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Learn Exercise 3.3 with Free Lessons & Tips

Question 1:

A fraction becomes $\frac{9}{11}$ , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$ . Find the fraction

Solution 1:

Let the fraction be
According to the question, $\frac{x+2}{y+2}=\frac{9}{11}$

$so, 11x-9y=-4$ ......................(i)

$\frac{x+3}{y+3}=\frac{5}{6}$

$6x-5y=-3$ ....................(ii)

From the eq (i), we get $x= \frac{-4+9y}{11}...................(iii)$

Substituting the value of x in (ii) y=9

therefore $x=\frac{-4+81}{11}=7$

the fraction obtained is therefore, $\frac{7}{9}$

Comments

Tasneem

Question 2:

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution 2:

Let the age of Jacob be x and the age of his son be y.

According to the question, $(x+5)=3(y+5)$

$x-3y=10..............(i)$

$(x-5)=7(y-5)$

$x-7y= -30$ ................(ii)

from (i), we get, $x=3y+10...............(iii)$

substituting this in (ii), y=10

therefore $x=3\times 10+10= 40$

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

Comments

Tasneem

Question 3:

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

(i) x + y = 14, x – y = 4 (ii) s – t = 3, $\frac{s}{3}+\frac{t}{2}=6$ (iii) 3x – y = 3, 9x – 3y = 9 (iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3 (v) $\sqrt{2}x+\sqrt{3}y=0, \sqrt{3}x-\sqrt{8}y=0$ (vi) $\frac{3x}{2}-\frac{5y}{3}=-2$ , $\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Solution 3:

(i) x + y = 14      ... (i)
x - y = 4        ... (ii)
From (i), we obtain:
x = 14 - y      ... (iii)
Substituting this value in equation (ii), we obtain:

$(14-y)-y=4$

14-2y=4

y=5

Substituting the value of y in equation (iii), we obtain: x=9

$(ii)s-t=3.....(i)$

$\frac{s}{3}+\frac{t}{c}=6$ $.....(ii)$

From (i), we obtain:
s = t + 3 ...(iii)

Substituting this value in equation (ii), we obtain:

$\frac{t+3}{3}+\frac{t}{2}=6$ , t=6

Substituting the value of t in equation (iii), we obtain:
s = 9 and t=6

(iii) 3x - y = 3        ... (i)
9x - 3y = 9        ... (ii)
From (i), we obtain
y = 3x - 3        ... (iii)
Substituting this value in equation (ii), we obtain:

$9x-3(3x-3)=9$

9=9

Thus, the given pair of equations has infinitely many solutions and the relation between these variables can be given by
y = 3x - 3
So, one of the possible solutions can x = 3, y = 6.

$(iv)0.2x+0.3y=1.3.......(i)$

$0.4x+0.5y=2.3............(ii)$

$x= \frac{1.3-0.3y}{0.2}.....(iii)$

substituting the value in equation (ii), we get y=3

Putting this in equation (i), we get x=2

$(v)\sqrt{2}x+\sqrt{3}y=0..........(i)$

$\sqrt{3}x-\sqrt{8}y=0$ $..........(ii)$

From equation (i) we obtain, $x= \frac{-\sqrt{3}y}{\sqrt{2}}$ $...........(iii)$

Substituting this value in equation (ii), we get, y=0

Substituting the value of y in equation (iii), we obtain: x=0

$(vi)$ $\frac{3}{2}x-\frac{5}{3}y=-2$ .............. (i)

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}................(ii)$

from equation (i), we obtain, $x= \frac{-12+10y}{9}......... (iii)$

Substituting this value in equation (ii) we get, x=2 and y=3

Comments

Tasneem

Question 4:

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

Solution 4:

$2x+3y=11.......(i)$

$2x-4y=-24....... (ii)$

From equaton (i), we have $y=\frac{11-2x}{3}$ ...........(iii)

Substituting the value of y in equation (ii)

we get x= -2

substituting value of x in equation (iii), we get y=5

Thus the solution is x=-2 and y=5

Now we have, $y=mx+3$

$5=m(-2)+3, m=-1$

Comments

Tasneem

Question 5:

Form the pair of linear equations for the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

Solution 5:

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

According to the question: $y=3x.......(i)$

$y=x=26..................(ii)$

Substituting the value of y in the (i), we get, $x= 13..............(iii)$

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,

From (1), we obtain
x = 180º - y (3)
Substituting this in equation (2), we obtain $180^{\circ}-y-y=18^{\circ}$

$162^{\circ}=2y$ , $y=81^{\circ}$ $..............(iv)$

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 question,

$7x+6y=3800..............(i)$

$3x+5y=1750..............(ii)$

From (i), we get, $y=\frac{3800-7x}{6}.................(iii)$

Substituting this in (ii), we get $3x+5\left ( \frac{3800-7x}{6} \right )=1750$

$x=500...............(iv)$

Substituting this (iii), $y= \frac{3800-7\times 500}{6}= \frac{300}{6}=50$

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

Comments

Tasneem

Question 6:

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Solution 6:

Let the fixed rate be Rs. x and per km change be Rs. y

according to the question, $x+10y=105$

$x+15y=155...............(ii)$

from (i), we get $x=105-10y.................(iii)$

substituting this in equation (ii), y=10

placing the value of y in equation (iii),we get x=5

Hence the fixed charges are Rs. 5 and the charges per km are Rs. 10

Therefore, charge for 25 km is x+25y = Rs. 255

Comments

Tasneem

All Exercises - Chapter 3 - Pair of Linear Equations in Two Variable

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Exercise 3.5

Exercise 3.6

Exercise 3.7

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Learn Exercise 3.3 with Free Lessons & Tips

All Exercises - Chapter 3 - Pair of Linear Equations in Two Variable

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