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NCERT exemplar Class 10 Maths solutions chapter 3 will help the students to solve pair of linear equations in two variables in an easy way. This chapter provides a detailed understanding of Linear equations in two variables. The NCERT exemplar Class 10 Maths chapter 3 solutions clarifies the difficulties faced by the students while practicing exemplar and prove to be sufficient material to study NCERT Class 10 Maths.
These NCERT exemplar Class 10 Maths chapter 3 solutions are highly detailed and explanatory which covers the concepts related to linear equations in two variables. The NCERT exemplar Class 10 Maths solutions chapter 3 explores the CBSE Syllabus for Class 10 of this specific chapter along with additional interesting practice problems of NCERT.
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Question:1
Choose the correct answer from the given four options:Graphically, the pair of equations 6x – 3y + 10 = 0, 2x – y + 9 =0 represents two lines which are
(A) intersecting at exactly one point.
(B) intersecting at exactly two points.
(C) coincident.
(D) parallel.
Answer:
Answer: D
Solution:
In equation
6x – 3y + 10 = 0
a1 = 6, b1 = –3, c1 = 10
In equation
2x – y + 9 = 0
a2 = 2, b2 = –1, c2 = 9
Here,
The lines are parallel.
Question:2
The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have
(A) a unique solution
(B) exactly two solutions
(C) infinitely many solutions
(D) no solution
Answer:
Answer: D
Solution:
In equation
x + 2y + 5 = 0
a1 = 1, b1 = 2, c1 = 5
In equation
–3x –6y + 1 = 0
a2 =–3, b2 = –6, c2 = 1
Here
Hence these lines are parallel to each other and have no solution.
Question:3
If a pair of linear equations is consistent, then the lines will be
(A) parallel
(B) always coincident
(C) intersecting or coincident
(D) always intersecting
Answer:
If the pair of linear equations is consistent then they have unique or infinitely many solutions.
In unique solution,
Lines are intersecting
In infinitely many solution,
Lines are coincident
Hence the pair of lines are intersecting or coincident.
Question:4
The pair of equations y = 0 and y = –7 has
(A) one solution
(B) two solutions
(C) infinitely many solutions
(D) no solution
Answer:
Solution:
Line y = 0 and y = –7 are parallel to each other.
Hence, they never intersect. Hence the pair of equations have no solution.
Question:5
The pair of equations x = a and y = b graphically represents lines which are
(A) parallel
(B) intersecting at (b, a)
(C) coincident
(D) intersecting at (a, b)
Answer:
Answer: D
Solution:
The lines x = a and y = b are intersecting at (a, b)
Question:6
For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = –16 represent coincident lines?
(A)
(B) –
(C) 2
(D) –2
Answer:
Answer: C
Solution:
Equations are
3x – y + 8 = 0
6x – ky + 16 = 0
In equation
3x – y + 8 = 0
a1 = 3, b1 = –1, c1 = 8
In equation
6x – ky + 16 =0
a2 = 6, b2 = –k, c2 = 16
Since lines are coincident
Hence
….(i)
1/k = 1/2 ( from equation (1))
k = 2 Hence the value of k is 2.
Question:7
If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is
(A)
(B)
(C)|
(D)
Answer:
Answer: C
Solution:
Equations are
3x + 2ky = 2
2x + 5y + 1 = 0
In equation
3x + 2ky – 2 = 0
a1 = 3, b1 = 2k, c1 = –2
In equation
2x + 5y + 1 = 0
a2 = 2, b2 = 5, c2 = 1
Since lines are parallel hence
……(1)
( from equation (1))
Hence the value of k is .
Question:8
The value of c for which the pair of linear equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is
(A) 3
(B) – 3
(C) –12
(D) no value
Answer:
Answer: D
Solution:
Equations are
cx – y – 2 = 0, 6x – 2y – 3 = 0
In equation
cx – y – 2 = 0
a1 = c, b1 = –1, c1 = –2
In equation
6x – 2y – 3 = 0
a2 = 6, b2 = –2, c2 = –3
Since equations have infinitely many solutions
Hence
Here we find that i.e., equations are inconsistent
So, we can not assign any value to c.
Question:9
One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be
(A) 10x + 14y + 4 = 0
(B) –10x – 14y + 4 = 0
(C) –10x + 14y + 4 = 0
(D) 10x – 14y = –4
Answer:
Answer: D
Solution:
Given equation is
–5x + 7y – 2 = 0
a1 = –5, b1 = 7, c1 = –2
For dependent linear equation
a2 = –5k, b2 = 7k, c2 = –2k
Put k = 2
a2 = –10, b2 = 14, c2 = –4
The revived equation is of the type a2x + b2y + c2 = 0
Put a2 = – 10, b2 = 14, c2= –4
–10x + 14y – 4 = 0
10x – 14 y = –4
Question:10
A pair of linear equations which has a unique solution x = 2, y = –3 is
(A) x + y = –1 ; 2x – 3y = –5
(B) 2x + 5y = –11 ; 4x + 10y = –22
(C) 2x – y = 1 ; 3x + 2y = 0
(D) x– 4y –14 = 0 ; 5x – y – 13 = 0
Answer:
Answer: B, D
Solution:
A pair which have unique solution x = 2, y = –3 will satisfy L.H.S. = R.H.S. if we put x = 2 and y = –3
Put x = 2, y = –3 in option (A)
x + y = –1 ; 2x – 3y = –5
2 – 3 = –1 ; 2(2) – 3 (–3) = –5
–1 = –1 (True) ; 4 + 9 = –5
13 = – 5 (False)
Put x = 2, y = –3 in option (B)
2x + 5y = –11 ; 4x + 10y = –22
2(2) + 5(–3) = –11 ; 4(2) + 10 (–3) = –22
4 – 15 = –11 ; 8 – 30 = –22
–11 = –11 (True) –22 = –22 (True)
Put x = 2, y = – 3 in option (C)
2x – y = 1 ; 3x + 2y = 0
2(2) – (–3) = 1 ; 3(2) + 3 (–3) = 0
4 + 3 = 1 ; 6 – 6 = 0
7 = 1 (False) ; 0 = 0 (True)
Put x = 2, y = – 3 in option (D)
x – 4y – 14 = 0 ; 5x – y – 13 = 0
2 – 4 (–3) – 14 = 0 ; 5(2) – (–3) – 13 = 0
2 + 12 – 14 = 0 ; 10 + 3 – 13 = 0
0 = 0 (True) ; 13 – 13 = 0
0 = 0 (True)
By putting the value of x = 2, y = –3 in A, B, C, D; two options B, D satisfy L.H.S = R.H.S
Hence both (B, D) are correct
Question:11
If x = a, y = b is the solution of the equations x – y = 2 and x + y = 4, then the values of a and b are, respectively
(A) 3 and 5
(B) 5 and 3
(C) 3 and 1
(D) –1 and –3
Answer:
Answer : (C)
Solution:
Equation are x – y = 2 … (1)
x + y = 4 … (2)
Add equation (1) and (2)
x – y = 2
x + y = 4
2x = 6
x = 3
Put x = 3 in equation (2)
3 + y = 4
y = 4 – 3
y = 1
it is given that a = x, b = y
Hence a = 3, b = 1
Question:12
Aruna has only Rs 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Rs 1 and Rs 2 coins are, respectively
(A) 35 and 15
(B) 35 and 20
(C) 15 and 35
(D) 25 and 25
Answer:
Answer:(D) 25 and 25
Solution:
Let Rs. 1 coins = x
Rs. 2 coins = y
As per the question:
x + y = 50 …(1)
x + 2y = 75 … (2)
Find value of x, y using elimination method.
x + y = 50
x + 2y = 75
y = 25 (subtract equation 1 from equation 2)
put y = 25 in equation (1)
x + 25 = 50
x = 25
x = 25, y = 25
Hence, Rs. 1 coins = 25
Rs. 2 coins = 25
Question:13
The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively
(A) 4 and 24
(B) 5 and 30
(C) 6 and 36
(D) 3 and 24
Answer:
Answer:(C) 6 and 36
Solution:
According to questions at present father’s age is six times than son’s age.
Let son age = x
Father age = y
y = 6x … (1)
After 4 years Father’s age will be four times than son’s age. That is
y + 4 = 4 (x + 4)
y + 4 = 4x + 16
4x – y + 12 = 0 …(2)
Put y = 6x in equation (2)
4x – 6x + 12 = 0
+ 2x = + 12
x = = 6
Put x = 6 in equation (1)
y = 6 (6) = 36
y = 36
Hence present age of father is 36 and son is 6.
Question:1
Do the following pair of linear equations have no solution? Justify your answer.
(i) 2x + 4y = 3 ; 12y + 6x = 6
(ii) x = 2y; y = 2x
(iii) 3x + y – 3 = 0 ;
Answer:
(i)Solution:
Equations are 2x + 4y = 3
12y + 6x = 6
In equation 2x + 4y – 3 = 0
a1 = 2, b1 = 4, c1 = –3
In equation 12y + 6x – 6 = 0
a2 = 6, b2 = 12, c2 = –6
For no solution also, Here we have
Hence the given equation have no solution.
(ii)Solution:
Equations are x – 2y = 0
–2x + y = 0
In equation
x – 2y = 0
a1 = 1, b1 = –2, c1 = 0
In equation
–2x + y = 0
a2 = –2, b2 = 1, c2 = 0
For no solution but here
Hence the given pair of equations a unique solution.
(iii)Solution:
Equations are 3x + y = 3
In equation
3x + y – 3 = 0
a1 = 3, b1 = 1, c1 = –3
In equation
a2 = 2, b2= , c2 = –2
For no solution but here
Hence the given pair of equations have infinity many solutions.
Question:2
Do the following equations represent a pair of coincident lines? Justify your answer.
(i); 7x + 3y = 7
(ii) –2x – 3y = 1 ; 6y + 4x = –2
(iii)
Answer:
(i)Solution:
Equation are 3x + y = 3
7x + 3y = 7
In equation
3x + y – 3 = 0
a1 = 3 ; b1 = ; c1 = –3
In equation
7x + 3y – 7 = 0
a2 = 7 ; b2 = 3; c2 = –7
For coincident lines but here
Hence the given pair of equations does not represent a pair of coincident lines.
(ii)Solution:
Equation are –2x – 3y = 1
4x + 6y = –2
In equation
–2x – 3y – 1 = 0
a1 = –2 ; b1 = –3 ; c1 = –1
In equation
4x + 6y + 2 = 0
a2 = 4 ; b2 = 6; c2 = 2
For coincident lines also here
Hence the given pair of equation represents a pair of coincident lines.
Solution: (iii)
Equation are
4x + 8y + = 0
In equation
a1 = 1/2 ; b1 = 1 ; c1 =
In equation
4x + 8y + = 0
a2 = 4 ; b2 = 8; c2 =
For coincident lines but here
Hence the given pair of equations does not represent a pair of coincident lines.
Question:3
Answer:
Solution: (i)
Equation are : –3x – 4y = 12
4y + 3x = 12
In equation
–3x – 4y – 12 = 0
a1 = –3; b1 = –4 ; c1 = –12
In equation
4y + 3x – 12 = 0
a2 = 3 ; b2 = 4; c2 = –12
For consistency either but here
Hence the pair of linear equations is not consistent.
Solution: (ii)
Equation are :
In equation
a1= ; b1 = –1 ; c1 =
In equation
a2 = 1/5 ; b2 = – 3; c2 = –1/6
For consistency either also here
Hence given equations are consistent.
Solution: (iii)
Equation are: 2ax + by = a
4ax + 2by = 2a
In equation
2ax + by – a = 0
a1 = 2a; b1 = b; c1 = –a
In equation
4ax + 2by – 2a = 0
a2 = 4a ; b2 = 2b; c2 = –2a
For consistency either
Also, here
Hence given equations are consistent
Solution: (iv)
Given, equations are x + 3 y = 11
4x + 12y =22
In equation
x + 3y = 11
a1 = 1; b1 = 3; c1 = –11
In equation
4x + 12y = 22
a2 = 4 ; b2 = 12; c2 = –22
For consistency either but here
Hence given equations are not consistent.
Question:4
Answer:
Solution:
Here equations are x + 3y = –7
2x + 6y = 14
a1 = , b1 = 3, c1 = 7
a2 = 2, b2 = 6, c2 = –14
The equation have infinitely many solution
Here we can see that
Hence no value of exist because it is given that equation has infinitely many solutions
Hence the statement is false.
Question:5
Answer:
Solution:
Equation are x – 2y = 8
5x – 10y = c
Here a1 = 1, b1 = –2, c1 = –8
a2 = 5, b2 = –10, c2 = –c
For unique solution but here
Hence the given statement is false.
Question:6
Answer:
Solution:
From the above graph it is clear that x = 7 is not parallel to x axis but it is parallel to y-axis.Hence the given statement is false.
Question:1
Answer:
(i)Solution:
The given equations are
x + y = , x + y = 1
In equation
x + y – = 0
a1 = , b1 = 1, c1 = –
In equation
x + y – 1 = 0
a2 =1, b2 = , c2 = –1
For no solution
=1,-1
Hence value of is -1
because in this case )
(ii)Solution:
The given equations are
x + y = , x + y = 1
In equation
x + y – = 0
a1 = , b1 = 1, c1 = –
In equation
x + y – 1 = 0
a2 =1, b2 = , c2 = –1
For infinite many solution
Only one value of l satisfy all the three equations, that is = 1
(iii) Solution:
The given equations are
x + y = , x + y = 1
In equation
x + y – = 0
a1 = , b1 = 1, c1 = -
In equation
x + y = 1
a2 =1, b2 = , c2 =–1
For unique solution
All real values of except
Question:2
For which value(s) of k will the pair of equations kx+ 3y = k – 3, 12x + ky= k have no solution?
Answer:
Solution:
The given equations are
kx + 3y = k –3
12x + ky = k
In equation
kx + 3y = k – 3
a1 = k, b1 = 3, c1 = –k + 3
In equation
12x + ky = k
a2 = 12, b2 = k, c2 = –k
We know that if the equations have no solution then
= 36
( because if , then which is not possible in no solution.)
Hence value of k is –6.
Question:3
Answer:
Solution:
The given equations are
x + 2y = 1
(a – b) x + (a + b)y = a + b – 2
In equation
x + 2y = 1
a1 = 1, b1 = 2, c1 = –1
In equation
(a – b)x (a + b)y = a + b – 2
a2 = (a – b), b2 = (a + b), c2 = – (a + b – 2)
For infinitely many solutions
|
|
Add equations (1) and (2)
–a + 3b + a + b = 0 + 4
4b = 4
b = 1
Put b = 1 in equation (1)
–a + 3 (1) = 0
a = 3
Hence a = 3 and b = 1.
Question:4
Answer:
(i)Solution:
In equation 3x – y – 5 = 0
a1 = 3, b1 = –1, c1 = –5
In equation 6x – 2y – p = 0
a2 = 6, b2 = –2, c2 = –p
If the lines are parallel
Any real value of p except 10.
(ii)Solution:
Here the equations are
–x + py = 1, px – y = 1
In equation
–x + py = 1
a1 = –1, b1 = p, c1 = –1
In equation
px – y = 1
a2 = p, b2 = –1, c2 = –1
For no solution
= 1
p = ± 1
p = + 1 ( because if p = –1 then )
(iii)Solution:
Here the equation are –3x + 5y = 7, 2px – 3y = 1
In equation –3x + 5y = 7
a1 = –3, b1 = 5, c1 = –7
In equation 2px – 3y = 1
a2 = 2p, b2 = –3, c2 = –1
For unique solution
Hence p can have any real value for unique solution except 0.9
(iv)Solution:
Here the equation are 2x + 3y – 5 = 0
px – 6y –8 = 0
In equation 2x + 3y – 5 = 0
a1 = 2, b1 = 3, c1 = –5
In equation px – 6y – 8 = 0
a2 = p, b2 = –6, c2 = –8
For unique solution
Hence the equations have unique solution for every real value of p except –4.
(v)Solution:
Here the given equations are 2x + 3y = 7
a1 = 2, b1 = 3, c1 = –7
In equation 2px + py = 28 – qy
2px + py + qy = 28
2px + y(p + q) = 28
a2 = 2p, b2 = p + q, c2 = –28
For infinite many solution
4 = p
p = 4
p + q = 12
Put p =4
q=12-4
q=8
For infinitely many solutions, p = 4, q = 8
Question:5
Answer:
Solution:
Equations are: x – 3y – 2 = 0
–2x + 6y – 5 = 0
Here a1 = 1, b1 = –3, c1 = –2
a2 = –2, b2 = 6, c2 = –5
As we know that two lines cross each other when but here
Hence the paths do not cross each other.
Question:6
Answer:
Solution:
Let the equations be
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Put x = –1 and y = 3 in given equations.
–a1 + 3b1 + c1 = 0 … (1)
–a2 + 3b2 + c2 = 0 … (2)
Since for different values of a1, b2, c1 and a2, b2, c2 we have different solutions.
Hence infinite many pairs having x = –1, y = 3 as solution.
Question:7
If 2x + y = 23 and 4x – y = 19, find the values of 5y – 2x and
Answer:
Solution:
Given equations are
2x + y = 23 … (1)
4x – y = 19 … (2)
Using elimination method in equation (1) and (2) we get
2x + y = 23
4x – y = 19
6x = 42
x =
Put x = 7 in equation (1) we get
2(7) + y = 23
y = 23 – 14
y = 9
Now 5y – 2x = 5(9) – 2 (7)
= 45 – 14
= 31
Question:8
Find the values of x and y in the following rectangle [see Fig. ].
Answer:
Solution:
We know that opposite sides of a rectangle are equal
3x + y = 7 … (1)
x + 3y = 13 … (2)
Find the value of x and y using elimination method in equation (1) and (2)
9x + 3y = 21 {Multiply (1) by 3}
x + 3y = 13
-___-___-_______
8x = 8
x = = 1
Put x = 1 in equation (1)
3(1) + y = 7
3 + y = 7
y = 7 – 3
y = 4
Question:9
Answer:
(i)Solution:
Given equations are x + y = 3.3 … (1)
0.6 = 2y – 3x …(2)
Using elimination method in equation (1) and (2) we get
2x + 2y = 6.6 [multiply eq. (1) by 2]
–3x + 2y = 0.6
+ – –
5x = 6
x = = 1.2
Put x = 1.2 in (1) we get
1.2 + y = 3.3
y = 3.3 – 1.2
y = 2.1
(ii)Answer: y = 6 ; y = 8
Solution:
Given equations are :
………(1)
……….(2)
Multiply eq. (1) by 12 and eq. (2) by 24 and then add them.
We get
4x + 3y + 20x -3y = 48 + 96
24x = 144
x = 6
Put x = 6 in eq. (1) we have
y = 8
(iii)Solution:
Equation are: … (1)
… (2)
Multiply equation (1) by 6 and equation (2) by 4 and then subtract equation (2) from (1)
24x + = 90
24x – 32/y = 56
– + –
68 = 34y
y =
y = 2
Put y = 2 in equation (1) we get
4x = 15 – 3
4x = 12
x = = 3
x = 3
(iv)Solution:
Let = u and = v and put in the given equations
u/2 – v = – 1 … (1)
u + = 8 … (2)
Multiply eq. (1) by 2 and subtract from eq. (2) we get
v + 4v = 10 × 2
5v = 20
v =
v = 4
v = 1/y = 4
y =
Put v = 4 in equation (1) we get
u/2= –1 + 4
u = 6
(because 1/x = u)
(v)Solution:
Equations are 43x + 67y = –24 … (1)
67x + 43y = 24 … (2)
Multiply equation (1) by 67 and equation (2) by 43 and then subtract equation (2) from (1)
2881x + 4489y = –1608
2881x + 1849y = 1032
– – –
2640y = –2640
y = –1
Put y = –1 in (1) we get
43x + 67(–1) = –24
43x = –24 + 67
43x = 43
x = 1
(vi)Solution:
Given equation are:
Using cross multiplication, we have
x = a2
y =
y = b2
(vii)Solution:
… (1)
… (2)
Inverse equation (1) & (2) then simplify them we get
… (3)
…(4)
Divide equation (4) by 2 we get
… (5)
Add (3) and (5) we get
2 + 1 = –2y
y =
Put y = in equation (4) we have
Question:10
Find the solution of the pair of equation and .
Hence, find , if y = x + 5.
Answer:
Solution:
x + 2y = 10 …(1)
3x + 4y = 360 … (2)
Multiply eq. (1) by 2 and subtract from eq. (2) we get
3x – 2x = 360 – 20.
x = 340
Put x = 340 in eq. 1 we get
340 + 2y = 10
2y = 10 – 340
y =
y = – 165
y = x + 5
Put x = 340 and y = –165 we get
–165 = (340) + 5
–170 = (340)
Question:11
Answer:
(i)Solution:
Given equations are 3x + y + 4 = 0
6x – 2y + 4 = 0
Here a1 = 3, b1 = 1, c1 = 4
a2 = 6, b2 = –2, c2 = 4
Hence the given pair of linear equations are intersecting at one point.
Hence given pair of linear equations is consistent
Let as plot the graph of given equations
In equation 3x + y + 4 = 0
x | 0 | -1 | -2 |
y | -4 | -1 | 2 |
In equation 6x – 2y + 4 = 0
x | 0 | -1 | -2 |
y | 2 | -1 | -4 |
Here two lines AB and CD intersect at only one point that is E.Hence given pair of linear equation is consistent.(ii)Solution:
Given equations are x –2y –6 = 0
3x – 6y = 0
Here a1 = 1, b1 = –2, c1 = –6
a2 = 3, b2 = –6, c2 = 0
Here
Here the given pair of linear equations is parallel. Therefore it has no solution Hence given pair of linear equations is inconsistent
(iii)Solution:
The given equations are x + y – 3 = 0
3x +3y – 9 = 0
Here a1 = 1, b1 = 1, c1 = –3
a2 = 3, b2 = 3, c2 = –9
Therefore the given pair of equations is coincident and have infinitely many solution.Hence, the given pair of linear equations is consistent Let us plot the graph of given equations
x + y = 3
x | 0 | 1 | 2 |
y | 3 | 2 | 1 |
3x + 3y = 9
x | 0 | 1 | 2 |
y | 3 | 2 | 1 |
Here lines AB and CD is coincident.Therefore the above linear equations have infinitely many solutions.
Question:12
Answer:
Solution:
Given equation are:
2x + y = 4
2x –y = 4
2x + y = 4
x | 0 | 1 | 2 |
y | 4 | 2 | 0 |
2x-y=4
x | 0 | 1 | 2 |
y | -4 | -2 | 0 |
Hence from the above graph it is clear that both the lines and y-axis makes triangle ABC.
Now, area of triangle ABC is
2(area of AOB)
2(4)
8 sq. units.
Hence, required area of triangle is 8 sq. units.
Question:13
Answer:
Solution:
Given equations are x + y = 2
2x – y = 1
x + y = 2
x | 1 | 2 | 3 |
y | 1 | 0 | -1 |
2x – y = 1
x | 1 | 2 | 3 |
y | 1 | 3 | 5 |
From the graph it is clear that two lines AB and CD intersect and point E (1, 1) Hence infinite lines can pass through the intersecting point E like 4 = x, 2x – y = 1 etc.
Question:14
If x + 1 is a factor of 2x3+ ax2+ 2bx + 1, then find the values of a and b given that 2a – 3b = 4.
Answer:
Solution:
It is given that (x + 1) is a factor of P(x) = 2x3 + ax2 + 2bx + 1 then according to remainder theorem P(–1) is equal to 0.
P(–1) = 2(–1)3 + a(–1)2 + 2b (–1) + 1
0 = –2 + a – 2b + 1
a – 2b = 1 … (1)
Given 2a – 3b = 4 … (2)
Solving eq. (1) and (2) by substitution method
2a – 4b = 2 {multiply eq. (1) by 2}
2a – 3b = 4
– + –
–b = –2
b = 2
Put b = 2 in equation (1)
a – 2(2) = 1
a – 4 = 1
a = 1 + 4
a = 5
Question:15
Answer:
Solution:
Given: x – y = …(1)
According to sum of interior angles of a triangle’ property
x + y + = 180
x + y = … (2)
Solve eq. (1) and eq. (2) using substitution method.
x – y =
x + y =
2x = 170
x =
x =
Put x = in eq. (1) we get
– y =
– = y
= y
Question:16
Answer:
Solution: Let x is the age of salim
y is the age of his daughter
According to question
(x – 2) = 3(y – 2)
x – 2 = 3y – 6
x – 3y = –4 …(1)
According to second condition
x + 6 = 2(y + 6) + 4
x + 6 = 2y + 12 + 4
x – 2y = 10 … (2)
Solve (1) and (2) by substitution method
x – 3y = – 4
x – 2y = 10
– + –
–y = – 14
y = 14
Put y = 14 in equation (1)
x – 3(14) = – 4
x – 42 = –4
x = –4 + 42
x = 38
Salim’s age = 38 years
His daughter’s age = 14 years
Question:17
Answer:
Solution:
Let x be the age of father y and z be the ages of children
According to the question:
x = 2 (y + z) … (1)
After 20 years
x + 20 = (y + 20) + (z + 20)
x + 20 = y + z + 40
x + 20 – 40 = y + z
x – 20 = y + z … (2)
Put the value of (y + z) from equation (2) in equation (1) we get
x = 2 (x – 20)
x = 2x – 40
x – 2x = – 40
–x = – 40
x = 40
Hence father’s age is 40 years
Question:18
Answer:
Solution:
Let x and y be two numbers
According to question
x : y = 5 : 6
x = … (1)
According to second condition
5x – 40 = 4y – 32
5x – 4y = –32 + 40
5x – 4y = 8 … (2)
Put value of x from equation (1) in (2) we get
25y – 24y = 48
y = 48
Put y = 48 in eq. (1)
x =
x = 40
Question:19
Answer:
Solution:
According to question:
Let x and y are the number of students in class A and B.
x – 10 = y + 10
x – y = 20 … (1)
Second condition is
(x + 20) = 2(y – 20)
x + 20 = 2y – 40
x – 2y = – 60 … (2)
Subtract (1) and (2) we get
y = 20 + 60
y = 80
Put y = 80 in (1)
x – 80 = 20
x = 20 + 80
x = 100
Hence in hall A there are 100 students and in hall B there are 80 students.
Question:20
Answer:
Solution:
Let x be the fixed charges and y be the extra charges for each day.
According to question equation for latika is
x + 4y = 22 … (1)
For Anand
x + 2y = 16 … (2)
Subtract equation (1) and (2) we get
4y – 2y = 22 – 16
2y = 6
y = = 3
Put y = 3 in equation (1)
x + 4 (3) = 22
x = 22 – 12
x = 10
The fixed charge is = 10 Rs.
Additional charges = 3 Rs.
Question:21
Answer:
Solution:
Let x be the number of correct answer and (120 – x) be the number of wrong answer .
According to question
x –
3x – 120 = 180
3x = 180 + 120
3x = 300
x = 100
Hence Jayanti answered 100 questions correctly.
Question:22
Answer:
Solution:
According to property of cyclic quadrilateral, sum of opposite angles = .
A + C =
6x + + x + y =
7x + y = –
7x + y = … (1)
Similarly B + D =
5x + 3y – =
5x + 3y = +
5x + 3y = … (2)
Solving eq. (1) and eq. (2) we get
21x + 3y = {Multiply eq. (1) by 3} … (3)
Now subtract equation (2) from (3) then we get
16x =
x =
Put x = in eq. (1) we get
7() + y =
y = –
y =
Hence A = 6x +
= 6 × +
= +
=
B = (5x)
= 5 × =
( + ) =
D = (3y – 10)°
= 3 × –
= –
=
Hence the value of four angles are:
, , and respectively.
Question:1
Answer:
Solution:
Here the equations are 2x + y = 6, 2x – y = –2
2x + y = 6
x | 0 | 3 |
y | 6 | 0 |
2x – y = –2
x | 0 | -1 |
y | 2 | 0 |
In graph we find that the lines are intersecting at (1, 4) Hence x = 1, y = 4 is the solution of the equations.The two triangles are ABC and DBE
Area of ABE = × Base × perpendicular
Area of ADC = × Base × perpendicular
ABE area : ADC Area
4 : 1
Question:2
Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8
Answer:
Solution:
y = x
y – x = 0
x | 0 | 1 |
y | 0 | 1 |
3y – x = 0
x | 0 | 3 |
y | 0 | 1 |
x + y = 8
x | 0 | 8 |
y | 8 | 0 |
The require triangle is ABC with vertices (0, 0), (4, 4), (2, 6)
Question:3
Answer:
Solution:
The given equations are x = 3, x = 5, 2x – y – 4 = 0
2x – y = 4
x | 2 | 0 |
y | 0 | -4 |
The quadrilateral ABCD is a trapezium.
Area of ABCD (trapezium) = × (a + b)h
a = AD, b = BC, h = AB
= × (AD + BC) × AB
= × (2 + 6) × 2
Area = 8 sq. units
Question:4
Answer:
Solution:
Let the cost of a pen is x and the cost of a pencil box is y.
As per the question
4x + 4y = 100 … (1)
3x = y + 15 … (2)
Divide equation (1) by 4.
x + y = 25 ………..(3)
By adding equation (1) and (3) we get
4x = 40
x = 10
Put x = 10 in eq. (1)
40 + 4y = 100
4y = 100 – 40
y = 60/4 = 15
x = 10, y = 15
Cost of pen = Rs. 10
Cost of pencil box = Rs.15
Question:5
Answer:
Solution:
Determine, algebraically, the vertices of the triangle formed by the lines
3x – y = 3 … (1)
2x – 3y = 2 … (2)
x + 2y = 8 ... (3)
For vertices of A, solve equation (1) and (3)
3x – y = 3
x + 2y = 8 × 3
3x + 6y = 24
3x + y = –3
7y = 21
y = 3
Put y = 3 in eq. (1)
3x – 3 = 3
3x = 6
x = 2
point A is (2, 3)
For vertices of B solve equation (1) and (2)
3x – y = 3 × 3
2x – 3y = 2
9x – 3y = 9
– + –
–7x = –7
x = 1
Put x = 1 in equation (1)
3(1) – y = 3
3 – 3 = y
y = 0
point B is (1, 0)
For vertices of C solve (2) and (3)
2x – 3y = 2
x + 2y = 8 × 2
2x + 4y = 16
2x – 3y = 2
– + –
7y = 14
y = 2
Put y = 2 in eq. (2)
2x – 3(2) = 2
2x = 2 + 6
x = = 4
Point C (4, 2)
Hence vertices of triangle ABC are A (2, 3), B (1, 0) and C (4, 2)
Question:6
Answer:
Solution:
We know that
Speed =
Time = Distance / Speed
Let the speed of rickshaw is x and of bus is y.
So, according to question
Solve eq. (1) and (2)
– – –
21y = 840
Put y = 40 in eq. (1)
40 = 4x
x = 10
Speed of rickshaw = 10 km/h
Speed of bus = 40 km/h
Question:7
Answer:
Solution:
According to question,
Person’s speed with upstream = (5 – x) km/h
Person’s speed with downstream = (5 + x) km/h
Time with upstream
Time with downstream
(5 + x)40 = 120 (5 – x)
200 + 40x = 600 - 120 x
160x = 400
x = 2.5
Speed of stream = 2.5 km/h
Question:8
Answer:
Solution:
Let speed of boat in still water = x km/h
Let speed of stream = y km/h
Speed of boat in upstream = (x - y) km/h
Speed of boat in downstream = (x + y) km/h
We know that time =
So, … (1)
… (2)
Put in (1), (2)
30u + 28v = 7 … (3)
21u + 21v = 5 … (4)
Solve equation (3), (4)
Multiply equation (3) by 21 and (4) by 30
630u + 588 v = 147
630u + 630v = 150
_-______-____-________
+42v = +3
Put in eq. (4)
21u =
We put
So,
|
x-y=6..(5) x+y=14...(6)
By adding equation (5) and (6)
2x = 20
x = 10
Put x = 10 in eq. (6)
10 + y = 14
y = 4
Hence speed of the boat in still water = 10 km/hr.
Speed of the stream = 4 km/hr.
Question:9
Answer:
Solution:
Let the first digit = x
Let the ten’s digit = y
Therefore the number = 10y + x
As per the question
8(x + y) –5 = 10y + x
8x + 8y – 10y – x = 5
7x – 2y = 5 … (1)
16(y – x) + 3 = 10y + x
16y – 16x – 10y – x = – 3
–17x + 6y = –3
17x – 6y = 3 … (2)
Multiply equation (1) by 3
21x – 6y = 15
17x – 6y = 3
– + –
4x = 12
x = 3
Put x = 3 in eq. (1)
7(3) –2y = 5
–2y = 5 –21
y = = 8
Hence the number is = 10y + x = 10(8) + 3 = 83
Question:10
Answer:
Solution:
Let the full ticket cost = y Rs.
Let reservation charge = x Rs.
According to questions
x + y = 2530 … (1)
(x + y) + (x + y/2) = 3810 …(2)
Here x + y is for full ticket and x + y/2 is for half.
Solve equation (1) and (2)
x + y + x + = 3810
2x + y + = 3810
4x + 2y + y = 3810 × 2
4x + 3y = 7620 … (3)
Multiply equation (1) by 3
3x + 3y = 7590
4x + 3y = 7620
– – –
–x = – 30
Put x = 30 in eq. (1)
30 + y = 2530
y = 2500
Hence Reservation charge = 30Rs.
Full First class charge = 2500 Rs.
Question:11
Answer:
Solution:
Let the cost price of saree = Rs. x
Let the cost price of sweater = Rs. y
Saree with 8% profit =
Sweater with 10% discount = =
Saree with 10% profit = x +
Sweater with 8% discount = y –
According to question:
108x + 90y = 100800
6x + 5y = 5600 … (1) (Divide by 18)
110x + 92y = 102800
55x + 56y = 51400 … (2) (Divide by 2)
Multiply equation (1) by 46 and eq. (2) by 5
276 x + 230 y = 257600
275 x + 230 y = 257000
– – –
x = 600
Put x = 600 in eq. (1)
6(600) + 5y = 5600
5y = 5600 –3600
5y = 2000
y = 400
Price of saree = Rs. 600
Price of sweater = Rs. 400
Question:12
Answer:
Solution:
Let money invested in A and B are x, y respectively.
So, according to question.
0.08x + 0.09y = 1860 … (1)
0.09x + 0.08y = 1880 … (2)
Multiply equation (1) by 9 and (2) by 8
0.72x + 0.81y = 16740
0.72x + 0.64y = 15040
– – –
0.17y = 1700
y =
Put y = 10000 in eq. (1)
0.08x + 900 = 1860
0.08x = 960
x = = 12000
x = 12000
He invested Rs. 12000 in A and Rs. 10000 in B.
Question:13
Answer:
Solution:
Let bananas in lot A = x
Bananas in lot B = y
According to question:
2x + 3y = 400 × 3
2x + 3y = 1200 … (1)
x(1) + (4) = 460
5x + 4y = 460 × 5
5x + 4y = 2300 … (2)
Multiply equation (1) by 5 and eq. (2) by 2
10x + 15y = 6000
10x + 8y = 4600
– – –
7y = 1400
Put y = 200 in eq. (1)
2x + 3(200) = 1200
2x + 600 = 1200
2x = 600
x = 300
Total number of bananas = x + y = 300 + 200 = 500
These Class 10 Maths NCERT exemplar chapter 3 solutions provide a detailed knowledge of linear equations in two variables. The knowledge of this chapter will be useful whether a student will pursue mathematics in higher classes or not. Even for exams like NEET the basic method of solving linear equations is a prerequisite. A good volume of practice problems available in the exemplar provides a student with ample opportunity to clear the concept of Pair of Linear Equations in Two Variables.
Class 10 Maths NCERT exemplar solutions chapter 3 Pair of Linear Equations in Two Variables is a good training practice for the students to solve problems of RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths, NCERT Class 10 Maths, et cetera.
NCERT exemplar Class 10 Maths solutions chapter 3 pdf download is the feature that makes these solutions accessible to the students in an offline scenario and can keep the doubts at bay while practicing NCERT exemplar Class 10 Maths chapter 3.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
No, a pair of linear equations may have a unique solution or no solution or infinite possible solutions.
For any linear equation in two variables if we try to draw the graph, it will be straight line
If two straight lines are parallel to each other it means they don’t have any common point. Therefore, equations representing these two straight lines will not have any solution
As these two straight lines are passing through origin that means (zero, zero) will satisfy both the equations, hence, the common solution will be X =0 and Y =0.
The chapter of Pair of Linear Equations in Two Variables is essential for the board exams as it includes around 10-12% weightage of the whole paper.
Generally, a total of 3-4 questions of different types appear in the board examinations from this chapter. NCERT exemplar Class 10 Maths solutions chapter 3 is adequate to correctly answer these questions from Pair of Linear equations in two variables.
Dear aspirant !
Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .
Hope you get it !
Thanking you
Hello aspirant,
The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.
To get the complete datesheet, you can visit our website by clicking on the link given below.
https://school.careers360.com/boards/cbse/cbse-date-sheet
Thank you
Hope this information helps you.
Hello aspirant,
The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.
You don't need to worry. The class 7th paper will be simple and made by your own school teachers.
Thank you
Hope it helps you.
The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.
That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.
However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.
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For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.
Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.
Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.
Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.
Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.
A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.
The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.
A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.
Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.
An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.
They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.
In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.
In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.
Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.
For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.
Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.
Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.
Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.
Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.
A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.
Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.
A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.
A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.
A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.
The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.
An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.
An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.
Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack
An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.
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