Important Questions for CBSE Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables 4 Mark Question

1. 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, 4x − 2y = 4
(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
Ans. (i) x + y = 5, 2x + 2y = 10
For equation x + y – 5 = 0, we have following points which lie on the line

x	0	5
y	5	0

For equation 2x + 2y – 10 = 0, we have following points which lie on the line

x	1	2
y	4	3

We can see that both of the lines coincide. Hence, there are infinite many solutions. Any point which lies on one line also lies on the other. Hence, by using equation (x+y– 5=0), we can say thatx=5−y
We can assume any random values for y and can find the corresponding value of x using the above equation. All such points will lie on both lines and there will be infinite number of such points.
(ii) x– y = 8, 3x − 3y = 16
For x – y = 8, the coordinates are:

x	0	8
y	-8	0

And for 3x – 3y = 16, the coordinates

x	0
y		0

Plotting these points on the graph,it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.
(iii) 2x + y = 6, 4x − 2y = 4
For equation 2x + y – 6 = 0, we have following points which lie on the line

x	0	3
y	6	0

For equation 4x – 2y – 4 = 0, we have following points which lie on the line

x	0	1
y	-2	0

We can clearly see that lines are intersecting at (2, 2) which is the solution.
Hence x = 2 and y = 2 and lines are consistent.
(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
For 2x – 2y – 2 = 0, the coordinates are:

x	2	0
y	0	-2

And for 4x – 4y – 5 = 0, the coordinates

x	0
y		0

Plotting these points on the graph, it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.

---

2. Solve the following pair of linear equations by the substitution method.
(i) x+ y = 14
x – y= 4
(ii) s – t = 3

(iii) 3x – y = 3
9x − 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) 

(vi) 

Ans. (i) x+y=14 … (1)
x– y=4 … (2)
x=4+y from equation (2)
Putting this in equation (1), we get
4+y+y=14
⇒ 2y=10
⇒ y=5
Putting value of y in equation (1), we get
x+5=14
⇒ x=14 – 5=9
Therefore, x=9 and y=5
(ii) s– t=3 … (1)
 … (2)
Using equation (1), we can say that s=3+t
Putting this in equation (2), we get

⇒ 
⇒ 5t+6=36
⇒ 5t=30
⇒ t=6
Putting value of t in equation (1), we get
s– 6=3
⇒ s=3+6=9
Therefore, t=6 and s=9
(iii) 3x– y=3 … (1)
9x−3y=9 … (2)
Comparing equation 3x– y=3 with a₁x+b₁y+c₁=0 and equation 9x−3y=9 with a₂x+b₂y+c₂=0,
We get a₁=3,b₁=−1, c₁=−3, a₂=9,b₂=−3 and c₂=−9
Here 
Therefore, we have infinite many solutions for x and y
(iv) 0.2x+0.3y=1.3 … (1)
0.4x+0.5y=2.3 … (2)
Using equation (1), we can say that
0.2x=1.3−0.3y
⇒ x=
Putting this in equation (2), we get
0.4+0.5y=2.3
⇒ 2.6−0.6y+0.5y=2.3
⇒ −0.1y=−0.3
⇒ y=3
Putting value of y in (1), we get
0.2x+0.3(3)=1.3
⇒ 0.2x+0.9=1.3
⇒ 0.2x=0.4
⇒ x=2
Therefore, x=2 and y=3
(v)  ……….(1)
 ……….(2)
Using equation (1), we can say that
x=
Putting this in equation (2), we get

⇒ 
⇒ 
⇒ y=0
Putting value of y in (1), we get x=0
Therefore, x=0 and y=0
(vi)  … (1)
 … (2)
Using equation (2), we can say that
x=
⇒ x=
Putting this in equation (1), we get

⇒ 
⇒ 
⇒ 
⇒ 
⇒ y=3
Putting value of y in equation (2), we get

⇒ 
⇒ 
⇒ x = 2
Therefore, x=2 and y=3

---

3. Form a 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 Rs 3800. Later, she buys 3 bats and 5 balls for Rs 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 Rs 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 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes . 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?
Ans. (i) Let first number be x and second number be y.
According to given conditions, we have
x– y=26 (assuming x>y) … (1)
x=3y (x>y) … (2)
Putting equation (2) in (1), we get
3y– y=26 ⇒ 2y=26 ⇒ y=13
Putting value of y in equation (2), we get
x=3y=3×13=39
Therefore, two numbers are 13 and 39.
(ii) Let smaller angle =xand let larger angle =y
According to given conditions, we have
y=x+18 … (1)
Also, x+y=180⁰ (Sum of supplementary angles) … (2)
Putting (1) in equation (2), we get
x+x+18=180
⇒ 2x=180 – 18=162
⇒ x=81⁰
Putting value of x in equation (1), we get
y=x+18=81+18=99⁰
Therefore, two angles are 81⁰ and 99⁰.
(iii) Let cost of each bat = Rs x and let cost of each ball = Rs y
According to given conditions, we have
7x+6y=3800 … (1)
And, 3x+5y=1750 … (2)
Using equation (1), we can say that
7x=3800−6
⇒ x=
Putting this in equation (2), we get
3+5y=1750
⇒ +5y=1750
⇒ 
⇒ 
⇒ 17y=850
⇒ y=50
Putting value of y in (2), we get
3x+250=1750
⇒ 3x=1500
⇒ x=500
Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50
(iv) Let fixed charge = Rs x and let charge for every km = Rs y
According to given conditions, we have
x+10y=105 … (1)
x+15y=155 … (2)
Using equation (1), we can say that
x=105−10y
Putting this in equation (2), we get
105−10y+15y=155
⇒ 5y=50
⇒ y=10
Putting value of y in equation (1), we get
x+10(10)=105
⇒ x=105 – 100=5
Therefore, fixed charge = Rs 5 and charge per km = Rs 10
To travel distance of 25 Km, person will have to pay = Rs (x+25y) = Rs (5+25 × 10) = Rs(5+250) = Rs 255
(v) Let numerator = x and let denominator = y
According to given conditions, we have
 … (1)
 … (2)
Using equation (1), we can say that
11(x+2)=9y+18
⇒ 11x+22=9y+18
⇒ 11x=9y– 4
⇒ x=
Putting value of x in equation (2), we get
6 =5(y+3)
⇒ 
⇒ 
⇒ 
⇒ y=9
Putting value of y in (1), we get
 ⇒ x+2=9 ⇒ x=7
Therefore, fraction =
(vi) Let present age of Jacob = x years
Let present age of Jacob’s son = y years
According to given conditions, we have
(x+5)=3(y+5) … (1)
And, (x−5)=7(y−5) … (2)
From equation (1), we can say that
x+5=3y+15
⇒ x=10+3y
Putting value of x in equation (2) we get
10+3y– 5=7y−35
⇒ −4y=−40
⇒ y=10 years
Putting value of y in equation (1), we get
x+5=3(10+5)=3×15=45
⇒ x=45 – 5=40 years
Therefore, present age of Jacob = 40 years and, present age of Jacob’s son = 10 years

---

4. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5, 2x – 3y = 4
(ii) 3x + 4y = 10, 2x – 2y = 2
(iii) 3x − 5y – 4 = 0, 9x = 2y + 7
(iv) 
Ans. (i) x+y=5 … (1)
2x – 3y=4 … (2)
Elimination method:
Multiplying equation (1) by 2, we get equation (3)
2x+2y=10 … (3)
2x−3y=4 … (2)
Subtracting equation (2) from (3), we get
5y=6
⇒ y=
Putting value of y in (1), we get
x+=5 ⇒ x=5−=
Therefore, x= and y=
Substitution method:
x+y=5 … (1)
2x−3y=4 … (2)
From equation (1), we get,
x=5−y
Putting this in equation (2), we get
2(5−y)−3y=4 ⇒ 10−2y−3y=4
⇒ 5y=6
⇒ y=
Putting value of y in (1), we get
x=5−=
Therefore, x= and y=
(ii) 3x+4y=10 … (1)
2x – 2y=2 … (2)
Elimination method:
Multiplying equation (2) by 2, we get (3)
4x−4y=4 … (3)
3x+4y=10 … (1)
Adding (3) and (1), we get
7x=14
⇒ x=2
Putting value of x in (1), we get
3(2)+4y=10
⇒ 4y=10 – 6=4
⇒ y=1
Therefore, x=2 and y=1
Substitution method:
3x+4y=10 … (1)
2x−2y=2 … (2)
From equation (2), we get
2x=2+2y
⇒ x=1+y … (3)
Putting this in equation (1), we get
3(1+y)+4y=10
⇒ 3+3y+4y=10
⇒ 7y=7
⇒ y=1
Putting value of y in (3), we getx=1+1=2
Therefore, x=2 and y=1
(iii) 3x−5y– 4=0 … (1)
9x=2y+7 … (2)
Elimination method:
Multiplying (1) by 3, we get (3)
9x−15y– 12=0 … (3)
9x−2y– 7=0 … (2)
Subtracting (2) from (3), we get
−13y– 5=0
⇒ −13y=5
⇒ y=
Putting value of y in (1), we get
3x– 5−4=0
⇒ 3x=4−=
⇒ x=
Therefore, x= and y=
Substitution Method:
3x−5y– 4=0 … (1)
9x=2y+7 … (2)
From equation (1), we can say that
3x=4+5y
⇒ x=
Putting this in equation (2), we get
9−2y=7
⇒ 12+15y−2y=7
⇒ 13y=−5
⇒ y=
Putting value of y in (1), we get
3x– 5=4
⇒ 3x=4−= 
⇒ x=
Therefore, x= and y=
(iv) …(1)
 … (2)
Elimination method:
Multiplying equation (2) by 2, we get (3)
 … (3)
 … (1)
Adding (3) and (1), we get

⇒ x=2
Putting value of x in (2), we get
2−= 3
⇒ y=−3
Therefore, x=2 and y=−3
Substitution method:
 … (1)
 … (2)
From equation (2), we can say that
Putting this in equation (1), we get

⇒ 
⇒ 5y+9=−6
⇒ 5y=−15
⇒ y=−3
Putting value of y in (1), we get

⇒ x= 2
Therefore, x=2 and y=−3

---

5. 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 ½ 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 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 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.
Ans. (i) Let numerator =xand let denominator =y
According to given condition, we have

⇒ x+1=y– 1 and 2x=y+1
⇒ x– y=−2 … (1) and 2x– y=1 … (2)
So, we have equations (1) and (2), multiplying equation (1) by 2 we get (3)
2x−2y=−4 … (3)
2x– y=1 … (2)
Subtracting equation (2) from (3), we get
−y=−5
⇒ y=5
Putting value of y in (1), we get
x– 5=−2
⇒ x=−2+5=3
Therefore, fraction =
(ii) Let present age of Nuri =x years and let present age of Sonu =y years
5 years ago, age of Nuri = (x – 5) years
5 years ago, age of Sonu = (y – 5) years
According to given condition, we have
(x−5)=3(y−5)
⇒ x– 5=3y– 15
⇒ x−3y=−10 … (1)
10 years later from present, age of Nuri =(x+10) years
10 years later from present, age of Sonu =(y+10) years
According to given condition, we have
(x+10)=2(y+10)
⇒ x+10=2y+20
⇒ x−2y=10 … (2)
Subtracting equation (1) from (2), we get
y=10−(−10)=20 years
Putting value of y in (1), we get
x– 3(20)=−10
⇒ x– 60=−10
⇒ x=50 years
Therefore, present age of Nuri =50 years and present age of Sonu =20 years
(iii) Let digit at ten’s place =xand Let digit at one’s place =y
According to given condition, we have
x+y=9 … (1)
And 9(10x+y)=2(10y+x)
⇒ 90x+9y=20y+2x
⇒ 88x=11y
⇒ 8x=y
⇒ 8x– y=0 … (2)
Adding (1) and (2), we get
9x=9
⇒ x=1
Putting value of x in (1), we get
1+y=9
⇒ y=9 – 1=8
Therefore, number =10x+y=10(1)+8=10+8=18
(iv) Let number of Rs 100 notes =xand let number of Rs 50 notes =y
According to given conditions, we have
x+y=25 … (1)
and 100x+50y=2000
⇒ 2x+y=40 … (2)
Subtracting (2) from (1), we get
−x=−15
⇒ x=15
Putting value of x in (1), we get
15+y=25
⇒ y=25 – 15=10
Therefore, number of Rs 100 notes =15 and number of Rs 50 notes =10
(v) Let fixed charge for 3 days = Rs x
Let additional charge for each day thereafter = Rs y
According to given condition, we have
x+4y=27 … (1)
x+2y=21 … (2)
Subtracting (2) from (1), we get
2y=6
⇒ y= 3
Putting value of y in (1), we get
x+4(3)=27
⇒ x=27 – 12=15
Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3

---

6. C 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
Ans. (i) x−3y– 3=0
3x−9y– 2=0
Comparing equation x−3y– 3=0 with a₁x+b₁y+c₁=0 and 3x−9y– 2=0 with a₂x+b₂y+c₂=0,
We get a₁=1,b₁= −3, c₁=−3, a₂=3,b₂=−9, c₂=−2
Here this means that the two lines are parallel.
Therefore, there is no solution for the given equations i.e. it is inconsistent.
(ii) 2x+y=5
3x+2y=8
Comparing equation 2x+y=5 with a₁x+b₁y+c₁=0 and 3x+2y=8 with a₂x+b₂y+c₂=0,
We get a₁= 2,b₁= 1, c₁=−5, a₂=3,b₂= 2, c₂=−8
Here this means that there is unique solution for the given equations.


⇒ 
 
⇒ x=2 and y=1
(iii) 3x−5y=20
6x−10y=40
Comparing equation 3x−5y=20 with a₁x+b₁y+c₁=0 and 6x−10y=40 with a₂x+b₂y+c₂=0,
We get a₁= 3,b₁= –5, c₁=−20, a₂= 6,b₂= –10, c₂=−40
Here 
It means lines coincide with each other.
Hence, there are infinite many solutions.
(iv) x−3y– 7=0
3x−3y– 15=0
Comparing equation x−3y– 7=0 with a₁x+b₁y+c₁=0 and 3x−3y– 15=0 with a₂x+b₂y+c₂=0,
We get a₁= 1,b₁= –3, c₁=−7, a₂= 3,b₂= –3, c₂=−15
Here this means that we have unique solution for these equations.


⇒ 
⇒ 
⇒ x= 4 and y= –1

---

7. 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 when 1 is subtracted from the numerator and it becomes  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.
Ans. (i) Let fixed monthly charge = Rs x and let charge of food for one day = Rs y
According to given conditions,
x + 20y = 1000 … (1),
and x + 26y = 1180 … (2)
Subtracting equation (1) from equation (2), we get
6y = 180
⇒ y = 30
Putting value of y in (1), we get
x + 20 (30) = 1000
⇒ x = 1000 – 600 = 400
Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30
(ii) Let numerator = xand let denominator = y
According to given conditions,

⇒ 3x – 3 = y … (1) 4x = y + 8 … (1)
⇒ 3x – y = 3 … (1) 4x – y = 8 … (2)
Subtracting equation (1) from (2), we get
4x – y − (3x − y) = 8 – 3
⇒ x = 5
Putting value of x in (1), we get
3 (5) – y = 3
⇒ 15 – y = 3
⇒ y = 12
Therefore, numerator = 5 and, denominator = 12
It means fraction = 
(iii) Let number of correct answers = x and let number of wrong answers = y
According to given conditions,
3x – y = 40 … (1)
And, 4x − 2y = 50 … (2)
From equation (1), y = 3x – 40
Putting this in (2), we get
4x – 2 (3x − 40) = 50
⇒ 4x − 6x + 80 = 50
⇒ −2x = −30
⇒ x = 15
Putting value of x in (1), we get
3 (15) – y = 40
⇒ 45 – y = 40
⇒ y = 45 – 40 = 5
Therefore, number of correct answers = x = 15and number of wrong answers = y = 5
Total questions = x + y = 15 + 5 = 20
(iv) Let speed of car which starts from part A = x km/hr
Let speed of car which starts from part B = y km/hr
According to given conditions,
(Assuming x > y)
⇒ 5x − 5y = 100
⇒ x – y = 20 … (1)
And, 
⇒ x + y = 100 … (2)
Adding (1) and (2), we get
2x = 120
⇒ x = 60 km/hr
Putting value of x in (1), we get
60 – y= 20
⇒ y = 60 – 20 = 40 km/hr
Therefore, speed of car starting from point A = 60 km/hr
And, Speed of car starting from point B = 40 km/hr
(v) Let length of rectangle = x units and Let breadth of rectangle = y units
Area =xy square units. According to given conditions,
xy– 9 = (x − 5) (y + 3)
⇒ xy– 9 = xy + 3x − 5y – 15
⇒ 3x − 5y = 6 … (1)
And, xy + 67 = (x + 3) (y + 2)
⇒ xy+ 67 = xy + 2x + 3y + 6
⇒ 2x + 3y = 61 … (2)
From equation (1),
3x = 6 + 5y
⇒ x = 
Putting this in (2), we get
2 + 3y = 61
⇒ 12 + 10y + 9y = 183
⇒ 19y = 171
⇒ y = 9 units
Putting value of y in (2), we get
2x + 3 (9) = 61
⇒ 2x = 61 – 27 = 34
⇒ x = 17 units
Therefore, length = 17 units and, breadth = 9 units

---

8. Solve the following pairs of equations by reducing them to a pair of linear equations:
 
(i) 

(ii) 

(iii) +3y=14
−4y=23
(iv) 

(v) 7x−2y=5xy
8x+7y=15xy
(vi) 6x+3y−6xy=0
2x+4y−5xy=0
(vii) 

(viii) 

Ans. (i)  … (1)
 … (2)
Let =p and =q
Putting this in equation (1) and (2), we get

⇒ 3p+2q=12 and 6(2p+3q)=13(6)
⇒ 3p+2q=12 and 2p + 3q =13
⇒ 3p+2q– 12=0 … (3) and 2p+3q– 13=0 … (4)


⇒ 
⇒ 
⇒ 
⇒ p=2 and q=3
But =p and =q
Putting value of p and q in this we get
x= and y=
(ii)  … (1)
 … (2)
Let =p and =q
Putting this in (1) and (2), we get
2p+3q=2 … (3)
4p−9q=−1 … (4)
Multiplying (3) by 2 and subtracting it from (4), we get
4p−9q+1 – 2(2p+3q−2)=0
⇒ 4p−9q+1−4p−6q+4=0
⇒ −15q+5=0
⇒ q=
Putting value of q in (3), we get
2p+1=2
⇒ 2p=1 ⇒ p= ½
Putting values of p and q in (=p and =q), we get
 and 
⇒ 
⇒ x=4 and y=9
(iii) +3y=14 … (1)
−4y=23 … (2) and Let =p … (3)
Putting (3) in (1) and (2), we get
4p+3y=14 … (4)
3p−4y=23 … (5)
Multiplying (4) by 3 and (5) by 4, we get
3(4p+3y– 14= 0) and, 4(3p−4y– 23= 0)
⇒ 12p+9y– 42=0 … (6) 12p−16y– 92=0 … (7)
Subtracting (7) from (6), we get
9y−(−16y) – 42−(−92)=0
⇒ 25y+50=0
⇒ y=50 – 25=−2
Putting value of y in (4), we get
4p+3(−2)=14
⇒ 4p– 6=14
⇒ 4p=20
⇒ p=5
Putting value of p in (3), we get
=5
⇒ x=
Therefore, x= and y=−2
(iv)  … (1)
 … (2)
Let 
Putting this in (1) and (2), we get
5p+q=2
⇒ 5p+q– 2=0 … (3)
And, 6p−3q=1
⇒ 6p−3q– 1=0 … (4)
Multiplying (3) by 3 and adding it to (4), we get
3(5p+q−2)+6p−3q– 1=0
⇒ 15p+3q– 6+6p−3q– 1=0
⇒ 21p– 7=0
⇒ p=
Putting this in (3), we get
5()+q– 2=0
⇒ 5+ 3q= 6
⇒ 3q = 6 – 5 = 1
⇒ q=
Putting values of p and q in (), we get

⇒ 3=x−1 and 3=y– 2
⇒ x=4 and y=5
(v) 7x−2y=5xy … (1)
8x+7y=15xy … (2)
Dividing both the equations by xy, we get


Let =p and =q
Putting these in (3) and (4), we get
7q−2p=5 … (5)
8q+7p=15 … (6)
From equation (5),
2p=7q– 5
⇒ p=
Putting value of p in (6), we get
8q+7()=15
⇒ 16q+49q– 35= 30
⇒ 65q= 30 +35=65
⇒ q=1
Putting value of q in (5), we get
7(1)−2p=5
⇒ 2p=2 ⇒ p=1
Putting value of p and q in (=p and =q), we getx=1 and y=1
(vi) 6x+3y−6xy=0 … (1)
2x+4y−5xy=0 … (2)
Dividing both the equations by xy, we get


Let =p and =q
Putting these in (3) and (4), we get
6q+3p– 6=0 … (5)
2q+4p– 5=0 … (6)
From (5),
3p=6−6q
⇒ p=2−2q
Putting this in (6), we get
2q+4(2−2q) – 5=0
⇒ 2q+8−8q– 5=0
⇒ −6q=−3
⇒ q= ½
Putting value of q in (p=2 – 2q), we get
p=2 – 2 (½)=2 – 1=1
Putting values of p and q in (=p and =q), we getx=1 and y=2
(vii)  … (1)
 … (2)
Let 
Putting this in (1) and (2), we get
10p+2q=4 … (3)
15p−5q=−2 … (4)
From equation (3),
2q=4−10p
⇒ q=2−5p … (5)
Putting this in (4), we get
15p– 5(2−5p)=−2
⇒ 15p– 10+25p=−2
⇒ 40p=8
⇒ p=
Putting value of p in (5), we get
q=2 – 5()=2 – 1=1
Putting values of p and q in (),we get

⇒ x+y=5 … (6) and x– y=1 … (7)
Adding (6) and (7), we get
2x=6
⇒ x=3
Putting x=3 in (7), we get
3 – y=1
⇒ y=3 – 1=2
Therefore, x=3 and y=2
(viii)  … (1)
 … (2)
Let 
Putting this in (1) and (2), we get
p+q= and 
⇒ 4p+4q=3 … (3) and 4p−4q=−1 … (4)
Adding (3) and (4), we get
8p=2
⇒ p= ¼
Putting value of p in (3), we get
4 (¼)+4q=3
⇒ 1+4q=3
⇒ 4q=3 – 1=2
⇒ q= ½
Putting value of p and q in , we get

⇒ 3x+y=4 … (5) and 3x– y=2 … (6)
Adding (5) and (6), we get
6x=6
⇒ x=1
Putting x=1 in (5) , we get
3(1)+y=4
⇒ y=4 – 3=1
Therefore, x=1 and y=1

---

9. Formulate the following problems as a part 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.
Ans. (i) Let speed of rowing in still water =x km/h
Let speed of current =y km/h
So, speed of rowing downstream =(x+y) km/h
And, speed of rowing upstream =(x−y) km/h
According to given conditions,

⇒ 2x+2y=20 and 2x−2y=4
⇒ x+y=10 … (1) and x– y=2 … (2)
Adding (1) and (2), we get
2x=12
⇒ x=6
Putting x=6 in (1), we get
6+y=10
⇒ y=10 – 6=4
Therefore, speed of rowing in still water =6 km/h
Speed of current =4 km/h
(ii) Let time taken by 1 woman alone to finish the work =x days
Let time taken by 1 man alone to finish the work =y days
So, 1 woman’s 1 day work =()th part of the work
And, 1 man’s 1 day work =()th part of the work
So, 2 women’s 1 day work =()th part of the work
And, 5 men’s 1 day work =()th part of the work
Therefore, 2 women and 5 men’s 1 day work =(+)th part of the work… (1)
It is given that 2 women and 5 men complete work in = 4 days
It means that in 1 day, they will be completing th part of the work … (2)
Clearly, we can see that (1) = (2)
⇒  … (3)
Similarly,  … (4)
Let 
Putting this in (3) and (4), we get
2p+5q= and 3p+6q=
⇒ 8p+20q=1 … (5) and 9p+18q=1 … (6)
Multiplying (5) by 9 and (6) by 8, we get
72p+180q=9 … (7)
72p+144q=8 … (8)
Subtracting (8) from (7), we get
36q=1
⇒ q=
Putting this in (6), we get
9p+18()=1
⇒ 9p= ½
⇒ p =
Putting values of p and q in , we get x=18 and y=36
Therefore, 1 woman completes work in =18 days
And, 1 man completes work in =36 days
(iii) Let speed of train =x km/h and let speed of bus =y km/h
According to given conditions,

Let 
Putting this in the above equations, we get
60p+240q=4 … (1)
And 100p+200q= … (2)
Multiplying (1) by 5 and (2) by 3, we get
300p+1200q=20 … (3)
300p+600q= … (4)
Subtracting (4) from (3), we get
600q=20−=7.5
⇒ q=
Putting value of q in (2), we get
100p+200()=
⇒ 100p+ 2.5 =
⇒ 100p=– 2.5
⇒ p=
But 
Therefore, x=km/h and y=km/h
Therefore, speed of train =60 km/h
And, speed of bus =80 km/h

---

10. Solve the following pair of linear equations:
(i) 

(ii) 

(iii) 

(iv) 

(v) 

Ans. (i)  … (1)
 … (2)
Multiplying equation (1) by p and equation (2) by q, we obtain:
 … (3)
 … (4)
Adding equations (3) and (4), we obtain:

 
  = 1
Substituting the value of in equation (1), we obtain:

 
 
Hence the required solution is x = 1 and y = –1.
(ii)  … (1)
 … (2)
Multiplying equation (1) by a and equation (2) by b, we obtain:
 … (3)
 … (4)
Subtracting equation (4) from equation (3),

 
Substituting the value of x in equation (1), we obtain:

 
 
 
 
 
(iii)    ……..(1)
 ……..(2)
Multiplying equation (1) and (2) by b and a respectively, we obtain:
 ……..(3)
 ……..(4)
Adding equations (3) and (4), we obtain:

 
 
Substituting the value of in equation (1), we obtain:

 
 
(iv)  … (1)

  ……..(2)
Subtracting equation (2) from (1), we obtain:

 
 
 
Substituting the value of in equation (1), we obtain:

 
 
 
(v) 152x – 378y = –74 … (1)
–378x + 152y = –604 … (2)
Adding the equations (1) and (2), we obtain:
–226x – 226y = –678
 x + y = 3 ………(3)
Subtracting the equation (2) from equation (1), we obtain:
530x – 530y = 530
 x – y = 1 ……..(4)
Adding equations (3) and (4), we obtain:
2x = 4
 x = 2
Substituting the value of x in equation (3), we obtain:
y = 1

---

11. Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi he takes 6 minute longer. Find the speed of the train and that of the taxi
Ans. 
According to question,

---

12. If in a rectangle the length is increased and breadth is decreased by 2 units each, the area is reduced by 28 square units, and if the length is reduced by 1 unit and breadth is increased by 2 units, the area increased by 33 square units. Find the dimensions of the rectangle.
Ans. 

---

13. On comparing the ratios , 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) 
9x − 10y = 14
(iv) 5x − 3y = 11
−10x + 6y = −2
Ans. (i) 3x+2y=5, 2x−3y=7
Comparing equation 3x+2y=5 with a₁x+b₁y+c₁=0and 2x−3y=7 with a₂x+b₂y+c₂=0,
We get, a₁= 3,b₁= 2,c₁= 5, a₂= 2,b₂= –3,c₂= –7
and
Here which means equations have unique solution.
Hence they are consistent.
(ii) 2x−3y=8, 4x−6y=9
Comparing equation 2x−3y=8 with a₁x+b₁y+c₁=0and 4x−6y=9 with a₂x+b₂y+c₂=0,
We get, a₁= 2,b₁= –3,c₁= –8, a₂= 4,b₂= –6,c₂= –9
Here because 
Therefore, equations have no solution because they are parallel.
Hence, they are inconsistent.
(iii)  9x − 10y = 14
Comparing equation with a₁x+b₁y+c₁=0and 9x − 10y = 14 with a₂x+b₂y+c₂=0,
We get, a₁=,b₁=,c₁= –14, a₂= 9,b₂= –10,c₂= –14
and
Here 
Therefore, equations have unique solution.
Hence, they are consistent.
(iv) 5x−3y=11, −10x+6y=−22
Comparing equation 5x−3y=11 with a₁x+b₁y+c₁=0and −10x+6y=−22 with a₂x+b₂y+c₂=0,
We get, a₁= 5,b₁= –3,c₁= –11, a₂= –10,b₂= 6,c₂= 22
and
Here 
Therefore, the lines have infinite many solutions.
Hence, they are consistent.

---

14. 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
Ans. (i) x−3y– 3=0
3x−9y– 2=0
Comparing equation x−3y– 3=0 with a₁x+b₁y+c₁=0 and 3x−9y– 2=0 with a₂x+b₂y+c₂=0,
We get a₁=1,b₁= −3, c₁=−3, a₂=3,b₂=−9, c₂=−2
Here this means that the two lines are parallel.
Therefore, there is no solution for the given equations i.e. it is inconsistent.
(ii) 2x+y=5
3x+2y=8
Comparing equation 2x+y=5 with a₁x+b₁y+c₁=0 and 3x+2y=8 with a₂x+b₂y+c₂=0,
We get a₁= 2,b₁= 1, c₁=−5, a₂=3,b₂= 2, c₂=−8
Here this means that there is unique solution for the given equations.


⇒ 
 
⇒ x=2 and y=1
(iii) 3x−5y=20
6x−10y=40
Comparing equation 3x−5y=20 with a₁x+b₁y+c₁=0 and 6x−10y=40 with a₂x+b₂y+c₂=0,
We geta₁= 3,b₁= –5, c₁=−20, a₂= 6,b₂= –10, c₂=−40
Here 
It means lines coincide with each other.
Hence, there are infinite many solutions.
(iv) x−3y– 7=0
3x−3y– 15=0
Comparing equation x−3y– 7=0 with a₁x+b₁y+c₁=0 and 3x−3y– 15=0 with a₂x+b₂y+c₂=0,
We get a₁= 1,b₁= –3, c₁=−7, a₂= 3,b₂= –3, c₂=−15
Here this means that we have unique solution for these equations.


⇒ 
⇒ 
⇒ x= 4 and y= –1

---

15. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x+5y=9
3x+2y=4
Ans. Substitution Method
8x+5y=9 … (1)
3x+2y=4 … (2)
From equation (1),
5y=9−8x
⇒ y=
Putting this in equation (2), we get
3x+2=4
⇒ 3x+=4
⇒ 3x−
⇒ 15x−16x=20 – 18
⇒ x=−2
Putting value of x in (1), we get
8(−2)+5y=9
⇒ 5y=9+16=25
⇒ y=5
Therefore, x=−2 and y=5
Cross multiplication method
8x+5y=9 … (1)
3x+2y=4 … (2)


⇒ 
⇒ 
⇒ x=−2 and y=5

---

16. In aABC, C = 3B = 2(A + B). Find three angles
Ans. C = 3B = 2(A + B)
Taking 3B = 2(A + B)
 B = 2A
 2A – B = 0 …….(1)
We know that the sum of the measures of all angles of a triangle is 180°.
A + B + C = 180°
 A + B + 3B = 180°
 A + 4B = 180° …….(2)
Multiplying equation (1) by 4, we obtain:
8A – 4B = 0 …….(3)
Adding equations (2) and (3), we get
9A = 180°
 A = 20°
From eq. (2), we get,
20° + 4B = 180°
 B = 40°
And C = 3 x 40°= 120°
Hence the measures of A, B and C are 20°, 40° and 120° respectively.