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Exercise 3.2 (Revised) - Chapter 3 Pair Of Linear Equations In Two Variables class 10 ncert solutions Maths - SaraNextGen [2024]


Exercise 3.2 (Revised) : Chapter 3 - Pair Of Linear Equations In Two Variables - Ncert Solutions class 10 - Maths

Question 1:

Solve the following pair of linear equations by the substitution method.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m5abd3f07.gif

Answer:

(i) x + y = 14 (1)

x − y = 4 (2)

From (1), we obtain

x = 14 − y (3)

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_397c34dd.gif

Substituting this in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_1b6953c1.gif

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_579081c6.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_31acc4f4.gif

From (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m2e4449a4.gif

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m50b1fbee.gif

Substituting in equation (3), we obtain

s = 9

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_4dd19828.gif s = 9, t = 6

(iii)3x − y = 3 (1)

9x − 3y = 9 (2)

From (1), we obtain

y = 3x − 3 (3)

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_mb756bcc.gif

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by

y = 3x − 3

Therefore, one of its possible solutions is x = 1, y = 0.

(iv) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m548baf5a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m48b6ba7e.gif

From equation (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_3696c842.gif

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_149779bc.gif

Substituting this value in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_36ac7c0.gif

(v) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m58bbbcc9.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m2f6ea114.gif

From equation (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_74269a3a.gif

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_5aaf976.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_m2b78ad47.gif

Substituting this value in equation (3), we obtain

x = 0

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_4dd19828.gif x = 0, y = 0

(vi) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_7d46119a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_50583027.gif

From equation (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_595abfcc.gif

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_4fc1d39a.gif

Substituting this value in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2096/Chapter%203_html_6e379720.gif

Hence, x = 2, y = 3

 

Question 2:

Solve 2x + 3y = 11 and 2− 4y = − 24 and hence find the value of ‘m’ for which y = mx + 3.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2099/Chapter%203_html_mb59bb18.gif

From equation (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2099/Chapter%203_html_m6105a407.gif

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2099/Chapter%203_html_562d4bd9.gif

Putting this value in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2099/Chapter%203_html_16cb0eca.gif

Hence, x = −2, y = 5

Also,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2099/Chapter%203_html_m2f277835.gif

 

Question 3:

Answer:

(i) Let the first number be x and the other number be y such that y > x.

According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m3530d169.gif

On substituting the value of y from equation (1) into equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_777b9319.gif

Substituting this in equation (1), we obtain

y = 39

Hence, the numbers are 13 and 39.

(ii) Let the larger angle be x and smaller angle be y.

We know that the sum of the measures of angles of a supplementary pair is always 180º.

According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m6f2f566b.gif

From (1), we obtain

x = 180º − y (3)

Substituting this in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_9ba7e91.gif

Putting this in equation (3), we obtain

x = 180º − 81º

= 99º

Hence, the angles are 99º and 81º.

(iii) Let the cost of a bat and a ball be x and y respectively.

According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m27073f02.gif

From (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m912cd04.gif

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_3b4b5c17.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_61b49d91.gif

Substituting this in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_3b463434.gif

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

(iv) Let the fixed charge be Rs and per km charge be Rs y.

According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_5632f728.gif

From (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m6f6255fe.gif

Substituting this in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m4d22fd06.gif

Putting this in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_4be108bc.gif

Hence, fixed charge = Rs 5

And per km charge = Rs 10

Charge for 25 km = x + 25y

= 5 + 250 = Rs 255

(v) Let the fraction be https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_1141e1f1.gif .

According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m3297eb45.gif

From equation (1), we obtain https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_6b918ebf.gif

Substituting this in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_5d2e8d7f.gif

Substituting this in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_13dc9195.gif

Hence, the fraction is https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m1e454493.gif .

(vi) Let the age of Jacob be and the age of his son be y.

According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_d18f958.gif

From (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_m736d40a6.gif

Substituting this value in equation (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_3dec185f.gif

Substituting this value in equation (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/130/2113/Chapter%203_html_f2c964b.gif

Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.

Also Read : Exercise-3.3-(Revised)-Chapter-3-Pair-Of-Linear-Equations-In-Two-Variables-class-10-ncert-solutions-Maths

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