and
Page No 3.56:
Answer:20
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Consider the following for x
Now consider the following for y
Hence we get the value of and
Page No 3.56:
Answer:21
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Consider the following for x
Now consider the following for y
Hence we get the value of and
Page No 3.56:
Answer:22
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
Now rewriting the given equation as
By cross multiplication method we get
Consider the following for u
Consider the following for v
We know that
Now adding eq. (3) and (4) we get
And after substituting the value of x in eq. (4) we get
Hence we get the value of and
Page No 3.56:
Answer:23
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
After rewriting equations
By cross multiplication method we get
For y consider the following
Hence we get the value of and
Page No 3.57:
Answer:24
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation, after rewriting equations
By cross multiplication method we get
Consider the following for y
Hence we get the value of and
Page No 3.57:
Answer:25
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
Let
Rewriting equations
Now, by cross multiplication method we get
For u consider the following
For y consider
We know that
Now
Hence we get the value of and
Page No 3.57:
Answer:26
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Now for y
Hence we get the value of and
Page No 3.57:
Answer:27
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
For y
Hence we get the value of and
Page No 3.57:
Answer:28
GIVEN:
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
By cross multiplication method we get
Hence we get the value of
Page No 3.71:
Answer:1
GIVEN:
To find: To determine whether the system has a unique solution, no solution or infinitely many solutions
We know that the system of equations
For unique solution
For no solution
For infinitely many solution
Here,
Since which means hence the system of equation has no solution.
Hence the system of equation has no solution
Page No 3.71:
Answer:2
GIVEN:
To find: To determine whether the system has a unique solution, no solution or infinitely many solutions
We know that the system of equations
For unique solution
For no solution
For infinitely many solution
Here,
Since which means hence the system of equation has infinitely many solution.
Hence the system of equation has infinitely many solutions
Page No 3.71:
Answer:3
GIVEN:
To find: To determine whether the system has a unique solution, no solution or infinitely many solutions
We know that the system of equations
For unique solution
For no solution
For infinitely many solution
Here,
Since which means hence the system of equation has infinitely many solution.
Hence the system of equation has infinitely many solutions
Page No 3.71:
Answer:4
GIVEN:
To find: To determine whether the system has a unique solution, no solution or infinitely many solutions
We know that the system of equations
For unique solution
For no solution
For infinitely many solution
Here,
Since which means hence the system of equation has no solution.
Hence the system of equation has no solution
Page No 3.71:
Answer:5
GIVEN:
To find: To determine to value of k for which the system has a unique solution.
We know that the system of equations
For unique solution
Here,
Hence for the system of equation has unique solution.
Page No 3.71:
Answer:6
GIVEN:
To find: To determine to value of k for which the system has a unique solution.
We know that the system of equations
For unique solution
Here,
Hence for the system of equation has unique solution
Page No 3.71:
Answer:7
GIVEN:
To find: To determine to value of k for which the system has a unique solution.
We know that the system of equations
For unique solution
Here,
Hence already for the system of equation to have unique solution but the value of k should be a real number
Hence for the system of equation has unique solution.
Page No 3.71:
Answer:8
GIVEN:
To find: To determine to value of k for which the system has a unique solution.
We know that the system of equations
For unique solution
Here,
Hence for the system of equation has unique solution
Page No 3.72:
Answer:9
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Hence for the system of equation have infinitely many solutions
Page No 3.72:
Answer:10
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Hence for the system of equation have infinitely many solutions.
Page No 3.72:
Answer:11
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Hence for the system of equation have infinitely many solutions.
Page No 3.72:
Answer:12
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Hence for the system of equation have infinitely many solutions
Page No 3.72:
Answer:13
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider the following
Now consider the following
Hence for the system of equation have infinitely many solutions.
Page No 3.72:
Answer:14
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Consider the following for k
Now consider the following
Hence for the system of equation have infinitely many solutions.
Page No 3.72:
Answer:15
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Hence for the system of equation have infinitely many solutions.
Page No 3.72:
Answer:16
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Consider the following relation to find k
Now consider the following
Hence for the system of equation have infinitely many solutions
Page No 3.72:
Answer:17
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Consider the following relation to find k
Now again consider the following
Hence for the system of equation have infinitely many solutions
Page No 3.72:
Answer:18
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Now again consider the following to find k
Hence for the system of equation have infinitely many solutions
Page No 3.72:
Answer:19
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here,
Consider the following to find out k
Now again consider the following relation
So the common solution is 7
Hence for the system of equation have infinitely many solutions
Page No 3.72:
Answer:20
GIVEN:
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
For no solution
Here,
Hence for the system of equation have infinitely many solutions.
Page No 3.72:
Answer:21
GIVEN:
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
For no solution
Here,
Hence for the system of equation has no solution
Page No 3.72:
Answer:22
GIVEN:
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
For no solution
Here,
Hence for the system of equation has no solution.
Page No 3.72:
Answer:23
GIVEN:
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
For no solution
Here,
Hence for the system of equation has no solution.
Page No 3.72:
Answer:24
GIVEN:
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
For no solution
Here,
Hence for the system of equation has no solution.
Page No 3.72:
Answer:25
GIVEN:
To find: To determine for what value of c the system of equation has no solution
We know that the system of equations
For no solution
Here,
Hence for the system of equation has no solution.
Page No 3.72:
Answer:26
GIVEN:
To find: To determine for what value of k the system of equation will be inconsistent
We know that the system of equations
For the system of equation to be inconsistent
Here,
Hence for the system of equation will be inconsistent.
Page No 3.72:
Answer:27
GIVEN:
$\alpha x+3y=\alpha -3\phantom{\rule{0ex}{0ex}}12x+\alpha y=\alpha $
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
For no solution
Here,
$\frac{\alpha}{12}=\frac{3}{\alpha}\ne \frac{\alpha -3}{\alpha}$
Consider the following for α
$\frac{\alpha}{12}=\frac{3}{\alpha}\phantom{\rule{0ex}{0ex}}{\alpha}^{2}=12\times 3\phantom{\rule{0ex}{0ex}}{\alpha}^{2}=36\phantom{\rule{0ex}{0ex}}\alpha =\pm 6$
Now consider the following
$\frac{3}{\alpha}\ne \frac{\alpha -3}{\alpha}\phantom{\rule{0ex}{0ex}}3\alpha \ne \alpha (\alpha -3)\phantom{\rule{0ex}{0ex}}3\alpha \ne {\alpha}^{2}-3\alpha \phantom{\rule{0ex}{0ex}}6\alpha \ne {\alpha}^{2}\phantom{\rule{0ex}{0ex}}\alpha \ne 6$
Hence the common value of α is − 6
Hence for α = -6 the system of equation has no solution
Page No 3.72:
Answer:28
GIVEN:
To find: To determine for what value of k the system of equation has
(1) Unique solution
(2) No solution
We know that the system of equations
(1) For Unique solution
Here,
Hence for the system of equation has unique solution.
(2) For no solution
Hence for the system of equation has no solution
Page No 3.73:
Answer:29
GIVEN:
To find: To determine for what value of c the system of equation has infinitely many solution
We know that the system of equations
For infinitely many solution
Here
Consider the following
Now consider the following for c
But it is given that c ≠ 0. Hence c = 6
Hence for the system of equation have infinitely many solutions.
Page No 3.73:
Answer:30
GIVEN:
To find: To determine for what value of k the system of equation has
(1) Unique solution
(2) No solution
(3) Infinitely many solution
We know that the system of equations
(1) For Unique solution
Here,
Hence for the system of equation has unique solution
(2) For no solution
Here,
Hence for the system of equation has no solution
(3) For infinitely many solution
Here,
But since here
Hence the system does not have infinitely many solutions.
Page No 3.73:
Answer:31
GIVEN:
To find: To determine for what value of k the system of equation will represents coincident lines
We know that the system of equations
For the system of equation to represent coincident lines we have the following relation
Here,
Hence for the system of equation represents coincident lines
Page No 3.73:
Answer:32
GIVEN:
To find: To determine the condition for the system of equation to have a unique equation
We know that the system of equations
For unique solution
Here
Hence for the system of equation have unique solution.
Page No 3.73:
Answer:33
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Again consider
Hence for and the system of equation has infinitely many solution.
Page No 3.73:
Answer:34
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider the following
5a − 4b + 21 = 0……(1)
Again
a + b + 6 = 0…… (2)
Multiplying eq. (2) by 4 and adding eq. (1)
Putting the value of a in eq. (2)
Hence for and the system of equation has infinitely many solution.
Page No 3.73:
Answer:35
GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
…… (1)
Again consider
…… (2)
Multiplying eq. (2) by 3 and subtracting from eq. (1)
Putting the value of q in eq. (2)
Hence for and the system of equation has infinitely many solution.
Page No 3.73:
Answer:36
(i) GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider
Again consider
Hence forand the system of equation has infinitely many solution.
(ii) GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider the following
Again consider
Hence for and the system of equation has infinitely many solution.
(iii) GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider the following
Again consider
Hence forand the system of equation has infinitely many solution.
(iv) GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider the following
…… (1)
Again consider
…… (2)
Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. 2
Substituting the value of ‘a’ in eq. (2) we get
Hence forand the system of equation has infinitely many solution.
(v) GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider the following
…… (1)
Again consider the following
…… (2)
Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. (2)
Substituting the value of b in eq. (2) we get
Hence forand the system of equation has infinitely many solution.
(vi) GIVEN:
To find: To determine for what value of k the system of equation has infinitely many solutions
Rewrite the given equations
We know that the system of equations
For infinitely many solution
Here
Consider the following
Hence for the system of equation have infinitely many solutions.
(vii) GIVEN :
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
For infinitely many solution
Here
Consider the following
Hence for the system of equation have infinitely many solutions.
(viii) Given:
$x+2y=1\phantom{\rule{0ex}{0ex}}\left(a-b\right)x+\left(a+b\right)y=a+b-2$
We know that the system of equations
has infinitely many solutions if
So,
$\frac{1}{a-b}=\frac{2}{a+b}=\frac{1}{a+b-2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle a-b}}=\frac{{\displaystyle 2}}{{\displaystyle a+b}}\mathrm{and}\frac{{\displaystyle 2}}{{\displaystyle a+b}}=\frac{{\displaystyle 1}}{{\displaystyle a+b-2}}\phantom{\rule{0ex}{0ex}}\Rightarrow a+b=2a-2b\mathrm{and}2a+2b-4=a+b\phantom{\rule{0ex}{0ex}}\Rightarrow a=3b\mathrm{and}a+b=4\phantom{\rule{0ex}{0ex}}\Rightarrow a-3b=0\mathrm{and}a+b=4$
Solving these two equations, we get
$-4b=-4\phantom{\rule{0ex}{0ex}}\Rightarrow b=1$
Putting b = 1 in a + b = 4, we get
a = 3
(ix) Given:
$2x+3y=7\phantom{\rule{0ex}{0ex}}2ax+ay=28-by$
$\Rightarrow 2ax+\left(a+b\right)y=28$
We know that the system of equations
has infinitely many solutions if
$\therefore \frac{2}{2a}=\frac{3}{a+b}=\frac{7}{28}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{a}=\frac{3}{a+b}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{a}=\frac{3}{a+b}\mathrm{and}\frac{3}{a+b}=\frac{1}{4}$
Now,
$\frac{1}{a}=\frac{3}{a+b}$
⇒ a + b = 3a
⇒ b = 2a .....(1)
Also, $\frac{3}{a+b}=\frac{1}{4}$
⇒ a + b = 12 .....(2)
Solving (1) and (2), we get
a = 4 and b = 8
Page No 3.77:
Answer:10
To find:
(1) Total amount of A.
(2) Total amount of B.
Suppose A has Rs x and B has Rs y
According to the given conditions,
x + 100 = 2(y − 100)
x + 100 = 2y − 200
x − 2y = −300 ....(1)
and
y + 10 = 6(x − 10)
y + 10 = 6x − 60
6x − y = 70 ....(2)
Multiplying equation (2) by 2 we get
12x − 2y = 140 ....(3)
Subtracting (1) from (3), we get
11x = 440
x = 40
Substituting the value of x in equation (1), we get
40 − 2y = −300
−2y = −340
y = 170
Hence A has and B has
Page No 3.77:
Answer:1
Given:
(i) 5 pens and 6 pencils together cost of Rs. 9.
(ii) 3 pens and 2 pencils cost Rs. 5.
To Find: Cost of 1 pen and 1 pencil.
Let
(i) The cost of 1 pen = Rs x.
(ii) The cost of 1 pencil = Rs y.
According to question
Thus we get the following system of linear equation
By using cross multiplication we have
Cost of one pen =
Cost of one pencil =
Page No 3.77:
Answer:2
Given:
(i) 7 Audio cassettes and 3 Video cassettes cost is 1110.
(ii) 5 Audio cassettes and 4 Video cassettes cost Rs. 1350.
To Find: Cost of 1 audio cassette and 1 video cassettes.
Let (i) the cost of 1 audio cassette = Rs. x.
(ii) the cost of 1 video cassette = Rs. y.
According to the given conditions, we have
Thus, we get the following system of linear equation,
By using cross multiplication, we have
Hence cost of 1 audio cassette =
Hence cost of 1 video cassette =
Page No 3.77:
Answer:3
Given:
(i) Total numbers of pens and pencils = 40.
(ii) If she has 5 more pencil and 5 less pens, the number of pencils would be 4 times the number of pen.
To find: Original number of pens and pencils.
Suppose original number of pencil = x
And original number of pen = y
According the given conditions, we have,
Thus we got the following system of linear equations
Substituting the value of y from equation 1 in equation 2 we get
Substituting the value of y in equation 1 we get
Hence we got the result number of pencils is and number of pens are
Page No 3.77:
Answer:4
Given:
(i) Cost of 4 tables and 3 chairs = Rs 2250.
(ii) Cost of 3 tables and 4 chairs = Rs 1950.
To find: The cost of 2 chairs and 1 table.
Suppose, the cost of 1 table = Rs x.
The cost of 1 chair = Rs y.
According to the given conditions,
4x + 3y = 2250,
4x + 3y − 2200 = 0 …… (1)
3x + 4y = 1950,
3x + 4y − 1950 = 0 …… (2)
Solving eq. (1) and Eq. (2) by cross multiplication
∴ cost of 1 chairs = Rs. 150.
Hence total cost of 2 chairs and 1 table =
Page No 3.77:
Answer:5
Given:
(i) Cost of 3 bags and 4 pens = Rs. 257.
(ii) Cost of 4 bags and 3 pens = Rs. 324.
To Find: Cost of 1 bag and 10 pens.
Suppose, the cost of 1 bag = Rs. x.
and the cost 1 pen = Rs. y.
According to the given conditions, we have
3x + 4y = 257,
3x + 4y − 257 = 0 …… (1)
4x + 3y = 324
4x +3y − 324 = 0 …… (2)
Solving equation 1 and 2 by cross multiplication
Total cost of 1 bag and 10 pens =
Hence total cost of 1 bag and 10 pens =
Page No 3.77:
Answer:6
Given:
(i) Cost of 5 books and 7 pens = Rs. 79.
(ii) Cost of 7 books and 5 pens = Rs. 77.
To find: Cost of 1 book and 2 pens.
Suppose the cost of 1 book = Rs x.
and the cost of 1 pen = Rs y.
According to the given conditions, we have
5x + 7y = 79
5x + 7y − 79 = 0 …… (1)
7x + 5y = 77,
5x + 7y − 77 = 0 …… (2)
Thus we get the following system of linear equation,
Hence, the cost of 1 book = Rs 6
and the cost of 1 pen = Rs 7.
Therefore the cost of 2 pen = Rs 14.
Total cost of 1 book and 2 pens = 14 + 6 = 20
Total cost of 1 book and 2 pens =
Hence total cost of 1 book and 2 pens =
Page No 3.77:
Answer:7
To find:
(1) Total mangoes of A.
(2) Total mangoes of B.
Suppose A has x mangoes and B has y mangoes,
According to the given conditions,
3x + 6y + 270 = 0 …… (3) and
Now adding eq.2 and eq.3
5y = 310
y =
Hence A has 34 mangoes and B has 62 mangoes.
Page No 3.77:
Answer:8
Given:
(i) On selling of a T.V. at 5% gain and a fridge at 10% gain, shopkeeper gain Rs.2000.
(ii) Selling T.V. at 10% gain and fridge at 5% loss. He gains Rs. 1500.
To find: Actual price of T.V. and fridge.
According to the question:
Hence we got the pair of equations
1x + 2y − 40000 = 0 …… (1)
2x − 1y − 30000 = 0 …… (2)
Solving the equation by cross multiplication method;
Cost of T.V. =
Cost of fridge =
Hence the cost of T.V. is and that of fridge is _{.}
Page No 3.77:
Answer:9
Given: (i) 7 bats and 6balls cost is Rs3800
(ii) 3 bats and 5balls cost is Rs1750
To find: Cost of 1 bat and 1 ball
Let (i) the cost of 1 bat = Rs. x.
(ii) the cost of 1 ball = Rs. y.
According to the given conditions, we have
Thus, we get the following system of linear equation,
7x + 6y − 3800 = 0 …… (1)
3x + 5y − 1750 = 0 …… (2)
By using cross multiplication, we have
Hence cost of 1 bat =
Hence cost of 1 ball =
Page No 3.77:
Answer:11
To find:
(1) the fixed charge
(2) The charge for each day
Let the fixed charge be Rs x
And the extra charge per day be Rs y.
According to the given conditions,
Subtracting equation 1 and 2 we get
Substituting the value of y in equation 1 we get
Hence the fixed charge is and the charge of each day
Page No 3.82:
Answer:1
Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The sum of the two numbers is 8. Thus, we have
The sum of the two numbers is four times their difference. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
Multiplying the first equation by 5 and then adding with the second equation, we have
Substituting the value of x in the first equation, we have
Hence, the numbers are 5 and 3.
Page No 3.82:
Answer:2
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The sum of the digits of the number is 13. Thus, we have
After interchanging the digits, the number becomes.
The difference between the number obtained by interchanging the digits and the original number is 45. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
Page No 3.82:
Answer:3
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The sum of the digits of the number is 5. Thus, we have
After interchanging the digits, the number becomes.
The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
Page No 3.82:
Answer:4
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The sum of the digits of the number is 15. Thus, we have
After interchanging the digits, the number becomes.
The number obtained by interchanging the digits is exceeding by 9 from the original number. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
Page No 3.83:
Answer:5
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The two digits of the number are differing by 2. Thus, we have
After interchanging the digits, the number becomes.
The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have
So, we have two systems of simultaneous equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
(i) First, we solve the system
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
(ii) Now, we solve the system
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
There are two such numbers.
Page No 3.83:
Answer:6
Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The sum of the two numbers is 1000. Thus, we have
The difference between the squares of the two numbers is 256000. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the numbers are 628 and 372.
Page No 3.83:
Answer:7
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The two digits of the number are differing by 3. Thus, we have
After interchanging the digits, the number becomes.
The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have
So, we have two systems of simultaneous equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
(i) First, we solve the system
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
(ii) Now, we solve the system
Adding the two equations, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
Note that there are two such numbers.
Page No 3.83:
Answer:8
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The number is 4 times the sum of the two digits. Thus, we have
After interchanging the digits, the number becomes.
If 18 is added to the number, the digits are reversed. Thus, we have
So, we have the systems of equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
Subtracting the first equation from the second, we have
Substituting the value of y in the first equation, we have
Hence, the number is.
Page No 3.83:
Answer:9
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The number is 3 more than 4 times the sum of the two digits. Thus, we have
After interchanging the digits, the number becomes.
If 18 is added to the number, the digits are reversed. Thus, we have
So, we have the systems of equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
Subtracting the first equation from the second, we have
Substituting the value of y in the first equation, we have
Hence, the number is.
Page No 3.83:
Answer:10
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The number is 4 more than 6 times the sum of the two digits. Thus, we have
After interchanging the digits, the number becomes.
If 18 is subtracted from the number, the digits are reversed. Thus, we have
So, we have the systems of equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
Multiplying the second equation by 5 and then subtracting from the first, we have
Substituting the value of y in the second equation, we have
Hence, the number is.
Page No 3.83:
Answer:11
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The number is 4 times the sum of the two digits. Thus, we have
After interchanging the digits, the number becomes.
The number is twice the product of the digits. Thus, we have
So, we have the systems of equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
Substituting in the second equation, we get
Or
Substituting the value of y in the first equation, we have
Hence, the number is.
Note that the first pair of solution does not give a two digit number.
Page No 3.83:
Answer:12
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The product of the two digits of the number is 20. Thus, we have
After interchanging the digits, the number becomes.
If 9 is added to the number, the digits interchange their places. Thus, we have
So, we have the systems of equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
Substituting from the second equation to the first equation, we get
Or
Substituting the value of y in the second equation, we have
Hence, the number is.
Note that in the first pair of solution the values of x and y are both negative. But, the digits of the number can’t be negative. So, we must remove this pair.
Page No 3.83:
Answer:13
Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The difference between the two numbers is 26. Thus, we have
One of the two numbers is three times the other number. Here, we are assuming that x is greater than or equal to y. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
Substitutingfrom the second equation in the first equation, we get
Substituting the value of y in the first equation, we have
Hence, the numbers are 39 and 13.
Page No 3.83:
Answer:14
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The sum of the two digits of the number is 9. Thus, we have
After interchanging the digits, the number becomes.
Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have
So, we have the systems of equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
Substituting from the second equation to the first equation, we get
Substituting the value of y in the second equation, we have
Hence, the number is.
Page No 3.83:
Answer:15
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
The difference between the two digits of the number is 3. Thus, we have
After interchanging the digits, the number becomes.
Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have
So, we have two systems of simultaneous equations
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
(i) First, we solve the system
Multiplying the first equation by 2 and then subtracting from the second equation, we have
Substituting the value of x in the first equation, we have
Hence, the number is.
(ii) Now, we solve the system
Multiplying the first equation by 2 and then subtracting from the second equation, we have
Substituting the value of x in the first equation, we have
But, the digits of the number can’t be negative. Hence, the second case must be removed.
Page No 3.85:
Answer:1
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
The numerator of the fraction is 4 less the denominator. Thus, we have
If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is 8 times the numerator. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
Subtracting the second equation from the first equation, we get
Substituting the value of x in the first equation, we have
Hence, the fraction is.
Page No 3.85:
Answer:2
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
If 2 is added to both numerator and the denominator, the fraction becomes. Thus, we have
If 3 is added to both numerator and the denominator, the fraction becomes. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.85:
Answer:3
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
If 1 is subtracted from both numerator and the denominator, the fraction becomes. Thus, we have
If 1 is added to both numerator and the denominator, the fraction becomes. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.85:
Answer:4
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes. Thus, we have
If 1 is added to the denominator, the fraction becomes. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.86:
Answer:5
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes. Thus, we have
If the denominator is doubled and the numerator is increased by 8, the fraction becomes. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.86:
Answer:6
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
If 3 is added to the denominator and 2 is subtracted from the numerator, the fraction becomes. Thus, we have
If 6 is added to the numerator and the denominator is multiplied by 3, the fraction becomes. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.86:
Answer:7
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
The sum of the numerator and the denominator of the fraction is. Thus, we have
If the denominator is increased by 2, the fraction reduces to. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.86:
Answer:8
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
If 2 is added to the numerator of the fraction, it reduces to. Thus, we have
If 1 is subtracted from the denominator, the fraction reduces to. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.86:
Answer:9
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have
If the numerator and denominator are increased by 3, they are in the ratio 2:3. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.86:
Answer:10
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
The sum of the numerator and denominator of the fraction is 3 less than twice the denominator. Thus, we have
If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.86:
Answer:11
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is
The sum of the numerator and denominator of the fraction is 12. Thus, we have
If the denominator is increased by 3, the fraction becomes. Thus, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the fraction is.
Page No 3.88:
Answer:1
Let the present age of father be x years and the present age of son be y years.
Father is three times as old as his son. Thus, we have
After 12 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of father isyears and the present age of son isyears.
Page No 3.88:
Answer:2
Let the present age of A be x years and the present age of B be y years.
After 10 years, A’s age will beyears and B’s age will beyears. Thus using the given information, we have
Before 5 years, the age of A wasyears and the age of B wasyears. Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of A isyears and the present age of B isyears.
Page No 3.89:
Answer:3
Let the present ages of A, B, F and S be x, y, z and t years respectively.
A is elder to B by 2 years. Thus, we have
F is twice as old as A. Thus, we have
B is twice as old as S. Thus, we have
The ages of F and S is differing by 40 years. Thus, we have
So, we have four equations
……(1)
……(2)
……(3)
……(4)
Here x, y, z and t are unknowns. We have to find the value of x.
By using the third equation, the first equation becomes
From the fourth equation, we have
Hence, we have
Using the second equation, we have
Hence, the age of A isyears.
Page No 3.89:
Answer:4
Let the present age of the man be x years and the present age of his son be y years.
After 6 years, the man’s age will beyears and son’s age will beyears. Thus using the given information, we have
Before 3 years, the age of the man wasyears and the age of son’s wasyears. Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of the man isyears and the present age of son isyears.
Page No 3.89:
Answer:5
Let the present age of father be x years and the present age of his son be y years.
After 10 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have
Before 10 years, the age of father wasyears and the age of son wasyears. Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of father isyears and the present age of son isyears.
Page No 3.89:
Answer:6
Let the present age of father be x years and the present age of his son be y years.
The present age of father is three years more than three times the age of the son. Thus, we have
After 3 years, father’s age will beyears and son’s age will beyears.
Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of father isyears and the present age of son isyears.
Page No 3.89:
Answer:7
Let the present age of father be x years and the present age of his son be y years.
The present age of father is three times the age of the son. Thus, we have
After 12 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of father isyears and the present age of son isyears.
Page No 3.89:
Answer:8
Let the present age of father be x years and the present ages of his two children’s be y and z years.
The present age of father is three times the sum of the ages of the two children’s. Thus, we have
After 5 years, father’s age will beyears and the children’s age will beandyears. Thus using the given information, we have
So, we have two equations
Here x, y and z are unknowns. We have to find the value of x.
Substituting the value offrom the first equation in the second equation, we have
By using cross-multiplication, we have
Hence, the present age of father isyears.
Page No 3.89:
Answer:9
Let the present age of father be x years and the present age of his son be y years.
After 2 years, father’s age will beyears and the age of son will beyears. Thus using the given information, we have
Before 2 years, the age of father wasyears and the age of son wasyears. Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of father isyears and the present age of son isyears.
Page No 3.89:
Answer:10
Let the present age of Nuri be x years and the present age of Sonu be y years.
After 10 years, Nuri’s age will be(x + 10) years and the age of Sonu will be(y + 10) years. Thus using the given information, we have
Before 5 years, the age of Nuri was(x – 5)years and the age of Sonu was(y – 5)years. Thus using the given information, we have
So, we have two equations
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
Hence, the present age of Nuri isyears and the present age of Sonu isyears.
Page No 3.89:
Answer:11
Let the present ages of Ani, Biju, Dharam and Cathy be x, y, z and t years respectively.
The ages of Ani and Biju differ by 3 years. Thus, we have
Dharam is twice as old as Ani. Thus, we have
Biju is twice as old as Cathy. Thus, we have
The ages of Cathy and Dharam differ by 30 years. Clearly, Dharam is older than Cathy. Thus, we have
So, we have two systems of simultaneous equations
(i)
(ii)
Here x, y, z and t are unknowns. We have to find the value of x and y.
(i) By using the third equation, the first equation becomes
From the fourth equation, we have
Hence, we have
Using the second equation, we have
From the first equation, we have
Hence, the age of Ani isyears and the age of Biju isyears.
(ii) By using the third equation, the first equation becomes
From the fourth equation, we have
Hence, we have
Using the second equation, we have
From the first equation, we have
Hence, the age of Ani isyears and the age of Biju isyears.
Note that there are two possibilities.
Page No 3.98:
Answer:1
We have to find the speed of car
Let and be two cars starting from pointsand respectively. Let the speed of car be x km/hr and that of carbe y km/hr.
Case I: When two cars move in the same directions:
Suppose two cars meet at point, Then,
Distance travelled by car
Distance travelled by car
It is given that two cars meet in 7 hours.
Therefore, Distance travelled by car X in hours = km
Distance traveled by car y in 7 hours = km
Clearly
Dividing both sides by common factor 7 we get,
Case II : When two cars move in opposite direction
Suppose two cars meet at point. Then,
Distance travelled by car,
Distance travelled by car.
In this case, two cars meet in 1 hour
Therefore Distance travelled by car X in1 hour km
Distance travelled by car Y in 1 hour km
From the above clearly,
...(ii)
By solving equation (i) and (ii), we get
Substituting in equation (ii) we get
Hence, the speed of car starting from point A is
The speed of car starting from point B is.
Page No 3.98:
Answer:2
Let the speed of the sailor in still water be x km/hr and the speed of the current be y km/hr
Speed upstream
Speed downstream
Now, Time taken to cover 8km down stream =
Time taken to cover 8km upstream=
But, time taken to cover 8 km downstream in 40 minutes or that is
Dividing both sides by common factor 2 we get
Time taken to cover 8km upstream in1hour
...(ii)
By solving these equation and we get
Substitute in equationwe get
Hence, the speed of sailor is
The speed of current is
Page No 3.98:
Answer:3
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr
Speed upstream =km/hr
Speed down stream =km/hr
Now,
Time taken to cover 30 km upstream = hrs
Time taken to cover 44 km down stream = hrs
But total time of journey is 10 hours
Time taken to cover 40 km upstream=
Time taken to cover 55 km down stream =
In this case total time of journey is given to be 13 hours
Therefore, ...(ii)
Putting = u and in equation and we get
Solving these equations by cross multiplication we get
and
Now,
By solving equation and we get ,
Substituting in equation we get ,
Hence, speed of the boat in still water is
Speed of the stream is
Page No 3.98:
Answer:4
We have to find the speed of the boat in still water and speed of the stream
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr then
Speed upstream
Sped down stream
Now, Time taken to cover km down stream =
Time taken to cover km upstream =
But, total time of journey is 6 hours
Time taken to cover km upstream =
Time taken to cover km down stream =
In this case total time of journey is given to or
...(ii)
By and in equation (i) and (ii) we get
Solving these equations by cross multiplication we get
and
and
Now,
and
By solving equation and we get,
By substituting in equation we get
Hence, the speed of the stream is
The speed of boat is
Page No 3.98:
Answer:5
Let the speed of Ajeet and Amit be x Km/hr respectively. Then,
Time taken by Ajeet to cover
Time taken by Amit to cover
By the given conditions, we have
If Ajeet doubles his speed, then speed of Ajeet is
Time taken by Ajeet to cover
Time taken by Amit to cover
According to the given condition, we have
...(ii)
Putting and, in equation (i) and (ii), we get
Adding equations (iii) and (iv), we get
Putting in equation (iii), we get
Now,
and
Hence, the speed of Ajeet is
The speed of Amit is
Page No 3.98:
Answer:6
Let the actual speed of the train be x Km/hr and the actual time taken by y hours. Then,
Distance covered=
If the speed is increased by, then time of journey is reduced by 1 hour i.e., when speed is , time of journey is
Distance covered= km
...(ii)
When the speed is reduced by, then the time of journey is increased by i.e., when speed is, time of journey is
Distance covered =
Thus we obtain the following equations
By using elimination method, we have
Putting the value in equation (iii) we get
Putting the value of x and y in equations (i) we get
Distance covered =
=
Hence, the distance is,
The speed of walking is .
Page No 3.98:
Answer:7
Let the speed of the train be x km/hour that of the car be y km/hr, we have the following cases
Case I: When Ramesh travels 760 Km by train and the rest by car
Time taken by Ramesh to travel 160 Km by train =
Time taken by Ramesh to travel (760-160) =600 Km by car =
Total time taken by Ramesh to cover 760Km = +
It is given that total time taken in 8 hours
Case II: When Ramesh travels 240Km by train and the rest by car
Time taken by Ramesh to travel 240 Km by train =
Time taken by Ramesh to travel (760-240) =520Km by car =
In this case total time of the journey is 8 hours 12 minutes
...(ii)
Putting and, , the equations and reduces to
Multiplying equation (iii) by 6 and (iv)by 20 the above system of equation becomes
Subtracting equation from we get
Putting in equation, we get
Now
and
Hence, the speed of the train is ,
The speed of the car is .
Page No 3.98:
Answer:8
Let the speed of the train be x km/hr that of the car be y km/hr, we have the following cases:
Case I: When a man travels 600Km by train and the rest by car
Time taken by a man to travel 400 Km by train =
Time taken by a man to travel (600-400) =200Km by car =hrs
Total time taken by a man to cover 600Km =
It is given that total time taken in 8 hours
Case II: When a man travels 200Km by train and the rest by car
Time taken by a man to travel 200 Km by train =
Time taken by a man to travel (600-200) = 400 Km by car
In this case, total time of the journey in 6 hours 30 minutes + 30 minutes that is 7 hours,
...(ii)
Putting and, , the equations and reduces to
Multiplying equation (iii) by 6 the above system of equation becomes
Substituting equation and, we get
Putting in equation, we get
Now
and
Hence, the speed of the train is,
The speed of the car is.
Page No 3.98:
Answer:9
Let x and y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of the car Y be y km/hr.
Case I: When two cars move in the same directions:
Suppose two cars meet at point Q, then,
Distance travelled by car X = AQ
Distance travelled by car Y = BQ
It is given that two cars meet in 8 hours.
Distance travelled by car X in 8 hours =km
AQ=
Distance travelled by car Y in 8 hours =km
BQ =
Clearly AQ-BQ = AB
Both sides divided by 8, we get
Case II: When two cars move in opposite direction
Suppose two cars meet at point P, then,
Distance travelled by X car X=AP
Distance travelled by Y car Y=BP
In this case, two cars meet in 1 hour 20 minutes, we can write it as 1 hour or
hours that is hours.
Therefore,
Distance travelled by car y in hours = km
Distance travelled by car y in hours = km
...(ii)
By solving (i) and (ii) we get,
By substituting in equation (ii), we get
Hence, speed of car X is_{ } , speed of car Y is.
Page No 3.98:
Answer:10
We have to find the speed of the boat in still water and speed of the stream
Let the speed of the boat in still water be km/hr and the speed of the stream be km/hr then
Speed upstream
Sped down stream
Now, Time taken to cover km upstream =
Time taken to cover km down stream =
But, total time of journey is 8 hours
Time taken to cover km upstream =
Time taken to cover km down stream =
In this case total time of journey is given to
...(ii)
By and in equation (i) and (ii) we get
Solving these equations by cross multiplication we get
and
and
Now,
and
By solving equation and we get,
By substituting in equation we get
Hence, the speed of boat in still water is ,
The speed of the stream is.
Page No 3.99:
Answer:11
Let the speed of the train be x km/hour that of the bus be y km/hr, we have the following cases
Case I: When Roohi travels 300 Km by train and the rest by bus
Time taken by Roohi to travel 60 Km by train =
Time taken by Roohi to travel (300-60) =240 Km by bus =
Total time taken by Roohi to cover 300Km= +
It is given that total time taken in 4 hours
Case II: When Roohi travels 100 km by train and the rest by bus
Time taken by Roohi to travel 100 Km by train =
Time taken by Roohi to travel (300-100) =200Km by bus =
In this case total time of the journey is 4 hours 10 minutes
...(i)
Putting and, , the equations and reduces to
Subtracting equation (iv) from (iii)we get
Putting in equation (iii), we get
Now
and
Hence, the speed of the train is,
The speed of the bus is.
Page No 3.99:
Answer:12
Let the speed of rowing in still water be x km/hr and the speed of the current be y km/hr
Speed upstream
Speed downstream
Now,
Time taken to cover km down stream =
Time taken to cover km upstream =
But, time taken to cover km downstream in
Time taken to cover km upstream in 2 hours
...(i)
By solving these equation (i) and (ii) we get
Substitute in equation (i)we get
Hence, the speed of rowing in still water is,
The speed of current is .4 km/ hr
Page No 3.99:
Answer:13
Let the speed of A and B be x Km/hr and y Km/hr respectively. Then,
Time taken by A to cover ,
And, Time taken by B to cover .
By the given conditions, we have
If A doubles his pace, then speed of A is
Time taken by A to cover ,
Time taken by B to cover .
According to the given condition, we have
Putting and, in equation (i) and (ii), we get
Adding equations (iii) and (iv), we get,
Putting in equation (iii), we get