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SEBA Class 10 Mathematics Chapter 3 Pair of Linear Equations in two Variables
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Pair of Linear Equations in two Variables
Chapter – 3
| Exercise 3.1 |
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.
Ans: Let the present age of Aftab’s daughter= x year and the present age of Aftab = y years (y > x) According to the given conditions Seven years ago,
From the graph, we find x = 12
and y = 42
Thus, the present age of Aftab’s daughter = 12 years and the present age of Aftab = 42 years.
2. The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900.
Later, another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Ans: Let the price of one bat = Rs. x
And the price of one ball According to the given conditions we have, From the graph, we find that From 3x+6y=3900 1x + 3y = 1300 x = 12 = Rs. y
We can write the equations as x + 2y = 1300
x + 3y = 1300
3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4kg apples and 2 kg. grapes is Rs. 300. Represent the situation algebraically and geometrically.
Ans: The basic assumption in this problem has to be made that the prices of apples and grapes per kg remain the same after one month.
Let the cost is 1kg. apples be and the cost of 1kg grapes be = Rs. x = Rs. y
According to the given situations, cost of 2 kg of apples and 1 kg of grapes = Rs. 160
Rs. x × 2 + Rs y x 1 = Rs. 160
Rightarrow 2x + y = 160 After one month
Cost of 4 kg of apples and 2 kg of grapes = Rs. 300
Rs. x × 4 + Rs y x 2 = Rs. 300
Rightarrow 4x + 2y = 300
| Exercise 3.2 |
1. Form the 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.
Ans: Let the number of boys be x and the number of girls be y.
According to the given conditions
x + y = 10 and y = x + 4
We get the required pair of linear equations as
x + y-10 = 0, x – y + 4 = 0
Graphical solution
x + y – 10 = 0 … (1)
| × | 2 | 5 |
| Y = 10 – × | 8 | 5 |
× – y + = …..(2)
| × | 2 | 4 |
| Y = × + 4 | 6 | 8 |
From the graph, we have x = 3, y=7
Common solution of the two linear equations.
Hence, the number of boys = 3
and the number of girls = 7
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Ans: Let the cost of one pencil be Rs. x and the cost of one pes be Rs. y
According to the given condition,
Cost of 5 pencils and 7 pens = Rs. 50
Cost of 7 pencils and 5 pens = Rs. 46
5x + 7y = 50 and 7x + 5y=46
Hence, the required linear equations are
5x + 7y = 50, 7x + 5y = 46
Graphical solution
5 × + y = 50 …(1)
| × | 2 | 3 | 6 |
| Y= 50 – 5× /7 | 5.7 | 5 | 2.8 |
7× + 5y = 46 …(2)
| × | 2 | 3 | 4 |
| Y = 46 – 7 × /5 | 6.4 | 5 | 3.6 |
From the graph, we
have the common
solution x = 3, y = 5
Hence, the cost of
one pencil = Rs. 3 and
the cost of one pen = Rs.
2. One compares the ratios a and find out whether the lines C2 representing the following pairs of linear equations intersect at a point, are parallel or coincident.
(i) 5x-4y+8=0… (1)
7x+6y-9=0… (2)
Ans: 5x-4y+8=0
7x+6y-9=0
a₁ /b₂ 5/7, b₁/b₂ = -4/6 = -4/6 = -⅔
⇒a₁/b₂ ≠ b₁/b₂
⇒Lines represented by (1) and (2) intersect at a point
(ii) 9x+3y+12=0 … (1)
18x+6y+ 24 = 0… (2)
Ans: 9× + 3y + 12 = 0
18× + 6 y + 24 = 0
a₁/a₂ = 8/19 = ½ , b₁/b₂= 3/6 = ½ ⇒a₁a₂ /b₁/b₂ = C₁ /C ₂
⇒ Liner represented by (i) ans (ii) are coincident
(iii) 6x-3y+10=0… (1)
2x-y+9=0… (2)
Ans: 6x-3y+10=0
2x-y+9=0
a₁/a₂ = 6/2 = 3, b₁/b₂ = -3/-1 = 3 ,C₁ /C ₂ = 10/9
a₁/a₂ = b₁/b₂ ≠ C₁ /C
⇒lines represented by (1) and (2) are parallel.
3. On the comparing the rations a₁/a₂ = b₁/b₂ and C₁ /C find out whether the following pairs of lines questions are consistent for inconsistent.
(i) 3x+2y=5; 2x-3y=7
Ans: 3× + 2y – 5 = 0
2 × – 3y – 7 = 0
a₁/a₂ = 3/2 ; b₁/b₂ = 2/-3 = -⅔ ⇒ a₁ / a₂ ≠ C₁ /C₂
⇒The equations have a unique Hence, consistent
(ii) 2x-3y = 8; 4x-6y=9
Ans: 2× – 3y – 8 = 0
4× – 6y – 9 = 0
a₁/a₂ = 2/4 = ½ ; b₁b₂ = -3 /-6 = ½; C₁ /C₂
= -8/-9 = 8/9 ⇒ a₁/a₂ = b₁b₂ ⇒ C₁ /C₂
⇒ The equation have no solution.
Hence , inconsistent.
(iii) 3/2 + ⅗y = 7; 9x-10y = 14
Ans: 3/2 + ⅗y – 7 = 0
9x-10y – 14 = 0
a₁/a₂ = (3/2)/9 = ⅙ ; ≠ = (5/3)/-10 = ⅙ ⇒
a₁/a₂ ≠ b₁b₂
⇒ The equation have no solution. Hence , inconsistent.
(iv) 5× – 3y – 11 = 0
10× + 6y -22 = 0
Ans: 5× – 3y – 11 ,= 0
10x – 6y – 22 = 0
(v) 4/3× 2y = 8; +3y = 12
Ans:
4. Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + = 10
Ans: x + y – 5 = 0 ……(i)
2x + ½ , b₁b₂ = ½, C₁ /C₂ = -5/-10 = ½
i.e a₁/a₂ = b₁b₂ = C₁ /C₂
Hence , the pair of liner equation is consistent ( 1) and (2) are same equations and hence
The graph is a consistent straight Liner.
| X | 1 | 3 |
| Y = 5 – x | 4 | 2 |
Graph of liner equations (1) and (2) is the coincident line passing through the points (1,4)
and (3,2) . every point on the line given a solution.
(ii) X- y = 8, 3X- 3y = 16
Ans: x-y -8= 0 (i)
3x- 3y -16 = 0…(ii)
a₁/a₂ = ⅓, b₁b₂= -1/-3, ⅓ C₁/C₂ = -8/-16,= ½ ; a₁/a₂ = b₁b₂ ≠ C₁/C₂
The pair of equations is inconsistent and the graph of two equations is a pair of parallel
straight lines.
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4x – 4y – 5 = 0
5. Half the perimeter of rectangular garden, whose length is 4m more than its width, is 36m. Find the dimensions of the garden. Solution: Length = x metres Width = y metres (x > y)
Ans: Given that x = y + 4 x + y (Half the perimeter)
= 36 m x – y – 4 = 0 (1)
x – y – 36 = 0 … (2)
6. Given the linear equation 2x + 3y – 8 = 0 write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) Intersecting lines incident lines.
Ans: 2x + 3y – 8 = 40 ( Given Equation)
3x + 2y + 4 = 0 ( New Equation)
a1/a2 ≠ b1/b2
Here,
Hence , the graph of the two equations will be two interesting lines
(ii) parallel lines.
Ans: 2x + 3y – 8 = 0 ( Given Equation)
4x + 6y + 8 = 0 ( New Equation)
a1/a2 = b1/b2 ≠ c1/c2
Here
Hence, the graph of the two equations will be a pair of parallel lines.
(iii) coincident lines.
Ans: 2x + 3y-8=0 (Given equation)
4x+6y-16=0 (New equation)
Here
Hence , the graph of the two equations will be coincident lines.
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.
Example 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.
Ans: Let s and I be the ages (in years) of Aftab and his daughter, respectively. Then, the pair of linear equations that represent the situation is s – 7 = 7(t – 7) i.e., s – 7t + 42 = 0 s + 3 = 3(t + 3) i.e., s – 3t = 6 and Using Equation (2) we, get s = 3t + 6 Putting this value of s in Equation (1), we get (3t + 6) – 7t + 42 = 0 4t = 48 i.e., which gives t = 12
Putting this value of t in Equation (2) we get, s = 3(12) + 6 = 42 So. Aftab and his daughter are 42 and 12 years old, respectively. Verify this answer by checking if it satisfies the conditions of the given problems.
Example 2: The cost of 2 pencils and 3 erasers is Rs. 9 and the cost of 4 pencils and 6 erasers is Rs. 18. Find the cost of each pencil and each eraser.
Ans: The pair of linear equations formed were:
2x + 3y = 9
4x + 6y = 18 … (1) … (2)
We first express the value of x in terms of y from the equation 2x +
This statement is true for all values of y. However, we do not get a specific value of y as a solution. Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions. Observe that we have obtained the same solution graphically also. We can not find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.
Example 3: Let us consider the Example 3 of section 3.2. Will the rails cross each other?
Ans: The pair of linear equations formed were:
x + 2y – 4 = 0
2x + 4y – 12 = 0
We express x in terms of y from Equation (1) to get
x = 4 – 2y Now, we substitute this value of x in equation (2) we get
2(4 – 2y) + 4y – 12 = 0
I.e. 8-120
I.e. -4=0
Which is a false statement.
Therefore, the equations do not have a common solution. So, the two rails will not cross each other.
| Exercise 3.3 |
1. Slove the following pair of linear equations by the substitution method
Ans: x + y = 14 … (1)
x – y = 4 … (2)
From (2) we get x = 4 + y……(3)
Substituting x from (3) in (1) we get
4 + y + y = 14
2y =10 ⇒ 5
From (3), we get x = 4 + 5 = 9
x = 9 y = 5
(ii) s – t = 3
Ans: s – t = 3 … (1)
s/3 + t/2 = 6…….(2)
From (1) s = t + 3 ……..(3)
(iii) 3x – y = 3
9 x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
Ans: 0.2x + 0.3y = 1.3
0.4x + 05y = 2.3…….(2)
From (1) 2x + 3 = 13 ⇒ 3 y = 13 – 2x ⇒ y = 13 – 2x / 3 …..(3)
Substituting y from (3) in (2) , we get
(v) √2x + √3y = 0
Ans: √2x + √3y = 0………..(1)
√3 – √3 = 0 …….(2)
From (2), √2, √8y = √3 i.e y., √3// √8…….(3)
Substituting y from (3) in (1) we get
(vi) 3x/2 – 5y/3 = -2
Ans: 3x/2 – 5y/3 = – 2 …..(1)
x/3 + y/2 = 13/6……(2)
From (2) y/2 = 13/6 – x/3 = 13-2x/6 i.e., y = 2 🗙( 13-2x)/6
I.e.., y = (13-2x)/3
Substituting y from (3) in (1) , we get
3x/2 – 5/3 🗙 (13-2x)/ -2
3x/2 – 5/9 (13 – 2) = – 2
⇒ 18 🗙 3x/2 – 18 🗙 5/9 (13 – 2x)=- 2 🗙 18
⇒ 7x-10(13-2x) = – 3627x-130 + 20x = -36
⇒27 x + 20 x = 130 – 3647 x = 94 ⇒ x =2
Substituting x = 2 in (3), we get, y=
13-2-2-/3 =9/3 = 3
Hence, x = 2, y = 3
2. Solve 2x + 3y = 11 and 2x-4y=-24 and hence find the value of ‘m’
for which y = mx +3
Ans: 2x+3y=11….(1)
2x-4y=-24…..(2)
From (2), 4y=2x+24
Substituting y from (3) in (1)
2x + 3(+12)=11 = 4x + 3(x+12)=11×12
⇒ 4x + 3x + 36-22 7x 22-36
⇒ 7x=-14⇒x=-2
From (3), y= -2+12/2i.e.,y=5
Substituting x=-2 and y = 5 in equation y = mx + 3, we get
5 mx(-2)+3i.e., 5=-2m+3
i.e., 2m 3-5=-2 i.e., m=-1
3. From 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.
Ans: Let the two numbers be x and (x > y)
We are given that x – y = 26 and x = 3y
The linear equations are x – y = 26 ….(1) x = 3y … (2)
Substituting x from (2) in (1), we get,
3y – y = 26 i.e., 2y = 26 i.e., y = 13
From (2), x = 3 * 13 = 39 i.e., x = 39
Hence the two numbers are 39 and 13
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Ans: (ii) Let the supplementary angles be x0 and y0 (x > y)
Now, x + y = 180 … (1)
and x – y = 18 …(2)
From (2), y = x – 18 …(3)
Substituting y from (3) in (1),
x+x-18=180 Rightarrow 2x = 198 , i.e., x = 99
Then from (3), y = 99 – 18 = 81 i.e, y = 81
Hence, the angles are 99º and 81°.
(iii) The coach of a cricket team buys 7 bats and 6 balls of Rs. 3800. Later, she buys 3 bats and 5 balls for 1750. Find the cost of each bat and each ball.
Ans: Let cost of one bat = Rs. x
cost of one ball = Rs. y
Cost of 7 bats and 6 balls = Rs. 3800 i.e., 7x + 6y = 3800 … (1)
Cost of 7 bats and 6 balls = Rs. 3800 i.e., 7x + 6y = 3800 … (1)
Cost of 3 bats and 6 balls = Rs. 1750 i.e., 3x + 5y = 1750 … (2)
From (2), 5y = 1750 – 3x i.e., y = (1750 – 3x)/5 ….(3)
Substituting y from (3) in (1), we get
7x + 6 ✖ (1750 – 3x)/5 = 3800
Hence, cost of one bat = Rs. 500
cost of one ball = Rs. 50 i.e., y = 50
(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 15km, 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 25km?
Ans: Let fixed charges be Rs. x
and charges per km be Rs. y
x + 10y = 105 … (1)
x + 15y = 155 … (2)
From (1), x = (105 – 10y)… (3)
Substituting x from (3) in (2)
105 – 10y + 15y = 155, i.e 5y = 155 – 105 = 50 i.e.. y = 10
Now from (3) x = 105 – 10 10 = 5
Hence, Fixed charges Rs. 5
Rate per km = Rs 10 Amount to be paid for travelling 25 km = Rs. 5 + Rs. 10 ✖ 25
= Rs..5 + Rs 250 = Rs .255
(v) A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 Find the fraction.
Ans: Let Now, x/y be the fraction where x and y are positive integers.
(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: Let x (in years) be the present age of Jocob’s son and y (in years) be the present age of Jacob. 5 years hence, it has relation
(y + 5) = 3(x + 5) i.e., y = 5 = 3x + 15 i.c., 3x – y + 10 = 0 (1)
5 years ago, it has relation
y – 5 = 7(x – 5) i.e., 7x – y – 30 = 0 (2)
From (i), y = 3x + 10
Substituting value of y in (2)
7x – (3x + 10) – 30 = 0 .i.e ., 4x – 40 = 0 ⇨ x = 10
Substituting x = 10 in (3), we get, y = 3 ✖ 10 + 10 Le.. y = 40 Hence present age of Jacob is 40 and the present age of his son is 10 years.
Elimination Method:
Example 1: The ratio of incomes of two persons is 9:7 and the ratio of their expenditures is 4: 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes.
Ans: Let us denote the incomes of the two persons by Rs. 9r and Rs. 7x and their expenditures by Rs. 4y and Rs. 3y respectively. Then the equations formed in the situation is given by: 9x – 4y = 2000 and (1)… (2)
7x – 3y = 2000
Step 1: Multiply Equation (1) by and 3 and Equation (2) by 4 to make the coefficients of y equal. Then we get the equations:
27x – 12y = 6000…(3)
28x – 12y = 8000 (4)
Step 2: Substract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So we get
(28x – 27x) – (12y – 12y) = 8000 – 6000 i.e. x = 2000
Step 3: Substituting this value of x in (1), we get
9(2000) – 4y = 200 i.e., y = 4000
So, the solution of the equations is x
= 2000 y = 4000 Therefore, the monthly
incomes of the persons are Rs. 18,000 and Rs. 14,000 respectively.
Verification: 18000/14000 = 9/7
Also, the ratio of their expenditures = 18000 – 2000/14000 – 2000
= 16000/12000 = 4:3
Example 2: Use elimination method to find all possible solutions of the following pair of linear equations:
2x + 3y = 8 …(1)
4x + 6y = 7 … (2)
Step 1: Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficient of x equal. Then we get the equations as:
4x + 6y = 16….(3)
4x + 6y = 7 … (4)
Step 3: Subtracting Equation (4) from Equation (3),
(4x – 4x) + (6y – 6y) = 16 – 7
i.e. 09, which is a false statement.
Therefore, the pair of equations has no solution.
Example 3: The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Ans: Let the ten’s and the unit’s digit in the first number be x and y, respectively. So, the first number may be written as 10x + y in the expanded from
(for example, 56 = 10 (5) + 6
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y +x| for example, when 56 is reversed, we get 65 = 10(6) +5.
If x – y = 2 then solving (1) and (2) by elimination, we get the number 42. we get x = 4 and y – x = in the cases we get the number 42
If x – y = 2 then solving (1) and (3) by elimination, we get x = 2 and
y = 4. In this case, we get the number 24. Thus, there are two such numbers 42 and 24
Verification: Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4-2-2
| Exercise 3.4 |
1. Solve the following pair of linear equations by the elimination method and the substitution method.
(i) x + y = 5 and 2x – 3y = 4
Ans:
Hence , x = 19/ 5 y = 6/5
Hence, x = 19/5 y = 6/5
Solution by Substitution Method:
x + y = 5 ……(1)
2x – 3y = 4 … (2)
From (1), y = 5 – x … (3)
Substituting y from (3) in (2),
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
Ans: By Elimination Method :
3x – 5y = 4 … (1)
9x – 2y = 7 … (2)
Multiplying (1) by 3 and (2) by 1, we get
9x – 15y = 12 … (3)
9x – 2y = 7 … (4)
Subtracting (4) from (3) (9x – 15y) – (9x – 2y) = 12 – 7 – 15y + 2y = 5
-13y = 5y y = – 5/13
(iv) x/2 + (2y)/3 = – 1 and x – y/3 = 3
Ans:
(v) (3y)/2 – (5x)/3 = – 2 and y/3 + x/3 = 13/16
Ans:
(vi) x – y = 3 and x/3 + y/2 = 3
Ans:
(vii) 8/x – 9/y = 1 and 10/x + 6/y = 7
Ans:
∴ 1/y = ⅓ ⇒ y = 3
Putting the value v in equation (1)
2. From 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 What is the fraction? 1/2 if we only add 1 to the denominator. What is the friction?
Ans: Let the fraction be x/y According to the given conditions-
(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?
Ans: Let assume that Nuri’s age is x years
Sonu’s age is y years
5 years ago,
Nuri’s age = (x – 5) years
Sonu’s age = (y – 5) years
According to the question:
⇒ (x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x = 3y – 10 …………………………. (1)
10 years later,
Nuri’s age = (x + 10) years
Sonu’s age = (y + 10) years
According to the question:
⇒ (x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x = 2y + 10 …………………………. (2)
From equation (1) and (2)
⇒ 3y – 10 = 2y + 10
⇒ y = 20 years
⇒ x = 3 × 20 – 10
⇒ x = 50 years
Nuri’s age = x = 50 years
Sonu’s age = y = 20 years
(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?
Ans: Let, present age of Nuri = x (in years) present age of Sonu = y (in year)
(iii) The sum of the digits of a 20 digit number is 9. Also nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Ans: Let x be the digit at unit place and y be the digit at tens place of the number. According to the condition, x + y = 9 … (1) and the number is 10 * y deg + x When we reverse the , the value of the new number is 10x + y
(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.
Ans: Let Number of Rs. 50 notes = x
Number of Rs. 100 notes = y
x + y = 25……(1)
Also, 50x + 100y = 2000 x + 2y = 40
Subtracting (1) from (2), (x + 2y) – (x + y) = 40-25
y = 15, then x = 10
(v) A lending library has a fixed charge for the first three days and 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: Let the fixed charges for the first three days be Rs. x.
Let the additional charge per day be Rs. y.
⇒ x + 4y = 27
(∴ Rs. 4y are to be paid for 4 extra days.) … (1)
In the case of Susy, x + 2y = 21
Subtracting (2) from (1), 2y = 27 – 21 ⇒ y = 3
From (1), x + 4 x 3 = 27 ⇒ x = 15
∴ Fixed charges for first three days = Rs. 15
Addition charges per extra day = Rs. 3
Cross-Multiplication Method:
Example 1: From a bus stand in Bangalore, if we buy 2 tickets to Malleswaram and 3 tickets to Yeshwanthpur, the total cost is Rs. 46; but if we buy 3 tickets to Malleshwaram and 5 tickets to Yeshwanthpur the total cost is Rs. 74. Find the fares from the bus stand to Malleswaram, and to Yeshwanthpur.
Ans: Let Rs. x be the fare from the bus stand in Bangalore to Malleswaram, and Rs. y to Yeshwanthpur. From the given information, we have
Hence, the fare from the bus stand in Bangalore to Malleswaram is Rs. and the fare to Yeshwanthpur is Rs. 10.
Verification: You can check from the problem that the solution we have got correct.
Example 2: For which values of p does the pair of equations given below has unique solution?
4x + py + 8 = 0
2x + 2y + 2 = 0
Therefore, for all values of p, except 4, the given pair of equations will have a unique solution.
Example 3: For what values of k will the following pair of linear equations have infinitely many solutions?
Kx + 3y – (k-3) = 0
12x + ky – k = 0
Ans: a1/ a2 = k/12, b1/b2 = 3/k c1/c2 = k – 3/k
For pair of linear equations to have infinitely many solutions
a1/a2 = b1/b2 = c1/2
So we need k/12 = 3/k = k-3/k
k/12 = 3
Which gives k2 = 36 , i.e k = ± 6
Also, 3/k = k – 3/k
gives 3k = k 2 i.e., 6k, which means k = 0 or k = 6 Therefore, the value of k, that satisfies both the conditions is k = 6 For this value, the pair of linear equations has infinitely many solutions.
| Exercise 3.5 |
1. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using a cross-multiplication method.
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0
Ans: x – 3y – 3 = 0……1
3x – 9y – 2 = 0……….2
a1/a2 = 1 /3 , b1/b2 =-3/-9 = 1/3 , c1/c2 = 3/2 ⇨ a1/a2 = b1/b2 ≠ c1/c2
(ii) 2x + y = 5
3x + 2y = 5
Ans: 2x + y = 5………1
3x + 2y = 5……………2
a/a2 = ⅔ , b1/b2 = ½ ,, c1/c2 = ⅝
∴ a1/a2 ≠ b1/b2
Here we have a unique solution.
2x + y = 5
3x + 2y = 5
(iii) 3x – 5y = 20
6x – 10y = 40
(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
Ans: Do yourself.
(v) 3x + 3y = 6
4x + 6y = 12
(vi) x – 2y = 6
3x – 6y = 0
(vii) 3a/x – 2b/y = 5
a/x + 3b/y = 2
(viii) 2x + y – 15 = 0
3x – y – 5 = 0
Ans: 2x + y – 15 = 0
3x – y – 5 = 0
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?
(iii) For what value of p the system equation px – y = 2 6x – 2y = 3 has only one solution?
(iv) Find the value of k so that the following system of linear equations has no solution.
(v) Find the value of m such that the following system of linear equation has infinite number of solutions.mx + 4y = m – 4 16x + my = m
3. Solve the following pair of linear equations by the substitution and cross-multiplication methods :
(i) 8x + 5y = 9
3x + 2y = 4
(ii) 4x – 3y = 23 3x + 4y = 11
(iii) 2x + 3y-11 = 0
4x-3y + 5 = 0
(iv) 5x + 7y = 19 3x + 2y = 7
4. From the pair of linear equation 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 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.
Ans: let, fixed hostel charges (Monthly) = Rs. y
cost of food for one day = Rs. x
In the case of A student: 20x + y = 1000 … (1)
In the case of B student: 26x + y = 1180 Rs. y = Rs .x …(2)
Subtracting (1) from (2), 26 x – 20 x = 1180 – 1000
x = 30
From (1), y = 1000 – 20 x = 1000 – 20 ✖ 30
1000-600 = 400
Hence, monthly fixed charges s = Rs. 400
Cost of food per day = Rs. 30
(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?
Ans: Let, Number of right answers = x
Number of wrong answers = y
Total number of questions = x + y
In the first case,
Marks awarded for x right answers = 3x
Marks lost for y wrong answers = y ✖ 1 = y
Then, 3x – y = 40 … (1) In second case, Marks awarded for x right answer = 4x
Marks lost for y wrong answer = 2y
4x – 2y = 50 … (2)
From (1), y = 3x – 40 … (3)
From (2) and (3) eliminating y, 4x – 2(3x – 40) = 50
⇒ 4x – 6x – 80 = 50
2x = 30
Then y = 3 or x 15 – 40 =5 ⇒ y = 5
Total number of question = 15 + 5 = 20
(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?
Ans: Let speed of car I = x km / hr
Speed of car II = y km / hr
Car I starts from point A and car II starts from point B.
(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: In the first case, the area is reduced by 9 square units.
When length = x – 5 units
and breadth = y + 3 units.
⇒ xy – (x – 5)(y – 3) = 9
⇒ xy – (xy + 3x – 5y – 15) = 9
– 3x + 5y + 15 = 9
⇒ 3x – 5y = 6 … (1)
Adding (3) and (4), 19x =323 ⇒ x=17
Substituting x = 17 in (2), we get, 34 + 3y =61 ⇒ y=9
Hence, length = 17 units and breadth = 9 units.
| Exercise 3.6 |
1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
(iii) 4/x + 3y = 14
3/x – 4y = 23
(iv) 5/x – 1 + 1/y – 2 = 2
6/x – 1 – 3/y – 2 = 1
Ans:
Ans:
Ans:
(vii)
Ans:
2. Formulate the following problems as a 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 speed of the current.
Ans: Let, speed of Ritu in still water = x km/hr
speed of current = y km / hr
(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, ans also that taken by 1 man alone.
Ans: Let 1 woman finish the work in x days
and let I man finish the work in y days.
(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: Let the speed of the train be = x km / hr and the speed of the bus be = y km / hour . In the first case, Roohi travels 60 km by train and 240 km by bus in 4
| Exercise 3.7 |
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 age of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
2. One says, “Give me a hundred, friend! I shall then become twice as
rich as you.” The other replies, “If yo give me ten, I shall be six times as
rich as you.” Tell me what is the amount of their (respective) capital?
[Hint: x + 100 = 2(y – 100) y + 10 = 6(x – 10)
3. A train covered a certain distance at a uniform speed. If the train would have been 10km/hr 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.
5. In a ∆ABC, ∠C = 3∠B = 2(∠A + B). Find the three angles.
Ans: We have ∠C = 3∠B = 2(ZA+∠B)… (1)We know that the sum of the measures of three angles of a triangle is 180°.
∠A + ∠B + ∠C = 180°………..(2)
6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3 Determine the coordinates of the vertices of the triangle formed by these lines and the y axis.
In the figure we observe that the coordinates of the vertices of the angle AEF and A(1, 0) E(0, – 5) and F(0, – 3)
7. Solve the following pair linear of equations:
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