Pa.Linear Eq Ex 3.4 Solutions
Question - 1 : - Graphically, solve thefollowing pair of equations:
2x + y =6
2x – y +2 = 0
Find the ratio of theareas of the two triangles formed by the lines representing these equationswith the x-axis and the lines with the y-axis.
Answer - 1 : -
Solution:
Given equations are 2x+ y = 6 and 2x – y + 2 = 0
Table for equation 2x+ y – 6 = 0, for x = 0, y = 6, for y = 0, x = 3.
Table for equation 2x– y + 2 = 0, for x = 0, y = 2, for y = 0,x = – 1
Let A1 andA2 represent the areas of triangles ACE and BDE respectively.
Let, Area of triangleformed with x -axis = T1
T1 =Area of △ACE = ½ × AC × PE
T1 = ½× 4 × 4 = 8
And Area of triangleformed with y – axis = T2
T1 =Area of △BDE = ½ × BD × QE
T1 = ½× 4 × 1 = 2
T1:T2 =8:2 = 4:1
Hence, the pair ofequations intersect graphically at point E(1,4)
i.e., x = 1 and y = 4.
Question - 2 : - Determine,graphically, the vertices of the triangle formed by the lines
y = x, 3y = x, x + y =8
Answer - 2 : -
Solution:
Given linear equationsare
y = x …(i)
3y = x …(ii)
and x + y = 8 …(iii)
Table for line y = x,
Table for line x = 3y,
Table for line x + y =8
Plotting the points A(1, 1), B(2,2), C (3, 1), D (6, 2), we get the straight lines AB and CD.
Similarly, plottingthe point P (0, 8), Q(4, 4) and R(8, 0), we get the straight line PQR.
AB and CD intersectsthe line PR on Q and D, respectively.
So, △OQD is formed by these lines. Hence, thevertices of the △OQD formed by thegiven lines are O(0, 0), Q(4, 4) and D(6,2).
Question - 3 : - Draw the graphs of theequations x = 3, x = 5 and 2x – y –4 = 0. Also find the area of the quadrilateral formed by the lines andthe x–axis.
Answer - 3 : -
Solution:
Given equation oflines x = 3, x = 5 and 2x-y-4 = 0.
Table for line 2x – y– 4 = 0
Plotting the graph, weget,
From the graph, weget,
AB = OB-OA = 5-3 = 2
AD = 2
BC = 6
Thus, quadrilateralABCD is a trapezium, then,
Area of QuadrileralABCD = ½ × (distance between parallel lines) = ½ × (AB) × (AD + BC)
= 8 sq units
Question - 4 : - The cost of 4 pens and4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than thecost of a pencil box. Form the pair of linear equations for the abovesituation. Find the cost of a pen and a pencil box.
Answer - 4 : -
Solution:
Let the cost of a penand a pencil box be Rs x and Rs y respectively.
According to thequestion,
4x + 4y = 100
Or x + y = 25 …(i)
3x = y + 15
Or 3x-y = 15 …(ii)
On adding Equation (i)and (ii), we get,
4x = 40
So, x = 10
Substituting x = 10,in Eq. (i) we get
y = 25-10 = 15
Hence, the cost of apen = Rs. 10
The cost of a pencilbox = Rs. 15
Question - 5 : - Determine,algebraically, the vertices of the triangle formed by the lines
3x – y =3
2x – 3y =2
x + 2y = 8
Answer - 5 : -
Solution:
3x – y = 2 …(i)
2x -3y = 2 …(ii)
x + 2y = 8 …(iii)
Let the equations ofthe line (i), (ii) and (iii) represent the side of a ∆ABC.
On solving (i) and(ii),
We get,
[First,multiply Eq. (i) by 3 in Eq. (i) and then subtract]
(9x-3y)-(2x-3y) = 9-2
7x = 7
x = 1
Substituting x=1 inEq. (i), we get
3×1-y = 3
y = 0
So, the coordinate ofpoint B is (1, 0)
On solving lines (ii)and (iii),
We get,
[First,multiply Eq. (iii) by 2 and then subtract]
(2x + 4y)-(2x-3y) =16-2
7y = 14
y = 2
Substituting y=2 inEq. (iii), we get
x + 2 × 2 = 8
x + 4 = 8
x = 4
Hence, the coordinateof point C is (4, 2).
On solving lines (iii)and (i),
We get,
[First,multiply in Eq. (i) by 2 and then add]
(6x-2y) + (x + 2y) = 6+ 8
7x = 14
x = 2
Substituting x=2 inEq. (i), we get
3 × 2 – y = 3
y = 3
So, the coordinate ofpoint A is (2, 3).
Hence, the vertices ofthe ∆ABC formed by the given lines are as follows,
A (2, 3), B (1, 0) andC (4, 2).
Question - 6 : - Ankita travels 14 kmto her home partly by rickshaw and partly by bus. She takes half an hour if shetravels 2 km by rickshaw, and the remaining distance by bus.
Answer - 6 : -
Solution:
Let the speed of therickshaw and the bus are x and y km/h, respectively.
Now, she has takentime to travel 2 km by rickshaw, t1 = (2/x) hr
Speed = distance/ time
she has taken time totravel remaining distance i.e., (14 – 2) = 12km
By bus t2 =(12/y) hr
By first condition,
t1 + t2 =½ = (2/x) + (12/y) … (i)
Now, she has takentime to travel 4 km by rickshaw, t3 = (4/x) hr
and she has taken timeto travel remaining distance i.e., (14 – 4) = 10km, by bus = t4 =(10/y) hr
By second condition,
t3 + t4 =½ + 9/60 = ½ + 3/20
(4/x) + (10/y) =(13/20) …(ii)
Let (1/x) = u and(1/y) = v
Then Equations. (i)and (ii) becomes
2u + 12v = ½ …(iii)
4u + 10v = 13/20…(iv)
[First,multiply Eq. (iii) by 2 and then subtract]
(4u + 24v) – (4u +10v) = 1–13/20
14v = 7/20
v = 1/40
Substituting the valueof v in Eq. (iii),
2u + 12(1/40) = ½
2u = 2/10
u = 1/10
x = 1/u = 10km/hr
y = 1/v = 40km/hr
Hence, the speed ofrickshaw = 10 km/h
And the speed of bus =40 km/h.
Question - 7 : - Solve the following pair of linear equations by the elimination method and the substitution method:
Answer - 7 : -
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2+ 2y/3 = -1 and x-y/3 = 3
Answer
(i) x + y = 5 and 2x – 3y = 4
By the method of elimination.
x + y = 5 ……………………………….. (i)
2x – 3y = 4 ……………………………..(ii)
When the equation (i) is multiplied by 2, we get
2x + 2y = 10 ……………………………(iii)
When the equation (ii) is subtracted from (iii) we get,
5y = 6
y = 6/5 ………………………………………(iv)
Substituting the value of y in eq. (i) we get,
x=5−6/5 = 19/5
∴x = 19/5 , y = 6/5
By the method of substitution.
From the equation (i), we get:
x = 5 – y………………………………….. (v)
When the value is put in equation (ii) we get,
2(5 – y) – 3y = 4
-5y = -6
y = 6/5
When the values are substituted in equation (v), we get:
x =5− 6/5 = 19/5
∴x = 19/5 ,y = 6/5
(ii) 3x + 4y = 10 and 2x – 2y = 2
By the method of elimination.
3x + 4y = 10……………………….(i)
2x – 2y = 2 ………………………. (ii)
When the equation (i) and (ii) is multiplied by 2, we get:
4x – 4y = 4 ………………………..(iii)
When the Equation (i) and (iii) are added, we get:
7x = 14
x = 2 ……………………………….(iv)
Substituting equation (iv) in (i) we get,
6 + 4y = 10
4y = 4
y = 1
Hence, x = 2 and y = 1
By the method of Substitution
From equation (ii) we get,
x = 1 + y……………………………… (v)
Substituting equation (v) in equation (i) we get,
3(1 + y) + 4y = 10
7y = 7
y = 1
When y = 1 is substituted in equation (v) we get,
A = 1 + 1 = 2
Therefore, A = 2 and B = 1
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By the method of elimination:
3x – 5y – 4 = 0 ………………………………… (i)
9x = 2y + 7
9x – 2y – 7 = 0 …………………………………(ii)
When the equation (i) and (iii) is multiplied we get,
9x – 15y – 12 = 0 ………………………………(iii)
When the equation (iii) is subtracted from equation (ii) we get,
13y = -5
y = -5/13 ………………………………………….(iv)
When equation (iv) is substituted in equation (i) we get,
3x +25/13 −4=0
3x = 27/13
x =9/13
∴x = 9/13 and y = -5/13
By the method of Substitution:
From the equation (i) we get,
x = (5y+4)/3 …………………………………………… (v)
Putting the value (v) in equation (ii) we get,
9(5y+4)/3 −2y −7=0
13y = -5
y = -5/13
Substituting this value in equation (v) we get,
x = (5(-5/13)+4)/3
x = 9/13
∴x = 9/13, y = -5/13
(iv) x/2 + 2y/3 = -1 and x-y/3 = 3
By the method of Elimination.
3x + 4y = -6 …………………………. (i)
x-y/3 = 3
3x – y = 9 ……………………………. (ii)
When the equation (ii) is subtracted from equation (i) we get,
-5y = -15
y = 3 ………………………………….(iii)
When the equation (iii) is substituted in (i) we get,
3x – 12 = -6
3x = 6
x = 2
Hence, x = 2 , y = -3
By the method of Substitution:
From the equation (ii) we get,
x = (y+9)/3…………………………………(v)
Putting the value obtained from equation (v) in equation (i) we get,
3(y+9)/3 +4y =−6
5y = -15
y = -3
When y = -3 is substituted in equation (v) we get,
x = (-3+9)/3 = 2
Therefore, x = 2 and y = -3
Question - 8 : - Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
Answer - 8 : -
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?
Solution:
Let the fraction be a/b
According to the given information,
(a+1)/(b-1) = 1
=> a – b = -2 ………………………………..(i)
a/(b+1) = 1/2
=> 2a-b = 1…………………………………(ii)
When equation (i) is subtracted from equation (ii) we get,
a = 3 …………………………………………………..(iii)
When a = 3 is substituted in equation (i) we get,
3 – b = -2
-b = -5
b = 5
Hence, the fraction is 3/5.
(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?
Solution:
Let us assume, present age of Nuri is x
And present age of Sonu is y.
According to the given condition, we can write as;
x – 5 = 3(y – 5)
x – 3y = -10…………………………………..(1)
Now,
x + 10 = 2(y +10)
x – 2y = 10…………………………………….(2)
Subtract eq. 1 from 2, to get,
y = 20 ………………………………………….(3)
Substituting the value of y in eq.1, we get,
x – 3.20 = -10
x – 60 = -10
x = 50
Therefore,
Age of Nuri is 50 years
Age of Sonu is 20 years.
(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.
Solution:
Let the unit digit and tens digit of a number be x and y respectively.
Then, Number (n) = 10B + A
N after reversing order of the digits = 10A + B
According to the given information, A + B = 9…………………….(i)
9(10B + A) = 2(10A + B)
88 B – 11 A = 0
-A + 8B = 0 ………………………………………………………….. (ii)
Adding the equations (i) and (ii) we get,
9B = 9
B = 1……………………………………………………………………….(3)
Substituting this value of B, in the equation (i) we get A= 8
Hence the number (N) is 10B + A = 10 x 1 +8 = 18
(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 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.
Solution:
Let the number of Rs.50 notes be A and the number of Rs.100 notes be B
According to the given information,
A + B = 25 ……………………………………………………………………….. (i)
50A + 100B = 2000 ………………………………………………………………(ii)
When equation (i) is multiplied with (ii) we get,
50A + 50B = 1250 …………………………………………………………………..(iii)
Subtracting the equation (iii) from the equation (ii) we get,
50B = 750
B = 15
Substituting in the equation (i) we get,
A = 10
Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. 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.
Solution:
Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.
According to the information given,
A + 4B = 27 …………………………………….…………………………. (i)
A + 2B = 21 ……………………………………………………………….. (ii)
When equation (ii) is subtracted from equation (i) we get,
2B = 6
B = 3 …………………………………………………………………………(iii)
Substituting B = 3 in equation (i) we get,
A + 12 = 27
A = 15
Hence, the fixed charge is Rs.15
And the Charge per day is Rs.3
Question - 9 : - Solve the following pair of linear equations by the elimination method and the substitution method:
Answer - 9 : -
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2+ 2y/3 = -1 and x-y/3 = 3
Solutions:
(i) x + y = 5 and 2x – 3y = 4
By the method of elimination.
x + y = 5 ……………………………….. (i)
2x – 3y = 4 ……………………………..(ii)
When the equation (i) is multiplied by 2, we get
2x + 2y = 10 ……………………………(iii)
When the equation (ii) is subtracted from (iii) we get,
5y = 6
y = 6/5 ………………………………………(iv)
Substituting the value of y in eq. (i) we get,
x=5−6/5 = 19/5
∴x = 19/5 , y = 6/5
By the method of substitution.
From the equation (i), we get:
x = 5 – y………………………………….. (v)
When the value is put in equation (ii) we get,
2(5 – y) – 3y = 4
-5y = -6
y = 6/5
When the values are substituted in equation (v), we get:
x =5− 6/5 = 19/5
∴x = 19/5 ,y = 6/5
(ii) 3x + 4y = 10 and 2x – 2y = 2
By the method of elimination.
3x + 4y = 10……………………….(i)
2x – 2y = 2 ………………………. (ii)
When the equation (i) and (ii) is multiplied by 2, we get:
4x – 4y = 4 ………………………..(iii)
When the Equation (i) and (iii) are added, we get:
7x = 14
x = 2 ……………………………….(iv)
Substituting equation (iv) in (i) we get,
6 + 4y = 10
4y = 4
y = 1
Hence, x = 2 and y = 1
By the method of Substitution
From equation (ii) we get,
x = 1 + y……………………………… (v)
Substituting equation (v) in equation (i) we get,
3(1 + y) + 4y = 10
7y = 7
y = 1
When y = 1 is substituted in equation (v) we get,
A = 1 + 1 = 2
Therefore, A = 2 and B = 1
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By the method of elimination:
3x – 5y – 4 = 0 ………………………………… (i)
9x = 2y + 7
9x – 2y – 7 = 0 …………………………………(ii)
When the equation (i) and (iii) is multiplied we get,
9x – 15y – 12 = 0 ………………………………(iii)
When the equation (iii) is subtracted from equation (ii) we get,
13y = -5
y = -5/13 ………………………………………….(iv)
When equation (iv) is substituted in equation (i) we get,
3x +25/13 −4=0
3x = 27/13
x =9/13
∴x = 9/13 and y = -5/13
By the method of Substitution:
From the equation (i) we get,
x = (5y+4)/3 …………………………………………… (v)
Putting the value (v) in equation (ii) we get,
9(5y+4)/3 −2y −7=0
13y = -5
y = -5/13
Substituting this value in equation (v) we get,
x = (5(-5/13)+4)/3
x = 9/13
∴x = 9/13, y = -5/13
(iv) x/2 + 2y/3 = -1 and x-y/3 = 3
By the method of Elimination.
3x + 4y = -6 …………………………. (i)
x-y/3 = 3
3x – y = 9 ……………………………. (ii)
When the equation (ii) is subtracted from equation (i) we get,
-5y = -15
y = 3 ………………………………….(iii)
When the equation (iii) is substituted in (i) we get,
3x – 12 = -6
3x = 6
x = 2
Hence, x = 2 , y = -3
By the method of Substitution:
From the equation (ii) we get,
x = (y+9)/3…………………………………(v)
Putting the value obtained from equation (v) in equation (i) we get,
3(y+9)/3 +4y =−6
5y = -15
y = -3
When y = -3 is substituted in equation (v) we get,
x = (-3+9)/3 = 2
Therefore, x = 2 and y = -3
Question - 10 : - Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
Answer - 10 : -
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?
Solution:
Let the fraction be a/b
According to the given information,
(a+1)/(b-1) = 1
=> a – b = -2 ………………………………..(i)
a/(b+1) = 1/2
=> 2a-b = 1…………………………………(ii)
When equation (i) is subtracted from equation (ii) we get,
a = 3 …………………………………………………..(iii)
When a = 3 is substituted in equation (i) we get,
3 – b = -2
-b = -5
b = 5
Hence, the fraction is 3/5.
(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?
Solution:
Let us assume, present age of Nuri is x
And present age of Sonu is y.
According to the given condition, we can write as;
x – 5 = 3(y – 5)
x – 3y = -10…………………………………..(1)
Now,
x + 10 = 2(y +10)
x – 2y = 10…………………………………….(2)
Subtract eq. 1 from 2, to get,
y = 20 ………………………………………….(3)
Substituting the value of y in eq.1, we get,
x – 3.20 = -10
x – 60 = -10
x = 50
Therefore,
Age of Nuri is 50 years
Age of Sonu is 20 years.
(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.
Solution:
Let the unit digit and tens digit of a number be x and y respectively.
Then, Number (n) = 10B + A
N after reversing order of the digits = 10A + B
According to the given information, A + B = 9…………………….(i)
9(10B + A) = 2(10A + B)
88 B – 11 A = 0
-A + 8B = 0 ………………………………………………………….. (ii)
Adding the equations (i) and (ii) we get,
9B = 9
B = 1……………………………………………………………………….(3)
Substituting this value of B, in the equation (i) we get A= 8
Hence the number (N) is 10B + A = 10 x 1 +8 = 18
(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 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.
Solution:
Let the number of Rs.50 notes be A and the number of Rs.100 notes be B
According to the given information,
A + B = 25 ……………………………………………………………………….. (i)
50A + 100B = 2000 ………………………………………………………………(ii)
When equation (i) is multiplied with (ii) we get,
50A + 50B = 1250 …………………………………………………………………..(iii)
Subtracting the equation (iii) from the equation (ii) we get,
50B = 750
B = 15
Substituting in the equation (i) we get,
A = 10
Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. 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.
Solution:
Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.
According to the information given,
A + 4B = 27 …………………………………….…………………………. (i)
A + 2B = 21 ……………………………………………………………….. (ii)
When equation (ii) is subtracted from equation (i) we get,
2B = 6
B = 3 …………………………………………………………………………(iii)
Substituting B = 3 in equation (i) we get,
A + 12 = 27
A = 15
Hence, the fixed charge is Rs.15
And the Charge per day is Rs.3