Exercise 3.1
Question 1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.”(Isn’t this interesting?) Represent this situation algebraically and graphically.
Sol. Let Aftab’s present age be x years. Daughter’s present age be y years.
Seven years ago
Aftab’s age=(x – 7) years.Daughter’sage=(y – 7)years.According to the question,
Aftab’s age=7 times the age of the daughter
x-7=7(y-7) x-7=7y-49 x-7y=-42 (1)
After three years
Aftab’s age=(x + 3)years.Daughters age=(y + 3)years.According to the questions,
Aftabs age =3 times the age of the daughter
(x+3)=3(y+3) x+3=3y+9 x-3y=6 (2)
Thus,algebraic representation of given situation is given by the equations
(1)and (2).Now, represent this situation graphically, we find three solutions of each equation.
From eq. (1):
X-7y=-42 X=-42+7y
Thus, in the above graph, the two lines representing the two equations intersecting at the point C(42, 12).
Hence,x=42 and y=12.
:.Aftab’s present age = 42 years.His daughter’s present age = 12 years. Also, when his daughter was born, Aftab’s age=42-12=30 years
Note: Every solution of the equation is a point on the line representing it.
Question 2. The coach of a cricket team buys 3 bats and 6 balls for 3900. Later she buys another bat and 3 more balls of the same kind for < 1300. Represent this situation algebraically and geometrically.
Sol. Let the cost of a bat be < x.
Let the cost of a ball be < y.
Given,coach buys 3 bats and 6 balls for < 3900.
:. 3x+6y=3900
Similarly, coach buys another bat and 3 more balls for 1300.
:. x+3y=l300
Thus, the algebraic representation is given as
3x+6y=3900 or x+3y=l300.
For geometrical representation:
We find three solutions of each of the equation. From eq. (1): 3x + 6y = 3900
Plot these points A(0, 650), B(600, 350), C(1300, 0), D(100, 400) and E(400, 300) on graph to get the lines.
Thus, equation of lines intersect at the point C(1300, 0).
Hence,x=1300, y=0 is the required solution of the pair of linear equations.
Note : A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations.
Question 3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be < 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is < 300. Represent the situation algebraically and geometrically.
Sol. Let the cost of apple be < x per kg. Let the cost of grapes be < y per kg.
Since, the cost of 2kg of apples and 1 kg of grapes on a day was found to be < 160.
:.2x+ y=l60 …(1)
Similarly, the cost of 4 kg of apples and 2 kg of grapes is < 300.
:. 4x + 2y = 300 …(2)
Thus, the equations (1) and (2) represents the situation algebraically.
For graphical representation,
We find three solutions of each of the equation. From eq (1):2x + y = 160 y = 160 – 2x
Note: A pair of linear equations in two variables, which has no solution is called an inconsistent pair of linear equations
Excercise 3.2
Question 1. From pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 Pencils and 7 pens together cost rs. 50, whereas 7 pencils and 5 pens together cost 46 rs.Find the cost of one pen pencil and that of one pen.
Since,both the lines are intersecting at point (3, 7)
.’.X=3, y=7.
(ii)Let the cost of a pencil be x and the cost of a pen bey.
According to the question, we have
5x + 7 y = 50 and 7x + 5y = 46
This is the required pair of equations. Geometrically :
Now, we find two solutions for each of these equations.
Solution of the equations are given in table.
5x + 7 y = 50 7x + 5y = 46
Question 2. On comparing the ratios a1/a2,b1/b2, and c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x-4y+8=0; 7x+6y-9=0
(ii) 9x+3y+12=0; 18x+6y+24= 0
(iii) 6x-3y+10=0; 2x-y+9=0
Question 3. On comparing the ratios al/a2,b1/b2 and cl/c2, find out whether the following pair of linear equations are consistent or inconsistent.
(I) 3x+2y=5; 2x-3y=7
(ii) 2x-3y=8; 4x-6y=9
(iii) 3/2x+5/3y=7; 9x-10y=14
(iv) 5x-3y=ll; -10x+6y=-22
(v) 4/3x+2y=8; 2x+3y=12
Note : A dependent pair of linear equations is always consistent.
Question 4. Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically.
(I) x+y=5, 2x+2y=10
(ii) x- y = 8, 3x-3y =16,
(iii) 2x+ y-6= 0, 4x-2y-4= 0
(iv) 2x-2y-2 = 0, 4x-4y-5= 0
Sol. (I) Given equations are
x+y=5 and 2x+2y=l0
On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get
a1/a2=b1/b2=c1/c2=1/2
Thus, pair of linear equations are consistent.
Graphically : We have x+ y = 5.
Now,when x=0,y=5.When y=0,x=5 We have the following table
Similarly, we have 2x + 2y = 10
:.The table is given as
Now,we plot the points A(5, 0), B(0, 5), C(l, 4) and D(2, 3) on the graph paper. Join all the points to obtain the graph we observe C and D both lie on A and B.
Thus,the graph of the two equations are coincident.
Hence,the system of equations has infinitely many solutions. i.e.,consistent.
(ii) x-y-8=0 …(1)
3x-3y-16= 0 …(2)
On comparing with a1x + b1y + c1 = 0 and a2x + bzY + c2 = 0, we have
a1/a2=1/3 b1/b2=-1/-3=1/3, c1/c2=-8/-6= 1/2
Therefore,
:. the pair of equations is inconsistent and the graph of the two equations is a pair of parallel straight lines.
Question 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Sol. Let l be the length and b be the width of rectangular garden.
According to the question: t = b + 4 …(1)
and 1/2[2(l+b)]=36 …(2)
l+b=36 b+4+b=36
2b = 32 b = 16 and l = 16 + 4 = 20 [From eqn (l)]
Hence, l = 20 m and b = 16 m are the required dimensions of the garden
Question 6. Given the linear equation 2x + 3 y- 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i)intersecting lines (ii) parallel lines (iii) coincident lines
Note : Answer can be different from person to person.
Question 7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Sol. Given equations are x-y+1=0 and 3x+2y-12=0
For x-y+1=0, we have the table as follows:
For 3x+2y=12, we have the table as follows:
Now, we plot the points A(0, 1),B(2, 3),C(0, 6)and D(4, 0) on the graph paper. Join all the points to obtain the triangle.
Clearly, we obtain a triangle BED formed by the given lines and the x-axis.The co-ordinates of the vertices of triangle are B(2, 3), E(- 1, 0) and D(4, 0).
Exercise 3.3
Question 1. Solve the following pair of linear equations by the substitution method.
Sol. (i) The given system of equations is
x+y=14 and x-y=4
From eqn(1), we have: x= 14-y Putting value of x in equation (2),we get l4-y-y=4 14-2y=4
=> -2y=-10 => y=5
Thus, x=l4-y=l4-5=9. So, x=9 and y=5
(ii) Given pair of linear equations are
s-t=3 …(1)
s/3 + t/2 = 6 …(2)
From equation (1), we have: s=3+t
Putting value of sin eqn (2), we get
3+t/3+t/2=6 => 3/3+t/3+t/2=6 => t/3+t/2=6-1 => 2t+3t/6 = 5
=> 5t=5×6 => t=6
Putting t = 6 in eq. (3), we get:s=3+t => s=3+6=9 :. s=9 and t=6.
(iii) The given system of equations is
3x-y=3 …(1)
and, 9x-3y=9 …(2)
From eq (1):
y = 3x- 3
Substituting y=3x-3 in eq.(2),we get
9x-3(3x-3)=9 => 9x-9x+9=9 =>9=9
This statement is true for all values of x. However, we do not get a specific value of x as a solution. So, we cannot obtain a specific value of y. This situation has arisen because both the given equations are the same.
:. Equations (1) and (2) have infinitely many solutions.
(iv) The given system of equations is 0.2x+0.3y=1.3 => 2x+3y=13 …(1)
and,0.4x+O.Sy=2.3 => 4x+5y=23 …(2)
From eq.(2): 5y= 23-4x => Y=23-4x/5
Substituting y = 23-4x/5 in(1), we get 2x+3(23-4x/5)=13
lOx+69-12x=65 => -2x=-4 x=2
Putting x=2 in eq.(1), we get:2×2+3y=13
3y=13-4=9 => y=9/3=3
Hence, the solution of the given system of equations is x=2,and y=3
Question 2. Solve 2x+3y=ll and 2x-4y=-24 and hence find the value for ‘m’ for which y=mx+3.
Question 3. Form the pair of linear equations for the following problems and find their solution by substitution method.
(I) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for 3800. Later, she buys 3 bats and 5 balls for 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is 105 and for a journey of 15 km, the charge paid is rs.155 . What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km ?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and denominator.If,3 is added to both the numerator and the denominator it becomes 6,Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son.Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Sol. (I) Let the required numbers be x and y,(x > y).
According to the question,
x-y= 26 …(1)
and x= 3y …(2)
Put value of x = 3y in equation (l),we get:
3y-y=26 => 2y=26 => y=l3
Thus,x=3xl3=39. [From (2)]
Hence, the numbers are 39 and 13.
Note: We can also find the numbers by considering the case y>x.
(ii) Let the larger supplementary angle be x0 and smaller be y0.
According to the question,
x=y+l8°…(1)
Also, x+ y =180° (Supplementary angles)…(2)
Putting value of x from equation (1), in equation(2), we get
y+l8°+y=l80° 2y=l62° y=81°
Thus, x=l80°-8l0=99°[From (2)]
Hence, required angles are 99° and 81°.
(iii) Let the cost of one bat and one ball be < x and < y respectively. According to the question, 7x + 6y = 3800 …(1) and 3x + Sy= 1750 …(2) From eq.(2): 5y= 1750-3x => Y =l750-3x/5
Substituting y =1750-3x/5 in eq. (1), we get:
7x+6(175-3)/5 = 3800
35x+10500-18x=19000 => l7x =19000-10500
=> 17x = 8500
=> x = 8500/17 = 500
Putting x = 500 in eq. (2), we get:3(500)+5y=1750
=> 5y=1750-1500
=> 5y=250 => y =250/5 = 50
Hence the cost of one bat is <500 and the cost of one ball is rs. 50. (iv)Let the fixed charges of taxi be 105+5y=l55 =>5y=l55-105 =>5y=50 => y=l0
Putting y=10 in eq.(1),we get:x+10×10=105
X = 105-100 => X=5
Total charges for travelling a distance of 25km
= X + 25y = t(5+25X10) = rs.255
Hence, the fixed charge is rs.5, the charge per km is rs.10 and the total charges for travelling a distance of 25 km is rs.255.
(vi) Let the present age of Jacob be x years and his son by years. Five years hence,
According to the question
x+5=3(y+5)
=>x=3y+IO
Five years ago, according to the question,
x-5=7(y-5) => x-7y=-30
Putting value of x from eq.(1)in eq(2),we get
3y+I0-7y=-30 =>-4y=-40 => y=1O
Thus, x=3y+1O=3×10+10=40
:.present age of Jacob=40years and present age of his son=10years.
Exercise 3.4
Question 1. Solve the following pair of linear equations by the elimination method and the substitution method
(I) x + y = 5 and 2x- 3 y = 4 (ii)3x+4y=10and 2x-2y=2
(iii) 3x-5y-4=0 and 9x=2y+7 (iv) x/2+y=-1 and x-1=3
Sol. (ii) By elimination method :
The given system of equations is
3x + 4y = 10 …(1)
and,2x-2y= 2 …(2)
Multiplying equation (2) by 2, we get :
4x-4y = 4 …(3)
Adding eq.(1) and(3),we get
(3x+4y)+(4x-4y)=10 + 4
7x=14 =>X = 2
Putting x=2 in(1),we get :
3(2)+4y=10 4y=10-6 = 4
Hence, x=2,y=1 y=1
By substitution method :
The given system of equations is
3x + 4y = 10 …(1)
and,2x-2y=2 => x-y=1 …(2)
From equation(2),y=x-1
Substituting y = x-1 in eq.(1),we get:
3x+4(x-1)=10 => 3x+4x-4=10 => 7x=14 =>x = 2
Putting x=2 in eq.(1),we get:
3(2)+4y=10 => 4y=10-6=4 =>y = 1
Hence, x=2,y=1
(iii) By elimination method:
The given system of equations is
3x-5y-4=0 => 3x-Sy=4 …(1)
and 9x=2y+7 => 9x-2y=7 …(2)
Multiplying equation (1) by 3, we get
9x-15y = 12
Subtracting equation(2)from(3),we get
(9x-15y)-(9x-2y)=12-7 => -13y=5 => y = -5/13
Putting y =-5/13 in eq.(1),we get
3x=5(-5/13)=4 => 3x=4-25/13 => 13 = 52-25/13
=>3x = 27/13 =>x=9/13
hence, x=9/13, y= -5/13
By substitution method :
The given system of equations is
3x-5y-4=0 => 3x-5y=4 …(1)
and,9x=2y+7 => 9x – 2y = 7 …(2)
From equation(1),2y=9x-7 => y = 9x-7/2 …(3)
Substituting y = 9x-7/2 in equation(1),we get
3x-5(9x-7)/2 = 4 => 6x-45x+35=8 => -39x=8-35
-39x= -27 => x = -27/-39
Putting x= 9/13 in equation(3),we get
3x-5(9x-7/2)=4 => 6x-45x+35=8 =>-39x=8-35
=> -39x = -27 => x=-27/-39=9/13
Putting x=9/13 in equation(3), we get
y=9x-7/2 => (9(9/13)-7)/2 => y=(81/13-7)/2 = 81-91/13×2 = -10/26 = -5/13
Hence, x = 9/13, y = -5/13
(iv) By elimination method:
The given system of equations is
x/2 + 2y/3 = -1 => 3x + 4y =-6 …(1)
and x-y/3 = 3 => 3x-y = 9 …(2)
Multiplying eq.(2)by 4 and adding to(1),we get:
4(3x-y)+(3x+4y)= 4(9)+(-6)
=> 12x-4y+3x+4y= 36-6
=> 15x=30 => X=2
Putting x=2 in eq.(2),we get :
3(2)-y=9 =>-y=9-6=3 =>y=-3
Hence, x=2,y=-3
By substitution method :
The given system of equations is
x/2 + 2y/3 = -1 => 3x + 4y = -6 …..(1)
and, x-y/3 = 3 => 3x-y = 9 …..(2)
From eq.(2): y= 3x-9
Putting y= 3x-9 in eq(1), we get:
3x+4(3x-9)=-6 => 3x+12x-36 =-6 => 15x = 30 => x=2
Putting x=2 in eq(2), we get
3(2)-y=9 => -y=9-6=3 => y=-3
Hence, x=2, y=-3
Note : While using elimination method, eliminate any one variable first, to get a linear equation in one variable.
Question 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
(I) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later,
Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went, to a bank to withdraw rs. 2000. She asked the cashier to give her rs.50 and rs.100 notes only. Meena got rs.25 notes in all. Find how many notes of rs.50 and rs.100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day there after. Saritha paid rs.27 for a book kept for seven days, while Susy paid rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Exercise 3.5
Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.
(i) x-3y-3=0; 3x-9y-2=0 (ii) 2x+y=5; 3x+2y=8
(iii) 3x-5y=20; 6x-10y=40 (iv) x-3y-7=0; 3x-3y-15=0
Sol. (i) x-3y-3 = 0 …(i)
3x-9y-2 = 0 …(ii)
On comparing with a1x+b1y+c1=0
and a2x+bzY+c2=0,we get
a1/a2=1/3, b1/b2=-3/-9=1/3, c1/c2=-3/-2=3/2
Hence, no solution exists.
(ii) The given system of equations may be written as
2x+y-5=0 and 3x+2y-8=0
On comparing with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get
=> x/-8+10 = y/-15+16 = 1/4-3
=> x/2= y/1=1/1 => x/2=1/1 and y/1=1/1
=> x=2, y=1
x = 2 and y = l
Hence, the given system of equations has a unique solution given by
X = 2,y = l.
(iii) The given system of equations may be written as 3x-5y-20=0 and
6x-10y-40=0
On comparing with a1x+b2y+c1=0 and a2x+b2y+c2=0,we get
a1/a2=3/6=1/2,b1/b2=-5/-10=1/2 and c1/c2=-20/-40=1/2
Clearly, a1/a2=b1/b2=c1/c2
So, the given system of equations has infinitely many solutions.
(iv) The given system of equations is x-3y-7=0 and 3x-3y-15=0
On comparing with a1x+b1y+c1=0 and a2x+b2y+c2=0,we get
Note: In cross-multiplication method, the arrows between the two methods indicate that they are to be multiplied and the second product is to be subtracted from the first
Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x+3y=7
(a-b)x+(a+b)y = 3a+b-2
(ii) For which value of k will the following pair of linear equations have no solution?
3x+ y=l
(2k-1)x+(k-1)y = 2k+ 1
Question 3. Solve the following pair of linear equations by the substitution and cross multiplication methods:
8x+Sy= 9
3x+2y=4
Sol. 8x+5y- 9 = 0 …(1)
3x+2y-4=0 …(2)
By cross-multiplication method :
Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist ) by any algebraic method :
(I) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Sol.(i) Let fixed hostel charges (monthly) = rs.c
and cost of food for one day = rs.x
According to the question
20x + C = 1000 …(1)
26x + C = 1180 …(2)
Subtracting eq.(1) from eq.(2),we get:
(ii) (26x + c)-(20x + c) = 1180-1000 => 6x = 180x=30
Putting x=30 in eq.(1),we get:20(30)+c=1000
=> C=1000-600=400
Hence, monthly fixed charges = rs.400 and cost of food per day =rs.30
(ii) Let the fraction be x/y
According to the question: x-1/y = 1/3 => 3x-y=3 …(1)
and x/y+8 = 1/4 => 4x-y= 8 …(2)
Subtracting eq.(1) from eq.(2):(4x -y)-(3x -y)= 8-3
=> 4x-3x-y+y=5 => x=5
Putting x=5 in eq.(1),we get:3(5)-y=3 =>y=15-3=12
Hence, the fraction is 5/12.
(iii) Let number of right answers = x
Number of wrong answers = y
Total number of questions =x+y
First case,
Marks awarded for x right answers =3x
Marks lost for y wrong answers = y x 1=y
:. 3x-y = 40 …(1)
Second case,
Marks awarded for x right answers = 4x
Marks lost for y wrong answers = 2y
:. 4x-2y=50 …(2)
From eq.(1), y=3x-40
Putting y=3x-40 in eq.(2),we get
4x-2(3x-40)=50 => 4x-6x+80=50 => 2x=30 =>x=15
Now,y=3x-40=3(15)-40= 45-40 =5
:. total number of questions= 15+5=20
(iv) Let speed of car I= x krn/hr. Speed of car II= y km/hr
First case:
Two cars meet at C after 5 hrs.
AC= Distance travelled by car I in 5 hrs= 5x km
BC= Distance travelled by car II in 5 hrs = Sy km
AC-BC=AB => 5x-5y= 100 (·: AB= lO0km.)
x-y=20 …..(1)
Second case :
Two cars meet at C after one hour.
x+y=l00 …(2)
Adding(1) and(2), we get:
X = 60
Then,y = 100-x = 100-60 =40
Hence, the speeds of the two cars are respectively 60 km/hr and 40 km/hr.
(v) Let the length and breadth be x units and y units respectively.
:. Area xy square units
First case:
length = (x—5)units
breadth = (y+3) units
:.New area = Original area = 9
=> (x-5)(y+3)=xy-9 (Area of Rect.= length x Breath)
=xy +3x-Sy—15=xy— 93x Sy= 6 ….(1)
Second case:
length =(x+3)
and breadth =(y+ 2) units
:.New area = Original area + 67
=>(x+3)(y+2)= xy+67 (Area of Rect.= length x Breath)
=> xy+6+2x+3y= xy+67 => 2x+3y=61 …(2)
Mutiplying (1) by 3 and (2) by 5,
9x-15y=18 …(3)
10x+15y=305 ….(4)
Adding (3) and (4), we get: 19x=323 = x=17
Substituting x = 17 in (2), we get: 2(17)+3y= 61
3y=61-34 => y=27/3=9
Hence, length = 17 units. Breadth= 9 units.
Exercise 3.6
Question 1. Solve the following pairs of equations by reducing them to a pair of linear equations:
Note: You can use any of the three methods i.e.elimination.substitution or cross multiplication method to solve the equations algebraically.
Question 2. Formulate the following problems as pair of equations, and hence find their solutions :
(I) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Sol.(i) 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= (x -y) km/hr Speed downstream =(x+y) km/hr
Time taken to cover 20 km downstream = 2 hrs.
20/x+y =2 (:. Time= Distance/Speed)
Exercise 3.7
Question 1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Question 2. One says,”Give me a hundred, friend! I shall then become twice as rich as you”.The other replies,”If you given me ten,I shall be six times as rich as you”.Tell me what is the amount of their(respective) capital? [From the Bijaganita of Bhaskara II]
Hint:[x+l00 = 2(y-100),y+l0 = 6(x-10)).
Sol. Let one friend has rs.x and second has rs.y.
According to the question
x+ 100=2(y-100) => x-2y=-300 …(1)
and y+l0=6(x-10) =>6x-y=70 ….(2)
Multiplying equation(2) by 2, we get:
12x – 2y = 140 … (3)
Subtracting equation (1) from equation (3), we get
(12x-2y-140)-(x-2y+300) = 0 => llx-440 = 0 x = 40
Putting x = 40 in equation (1), we get
40 – 2y = – 300 => -2y = -340 => y = 170
Hence, one friend has rs.40 and second has rs.170.
Question 3. A train covered a certain distance at a uniform speed If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Sol. Let the speed of the train be x km/hr and the time taken to complete the journey be y hours
Question 4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less.If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Question 5. In a Triangle ABC, ang.C=3ang.B=2(ang.A+ang.B). Find the three angles.
Question 6. Draw the graphs of the equations 5x-y = 5 and 3x-y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.
Question 7. Solve the following pair of linear equations:
Question 8. ABCD is a cyclic quadrilateral(see fig.). Find the angles of the cyclic quadrilateral.
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