Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.
10th Standard Maths 1 Problem Set 1 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board
Class 10 Maths Part 1 Problem Set 1 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board
Choose correct alternative for each of the following questions.
Question 1.
To draw graph of 4x + 5y = 19, find y when x = 1.
(a) 4
(b) 3
(c) 2
(d) -3
Answer:
(b)
Question 2.
For simultaneous equations in variables x and y, D
x
= 49, Dy = β 63, D = 7 then what is x?
(a) 7
(b) -7
(c) \(\frac { 1 }{ 7 } \)
(d) \(\frac { -1 }{ 7 } \)
Answer:
(a)
Question 3.
Find the value of
(a) -1
(b) -41
(c) 41
(d) 1
Answer:
(d)
Question 4.
To solvex + y = 3; 3x β 2y β 4 = 0 by determinant method find D.
(a) 5
(b) 1
(c) -5
(d) -1
Answer:
(c)
Question 5.
ax + by = c and mx + n y = d and an β bm then these simultaneous equations have-
(a) Only one common solution
(b) No solution
(c) Infinite number of solutions
(d) Only two solutions.
Answer:
(a)
Question 2.
Complete the following table to draw the graph of 2x β 6y = 3.
Answer:
Question 3.
Solve the following simultaneous equations graphically.
i. 2x + 3y = 12 ; x β y = 1
ii. x β 3y = 1 ; 3x β 2y + 4 = 0
iii. 5x β 6y + 30 = 0; 5x + 4y β 20 = 0
iv. 3x β y β 2 = 0 ; 2x + y = 8
v. 3x + y= 10 ; x β y = 2
Answer:
i. The given simultaneous equations are
The two lines interest at point (3,2).
β΄ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x β y = 1.
ii. The given simultaneous equations are
The two lines intersect at point (-2, -1).
β΄ x = -2 and y = -1 is the solution of the simultaneous equations x β 3y = 1 and 3x β 2p + 4 = 0.
iii. The given simultaneous equations are
The two lines intersect at point (0, 5).
β΄ x = 0 and y = 5 is the solution of the simultaneous equations 5x β 6y + 30 = 0 and 5x + 4y β 20 = 0.
iv. The given simultaneous equations are
The two lines intersect at point (2, 4).
β΄ x = 2 and y = 4 is the solution of the simultaneous equations 3x β y β 2 = 0 and 2x + y = 8.
v. The given simultaneous equations are
The two lines intersect at point (3, 1).
β΄ x = 3 and y = 1 is the solution of the simultaneous equations 3x + y = 10 and x β y = 2.
Question 4.
Find the values of each of the following determinants.
Solution:
Question 5.
Solve the following equations by Cramerβs method.
Solution:
i. The given simultaneous equations are
6x β 3y = -10 β¦(i)
3x + 5y β 8 = 0
β΄ 3x + 5y = 8 β¦(ii)
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with a
1
x + b
1
y = c
1
and a
2
x + b
2
y = c
2
, we get
a
1
= 6, b
1
= -3, c
1
= 10 and
a
2
= 3, b
2
= 5, c
2
= 8
ii. The given simultaneous equations are
4m β 2n = -4 β¦(i)
4m + 3n = 16 β¦(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with a
1
m + b
1
n = c
1
and a
2
m + b
2
n = c
2
, we get
a
1
= 4, b
1
= -2, c
1
= -4 and
a
2
= 4, b
2
= 3, c
2
= 16
β΄ (m, n) = (1, 4) is the solution of the given simultaneous equations.
iii. The given simultaneous equations are
iv. The given simultaneous equations are
7x + 3y = 15 β¦(i)
12y β 5x = 39
i.e. -5x + 12y = 39 β¦(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a
1
x + b
1
y = c
1
and a
2
x + b
2
y = c
2
, we get
a
1
= 7, b
1
= 3, c
1
= 15 and
a
2
= -5, b
2
= 12, c
2
= 39
v. The given simultaneous equations are
β΄ 4(x + y β 8) = 2(3x β y)
β΄ 4x + 4y β 32 = 6x β 2y
β΄ 6x β 4x β 2y β 4y = -32
β΄ 2x β 6y = -32
β΄ x β 3y = -16 β¦(ii)[Dividing both sides by 2]
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with
a
1
x + b
1
y = c
1
and a
2
x + b
2
y = c
2
, we get
a
1
= 1, b
1
= -1, c
1
= -4 and
a
2
= 1, b
2
= -3, c
2
= -16
β΄ (x, y) = (2, 6) is the solution of the given simultaneous equations.
Question 6.
Solve the following simultaneous equations:
Answer:
i. The given simultaneous equations are
Subtracting equation (iv) from (iii), we get
β΄ (x, y) = (6, β 4) is the solution of the given simultaneous equations.
ii. The given simultaneous equations are
Adding equations (v) and (vi), we get
iii. The given simultaneous equations are
β΄ (x, y) = (1, 2) is the solution of the given simultaneous equations.
iv. The given simultaneous equations are
β΄ Equations (i) and (ii) become 7q β 2p = 5 β¦(iii)
8q + 7p = 15 β¦(iv)
Multiplying equation (iii) by 7, we get
49q β 14p = 35 β¦(v)
Multiplying equation (iv) by 2, we get
16q + 14p = 30 β¦(vi)
Adding equations (v) and (vi), we get
Substituting q = 1 in equation (iv), we get
8(1) + 7p = 15
β΄ 8 + 7p = 15
β΄ 7p = 15 β 8
β΄ 7p = 7
β΄ p = \(\frac { 7 }{ 7 } \) = 1
β΄ (P, q) = (1,1)
Resubstituting the values of p and q, we get
1 = \(\frac { 1 }{ x } \) and 1 = \(\frac { 1 }{ y } \)
β΄ x = 1 and y = 1
β΄ (x, y) = (1, 1) is the solution of the given simultaneous equations.
v. The given simultaneous equations are
Resubstituting the values of p and q, we get
β΄ 3x + 4y = 10 β¦(v)
and 2x β 3y = 1 β¦(vi)
Multiplying equation (v) by 3, we get
9x + 12y = 30 β¦(vii)
Multiplying equation (vi) by 4, we get
8x β 12y = 4 β¦(viii)
Adding equations (vii) and (viii), we get
Substituting x = 2 in equation (v), we get
3(2) + 4y = 10
β 6 + 4y = 10
β 4y = 10 β 6
β y = 4/4 = 1
β΄ y = 1
β΄ (x, y) = (2, 1) is the solution of the given simultaneous equations.
Question 7.
Solve the following word problems, i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unitβs place is 3 more than the digit in the tenβs place. Find the original number.
Solution:
Let the digit in unitβs place be x
and that in the tenβs place be y.
ii. Kantabai bought 1 \(\frac { 1 }{ 2 } \) kg tea and 5 kg sugar from a shop. She paid βΉ 50 as return fare for rickshaw. Total expense was βΉ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So, next month she placed the order online for 2 kg tea and 7 kg sugar. She paid βΉ 880 for that. Find the rate of sugar and tea per kg.
Solution:
Let the rate of tea be βΉ x per kg and that of sugar be βΉ y per kg.
According to the first condition,
cost of 1 \(\frac { 1 }{ 2 } \) kg tea + cost of 5 kg sugar + fare for rickshaw = total expense
According to the second condition,
cost of 2 kg tea + cost of 7 kg sugar = total expense
2x + 7y = 880 β¦(ii)
Multiplying equation (i) by 2, we get
6x + 20y = 2600 β¦(iii)
Multiplying equation (ii) by 3, we get
6x + 21y = 2640 β¦(iv)
Subtracting equation (iii) from (iv), we get
β΄ The rate of tea is βΉ 300 per kg and that of sugar is βΉ 40 per kg.
iii. To find number of notes that Anushka had, complete the following activity.
Solution:
Anushka had x notes of βΉ 100 and y notes of βΉ 50.
According to the first condition,
100x + 50y = 2500
β΄ 2x + y = 50 β¦(i) [Dividing both sides by 50]
According to the second condition,
100y + 50x = 2000
β΄ 2y + x = 40 β¦ [Dividing both sides by 50]
i.e. x + 2y = 40
β΄ 2x + 4y = 80 β¦(ii) [Multiplying both sides by 2]
Subtracting equation (i) from (ii), we get
β΄ Anushka had 20 notes of βΉ 100 and 10 notes of βΉ 50.
iv. Sum of the present ages of Manish and Savita is 31, Manishβs age 3 years ago was 4 times the age of Savita. Find their present ages.
Solution:
Let the present ages of Manish and Savita be x years and y years respectively.
According to the first condition,
x + y = 31 β¦(i)
3 years ago,
Manishβs age = (x β 3) years
Savitaβs age = (y β 3) years
According to the second condition,
(x β 3) = 4 (y β 3)
β΄ x β 3 = 4y β 12
β΄ x β 4y = -12 + 3
β΄ x β 4y = -9 β¦(ii)
Subtracting equation (ii) from (i), we get
β΄ x = 31 β 8
β΄ x = 23
The present ages of Manish and Savita are 23 years and 8 years respectively.
v. In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is βΉ 720. Find daily wages of skilled and unskilled workers.
Solution:
Let the daily wages of skilled workers be βΉ x
that of unskilled workers be βΉ y.
According to the first condition,
β΄ The daily wages of skilled workers is βΉ 450 and that of unskilled workers is βΉ 270.
vi. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A), Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Solution:
Let the speeds of Hamid and Joseph be x km/hr andy km/hr respectively.
Distance travelled by Hamid in 20 minutes
According to the first condition,
β΄ The speeds of Hamid and Joseph 50 km/hr and 40 km/hr respectively.