Mathematics Part i Solutions Solutions for Class 9 Maths Chapter 5 - Linear Equations In Two Variables

Answer:

Linear equations in two variables x and y will be
$$\begin{gathered} x+y=1 \\ 2x+3y=8 \\ x-y=6 \\ 5x-2y=3 \\ x+2y=3 \end{gathered}$$

Answer:

The given equation is x + y = 7.
The possible solution are
 

Answer:

Answer:

Disclaimer: There is an error in the given question. Instead of "Number of 5 rupee notes are less by 10 than number of 10 rupee notes" 
the sentence should have been "Number of 5 rupee notes are greater by 10 than number of 10 rupee notes". 

Let the number of Rs 5 notes be x and the number of Rs 10 notes be y. 
$$\begin{gathered} 5x+10y=350 \\ \Rightarrow x+2y=70 .....\left(\mathrm{I}\right) \end{gathered}$$
Number of 5 rupee notes are greater by 10 than number of 10 rupee notes
So, $x-y=10$                      
$\Rightarrow x=10+y$                       .....(II)
From (I) and (II) we have
$$\begin{gathered} 10+y+2y=70 \\ \Rightarrow 10+3y=70 \\ \Rightarrow 3y=60 \\ \Rightarrow y=20 \end{gathered}$$
Thus, x = 10 + 20 = 30
Number of Rs 5 notes = 30, number of Rs 10 notes = 20
 

Answer:

Disclaimer: There is an error in the given question. With the given statements, the two linear equations formed will be the same. 

Answer:

Let the age of Priyanka be x years and that of Deepika be y years. 
The sum of ages of Priyanka and Deepika is 34 years.
$x+y=34$                  .....(I)
Priyanka is elder to Deepika by 6 years.
$x-y=6$                  .....(II)
Adding (I) and (II) we have
$$\begin{gathered} 2x=40 \\ \Rightarrow x=20 \end{gathered}$$
Putting the value of x in (I) we have
y = 14
Thus, Priyanka's age = 20 years and Deepika's age = 14 years.
 

Answer:

Let the number of lions be x and the number of peacocks be y. 
The total number of lions and peacocks in a certain zoo is 50. 
So, $x+y=50$                     .....(I)
The total number of their legs is140. 
Lion has 4 legs while peacock has 2.
$$\begin{gathered} 4x+2y=140 \\ \Rightarrow 2x+y=70 .....\left(\mathrm{II}\right) \end{gathered}$$                      
Subtracting (II) from (I) we have
$$\begin{gathered} x=20 \\ y=30 \end{gathered}$$
Thus, the number of lions = 20 and the number of peacocks = 30. 


 

Answer:

Let the fixed monthly income be Rs x. 
Annual increament be Rs y. 
After 4 years, his monthly salary was Rs. 4500 
Monthly salary + annual increament of 4 years = 4500
$x+4y=4500$                        .....(I)
After 10 years his monthly salary became 5400 rupees
Monthly salary + annual increament of 10 years = 5400
$x+10y=5400$                        .....(II)
(II) $-$ (I)
6y = 900
y = 150
x = 4500 $-$ 150 $\times$ 4 = 3900
Thus, the monthly salary = Rs 3900
Annual increament = Rs 150  
 

Answer:

Let the price of 1 chair be Rs x and the price of 1 table be Rs y. 
The price of 3 chairs and 2 tables is 4500 rupees
$3x+2y=4500 .....\left(\mathrm{I}\right)$
Price of 5 chairs and 3 tables is 7000 rupees
$5x+3y=7000 .....\left(\mathrm{II}\right)$
Multiply (I) with 3 and (II) with 2
$$\begin{gathered} 9x+6y=13500 .....\left(\mathrm{III}\right) \\ 10x+6y=14000 .....\left(\mathrm{IV}\right) \\ \left(\mathrm{IV}\right)-\left(\mathrm{III}\right) \\ x=500 \\ \mathrm{Putting} \mathrm{this} \mathrm{value} \mathrm{of} x \mathrm{in} \left(\mathrm{I}\right) \\ 3\times 500+2y=4500 \\ \Rightarrow 1500+2y=4500 \\ \Rightarrow 2y=4500-1500=3000 \\ \Rightarrow y=1500 \end{gathered}$$
Price of 2 chairs and 2 tables will be
$$\begin{gathered} =2x+2y \\ =2\left(x+y\right) \\ =2\left(500+1500\right) \\ =2\times 2000 \\ =4000 \end{gathered}$$
 Thus, the price of 2 chairs and 2 tables will be Rs 4000

Answer:

Let the number at the ten's place be x and that at the one's place be y. 
Number obtained = 10x + y
The sum of the digits in a two-digits number is 9.
$x+y=9$                      .....(I)
The number obtained by interchanging the digits exceeds the original number by 27.
$$\begin{gathered} 10y+x-\left(10x + y\right)=27 \\ \Rightarrow 10y+x-10x-y=27 \\ \Rightarrow 9y-9x=27 \\ \Rightarrow y-x=3 .....\left(\mathrm{II}\right) \end{gathered}$$
Adding (I) and (II)
$$\begin{gathered} 2y=12 \\ \Rightarrow y=6 \end{gathered}$$
x = 3
Thus, the number is 36.  

Answer:

$\angle$A = $\angle$B + $\angle$C               .....(I)
$$\begin{gathered} \frac{\angle B}{\angle C}=\frac{4}{5} \\ \Rightarrow 5\angle B=4\angle C .....\left(\mathrm{II}\right) \end{gathered}$$
We know by angle sum property, 
$\angle$A + $\angle$B + $\angle$C = 180$°$
From (I) 
$\angle$B + $\angle$C +$\angle$B + $\angle$C = 180$°$
$$\begin{gathered} \Rightarrow 2\left(\angle \mathrm{B}+\angle \mathrm{C}\right)=180° \\ \Rightarrow 2\left(\angle \mathrm{B}+\frac{5\angle \mathrm{B}}{4}\right)=180° \\ \Rightarrow \angle \mathrm{B}+\frac{5\angle \mathrm{B}}{4}=90° \\ \Rightarrow 4\angle \mathrm{B}+5\angle \mathrm{B}=360° \\ \Rightarrow 9\angle \mathrm{B}=360° \\ \Rightarrow \angle \mathrm{B}=40° \end{gathered}$$
$$\begin{gathered} 5\times 40°=4\angle \mathrm{C} \\ \Rightarrow \angle \mathrm{C}=50° \end{gathered}$$
And $\angle$A = $\angle$B + $\angle$C = $40°+50°=90°$.
 

Answer:

Let the smaller part be x cm and larger part be y cm. 
$x+y=560$                    .....(I)
Twice the length of the smaller part is equal to $\frac{1}{3}$ of the larger part.
$$\begin{gathered} x=\frac{1}{3}y \\ \Rightarrow 3x=y \end{gathered}$$
Putting the value of y in (I) we have
$$\begin{gathered} x+3x=560 \\ \Rightarrow 4x=560 \\ \Rightarrow x=140 \end{gathered}$$
And thus, $y=3\times 140=420$
Thus, the larger part = 420 cm and smaller part is 140 cm.

 

Answer:

Let the number of correct answers be x and the number of incorrect answers be y. 
Total answers = Number of correct + number of incorrect = 60
$x+y=60$                      .....(I)
Total marks obtained = 90
$2x-y=90$                     .....(II)
Adding (I) and (II) we have
$$\begin{gathered} 3x=150 \\ \Rightarrow x=50 \\ y=10 \end{gathered}$$
Number of questions he got wrong is 10. 

Answer:

(i) 3x + 5y = 9             .....(I)
5x + 3y= 7                   .....(II)
Adding (I) and (II) we have
$$\begin{gathered} 8x+8y=16 \\ \Rightarrow x+y=2 \end{gathered}$$
Hence, the correct answer is option (A). 

(ii) Let the length be x and the breadth be y. 
Perimeter of the rectangle = 2(l + b) = 26
2(l + b) = 26
$$\begin{gathered} \Rightarrow 2\left[\left(x-5\right)+\left(y-5\right)\right]=26 \\ \Rightarrow 2\left[x+y-10\right]=26 \\ \Rightarrow x+y=23 \end{gathered}$$
Hence, the correct answer is option (C). 

(iii) Let Ajay's age be x years and Vijay's age be y years.
Ajay is younger than Vijay by 5 years.
$y-x=5 .....\left(\mathrm{I}\right)$
Sum of their ages is 25 years.
$x+y=25 .....\left(\mathrm{II}\right)$
Adding (I) and (II) we have
$$\begin{gathered} 2y=30 \\ \Rightarrow y=15 \\ x=10 \end{gathered}$$
Thus, Ajay's age is 10 years.
Hence, the correct answer is option (C). 


 

Answer:

Answer:

Answer:

(i)  $\frac{x}{3} + \frac{y}{4} = 4$            .....(I)
$\frac{x}{2} - \frac{y}{4}= 1$                   .....(II)
Multiplying (I) with LCM of 3 and 4 which is 12 we get
$4x+3y=48$                     .....(III)
Multiplying (II) with LCM of 2 and 4 which is 4 we get
$2x-y=4$                     .....(IV)
Multiplying (IV) with 2
$4x-2y=8$                    .....(V)
Subtracting (V) from (III)
$$\begin{gathered} 5y=40 \\ \Rightarrow y=8 \end{gathered}$$
Putting this value of y in (IV) we get
$$\begin{gathered} 2x-8=4 \\ \Rightarrow 2x=12 \\ \Rightarrow x=6 \end{gathered}$$

(ii)  
$$\begin{gathered} \frac{x}{3} + 5y = 13 .....\left(\mathrm{I}\right) \\ 2x + \frac{y}{2} =19 .....\left(\mathrm{II}\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow x+15y=39 .....\left(\mathrm{III}\right) \\ 4x+y=38 .....\left(\mathrm{IV}\right) \end{gathered}$$
Multiplying (III) with 4 we get
$4x+60y=156$            .....(V)
Subtracting (IV) from (V)
$$\begin{gathered} 59y=118 \\ \Rightarrow y=2 \end{gathered}$$
Putting the value of y in (IV) we get
$$\begin{gathered} 4x+2=38 \\ \Rightarrow 4x=36 \\ \Rightarrow x=9 \end{gathered}$$

(iii)  $\frac{2}{x} +\frac{ 3}{y} = 13 ; \frac{5}{x} - \frac{4}{y} = -2$
Let $\frac{1}{x}=u \mathrm{and} \frac{1}{y}=v$
So, the equations obtained are
$$\begin{gathered} 2u+3v=13 .....\left(\mathrm{I}\right) \left(\times 5\right) \\ 5u-4v=-2 .....\left(\mathrm{II}\right) \left(\times 2\right) \\ 10u+15v=65 .....\left(\mathrm{III}\right) \\ 10u-8v=-4 .....\left(\mathrm{IV}\right) \\ \mathrm{Subtracting} \left(\mathrm{IV}\right) \mathrm{from} \left(\mathrm{III}\right) \\ 23v=69 \\ \Rightarrow v=3 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} v \mathrm{in} \left(\mathrm{I}\right) \\ 2u+3\times 3=13 \\ \Rightarrow 2u=4 \\ \Rightarrow u=2 \\ \frac{1}{x}=u\Rightarrow x=\frac{1}{u}=\frac{1}{2} \\ \frac{1}{y}=v\Rightarrow y=\frac{1}{v}=\frac{1}{3} \end{gathered}$$
 

Answer:

Let the two digit number be 10x + y. 
A two digit number is 3 more than 4 times the sum of its digits.
$$\begin{gathered} 10x+y=4\left(x+y\right)+3 \\ \Rightarrow 10x+y=4x+4y+3 \\ \Rightarrow 2x-y=1 .....\left(\mathrm{I}\right) \end{gathered}$$
If 18 is added to this  number, the sum is equal to the number obtained by interchanging the digits.
$$\begin{gathered} 10x+y+18=10y+x \\ \Rightarrow 9x-9y=-18 \\ \Rightarrow x-y=-2 .....\left(\mathrm{II}\right) \end{gathered}$$
(I) $-$ (II)
$x=3,y=5$
Thus, the number obtained is 35. 

Answer:

Disclaimer: There is an error in the given question. Instead of 7 books and 5 pens there should be 7 books and 6 pens. 

Let the cost of 1 book be x and the cost of 1 pen be y. 
The total cost of 6 books and 7 pens is 79 rupees. 
$6x+7y=79 .....\left(\mathrm{I}\right)$
Total cost of 7 books and 6 pens is 77 rupeess.
$7x+6y=77 .....\left(\mathrm{II}\right)$
Adding (I) and (II)
$$\begin{gathered} 13x+13y=156 \\ \Rightarrow x+y=12 .....\left(\mathrm{III}\right) \end{gathered}$$
Subtracting (II) from (I)
$x-y=-2 .....\left(\mathrm{IV}\right)$
Adding (III) and (IV)
$$\begin{gathered} 2x=10 \\ \Rightarrow x=5 \end{gathered}$$
y = 7
Cost of 1 book and 2 pens is 5 + 14 = Rs 19.

Answer:

Let the incomes of the two persons be x and y. 
$$\begin{gathered} \frac{x}{y}=\frac{9}{7} \\ \Rightarrow 7x=9y \\ \Rightarrow 7x-9y=0 .....\left(\mathrm{I}\right) \end{gathered}$$
Savings of the two persons = Rs 200
Expenses = Income $-$ Savings 
$$\begin{gathered} \frac{x-200}{y-200}=\frac{4}{3} \\ \Rightarrow 3x-600=4y-800 \\ \Rightarrow 3x-4y=-200 .....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (I) with 3 and (II) with 7.
$$\begin{gathered} 21x-27y=0 .....\left(\mathrm{III}\right) \\ 21x-28y=-1400 .....\left(\mathrm{IV}\right) \\ \left(\mathrm{IV}\right)-\left(\mathrm{III}\right) \\ -y=-1400 \\ \Rightarrow y=1400 \end{gathered}$$
Putting the value y in (I) we get
x = 1800
Thus, the income of the two persons is Rs 1800 and Rs 1400. 

Answer:

Let the length be l and the breadth be b. 
Area of rectangle = lb
Length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. 
$$\begin{gathered} \left(l-5\right)\left(b+3\right)=lb-8 \\ \Rightarrow lb-5b+3l-15=lb-8 \\ \Rightarrow 3l-5b=7 .....\left(\mathrm{I}\right) \end{gathered}$$
If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units.
$$\begin{gathered} \left(l-3\right)\left(b+2\right)=lb+67 \\ \Rightarrow lb-3b+2l-6=lb+67 \\ \Rightarrow 2l-3b=73 .....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (I) with 2 and (II) with 3
$$\begin{gathered} 6l-10b=14 .....\left(\mathrm{III}\right) \\ 6l-9b=219 .....\left(\mathrm{IV}\right) \\ \left(\mathrm{III}\right)-\left(\mathrm{IV}\right) \\ b=205 \end{gathered}$$
Putting this value in (I) we get
$$\begin{gathered} 3l-5\left(205\right)=7 \\ \Rightarrow 3l=7+1025=1032 \\ \Rightarrow l=344 \end{gathered}$$
Thus, length = 344 units and breadth = 205 units. 

Answer:

Let the speed of the faster car be x km/h and that of the slower car be y km/h.
Total distance between them is 70 km. 
$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$
When both cars travel in the same direction, they meet in 7 hours. 
Total speed = x − y
$$\begin{gathered} x-y=\frac{70}{7} \\ \Rightarrow x-y=10 .....\left(\mathrm{I}\right) \end{gathered}$$
When both cars move towards each other,
Total speed = x + y
$$\begin{gathered} x+y=\frac{70}{1} \\ \Rightarrow x+y=70 .....\left(\mathrm{II}\right) \end{gathered}$$
Adding (I) and (II)
$$\begin{gathered} 2x=80 \\ \Rightarrow x=40 \end{gathered}$$
Putting the value of x in (I)
$$\begin{gathered} 40-y=10 \\ \Rightarrow y=40-10=30 \end{gathered}$$
Thus, speed of the cars are 40 km/h and 30 km/h. 

 

Answer:

Let the two digit number be 10x + y. 
Number obtained on interchanging the digits = 10y + x. 
The sum of a two digit number and the number obtained by interchanging its digits  is 99.
10x + y + 10y + x = 99
$$\begin{gathered} \Rightarrow 11x+11y=99 \\ \Rightarrow x+y=9 \end{gathered}$$
With the given information, only one equation can be formed. So, the number can be
18, 81, 54, 45, 27, 72, 36, 63.