NCERT Solutions are very beneficial for students while preparing for board examinations. NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Variables is an important topic for the board examinations and should be dealt with in complete detail. The solutions to the problems provided in the chapter are given in the NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.4. The NCERT Solutions Class 10 Maths contain stepwise solutions to all the Maths problems provided in the textbook or otherwise.
The students should practise the exercises repeatedly to score well in the board examinations. NCERT Solutions contain detailed study material, including all the important topics. The subject experts have come up with accurate NCERT solutions after a lot of research and brainstorming.
Download the PDF for NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.4
Access other exercise solutions of Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables
Exercise 3.1 Solutions – 3 Questions
Exercise 3.2 Solutions – 7 Questions
Exercise 3.3 Solutions – 3 Questions
Exercise 3.5 Solutions – 4 Questions
Exercise 3.6 Solutions – 2 Questions
Exercise 3.7 Solutions – 8 Questions
Access answers of Maths NCERT Class 10 Chapter 3 – Pair of Linear Equations in Two Variables 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
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2+ 2y/3 = -1 and x-y/3 = 3
Solutions:
(i) x + y = 5 and 2x – 3y = 4
By the method of elimination.
x + y = 5 ……………………………….. (i)
2x – 3y = 4 ……………………………..(ii)
When the equation (i) is multiplied by 2, we get
2x + 2y = 10 ……………………………(iii)
When equation (ii) is subtracted from (iii), we get
5y = 6
y = 6/5 ………………………………………(iv)
Substituting the value of y in eq. (i) we get
x=5−6/5 = 19/5
∴ x = 19/5 , y = 6/5
By the method of substitution.
From equation (i), we get
x = 5 – y………………………………….. (v)
When the value is put in equation (ii), we get
2(5 – y) – 3y = 4
-5y = -6
y =Â 6/5
When the values are substituted in equation (v), we get
x =5− 6/5 = 19/5
∴ x = 19/5 ,y = 6/5
Â
(ii) 3x + 4y = 10 and 2x – 2y = 2
By the method of elimination.
3x + 4y = 10……………………….(i)
2x – 2y = 2 ………………………. (ii)
When the equation (i) and (ii) are multiplied by 2, we get
4x – 4y = 4 ………………………..(iii)
When equations (i) and (iii) are added, we get
7x = 14
x = 2 ……………………………….(iv)
Substituting equation (iv) in (i), we get
6 + 4y = 10
4y = 4
y = 1
Hence, x = 2 and y = 1
By the method of substitution
From equation (ii), we get
x = 1 + y……………………………… (v)
Substituting equation (v) in equation (i), we get
3(1 + y) + 4y = 10
7y = 7
y = 1
When y = 1 is substituted in equation (v), we get
A = 1 + 1 = 2
Therefore, A = 2 and B = 1
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By the method of elimination.
3x – 5y – 4 = 0 ………………………………… (i)
9x = 2y + 7
9x – 2y – 7 = 0 …………………………………(ii)
When the equation (i) is multiplied by 3, we get,
9x – 15y – 12 = 0 ………………………………(iii)
When equation (iii) is subtracted from equation (ii), we get
13y = -5
y = -5/13 ………………………………………….(iv)
When equation (iv) is substituted in equation (i), we get
3x +25/13 −4=0
3x = 27/13
x =9/13
∴ x = 9/13 and y = -5/13Â
By the method of substitution.
From equation (i), we get
x = (5y+4)/3 …………………………………………… (v)
Putting the value (v) in equation (ii), we get
9(5y+4)/3 −2y −7=0
13y = -5
y = -5/13
Substituting this value in equation (v), we get
x = (5(-5/13)+4)/3
x = 9/13
∴ x = 9/13, y = -5/13
(iv) x/2 + 2y/3 = -1 and x-y/3 = 3
By the method of elimination.
3x + 4y = -6 …………………………. (i)
x-y/3 = 3
3x – y = 9 ……………………………. (ii)
When equation (ii) is subtracted from equation (i), we get
5y = -15
y = -3 ………………………………….(iii)
When equation (iii) is substituted in (i), we get
3x – 12 = -6
3x = 6
x = 2
Hence, x = 2 , y = -3
By the method of substitution.
From equation (ii), we get
x = (y+9)/3…………………………………(v)
Putting the value obtained from equation (v) in equation (i), we get
3(y+9)/3 +4y =−6
5y = -15
y = -3
When y = -3 is substituted in equation (v), we get
x = (-3+9)/3 = 2
Therefore, x = 2 and y = -3
2. 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 if we only add 1 to the denominator. What is the fraction?
Solution:
Let the fraction be a/b
According to the given information,
(a+1)/(b-1) = 1
=> a – b = -2 ………………………………..(i)
a/(b+1) = 1/2
=> 2a-b = 1…………………………………(ii)
When equation (i) is subtracted from equation (ii), we get
a = 3 …………………………………………………..(iii)
When a = 3 is substituted in equation (i), we get
3 – b = -2
-b = -5
b = 5
Hence, the fraction is 3/5.
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution:
Let us assume the present age of Nuri is x.
And the present age of Sonu is y.
According to the given condition, we can write as
x – 5 = 3(y – 5)
x – 3y = -10…………………………………..(1)
Now,
x + 10 = 2(y +10)
x – 2y = 10…………………………………….(2)
Subtract eq. 1 from 2 to get
y = 20 ………………………………………….(3)
Substituting the value of y in eq.1, we get
x – 3.20 = -10
x – 60 = -10
x = 50
Therefore,
Age of Nuri is 50 years
Age of Sonu is 20 years
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution:
Let the unit digit and tens digit of a number be x and y, respectively.
Then, number (n) = 10B + A
N after reversing the order of the digits = 10A + B
According to the given information, A + B = 9…………………….(i)
9(10B + A) = 2(10A + B)
88 B – 11 A = 0
-A + 8B = 0 ………………………………………………………….. (ii)
Adding the equations (i) and (ii), we get
9B = 9
B = 1……………………………………………………………………….(3)
Substituting this value of B in equation (i), we get A= 8
Hence, the number (N) is 10B + A = 10 x 1 +8 = 18
(iv) Meena went to a bank to withdraw Rs.2,000. She asked the cashier to give her Rs.50, and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.
Solution:
Let the number of Rs.50 notes be A and the number of Rs.100 notes be B.
According to the given information,
A + B = 25 ……………………………………………………………………….. (i)
50A + 100B = 2000 ………………………………………………………………(ii)
When equation (i) is multiplied by (ii), we get
50A + 50B = 1250 …………………………………………………………………..(iii)
Subtracting equation (iii) from equation (ii), we get
50B = 750
B = 15
Substituting in equation (i), we get
A = 10
Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.
According to the information given,
A + 4B = 27 …………………………………….…………………………. (i)
A + 2B = 21 ……………………………………………………………….. (ii)
When equation (ii) is subtracted from equation (i), we get
2B = 6
B = 3 …………………………………………………………………………(iii)
Substituting B = 3 in equation (i), we get
A + 12 = 27
A = 15
Hence, the fixed charge is Rs.15.
And the charge per day is Rs.3.
This is the fourth exercise of the NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables. It contains 2 questions, the solutions to which are provided in the NCERT Solutions Class 10 Maths. The students can refer to these solutions in case of any doubts.
Key Features of NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.4
- The answers to the questions are accurate.
- The solutions to all the questions in the textbook are provided here.
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- NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.4 can help the students to score well in the Maths examinations.
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