Sunday, May 31, 2020

NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.4 mathematics

EXERCISE 3.4




Q 1(i), Ex 3.4 - Linear Equations - Chapter 3 - Maths Class 10th - NCERT

Page No 56:

Question 1:

Solve the following pair of linear equations by the elimination method and the substitution method:
 

Answer:

(i) By elimination method


Multiplying equation (1) by 2, we obtain

Subtracting equation (2) from equation (3), we obtain

Substituting the value in equation (1), we obtain

By substitution method
From equation (1), we obtain
 (5)
Putting this value in equation (2), we obtain

−5y = −6

Substituting the value in equation (5), we obtain

(ii) By elimination method


Multiplying equation (2) by 2, we obtain

Adding equation (1) and (3), we obtain

Substituting in equation (1), we obtain

Hence, x = 2, y = 1
By substitution method
From equation (2), we obtain
 (5)
Putting this value in equation (1), we obtain

7y = 7

Substituting the value in equation (5), we obtain

(iii) By elimination method


Multiplying equation (1) by 3, we obtain

Subtracting equation (3) from equation (2), we obtain

Substituting in equation (1), we obtain

∴ 
By substitution method
From equation (1), we obtain
(5)
Putting this value in equation (2), we obtain

Substituting the value in equation (5), we obtain


∴ 
(iv)By elimination method


Subtracting equation (2) from equation (1), we obtain

Substituting this value in equation (1), we obtain

Hence, x = 2, y = −3
By substitution method
From equation (2), we obtain
(5)
Putting this value in equation (1), we obtain

5y = −15

Substituting the value in equation (5), we obtain

∴ x = 2, y = −3


Q 2(i), Ex 3.4 - Linear Equations - Chapter 3 - Maths Class 10th - NCERT

Q 2(ii), Ex 3.4 - Linear Equations - Chapter 3 - Maths Class 10th - NCERT

Q 2(iii), Ex 3.4 - Linear Equations - Chapter 3 - Maths Class 10th - NCERT

Q 2(iv), Ex 3.4 - Linear Equations - Chapter 3 - Maths Class 10th - NCERT

Q 2(v), Ex 3.4 - Linear Equations - Chapter 3 - Maths Class 10th - NCERT


Page No 57:

Question 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?
(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 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.

Answer:

(i)Let the fraction be .
According to the given information,

Subtracting equation (1) from equation (2), we obtain
= 3 (3)
Substituting this value in equation (1), we obtain

Hence, the fraction is .
(ii)Let present age of Nuri = x
and present age of Sonu = y
According to the given information,

Subtracting equation (1) from equation (2), we obtain
y = 20 (3)
Substituting it in equation (1), we obtain

Hence, age of Nuri = 50 years
And, age of Sonu = 20 years
(iii)Let the unit digit and tens digits of the number be and y respectively. Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the given information,
x + = 9 (1)
9(10y + x) = 2(10x + y)
88y − 11= 0
− x + 8=0 (2)
Adding equation (1) and (2), we obtain
9y = 9
y = 1 (3)
Substituting the value in equation (1), we obtain
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18
(iv)Let the number of Rs 50 notes and Rs 100 notes be x and respectively.
According to the given information,

Multiplying equation (1) by 50, we obtain

Subtracting equation (3) from equation (2), we obtain

Substituting in equation (1), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.
(v)Let the fixed charge for first three days and each day charge thereafter be Rs and Rs respectively.
According to the given information,

Subtracting equation (2) from equation (1), we obtain

Substituting in equation (1), we obtain

Hence, fixed charge = Rs 15
And Charge per day = Rs 3

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